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188400 | https://courses.lumenlearning.com/slcc-elementaryalgebra/chapter/4-3-solving-a-2x2-system-of-linear-equations-by-elimination/ | 4.3: Solving a 2×2 System of Linear Equations by Elimination
Section 4.3 Learning Objectives
4.3: Solving a 2×2 System of Linear Equations by Addition/Elimination
Solve systems of linear equations using elimination
Recognize when systems of linear equations have no solution or an infinite number of solutions
Solve a system of equations using the elimination method
The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation to eliminate one of the variable terms. In this method, you may or may not need to multiply the terms in one equation by a number first. We will first look at examples where no multiplication is necessary to use the elimination method. Then you will see examples using multiplication after you are familiar with the idea of the elimination method.
It is easier to show rather than tell with this method, so let’s dive right into some examples.
If you add the two equations,
x–y=−6 and x+y=8 together, watch what happens.
x−y=−6+x+y=8––––––––––––––2x+0=2
You have eliminated the y-term, and this equation can be solved using the methods for solving equations with one variable.
Let’s see how this system is solved using the elimination method.
Example 1
Use elimination to solve the system.
x–y=−6x+y=8
Show Solution
Add the equations.
x−y=−6+x+y=8––––––––––––––2x=2
Solve for x.
2x2=22x=1
Substitute x=1 into one of the original equations and solve for y.
x+y=81+y=8−1−1–––––––––––––y=7
Be sure to check your answer in both equations!
x–y=−61–7=−6−6=−6TRUEx+y=81+7=88=8TRUE
The answers check.
Answer
The solution is (1, 7).
Unfortunately not all systems work out this easily. How about a system like 2x+y=12 and −3x+y=2. If you add these two equations together, no variables are eliminated.
2x+y=12−3x+y=2–––––––––––––––−x+2y=14
But you want to eliminate a variable. So let’s add the opposite of one of the equations to the other equation. This means multiply every term in one of the equations by −1, so that the sign of every term is opposite.
2x+y=12→2x+y=12→2x+y=12−3x+y=2→−(−3x+y)=−(2)→3x–y=−25x+0y=10
You have eliminated the y-variable, and the problem can now be solved. The complete solution is shown in the example after the video.
The following video describe a similar problem where you can eliminate one variable by adding the two equations together.
Caution! When you add the opposite of one entire equation to another, make sure to change the sign of EVERY term on both sides of the equation. This is a very common mistake to make.
Example 2
Use elimination to solve the system.
2x+y=12−3x+y=2
Show Solution
You can eliminate the y-variable if you add the opposite of one of the equations to the other equation.
2x+y=12−3x+y=2
Rewrite the second equation as its opposite. In other words, multiply by −1.
Add. Solve for x.
2x+y=123x–y=−2––––––––––––––5x=10
5x5=105
x=2
Substitute x=2 into one of the original equations and solve for y.
2(2)+y=124+y=12−4−4–––––––––––––y=8
Be sure to check your answer in both equations!
2x+y=122(2)+8=124+8=1212=12TRUE−3x+y=2−3(2)+8=2−6+8=22=2TRUE
The answers check.
Answer
The solution is (2, 8).
The following are two more examples showing how to solve linear systems of equations using elimination.
Example 3
Use elimination to solve the system.
3y=2x−12x=5(5−y)
Show Solution
First, note that we will find the elimination process easier if the equations are in the standard form, Ax+By=C. So, we will start by subtracting 2x from both sides in the first equation.
3y=2x−1
−2x+3y=−1
Then we will distribute and add 5y to both sides in the second equation.
2x=5(5−y)
2x=25−5y
2x+5y=25
This results in the system shown below.
−2x+3y=−12x+5y=25
Notice the coefficients of each variable in each equation. If you add these two equations, the x-term will be eliminated since −2x+2x=0. Thus, we can add the equations and solve for y.
−2x+3y=−12x+5y=25–––––––––––––––8y=24
8y8=248
y=3
Substitute y=3into one of the original equations.
2x+5y=252x+5(3)=252x+15=25−15−15––––––––––––––2x=10
2x2=102
x=5
Check solutions.
−2x+3y=−1−2(5)+3(3)=−1−10+9=−1−1=−1TRUE2x+5y=252(5)+5(3)=2510+15=2525=25TRUE
The answers check.
Answer
The solution is (5, 3).
Example 4
Use elimination to solve for xand y.
4x+2y=145x+2y=16
Show Solution
Notice the coefficients of each variable in each equation. You will need to add the opposite of one of the equations to eliminate the variable y, as 2y+2y=4y, but 2y+(−2y)=0.
Change one of the equations to its opposite (i.e. multiply by −1), add, and solve for x. Let’s multiply the second equation by −1.
4x+2y=14−5x–2y=−16–––––––––––––––––––−x=−2
−x−1=−2−1
x=2
Substitute x=2 into one of the original equations and solve for y.
4x+2y=144(2)+2y=148+2y=14−8−8–––––––––––––––2y=6
2y2=62
y=3
We will leave it to you to check this solution.
Answer
The solution is (2, 3).
Go ahead and check this last example—substitute (2,3) into both equations. You get two true statements: 14=14 and 16=16.
Notice that you could have used the opposite of the first equation rather than the second equation and the result would be the same.
Recognize systems that have no solution or an infinite number of solutions
Just as with the substitution method, the elimination method will sometimes eliminate both variables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.
Let’s look at an example.
Example 5
Solve for xand y.
−x–y=−4x+y=2
Show Solution
Add the equations to eliminate the x-term.
−x–y=−4x+y=2––––––––––––0=−2
This false statement leads us to conclude that the system has no solution.
Answer
No Solution
Graphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.
If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.
Example 6
Solve for xand y.
x+y=2−x−y=−2
Show Solution
Add the equations to eliminate the x-term.
x+y=2−x−y=−2––––––––––––––0=0
This true statement leads us to conclude that there are an infinite number of solutions, given by all points on the line. Recall that we learned the proper way of expressing such an answer earlier in this chapter.
Answer
There are an infinite number of solutions, given by
{(x,y)|x+y=2}
Graphing these two equations will help to illustrate what is happening.
In the following video, a system of equations which has no solutions is solved using the method of elimination.
Solve a system of equations when multiplication is necessary to eliminate a variable
Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.
3x+4y=525x+y=30
If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate the same variable in the other equation.
We do this with multiplication. Notice that the first equation contains the term 4y, and the second equation contains the term y. If you multiply the second equation by −4, when you add both equations the y–variable terms will add up to 0.
The following example takes you through all the steps to find a solution to this system.
Example 7
Solve for xand y.
3x+4y=525x+y=30
Show Solution
Look for terms that can be eliminated. In this case, neither the x or yterms will cancel.
3x+4y=525x+y=30
Multiply the second equation by −4 so they do have the same coefficient.
−4(5x+y)=−4(30)
This gives
−20x−4y=−120
Rewrite the system and add the equations.
3x+4y=52−20x–4y=−120––––––––––––––––––––−17x=−68
Solve for x.
−17x−17=−68−17
x=4
Substitute x=4 into one of the original equations to find y.
3x+4y=523(4)+4y=5212+4y=52−12−12––––––––––––––––4y=40
4y4=404
y=10
Check your answer.
3x+4y=523(4)+4(10)=5212+40=5252=52TRUE5x+y=305(4)+10=3020+10=3030=30TRUE
The answers check.
Answer
The solution is (4, 10).
Caution! When you use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. Forgetting to multiply every term is a common mistake.
There are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both equations by different numbers.
Let’s eliminate the variable xthis time. We can achieve this by multiplying the first equation by 5 and the second equation by −3.
Example 8
Solve for x and y.
3x+4y=525x+y=30
Show Solution
Look for terms that can be eliminated. The equations do not have any x or yterms with opposite coefficients.
3x+4y=525x+y=30
In order to use the elimination method, you have to create variables that have the same coefficient in absolute value, but opposite signs—then you can eliminate them by adding the equation. Multiply the top equation by 5.
5(3x+4y)=5(52)15x+20y=260
Now multiply the bottom equation by −3.
−3(5x+y)=−3(30)−15x–3y=−90
Next add the equations, and solve for y.
15x+20y=260−15x–3y=–90––––––––––––––––––––17y=170
17y17=17017
y=10
Substitute y=10 into one of the original equations to find x.
3x+4y=523x+4(10)=523x+40=52−40−40––––––––––––––3x=12
3x3=123
x=4
You arrive at the same solution as before.
Answer
The solution is (4, 10).
These equations were multiplied by 5 and −3 respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.
In the following video, you will see an example of using the elimination method for solving a system of equations.
The next example reminds us that sometimes we must deal with fractions.
Example 9
Use elimination to solve the system of equations.
12x−34y=134x−32y=138
Show Solution
While we could try to jump into the elimination process immediately, you may find it easier to first clear out the fractions by multiplying each equation by its LCD.
Multiply the first equation by 4.
4⋅12x−4⋅34y=4⋅1
2x−3y=4
And multiply the second equation by 8.
8⋅34x−8⋅32y=8⋅138
6x−12y=13
So, we are trying to solve the equivalent system below.
2x−3y=46x−12y=13
Using elimination, we can multiply the top equation by −3 and add.
−6x+9y=−126x−12y=13––––––––––––––––−3y=1
−3y−3=1−3
y=−13
So, coincidentally, we also obtain a fractional answer. We now plug this value in to find x.For simplicity, we can plug into one of the two equations we obtained after eliminating the fractions.
2x−3y=4
2x−3(−13)=4
2x+1=4−1−1––––––––––––2x=3
2x3=32
x=32
We conclude that the solution to the system is the ordered pair (32,−13).
Answer
The solution is (32,−13).
It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as we saw in simpler systems earlier. In the following example, you will see a system that has infinitely many solutions.
Example 10
Solve the system of equations.
x−3y=−2−2x+6y=4
Show Solution
Look for terms that can be eliminated. The equations do not have any x or y terms with opposite coefficients.
x−3y=−2−2x+6y=4
Multiply the first equation by 2 so that the x terms will cancel out.
2(x−3y)=2(−2)
2x−6y=−4
Rewrite the system and add the equations.
2x−6y=−4−2x+6y=4–––––––––––––––0=0
Does this kind of solution look familiar? This implied that there were an infinite number of solution. If we solve both of these equations for y, you will see that they are the same equation.
Solve the first equation for y:
x−3y=−2−3y=−x−2y=13x+23
Solve the second equation for y:
−2x+6y=46y=2x+4y=13x+23
Both equations are the same when written in slope intercept form, and therefore the solution set consists of all ordered pairs that lie on the line.
Answer
There are an infinite number of solutions, given by
{(x,y)|x−3y=−2}
In the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a negative one first. Additionally, this system has an infinite number of solutions.
Summary
Combining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method. Once one variable is eliminated, it becomes much easier to solve for the other one.
Multiplication can be used to set up terms that can be eliminated in the equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate. Solving using the elimination method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions). |
188401 | https://artofproblemsolving.com/wiki/index.php/Law_of_Sines?srsltid=AfmBOoqJqHxUet1e7YrfARpW_EmhBr4bGWT7MRN2WwnTj0NnLtRB79AP | Art of Problem Solving
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Law of Sines
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.
Contents
[hide]
1 Statement
2 Proof
2.1 Method 1
2.2 Method 2
3 Method 3
4 Problems
4.1 Introductory
4.2 Intermediate
4.3 Olympiad
5 See Also
Statement
In triangle , where is the side opposite to , opposite to , opposite to , and where is the circumradius:
Proof
Method 1
In the diagram above, point is the circumcenter of . Point is on such that is perpendicular to . Since , and . But making . We can use simple trigonometry in right triangle to find that
The same holds for and , thus establishing the identity.
Method 2
This method only works to prove the regular (and not extended) Law of Sines.
The formula for the area of a triangle is .
Since it doesn't matter which sides are chosen as , , and , the following equality holds:
Assuming the triangle in question is nondegenerate, . Multiplying the equation by yields:
Method 3
We can circumvent some of the work in Method 1 by setting up the circle in a different way. Let be a diameter and be the center of the circle, and let be on . Furthermore, let , and let , , and .
We have that is a right angle, as is a diameter. Therefore, , so, rearranging, we have , or . Likewise, . Finally, we observe that , so evidently . Combining all three equalities,
Problems
Introductory
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(Source)
Intermediate
Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circlecircumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .
(Source)
Olympiad
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .
(Source)
See Also
Trigonometry
Trigonometric identities
Geometry
Law of Cosines
Retrieved from "
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Theorems
Trigonometry
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188402 | https://documents.thermofisher.com/TFS-Assets/CMD/Application-Notes/TN-152-Tips-Tricks-Fluorescence-Detection-Extended-Lifetime-Xenon-Flash-Lamp-TN70987-E.pdf | Tips and Tricks for Fluorescence Detection: Extend the Lifetime of Your Xenon Flash Lamp Melanie Neubauer and Holger Franz Thermo Fisher Scientific, Germering, Germany Technical Note 152 Summary The lifetime of xenon flash lamps is related to their operation time and flash frequency. The Thermo Scientific™ Dionex™ UltiMate™ 3000 FLD-3000 Fluorescence Detector Series offers three different lamp modes. These can be used in an effective and efficient way to achieve both the best limit of detection (LOD) and an extended lamp lifetime.
Background Two types of lamps are commonly used as a light source for fluorescence detection: continuous lamps and flash lamps. The FLD-3000 Series detectors use a xenon flash lamp that requires only a few seconds for warm-up. Therefore, the operation time of the lamp is equal to the data acquisition time. In contrast, continuous mercury and xenon arc lamps typically need a warm-up time of more than 1 hour to reach a thermal equilibrium. To ensure that the lamp is ready for each sample, it stays usually turned on after a run and after the end of a sequence, respectively. Leaving the lamp on between sequences reduces the number of samples that can be analyzed during the lifetime of the continuous lamp. A flash lamp with an average lifetime of 4000 hours (operation at 100 Hz) can therefore process a lot more than four times the samples of a 1000 hour continuous lamp.
FLD-3000 xenon flash lamps can be operated in three different flash frequency modes. The highest frequency results in the best LOD; the lowest flash frequency achieves the longest lamp lifetime. A smart use of the lamp operation modes achieves both best detection sensitivity and extended lifetime of the flash lamp.
Solution The FLD-3000 Fluorescence Detector Series is equipped with xenon flash lamps that offer three operation modes with different frequencies: • HighPower mode with a frequency of 300 Hz • Standard mode offers a flash rate of 100 Hz • LongLife mode that provides a frequency of 20 Hz Keywords Xenon, Flash Lamp, Fluorescence Detection, Lamp Frequency, LOD A change of the flash frequency has an impact on the lifetime of the lamp and on the observed baseline noise. The latter decreases with high flash frequencies, but the peak height remains constant for all lamp operation modes as shown in Figure 2.
Figure 2. Influence of the flash lamp frequency on the signal-to-noise.
3.65 4.15 0 HighPower Standard LongLife 10 × 106 Minutes Response (counts) 29114 17 × 104 15 × 104 Figure 1. UltiMate 3000 FLD-3000 Series Fluorescence Detector.
TN70987_E 07/16S Africa +43 1 333 50 34 0 Australia +61 3 9757 4300 Austria +43 810 282 206 Belgium +32 53 73 42 41 Brazil +55 11 3731 5140 Canada +1 800 530 8447 China 800 810 5118 (free call domestic) 400 650 5118 Denmark +45 70 23 62 60 Europe-Other +43 1 333 50 34 0 Finland +358 9 3291 0200 France +33 1 60 92 48 00 Germany +49 6103 408 1014 India +91 22 6742 9494 Italy +39 02 950 591 Japan +81 6 6885 1213 Korea +82 2 3420 8600 Latin America +1 561 688 8700 Middle East +43 1 333 50 34 0 Netherlands +31 76 579 55 55 New Zealand +64 9 980 6700 Norway +46 8 556 468 00 Russia/CIS +43 1 333 50 34 0 Singapore +65 6289 1190 Sweden +46 8 556 468 00 Switzerland +41 61 716 77 00 Taiwan +886 2 8751 6655 UK/Ireland +44 1442 233555 USA +1 800 532 4752 www.thermofisher.com/dionex ©2016 Thermo Fisher Scientific Inc. All rights reserved. All trademarks are the property of Thermo Fisher Scientific and its subsidiaries. This information is presented as an example of the capabilities of Thermo Fisher Scientific products. It is not intended to encourage use of these products in any manners that might infringe the intellectual property rights of others. Specifications, terms and pricing are subject to change. Not all products are available in all countries. Please consult your local sales representative for details.
Hence, a change of the operation mode influences the signal-to-noise ratio (S/N) and not the signal height itself. The trend of the S/N in relation to the lamp mode is illustrated in Figure 3. The higher the flash frequency, the better is the resulting S/N. This relation can be roughly estimated using Equation 1: with F1= Previous flash lamp frequency, F2= Planned flash lamp frequency.
Caused by photon shot noise, the baseline noise hence increases approximately by a factor of √3 (≈ 1.7) when changing the lamp mode from HighPower (300 Hz) to Standard mode (100 Hz). The noise rises by a factor of about 2.24 when the mode is switched from Standard to LongLife (20 Hz).
The FLD-3000 Series enables changes of the lamp mode during a chromatographic run. Figure 4 illustrates our recommendation for a smart use of the lamp modes. This approach combines both, an increased lamp lifetime and best S/N wherever required.
In quiet areas of the chromatogram, where no peaks of interest elute, the LongLife mode can be used. It provides sufficient S/N for monitoring the baseline, for example before the first peak elutes, or during the re-equilibration of the column (colored areas of the chromatogram). Switch to Standard or HighPower mode when peaks elute to obtain the best S/N (white part of the chromatogram).
The number of flashes during a chromatographic run is significantly reduced with this technique. In our example, the flash count is decreased by a factor of two compared to HighPower mode. This results in 100% more lamp lifetime without any drawbacks on the LOD.
Useful Links Find assistance on method development in the Fluorescence Method Development Handbook and get more information about fluorescence detection in UHPLC in Dionex, now a part of Thermo Fisher Scientific, Technical Note 92: Combining Fluorescence Detection with UHPLC: An Overview of the Technical Requirements.
For more information on our product portfolio, visit www.thermoscientific.com/liquidchromatography.
Figure 4. Smart use of the lamp modes extends the lifetime of the xenon flash lamp.
Figure 3. Signal-to-noise ratio of the peak shown in Figure 2 obtained with HighPower, Standard and LongLife modes.
Technical Note 152 References 1. Franz, H., Jendreizik, V. Fluorescence Method Development Handbook, 2013. [Online] (accessed Feb 13, 2014).
2. Lakowicz, J.R. Principles of Fluorescence Spectroscopy, 3rd er.; Springer Science+Business Media, LLC: New York, 2006 9482 29115 3632 6178 S/N 8000 7000 6000 5000 4000 3000 2000 1000 0 9000 10000 HighPower LongLife Standard 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 7.5 0 HighPower, 300 Hz LongLife, 20 Hz 4.5 × 106 Minutes Response (counts) 29116 LongLife, 20 Hz Noise (F2) = · Noise (F1), F1 F2 |
188403 | https://coolconversion.com/temperature/3500-C-to-Kelvin | 3500 C in Kelvin
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3500 Celsius equals 3773.2 Kelvin
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3500 degrees Celsius equals 3773.2 Kelvin
Use the folowing formula to convert from Celsius to Kelvin:
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How convert 3500 Celsius to Kelvin?
Use the formula below to convert from Celsius to Kelvin:
[K] = [°C] + 273.15
Thus, adding 273.15 to the value '3500' in Celsius we get:
3500 Celsius = 3500 + 273.15 = 3773.2 Kelvin.
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From Fahrenheit (F)F(F - 32) × 5/9(F - 32) × 5/9 + 273.15
From Celsius (C or o)(C × 9/5) + 32 C C + 273.15
From Kelvin (K)(K - 273.15) × 9/5 + 32 K - 273.15 K
Sample temperature conversions
4134 C to degrees Fahrenheit
3829 degrees Fahrenheit to Kelvins
3727 Kelvins to degrees Fahrenheit
158 degrees Fahrenheit to K
2122 degrees Celsius to K
4504 degrees Celsius to degrees Fahrenheit
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188404 | https://answers.everydaycalculation.com/simplify-fraction/4-52 | 4/52 simplified, Reduce 4/52 to its simplest form
Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Fraction simplifier
Reduce 4/52 to lowest terms
The simplest form of 4/52 is 1/13.
Steps to simplifying fractions
Find the GCD (or HCF) of numerator and denominator
GCD of 4 and 52 is 4
2. Divide both the numerator and denominator by the GCD
4 ÷ 4/52 ÷ 4
3. Reduced fraction: 1/13
Therefore, 4/52 simplified to lowest terms is 1/13.
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Equivalent fractions: 8/1042/2612/15620/26028/364
More fractions: 8/524/10412/524/1565/524/533/524/51
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188405 | https://www.fda.gov/food/alerts-advisories-safety-information/fda-advises-restaurants-and-retailers-not-serve-or-sell-and-consumers-not-eat-certain-shellfish | FDA Advises Restaurants and Retailers Not to Serve or Sell and Consumers Not to Eat Certain Shellfish from Oregon and Washington Potentially Contaminated with Paralytic Shellfish Toxins | FDA
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FDA Advises Restaurants and Retailers Not to Serve or Sell and Consumers Not to Eat Certain Shellfish from Oregon and Washington Potentially Contaminated with Paralytic Shellfish Toxins
Alerts, Advisories & Safety Information
FDA Advises Restaurants and Retailers Not to Serve or Sell and Consumers Not to Eat Certain Shellfish from Oregon and Washington Potentially Contaminated with Paralytic Shellfish Toxins
Alerts, Advisories & Safety Information
Audience
Restaurants and food retailers in Arizona (AZ), California (CA), Colorado (CO), Hawaii (HI), Nevada (NV), New York (NY), Oregon (OR), and Washington (WA) that have recently purchased oysters and bay clams from growing areas in Netarts Bay and Tillamook Bay, OR harvested on or after 5/28/24, and all shellfish species from growing areas in Willapa Bay, WA: Stony Point, harvested between 5/26/24 and 5/30/24; Bay Center, harvested between 5/29/24 and 5/30/24; and Bruceport, harvested between 5/29/24 and 5/30/24.
Consumers in AZ, CA, CO, HI, NV, NY, OR, and WA who have recently purchased oysters and bay clams harvested from growing areas in Netarts Bay and Tillamook Bay, OR harvested on or after 5/28/24, and all shellfish species from growing areas in Willapa Bay, WA: Stony Point, harvested between 5/26/24 and 5/30/24; Bay Center, harvested between 5/29/24 and 5/30/24; and Bruceport, harvested between 5/29/24 and 5/30/24.
Product
Certain oysters and bay clams harvested from OR growing areas in Netarts Bay and Tillamook Bay, harvested on or after 5/28/24, and shellfish species from growing areas in Willapa Bay, WA: Stony Point, harvested between 5/26/24 and 5/30/24; Bay Center, harvested between 5/29/24 and 5/30/24; and Bruceport, harvested between 5/29/24 and 5/30/24. The shellfish were distributed to restaurants and retailers in AZ, CA, CO, HI, NV, NY, OR, and WA and may have been distributed to other states as well.
Purpose
The FDA is advising restaurants and food retailers not to serve or sell and to dispose of, and consumers not to eat, oysters and bay clams harvested from growing areas in Netarts Bay and Tillamook Bay, OR harvested on or after 5/28/24, and all shellfish species from growing areas in Willapa Bay, WA: Stony Point, harvested between 5/26/24 and 5/30/24; Bay Center, harvested between 5/29/24 and 5/30/24; and Bruceport, harvested between 5/29/24 and 5/30/24, and distributed to AZ, CA, CO, HI, NV, NY, OR, and WA because they may be contaminated with the toxins that cause paralytic shellfish poisoning (PSP).
Molluscan shellfish contaminated with natural toxins from the water in which they lived can cause consumer illness. Most of these toxins are produced by naturally occurring marine algae (phytoplankton). Molluscan shellfish consume the algae which causes the toxins to accumulate in the shellfish’s flesh. Typically, contamination occurs following blooms of the toxic algal species; however, toxin contamination is possible even when algal concentrations are low in certain instances. One of the recognized natural toxin poisoning syndromes that can occur from consuming contaminated molluscan shellfish is paralytic shellfish poisoning (PSP).
PSP is caused by neurotoxins also referred to as saxitoxins or paralytic shellfish toxins (PSTs). Shellfish can retain the toxin for different lengths of time. Some species cleanse themselves of toxins rapidly, whereas others are much slower to remove the toxins. This lengthens the period of time they pose a human health risk from consumption.
Food containing PSTs may look, smell, and taste normal. These toxins cannot be removed by cooking or freezing. Consumers of these products who are experiencing symptoms of illness should contact their healthcare provider and report their symptoms to their local Health Department.
Symptoms of Paralytic Shellfish Poisoning
Most people with PSP will begin to develop symptoms within 30 minutes of consuming contaminated seafood. Effects of PSP intoxication can range from tingling of the lips, mouth, and tongue to respiratory paralysis and may include these other symptoms: numbness of arms and legs, “pins and needles” sensation, weakness, loss of muscle coordination, floating feeling, nausea, shortness of breath, dizziness, vomiting, and headache. Medical treatment consists of providing respiratory support and fluid therapy. For patients surviving 24 hours, with or without respiratory support, the prognosis is considered good, with no lasting side effects. In fatal cases, death is typically due to asphyxiation.
Due to the range in severity of illness, people should consult their healthcare provider if they suspect that they have developed symptoms that resemble paralytic shellfish poisoning.
Summary of Problem and Scope
On 5/30/2024 the Oregon Department of Agriculture advised the FDA of the recall of certain oysters and bay clams due to elevated PSP levels. The shellfish were harvested from Netarts Bay, OR and Tillamook Bay, OR, on or after 5/28/24. The Oregon product was shipped to OR and NY may have been distributed to other states as well.
On 5/30/2024 Washington State Department of Health advised the FDA that the state of Washington is conducting a recall for all shellfish species harvested from the following growing areas in Willapa Bay, WA due to elevated PSP levels: Stony Point, harvested 5/26/24 to 5/30/24; Bay Center, harvested 5/29/24 to 5/30/24; and Bruceport, harvested 5/29/24 to 5/30/24. The Washington product was shipped to AZ, CA, CO, HI, NV, OR, and WA, may have been distributed to other states as well.
FDA Actions
The FDA is issuing this alert advising restaurants and food retailers not to serve or sell and advising consumers not to eat oysters and bay clams from OR growing areas in Netarts Bay and Tillamook Bay, harvested on or after 5/28/2024 and all shellfish from growing areas in Willapa Bay, WA: Stony Point, harvested between 5/26/24 and 5/30/24; Bay Center, harvested between 5/29/24 and 5/30/24; and Bruceport, harvested between 5/29/24 and 5/30/24 due to possible contamination with paralytic shellfish toxins.
The FDA is awaiting further information on distribution of the shellfish harvested and will continue to monitor the investigation and provide assistance to state authorities as needed. As new information becomes available, the FDA will update the safety alert.
Recommendations for Restaurants and Retailers
Restaurants and retailers should not serve or sell the potentially contaminated shellfish. Restaurants and retailers should dispose of any products by throwing them in the garbage or contacting their distributor for return and destruction.
Restaurants and retailers should also be aware that shellfish may be a source of pathogens and should control the potential for cross-contamination of food processing equipment and the food processing environment. They should follow the steps below:
Wash hands with warm water and soap following the cleaning and sanitation process.
Retailers, restaurants, and other food service operators who have processed and packaged any potentially contaminated products need to be concerned about cross-contamination of cutting surfaces and utensils through contact with the potentially contaminated products.
Retailers that have sold bulk product should clean and sanitize the containers used to hold the product.
Regular frequent cleaning and sanitizing of food contact surfaces and utensils used in food preparation may help to minimize the likelihood of cross-contamination.
Recommendations for consumers
Consumers should not eat the potentially contaminated shellfish. Consumers who have symptoms of paralytic shellfish poisoning should contact their health care provider to report their symptoms and receive care.
To report a complaint or adverse event (illness or serious allergic reaction),visit Industry and Consumer Assistance.
Additional Information
Notice of Illness Outbreaks, Shellfish Closures, Reopenings, & Recalls | ISSCExternal Link Disclaimer
Food Poisoning from Marine Toxins | CDC Yellow Book 2024
Fish and Fishery Products Hazards and Controls
FDA Bad Bug Book (2nd Edition)External Link Disclaimer
Handwashing: A Healthy Habit in the Kitchen | CDC
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06/05/2024
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188406 | https://www.intrans.iastate.edu/wp-content/uploads/sites/15/2020/03/2F-2.pdf | 2F-2 Design Manual Chapter 2 - Stormwater 2F - Open Channel Flow Open Channel Flow 1 Revised: 2013 Edition A. Introduction The beginning of any channel design or modification is to understand the hydraulics of the stream. The procedures for performing uniform flow calculations aid in the selection or evaluation of appropriate depths and grades for natural or man-made channels. Allowable velocities are provided, along with procedures for evaluating channel capacity using Manning’s equation. All the methods described herein will be based on the conservation of mass, momentum and energy (in the form of Bernoulli’s theorem), and the Manning formula for frictional resistance. Steady uniform flow and steady non-uniform flow are the types of flow addressed in this section. B. Definitions Critical Flow: The variation of specific energy with depth at a constant discharge shows a minimum in the specific energy at a depth called critical depth at which the Froude number has a value of one. Critical depth is also the depth of maximum discharge, when the specific energy is held constant. Froude Number: The Froude number is an important dimensionless parameter in open-channel flow. It represents the ratio of inertia forces to gravity forces. This expression for Froude number applies to any single-section channel of nonrectangular shape. Hydraulic Jump: Hydraulic jumps occur at abrupt transitions from supercritical to subcritical flow in the flow direction. There are significant changes in the depth and velocity in the jump, and energy is dissipated. For this reason, the hydraulic jump is often employed to dissipate energy and control erosion at stormwater management structures. Kinetic Energy Coefficient: As the velocity distribution in a river varies from a maximum at the design portion of the channel to essentially zero along the banks, the average velocity head. Normal Depth: For a given channel geometry, slope, and roughness, and a specified value of discharge Q, a unique value of depth occurs in a steady uniform flow. It is called the normal depth. The normal depth is used to design artificial channels in a steady, uniform flow and is computed from Manning’s equation. Specific Energy: Specific energy (E) is the energy head relative to the channel bottom. If the channel is not too steep (slope less than 10%), and the streamlines are nearly straight and parallel (so that the hydrostatic assumption holds), the specific energy E becomes the sum of the depth and velocity head. The kinetic energy correction coefficient is taken to have a value of one for turbulent flow in prismatic channels but may be significantly different from one in natural channels. Steady and Unsteady Flow: A steady flow is when the discharge passing a given cross-section is constant with respect to time. When the discharge varies with time, the flow is unsteady. The maintenance of steady flow requires that the rates of inflow and outflow be constant and equal. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 2 Revised: 2013 Edition Subcritical Flow: Depths of flow greater than critical depths, resulting from relatively flat slopes. Froude number is less than one. Flow of this type is most common in flat streams. Supercritical Flow: Depths of flow less than critical depths resulting from relatively steep slopes. Froude number is greater than one. Flow of this type is most common is steep streams. Total Energy Head: The total energy head is the specific energy head plus the elevation of the channel bottom with respect to some datum. The curve of the energy head from one cross-section to the next defines the energy grade line. Uniform Flow and Non-uniform Flow: A non-uniform flow is one in which the velocity and depth vary over distance, while they remain constant in uniform flow. Uniform flow can occur only in a channel of constant cross-section, roughness, and slope in the flow direction; however, non-uniform flow can occur in such a channel or in a natural channel with variable properties. C. Uniform Flow (Manning’s Equation) 1. Manning’s Equation: The normal depth is used to design artificial channels in a steady, uniform flow and is computed from Manning’s equation: 𝑄= 𝐴𝑉= 1.486 𝑛 (𝐴𝑅 2 3 ⁄ ) (𝑠 1 2 ⁄ ) Equation 2F-2.01 where: V = Channel velocity, ft/s (see Tables 2F-2.03 and 2F-2.04 for permissible velocities) Q = Discharge, cfs A = Cross-sectional area of flow, ft2 n = Manning’s roughness coefficient (see Section 2B-3) R = hydraulic radius, ft = A/P P = wetted perimeter, ft s = slope of hydraulic grade line (pipe/channel slope), ft/ft The selection of Manning’s n is generally based on observation; however, considerable experience is essential in selecting appropriate n values. If the normal depth computed from Manning’s equation is greater than critical depth, the slope is classified as a mild slope, while on a steep slope, the normal depth is less than critical depth. Thus, uniform flow is subcritical on a mild slope and supercritical on a steep slope. Strictly speaking, uniform flow conditions seldom, if ever, occur in nature because channel sections change from point to point. For practical purposes in highway engineering, however, the Manning equation can be applied to most streamflow problems by making judicious assumptions. When the requirements for uniform flow are met, the depth (dn) and the velocity (Vn) are said to be normal and the slopes of the water surface and channel are parallel. For practical purposes, in open channel design, minor undulations in streambed or minor deviations from the mean (average) cross-section can be ignored as long as the mean slope of the channel can be represented as a straight line. The Manning equation can readily be solved either graphically or mathematically for the average velocity in a given channel if the normal depth is known, because the various factors in the equation are known or can be determined (the hydraulic radius can be computed from the normal depth in a given channel). Discharge (Q) is then the product of the velocity and the area of flow (A). Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 3 Revised: 2013 Edition 2. Continuity Equation: The continuity equation is the statement of conservation of mass in fluid mechanics. For the special case of steady flow of an incompressible fluid, it assumes the simple form: 𝑄= 𝐴1𝑉 1 = 𝐴2𝑉 2 Equation 2F-2.02 where: A = flow cross-sectional area, ft2 V = mean cross-sectional velocity, ft/s (measured perpendicular to cross-section) The subscripts 1 and 2 refer to successive cross-sections along the flow path. The continuity equation can be used with Manning’s equation to obtain steady uniform flow velocity as: 𝑉= 𝑄 𝐴= 1.49 (𝑅 2 3 ⁄ ) (𝐴 1 2 ⁄ ) 𝑛 Equation 2F-2.03 D. Energy Flow Flowing water contains energy in two forms, potential and kinetic. The potential energy at a particular point is represented by the depth of the water plus the elevation of the channel bottom above a convenient datum plane. The kinetic energy, in feet, is represented by the velocity head: Kinetic energy = 𝑉2 2𝑔 Equation 2F-2.04 In channel flow problems it is often desirable to consider the energy content with the channel bottom. This is called the specific energy or specific head and is equal to the depth of water plus the velocity head: Specific energy = 𝑑+ 𝑉2 2𝑔 Equation 2F-2.05 At other times it is desirable to use the total energy content (total head), which is the specific head plus the elevation of the channel bottom above a selected datum. For example, total head may be used in applying the energy equation, which states that the total head (energy) at one point in a channel carrying a flow of water is equal to the total head (energy) at any point downstream plus the energy (head) losses occurring between the two points. The energy (Bernoulli) equation is usually written: 𝑑1 + 𝑉 1 2 2𝑔+ 𝑍1 = 𝑑2 + 𝑉 2 2 2𝑔+ 𝑍2 + ℎ𝑙𝑜𝑠𝑠 Equation 2F-2.06 In this equation, cross-section 2 (subscript 2) is downstream from cross-section 1 (subscript 1), Z is the elevation of channel bottom, and hloss represents loss of head between cross-sections 1 and 2. A convenient way of showing specific head is to plot the water surface and the specific head lines above a profile of the channel bottom (see Figure 2F-2.01). Note in Figure 2F-2.01 that the line obtained by plotting velocity head above the water surface is the same line as that obtained by plotting specific head above the channel bottom. This line represents the total energy, potential and kinetic, of the flow in the channel, and is called the “total head line” or “total energy line.” Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 4 Revised: 2013 Edition The slope (gradient) of the energy line is a measure of the friction slope or rate of energy head loss due to friction. Under uniform flow, the energy line is parallel to the water surface and to the streambed. For flow to occur in a channel, the total head or energy line must slope negatively (downward) in the direction of flow. Figure 2F-2.01: Channel Flow Terms Figure 2F-2.02: Definition Sketch of Specific Head Source: Design Charts for Open-Channel Flow, FHWA 1. Critical Flow: The relative values of the potential energy (depth) and the kinetic energy (velocity head) are important in the analysis of open-channel flow. Consider, for example, the relation of the specific head, 𝑑+ 𝑉2 2𝑔 , and the depth of a given discharge in a given channel that can be placed on various slopes. Plotting values of specific head as ordinates and of the corresponding depth as abscissa will result in a specific-head curve such as that shown in Figure 2F-2.02. The straight diagonal line is drawn through points where depth and specific head are equal. The line thus represents the potential energy, and the ordinate interval between this line and the specific head curve is the velocity head for the particular depth. A change in the discharge or in the channel size or shape will change the position of the curve, but its general shape and location above and to the left of the diagonal line will remain the same. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 5 Revised: 2013 Edition Note that the ordinate at any point on the specific head curve represents the total specific energy at that point. The lowest point on the curve represents flow with the minimum energy content. The depth at this point is known as critical depth (dc) and the corresponding velocity is the critical velocity (Vc). With uniform flow, the channel slope at which critical depth occurs is known as the critical slope (Sc). The magnitude of critical depth depends only on the discharge and the shape of the channel, and is independent of the slope or channel roughness. Thus, in any given size and shape of channel, there is only one critical depth for a particular discharge. Critical depth is an important value in hydraulic analysis because it is a control in reaches of non-uniform flow whenever the flow changes from subcritical to supercritical. Typical occurrences of critical depths are: a. Entrance to a restrictive channel, such as a culvert or flume, on a steep slope b. At the crest of an overflow dam or weir c. At the outlet of a culvert or flume discharging with a free fall or into a relatively wide channel or a pond in which the depth is not enough to submerge critical depth in the culvert or flume. 2. Critical Depth Calculations: a. The general equation for determining critical depths on the discharge rate and channel geometry is: 𝑄2 𝑔= 𝐴3 𝑇 Equation 2F-2.07 where: g = acceleration of gravity, ft/s2 (32.2) A = cross-sectional area, ft2 T = top width of water surface, ft A trial and error procedure is needed to solve Equation 2F-2.07. The following guidelines are presented for evaluating critical flow conditions of open channel flow: 1) A normal depth of uniform flow within about 10% of critical depth is unstable (relatively large depth changes are likely for small changes in roughness, cross-sectional area, or slope) and should be avoided in design, if possible. 2) If the velocity head is less than one-half the mean depth of flow, the flow is subcritical. 3) If the velocity head is equal to one-half the mean depth of flow, the flow is critical. 4) If the velocity head is greater than one-half the mean depth of flow, the flow is supercritical. 5) If an unstable critical depth cannot be avoided in design, the least favorable type of flow should be assumed for the design. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 6 Revised: 2013 Edition b. The Froude number, Fr, calculated by the flowing equation, is useful for evaluating the type of flow conditions in an open channel: 𝐹𝑟= 𝑉 (𝑔𝐴 𝑇) 1 2 ⁄ Equation 2F-2.08 where: Fr = Froude number (dimensionless) V = velocity of flow, ft/s g = acceleration of gravity, ft/s2 (32.2) T = top width of flow, ft If Fr is greater than 1.0, flow is supercritical; if it is under 1.0, flow is subcritical. Fr is 1.0 for critical flow conditions. 3. Critical Slope: Critical slope is that channel slope for a particular channel and discharge, at which the normal depth for uniform flow will be the same as the critical depth. Critical slope varies with both the roughness and geometric shape of the channel and with the discharge. For large circular cross-section pipes, and for pipe-arch and oval pipe sections, a direct reading can be made on the part-full flow charts for critical depth, specific head, and critical slope (for certain values of n). 4. Supercritical Flow: Points on the left of the flow point of the specific head curve [Figure 2F-2.02 (B)] are for channel slopes steeper than critical (supercritical or steep slopes), and indicate relatively shallow depths and high velocities [Figure 2F-2.02 (A)]. Such flow is called supercritical flow. It is difficult to handle because violent wave action occurs when either the direction of flow or the cross-section is changed. Flow of this type is common in steep streams. In supercritical flow, the depth of flow at any point is influenced by a control upstream, usually critical depth. 5. Subcritical Flow: Points on the right of the low point of the specific head curve [Figure 2F-2.02 (B)] are for slopes flatter than critical (subcritical or mild slopes) and indicate relatively large depths with low velocities [Figure 2F-2.02 (C)]. Such flow is called subcritical flow. It is relatively easy to handle through transitions because the wave actions are tranquil. In subcritical flow, the depth at any point is influenced by a downstream control, which may be either critical depth or the water surface elevation in a pond or larger downstream channel. Figures 2F-2.02 (A) and 2F-2.02 (C) indicate the relationship of supercritical and subcritical flows, respectively, to the specific head curve. E. Non-uniform Flow Flow that varies in depth and velocity along the channel is called non-uniform. Truly uniform flow rarely exists in either natural or man-made channels, because changes in channel section, slope, or roughness cause the depths and average velocities of flow to vary from point to point along the channel, and the water surfaces will not be parallel to the streambed. Although moderate non-uniform flow actually exists in a generally uniform channel, it is usually treated as uniform flow in such cases. Uniform flow characteristics can readily be computed and the computed values are usually close enough to the actual for all practical purposes. The types of non-uniform flow are innumerable, but certain characteristic types are more common. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 7 Revised: 2013 Edition With subcritical flow, a change in channel shape, slope, or roughness affects the flow for a considerable distance upstream, and thus the flow is said to be under downstream control. If an obstruction, such as a culvert, causes ponding, the water surface above the obstruction will be a smooth curve asymptotic to the normal water surface upstream and to the pool level downstream (see Figure 2F-2.03). Another example of downstream control occurs where an abrupt channel enlargement, as at the end of a culvert not flowing full, or a break in grade from a mild to a steep slope, causes a drawdown in the flow profile to critical depth. The water surface profile upstream from a change in section or a break in channel slope will be asymptotic to the normal water surface upstream, but will drop away from the normal water surface on approaching the channel change or break in slope. In these two examples, the flow is non-uniform because of the changing water depth caused by changes in the channel slope or channel section. Direct solution of open-channel flow by the Manning equation or by the charts in this section is not possible in the vicinity of the changes in the channel section or channel slope. With supercritical flow, a change in the channel shape, slope, or roughness cannot be reflected upstream except for very short distances. However, the change may affect the depth of flow at downstream points; thus, the flow is said to be under upstream control. Most problems in highway drainage do not require the accurate computation of water surface profiles. However, the designer should know that the depth in a given channel may be influenced by conditions either upstream or downstream, depending on whether the slope is steep (supercritical) or mild (subcritical). Figure 2F-2.03 shows a channel on a mild slope, discharging into a pool. The vertical scale is exaggerated to illustrate the case more clearly. Cross-section 1 is located at the end of uniform channel flow in the channel and cross-section 2 is located at the beginning of the pool. Depth 2 is located at the beginning of the pool. The depth of flow (d) between sections 1 and 2 is changing and the flow is non-uniform. The water surface profile between the sections is known as backwater curve and is characteristically very long. Figure 2F-2.03: Water Surface Profile in Flow from a Channel to a Pool Source: Design Charts for Open-Channel Flow, FHWA Figure 2F-2.04 shows a channel in which the slope changes from subcritical to supercritical. The flow profile passes through critical depth near the break in slope (section 1). This is true whether the upstream slope is mild, as in the sketch, or whether the water above section 1 is ponded, as would be the case if section 1 were the crest of the spillway of a dam. If, at section 2, the total head were computed, assuming normal depth on the steep slope, it would plot (point a on the sketch) above the elevation of the total head at section 1. This is physically impossible, because the total head line must slope downward in the direction of flow. The actual total head line will take the position shown, and have a slope approximately equal to Sc at section 1 and approaching slope S0 farther downstream. The drop in the total head line hf between sections 1 and 2 represents the loss in energy due to Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 8 Revised: 2013 Edition friction. At section 2 the actual depth d2 is greater than dn because sufficient acceleration has not occurred and the assumption of normal depth at this point would clearly be in error. As section 2 is moved downstream so that total head for the normal depth drops below the pool elevation above section 1, the actual depth quickly approaches the normal depth for the steep channel. This type of water surface curve (section 1 to section 2) is characteristically much shorter than the backwater curve discussed in the previous paragraph. Another common type of non-uniform flow is the drawdown curve to critical depth which occurs upstream from section 1 (Figure 2F-2.04) where the water surface passes through the critical depth. The depth gradually increases upstream from critical depth to normal depth, provided the channel remains uniform through a sufficient length. The length of the drawdown curve is much longer than the curve from critical depth to normal depth in a steep channel. Figure 2F-2.04: Water Surface Profile in Changing from Subcritical to Supercritical Channel Slope Source: Design Charts for Open-Channel Flow, FHWA Figure 2F-2.05 shows a special case for a steep channel discharging into a pool. A hydraulic jump makes a dynamic transition from the supercritical flow in a pool. This situation differs from that shown in Figure 2F-2.03 because the flow approaching the pool in Figure 2F-2.05 is supercritical and the total head in the approach channel is large relative to the pool depth. In general, the supercritical flow can be changed to subcritical flow only by passing through a hydraulic jump. The violent turbulence in the jump dissipates energy rapidly, causing a sharp drop in the total head line between the supercritical and subcritical states of flow. A jump will occur whenever the ratio of the depth d1in the approach channel to the depth d2 in the downstream channel reaches a specific value. Note in Figure 2F-2.05 that normal depth in the approach channel persists well beyond the point where the projected pool level would intersect the water surface of the channel at normal depth. Normal depth can be assumed to exist on the steep slope upstream from section 1, which is located about at the toe of the jump. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 9 Revised: 2013 Edition Figure 2F-2.05: Water Surface Profile Illustrating Hydraulic Jump Source: Design Charts for Open-Channel Flow, FHWA F. Hydraulic Jump 1. General: The hydraulic jump consists of an abrupt rise of the water surface in the region of impact between rapid and tranquil flows. Flow depths before (supercritical depth, d1) and after (subcritical depth, d2) the jump are less than and greater than critical depth, respectively. The depth d1 is calculated based on the hydraulics of the channel. The depth d2 is calculated as shown in part 2. The zone of impact of the jump is accompanied by large-scale turbulence, surface waves, and energy dissipation. The hydraulic jump in a channel may occur at locations such as: a. The vicinity of a break in grade where the channel slope decreases from steep to mild. b. A short distance upstream from channel constrictions such as those caused by bridge piers. c. A relatively abrupt converging transition. d. A channel junction where rapid flow occurs in a tributary channel and tranquil flow in the main channel. e. Long channels where high velocities can no longer be sustained on a mild slope. Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 10 Revised: 2013 Edition Figure 2F-2.06: Hydraulic Jump and Depth of Flow 2. Hydraulic Jump Computations: The method for calculating the length of the hydraulic jump and the resulting flow depth and velocity downstream of the jump is discussed in detail in FHWA’s Hydraulic Engineering Circular No. 14 (HEC-14), Hydraulic Design of Energy Dissipators for Culverts and Channels. Due to the complex energy calculations required to analyze hydraulic jumps, the use of appropriate hydraulic design software is encouraged. Table 2F-2.03: Permissible Velocities for Channels with Erodible Linings, Based on Uniform Flow in Continuously Wet, Aged Channels Soil Type or Lining (earth; no vegetation) Maximum Permissible Velocities for… Clear Water (fps) Water Carrying Fine Silts (fps) Water Carrying Sand and Gravel (fps) Fine sand (non-colloidal) 1.5 2.5 1.5 Sandy loam (non-colloidal) 1.7 2.5 2.0 Silt loam (non-colloidal) 2.0 3.0 2.0 Ordinary firm loam 2.5 3.5 2.2 Volcanic ash 2.5 3.5 2.0 Fine gravel 2.5 5.0 3.7 Stiff clay 3.7 5.0 3.0 Graded, loam to cobbles (non-colloidal) 3.7 5.0 5.0 Graded, silt to cobbles (colloidal) 4.0 5.5 5.0 Alluvial silts (non-colloidal) 2.0 3.5 2.0 Alluvial silts (colloidal) 3.7 5.0 3.0 Coarse gravel (non-colloidal) 4.0 6.0 6.5 Cobbles and shingles 5.0 5.5 6.5 Shales and hard pans 6.0 6.0 5.0 Fabric and excelsior mat 7.0 7.0 7.0 Dry rip rap/gabions 10.0 10.0 10.0 Concrete pilot channel Use grass permissible velocity - Table 2F-2.04 Chapter 2 - Stormwater Section 2F-2 - Open Channel Flow 11 Revised: 2013 Edition Table 2F-2.04: Permissible Velocities for Channels Lined with Uniform Stands of Various Grass Covers, Well Maintained1 Cover Slope Range (percent) Permissible Velocity on… Erosion Resistant Soils (fps) Easily Eroded Soils (fps) Bermudagrass 0 to 5 8 6 5 to 10 7 5 Over 10 6 4 Buffalograss Kentucky bluegrass Smooth brome Blue grama 0 to 5 7 5 5 to 10 6 4 Over 10 5 3 Grass mixture 0 to 5 5 4 5 to 10 4 3 Lespedeza sericea Weeping lovegrass Yellow bluestem Kudzu Alfalfa Crabgrass 0 to 5 3.5 2.5 Common lespedeza2 Sudangrass 0 to 53 3.5 2.5 1 Use velocities of 5 fps only where good covers and proper maintenance can be obtained. 2 Annuals, used on mild slopes or as temporarily protection until permanent covers are established. 3 Use on slopes steeper than 5% is not recommended. Source: From Handbook of Channel Design for Soil and Water Conservation G. References U.S. Department of Transportation. Design Charts for Open-Channel Flow. Hydraulic Design Series No. 3. 1961. U.S. Soil Conservation Service. Handbook of Channel Design for Soil and Water Conservation. 1947. |
188407 | https://www.youtube.com/watch?v=YsVsuhdGRJk | Trigonometric Integrals Powers of Tangent and Secant 5 Examples Calculus 2 BC
ProfRobBob
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Examples at 6:21 10:58 13:14 20:34
Example involving Integration by Parts 22:38
Trigonometric Integrals Powers of Sine and Cosine
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Transcript:
Introduction BAM!!! Mr. Tarrou In part 2 of looking at
trigonometric integrals involving powers of secant and tangent we are going to be doing
five examples. Now in the description of this lesson not only will you will find time stamps
to those five examples if you want to skip ahead and you don't need to hear this entire
lesson, you will also find a link to part 1 where I talk about trigonometric integrals
involving powers of sine and cosine. Now the derivative of sine is cosine and the derivative
of cosine is negative sine. So, you right from one trig function to the other regardless
of which trig function you start with sine or cosine. Those problems tended to be fairly
straight forward. Of course in Calculus even stuff that seems straight forward at first
will be a little bit confusion for some of us. When you deal with secant and tangent
the derivative of tangent is secant squared and the derivative of secant is secant of
x times tangent of x. So, that little bit of um, an issue there where the derivative
of secant has in it's derivative secant and tangent, will with some of these problems
we will have a choice in how we start our problem. We are going to be talking about
in the third example, I think it is the third or maybe the second example, where we are
going to do that example two different ways. We are going to get two different answers
that look completely different, but when you graph them you will have the exact same graph
showing you that they are equivalent. Depending on the approach that you want to go through,
you will find that one way is easier than another and the choice that you make is dependent
on a particular problem that you are doing in your homework. So let's get to those five
examples shall we. We we have the guidelines for doing these trigonometric integrals which
you will probably find in your textbook. We are going to try and make sense of these and
then work through and clarify what is going on with our five examples. If the power of
the secant is even and positive save a secant squared factor and convert the remaining factors
to tangent, expand, and integrate. So we have the secant raised to a 2k power of x and the
tangent of x raised to an n power dx. Here we have this exponent of 2k. Any time you
take a number and you double it, of course you get even answers. We are looking at that
even power on our factor which is secant. We are going to pull one of those out so we
see a secant squared raised to the k-1 power and a secant squared over here. Of course
a power to power means that this would be 2k-2. Here we have a secant squared and when
you multiply like bases you add the exponents and 2k-2 plus 2 gets us back to that original
secant with the even power. I am just trying to explain a little bit what is going on this
expansion notation that you see in your textbook. Now we are going to take this secant squared
with the Pythagorean Identity which is of course 1 plus tangent squared of x is equal
to secant squared x, and we are going to do the expansion and have a bunch...it depends
on how you write your problem. I am going to give each of the terms their own integral
notation. What we will eventually see is that we have powers of tangent and then that u
prime or that du here, our derivative of tangent is equal to secant squared of x, and that
will allow us to go through the integral using the Power Rule. If the power of the tangent
is odd and positive, save a secant tangent factor and convert the remaining factors to
secant, expand, and then integrate. We have that notation over here. If there are no secant
factors and the power of tangent is even and positive, convert a tangent squared factor
to a secant squared factor so that you have that du or that u prime, and expand and repeat
if needed. This repetition here, I want to make up my own examples and not just use the
textbook examples or the textbook problems to not have any copyright issues with the
books. And why do something in our textbook already. This guideline about there are no
secant factors and the power of tangent is even and positive, my example is going to
have no secant factors and a power on tangent which is odd. But we are still going to basically
go through this process. We will just have an extra factor of tangent which we will have
to deal with by just using the basic integration rule for tangent. That the integral of tangent
is equal to negative natural log of the absolute value of cosine. If the integral is in the
form of secant raised to some power, secant of x raised to some power of m dx, where m
is odd and positive, use integration by parts. That will be our last example. We will just
do a basic one where we integrate the secant of x cubed. If all else fails, try converting
your integrand into sine and cosine. Kind of like when you were doing trig proofs and
all else failed trying to get the two sides of the identity to match. You would go into
the parent functions for the trig functions which is sine and cosine and see what would
happen there. We are not going to do any examples like this of our five examples. We are not
going to do need to do this last step We are going to save that for part three, our last
lesson helping you work with trig identities or excuse me these trigonometric integrals
that have some kind of powers of sine, cosine, secant, cosecant. What we are going to learn
here in this lesson dealing with powers of secant and tangent of course are going to
apply to integrands that powers of cosecant and cotangent. Because the derivative rules
work very very similar. The derivative of tangent is secant squared. The derivative
of cotangent is negative cosecant squared. Of course we should know what the derivative
of secant and the derivative of cosecant is, and see how similar those derivatives are.
So, let's get on to our first example right now:D So starting with our simplest examples
in the beginning and working through and increasing Example 1 Indefinite Integral the difficulty through our examples. We have
example number one, the indefinite integral of the secant of 4x raised to the power of
6 dx. Now if it was just integrating secant of 4x, we could use our basic integration
rule for secant and write the answer, just about. We have an inside function there of
4x. But, because the secant function is inside the power, then this secant is inside a power
rule...the power of six. We would need a du. We would need to be able to have a derivative
of that inside function. What we want to do, is we want to rewrite that integral to have
a tangent function and a secant function. With this being an even power we can pull
out a secant squared, be left with a secant to the 4th power, and then using one of those
Pythagorean Identities 1 plus tangent squared is equal to secant squared, convert those
remaining four powers of secant into tangents. Then we would have that inside function for
the power would be the tangent and that du or u prime would be the secant squared that
we are going to leave out. So we have the indefinite integral of secant of 4x to the
fourth power times secant of 4x squared dx. I always wrap my angles in a set of parenthesis
just to really make that notation as clear as possible. This is going to be our du, this
is going to be our u prime, this is going to be our derivative of our inside function.
That function inside the power which we are going to covert into tangent. The Pythagorean
Identities of course are not for trig function and powers of three, or four, or five, or
six, or whatever. They are for powers of two. So this is going to be rewritten as the definite
integral of secant squared of 4x then raised to a second power times secant squared of
4x dx. One plus tangent squared is equal to secant squared so we have the indefinite integral
of 1 plus tangent squared of 4x raised to the second power times secant squared of 4x
dx. Now for just the speed of the video and make this as efficient as possible and not
quite so boring for you to watch, I am going to step off. We are going to take this binomial,
raise it to a second power, have a trinomial. We will distribute that secant squared of
4x into each of those three terms. I am going to give each of those terms their own integral
just for maximum clarity of my work. You might find that you don't need to do that as you
practice these problems more and more. Maybe you want to streamline your work a little
bit. But I am going to break this down into those small parts as I can, as small of steps
as I can. Nanananana.... So there we go. Of course we have that binomial expanded. Again
like I said, multiplying that secant of 4x squared dx through each of these three terms.
Setting up a separate integral for each those integrands, and the integral of secant squared
of course is going to be tangent. I am just completing the chain rule here so that we
can recognize that pattern and then just integrate using the power rule. We have got the derivative
of tangent of 4x is going to be 4 times secant squared of 4x. Because of course you have
to take the derivative of that inside function which is going to be four. Then I am balancing
out those multiplications of four with divisions of four in each of those terms. Then finishing
up of course, don't forget your constant of integration and there is our final answer.
For our second example, it is going to follow that third guideline where it said if we had
no factors of secant and we had a tangent just simply raised to an even power, pull
out of two out, convert to secant and so on and repeat if necessary. We are going to do
that except it is going to have an odd power. That is just going to leave us, like I said,
an extra factor of tangent. When we are done we are just going to integrate that tangent
with a basic integration rule. So we have Example 2 Indefinite Integral started example two, the indefinite integral
of tangent of x raised to the second power dx. Now, just to be different from problems
that are in the textbook that I teach from, I got a single angle and I have a power of
seven. That means that whole idea of repeat if necessary is going to take a few repetitions.
More than likely your book probably has something like a tangent of 3x or 5x or some kind of
multiple angle and then it is just raised to the fifth power so you only have to repeat
this process twice. We are going to have to repeat it three times. But, the whole point
of taking a tangent squared factor out of that whatever power of tangent that you have
in your problem, is so that you can take the Pythagorean Identity of course and you get
a secant squared factor into your integrand so you can have a derivative of your inside
function. You have a derivative of that tangent which is secant squared. So far we are here
and we have the indefinite integral of tangent of x to the fifth power times secant squared
of x dx. Well this is our u, tangent of x, and our du is this secant of x squared dx.
So we are going to just use the power rule to integrate this first integral. Then continue
the expansion, again two more times, with this indefinite integral of the tangent of
the fifth power. See we are breaking this down in steps of two. So, I am going to step
off, reveal the rest of this example until we get to the very end and we have to integrate
that last odd power of tangent that we have. Remember that the indefinite integral of tangent
of x is negative natural log of the absolute value of cosine of x plus c. BAM!!! Alrighty.
Example number three. We are going to find Example 3 Indefinite Integral the indefinite integral of the tangent of
3x raised to the fifth power times the secant of 3x raised to the fourth power dx. Now to
remind us of what those guidelines said, it says if the power of the secant is even and
positive...check... save a secant squared factor and convert the remaining factors to
tangent, expand, and then integrate. If the power of the tangent is odd and positive...check...
then save a secant tangent factor and convert the remaining factors to secant, expand, and
integrate. Now, we haven't gone to powers of secant yet. Remembering that the derivative
of secant is secanttangent so we are going to do.... Basically we have a problem. An
integrand that follows both of those guidelines. Now we are going to do this again first going
into powers of secant and finish the power. Then we are going to look at how we could
go the other way, noticing that the secant factor has an even power as well. So we are
going to take this tangent and we are going to pull one of these out because we have noted
that the power of the tangent is odd. So save a secant tangent factor. Let's focus on getting
that extra secant out there in a second and get our factor of tangent. Get it into some
powers of secant. We are going to write this as the indefinite integral of tangent of 3x
times tangent to the...not squared...but fourth power of 3x times secant of 3x raised to the
fourth power dx. Now, really we are going to.... I am going to go to the line I was
about to write anyway and write this tangent squared squared. This will save a line of
work. Let's just do this. Excellent. Now, Pythagorean Identity tangent squared of an
angle is equal to secant squared of that same angle minus one. Now I am going to step off.
We are going to expand that binomial, distribute the tangent and the secant 3x to the fourth
power to each of those terms and come right back. Nanananana We have that binomial expanded,
multiplied the tangent of 3x and the secant of 3x to the fourth power through each of
those three terms. Now I am going to go ahead and show how we are saving that secant tangent
factor. Because we see that we have a secant raised to a high power in each of these three
terms and the derivative of secant is secant tangent. We are going to pull one of these
powers of secant out. I am going to also put the powers first just so that we have the
du at the end. We are going to have the indefinite integral of secant to the 7th power, or secant
of 3x to the seventh power. Now we are going to have that secant tangent, and inside of
our trig function we have a 3x. Well, the derivative of 3x is 3 so I am also going to
for purposes of completing that chain rule write this as 3 times secant of 3x tangent
of 3x dx. To balance that multiplication of 3 for the derivative of that inside function
of 3x, we are going to balance that of course with a coefficient or a factor of 1/3...a
division of three. I am going to do that with each of these three terms and finish up the
problem. Then we will look at this another way, where instead of putting this into powers
of secant we are going to put into powers of tangent pulling out a secant squared or
secant of 3x squared. And, show you why that version of this problem is going to be less
work. BAM! There is our final answer all in powers of secant. Now what we are going to
is, we are going to go and use the other guideline which is not the guideline about the power
on tangent is odd, but if the power of secant is even. We will rework this problem in terms
of power of tangent instead of powers of secant like we did in this particular example. It
is going to be less work, this second was we do the problem. It probably is a little
bit less work to go into powers of tangent just because the derivative of tangent is
secant squared as opposed to having the derivative of secant be secant and tangent. Just being
a little bit less writing. But, the main reason why this second version of the solution is
going to be less work is because we have a power of four which is closer to and easier
to get down into a power of two using that Pythagorean Identity. And the fact that the
derivative of tangent is secant squared. We have a power of two in that derivative as
opposed to the derivative of secant which is secant tangent. Not only do we have two
factors, but we have powers of one as opposed to powers of five, or four, or two. So, let's
break this off and hopefully you can see the difference between the yellow and white chalk.
I think I can fit the whole second version of the solution over here. We are going to
write this as the indefinite integral of tangent to the fifth power, or the tangent of 3x to
the fifth power times the secant of 3x squared times secant of 3x squared dx. Like you saw
do in the first two examples, I am going to step off just to make sure that this as neat
as possible and speed up the video a little bit. We are going to take this secant squared
and write it as one plus tangent squared and basically now finish this similar to examples
one and two. You will see that at least in this example, going into powers of tangent
allows us to more quickly solve this problem. BAM!!! There is a perfect example of how that
question can have two solutions that look so different. Not only in the amount of work
that it takes to complete those problems, but if you put your calculator in radian mode
and type in this equation y equals all of this. You can not worry about the plus c there.
And you type in y is equal to 1/18 tangent of 3x to the sixth power plus 1/24 tangent
of 3x to the eight power. You will get, as different as these equations look, the exact
same graph. Now let's move on to example number four which is going to be much quicker than
example number three. We will only do it once:) Example 4 Indefinite Integral And for example number four we have the indefinite
integral of tangent cubed, tangent of x cubed times secant of x cubed dx. A very similar
problem to the previous example. It is just with this one you don't have very good options,
you just want to pay attention to the fact that the tangent of x has an odd power on
it. So, we are going to save a factor of secant of x tangent of x and put this into powers
of secant. That is all we have to do with that problem. Taking out our factor of secant
tangent. Now we have the tangent of x being squared, using the Pythagorean Theorem to
convert that into secant squared minus one. Now distributing secant squared x secant x
tangent x through those parenthesis giving each of those terms their own integral notation,
and we have our u which is secant of x and du being secant of x tangents of x dx. Now
our last example is just going to fourth guideline which said that if secant has an odd power
to integrate using integration by parts. We are just going to do a simple indefinite integral
of secant of x cubed dx. It's our last example. Example 5 Indefinite Integral Example number 5. We have the indefinite integral
of secant of x cubed dx. Now this is a pretty innocent example. It looks very similar to
the one that we did earlier that was just having a factor of tangent being raised to
an odd power. You saw that really worked pretty easily, except for the fact it had a seventh
power and we had to do a couple of repetitions of the same type of process. But, it was not
a big deal. Here it is going to be a little bit more of a big deal. If you try and write
this as secant squared x times secant of x, and then take your secant squared and write
it as one plus tangent squared x, you are going to kind of just go in a circle and get
right back to where you started from. We are going to have a little back of that issue
well going through the integration by parts. You will see something come up that would
have shown up in that lesson when you were first learning integration by parts. Sometimes
you had to go through it multiple times and end up through the process of going through
integration by parts, getting one of your terms that matches the original problem..and
you are going to see that happen here. Our integration by parts is going to be the indefinite
integral of u dv is equal to.... Except that is not a u, or that I wrote a u and not a
v. Let's try that again, the indefinite integral of u dv is equal to uv minus the indefinite
integral of v du. Ok. So we want to, with integration by parts, look for a factor...the
most complicated factor that can be integrated using a basic integration rule. Well, right
now we are just looking at indefinite integral of secant of x cubed. So we are going to split
this into two parts like mentioned just a second ago and say that this is equal to the
indefinite integral of secant of x times secant squared x dx. This is the most complicated
factor that can be integrated using a basic integration rule. We are going to let dv equal
secant squared x dx. That means that v is the indefinite integral of secant squared
x dx. Of course that is going to be tangent. Now we got our v, we have our dv, we need
to know what u is. U is just whatever factor we have not used yet in our integrand which
is going to be secant. So u is equal to the secant of x and du is going to be the derivative
of secant is secant tangent of x dx. So we now we are just going to plug all of this
into our integration by parts formula. We have u and v, so we have secant of x tangent
of x. We can write that over here. This minus the indefinite integral of v which we have
here is tangent of x, du and du is secant of x tangent of x dx. Ok, now fantastic. Here
well... We have... We have a lot going on here. If we look at secant of x as our u,
then du would be secant of x tangent of x. We don't have another secant factor in here.
The derivative of tangent is secant squared. It seems like we are not going be able to
easily work this integrand, or integrate this expression. We are going to put the tangents
together and write that we have secant of x tangent of x minus... Ok... But whatever.
The same issue that we had here. We don't have a good u and du in here so we are going
to go ahead and convert tangent squared x into secant squared of x minus one. Distribute
that secant of x. Ok, now I am giving each of these terms their own integral notation.
Secant of x times secant of x squared of course secant cubed x. You just saw me do this in
my previous example. We have that negative there. Now I am going to do secant of x times
negative one which is negative secant of x dx, but then that negative is coming over
here so we have plus the indefinite integral of secant of x dx. Now, at this point see
we started off trying to integrate the indefinite integral of secant x cubed. We have another
secant x cubed. We had this happen when we first went through the process of learning
how to work with integration by parts. Sometimes we circled back to our original integrand.
I am going to go ahead here and the indefinite integral of secant of x is the natural log
of the absolute value of the secant of x plus tangent of x, get that plus c in there. We
still have an indefinite integral in here, but what I am going to do is add that to the
other side. Let's go ahead and do that at the same time. The indefinite integral of
secant of x is the natural log of the absolute values of secant of x plus tangent of x. I
am going to add both sides of this equation by the indefinite integral of the secant of
x cubed. So, one plus one is going to be two. Stretch out my spacing there a little bit.
Now I am going to divide both sides by two and get the indefinite integral of secant
cubed x dx is equal to 1/2 secant of x tangent of x plus 1/2 the natural log of absolute
value of the secant of x plus tangent of x plus c. You know c, and unknown constant divided
by two is just giving another unknown constant. Subscript of 1 or just leave it as another
unknown constant of integration. Let me double check and make sure that I did not do something
silly. We have got...there we go... 1/2 secant tangent plus 1/2 natural log. Excellent. That
is the end of my last example and the end of this particular lesson. I am Mr. Tarrou.
BAM!!! Go Do Your Homework:D Or, if you want to see even more examples I will have a part
three coming up and I will put a link in the description just as soon as it is done. |
188408 | https://www.reddit.com/r/learnmath/comments/1ci8n0g/derive_alternating_arithmetic_series_formula/ | Derive Alternating Arithmetic Series Formula : r/learnmath
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Derive Alternating Arithmetic Series Formula
Looking for a nice way to show that sum_(k=1)^n (-1)^(k + 1) k = 1/4 (-2 (-1)^n n + (-1)^(n + 1) + 1). Not a proof by induction, but rather a derivation.
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188409 | https://www.ncbi.nlm.nih.gov/books/NBK482230/ | Lisinopril - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Lisinopril
Edgardo Olvera Lopez; Mayur Parmar; Venkata Satish Pendela; Jamie M. Terrell.
Author Information and Affiliations
Authors
Edgardo Olvera Lopez 1; Mayur Parmar 2; Venkata Satish Pendela 3; Jamie M. Terrell 4.
Affiliations
1 Haysmed, The University of Kansas Health System
2 Nova Southeastern University
3 Rochester General Hospital
4 University of LA Monroe
Last Update: October 5, 2024.
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Continuing Education Activity
Lisinopril is an angiotensin-converting enzyme (ACE) inhibitor that has been prescribed for nearly 30 years to manage hypertension and reduce cardiovascular strain. As a competitive ACE inhibitor, lisinopril prevents the conversion of angiotensin I to angiotensin II, a potent vasoconstrictor. This activity reviews lisinopril's mechanism of action, highlighting its distinct pharmacokinetic parameters and half-life compared to other ACE inhibitors. The discussion also includes lisinopril's FDA-approved indications, off-label uses, contraindications, drug-drug interactions, and potential toxicity. Additionally, dosage recommendations, warnings, and necessary patient monitoring protocols are outlined to support safe administration. This activity also emphasizes the critical role of an interprofessional healthcare team in managing hypertension with lisinopril.
Objectives:
Determine the mechanism of action of lisinopril in the context of hypertension management.
Identify the various indications for prescribing lisinopril therapy.
Assess potential adverse effects associated with lisinopril and strategies for monitoring and managing these effects.
Develop collaboration and communication among interprofessional team members to improve the outcomes and treatment efficacy for patients receiving lisinopril therapy.
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Indications
Lisinopril is classified as an angiotensin-converting enzyme inhibitor and has been available for nearly 3 decades. The following are indications approved by the United States Food and Drug Administration (FDA) and conditions that lisinopril may effectively treat but are not FDA-approved.
FDA-Approved Indications
Hypertension
Heart failure (adjunctive therapy)
ST-segment elevation myocardial infarction (STEMI) (if given within 24 hours)
Off-Label Uses
Diabetic nephropathy
Proteinuria, particularly in patients with immunoglobulin A (IgA) nephropathy
Post-transplant erythrocytosis
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Mechanism of Action
Lisinopril is a competitive inhibitor of the angiotensin-converting enzyme (ACE) and prevents the conversion of angiotensin I to angiotensin II, a potent vasoconstrictor. A reduction in angiotensin II levels subsequently suppresses aldosterone secretion, which reduces sodium reabsorption in the collecting duct and potassium excretion. This process may result in a slight increase in serum potassium. By inhibiting the negative feedback of angiotensin II, lisinopril increases serum renin activity.The beneficial effects in patients with hypertension derive from inhibiting the renin-angiotensin-aldosterone system, resulting in reduced vasopressor and aldosterone activity even in patients with low renin levels. However, ACE also degrades bradykinin. Therefore, ACE inhibitors may increase the risk of angioedema.
Pharmacokinetics
Absorption:Lisinopril absorption is unchanged by food. After oral intake, it has low bioavailability, ranging from 10% to 30%. The time to peak concentration can vary from 6 to 8 hours.
Distribution:Lisinopril does not bind to albumin or other proteins, and its distribution in patients with heart failure is poor.
Metabolism: Unlike other ACE inhibitors (eg, enalapril, captopril), lisinopril has a long half-life, is hydrophilic, and is not broken down by the liver.
Elimination:Lisinopril is excreted unchanged in the urine.
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Administration
Available Dosages and Strengths
Lisinopril is available as 2.5 mg, 5mg, 10 mg, 20 mg, 30 mg, and 40 mg oral tablets and as a 1 mg/mL oral solution.
Adult Dosage
The standard adult dosage ranges from 2.5 to 40 mg daily, depending on the indication.Dosing and administration adjustments are recommended for patients with various conditions,such as kidney disease.
Hypertension
According to the 2017 guidelines, the American College of Cardiology/American Heart Association (ACC/AHA) recommends ACE inhibitors as first-line agents for managing hypertension.
The recommended initial dose is 10 mg daily, which is increased to 40 mg daily.
If adequate blood pressure control is not achieved with lisinopril alone, a low-dose diuretic can be added. In these cases, the lisinopril dose can be reduced. The recommended starting dose for adults with hypertension receiving diuretics is 5 mg once daily.
Heart failure
According to the 2013 ACCF/AHA Guidelines for the Management of Heart Failure(HF), ACE inhibitors are recommended for all patients with HF with reduced ejection fraction to minimize morbidity and mortality.
The recommended initial dose is 2.5 mg daily, with a maximum daily dose of 40 mg.
During the SOLVD trial,patients in the high-dose group had an overall 8%reduction in mortality compared to the low-dose group. The findings of this study have resulted in an inclination towards higher doses.
ST-Elevation Myocardial Infarction
The 2013 ACCF/AHA guidelines strongly recommend administering lisinopril within the first 24 hours for all hemodynamically stable patients with anterior ST-elevation myocardial infarction, HF, or ejection fraction ≤40%, unless contraindicated.
The recommended initial dose is 2.5 to 5 mg daily, with a slow titration to 40 mg daily, or the maximum tolerated dose.
Diabetes and hypertension
The American Diabetes Association (ADA) recommends ACE inhibitors as first-line agents for hypertension in patients with diabetes and a urinary albumin-to-creatinine ratio ≥30 mg/g creatinine.
The recommended initial dose is 2.5 to 10 mg daily, depending on blood pressure, and slowly titrated to a maximum daily dose of 40 mg. The target proteinuria is less than 1 g/day, as per KDIGO-2013.
Specific Patient Populations
Hepatic impairment:No dose adjustments are necessary for patients with hepatic impairment.
Renal impairment:The manufacturer recommends the following dose adjustments based on creatinine clearance (CrCl):
No dose adjustment is required in patients with a CrCl >30 mL/min.
For patients with a CrCl of 10 to 30 mL/min, the recommended initial dose of lisinopril should be reduced by 50%.
For patients with a CrCl <10 mL/min, the recommended initial dose is 2.5 mg once daily. This is also the recommended initial dose for patients on dialysis.
Pregnancy considerations: Lisinopril is pregnancy category class D due to its teratogenic effects (eg, impaired fetal renal function, oligohydramnios, lung hypoplasia, skeletal malformations, fetal/neonatal death). Thus, its use is contraindicated for women who are pregnant or may become pregnant and are not receiving appropriate contraception.
Breastfeeding considerations:Manufacturers recommend against using lisinopril in breastfeeding women because the amount secreted in breast milk and its effects in the breastfed infant is unknown.
Pediatric patients:For children 6 years and older, the initial dose is 0.07 to 0.1 mg/kg once daily, with a maximum initial dose of 5 mg daily.This dose may be increased every 1 to 2 weeks to a maximum tolerated dose of 0.6 mg/kg/day or 40 mg/day.
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Adverse Effects
The primary adverse reactions associated with ACE inhibitors include hyperkalemia, dry cough, angioedema, hypotension, dizziness, headache, and renal insufficiency. These effects may be more common in patients with renal, autoimmune, or collagen vascular diseases.The AHA/ACCF recommends caution when prescribing lisinopril for patients with cardiomyopathy with outflow obstruction, as the drug may exacerbate symptoms.Historically, ACE inhibitors have been associated with increased morbidity and mortality in patients with aortic stenosis. However, recent studies suggest that ACE inhibitors may be safer than originally thought and even provide some benefits in certain patients.
ACE inhibitor-induced cough is a dry, nonproductive, hacking cough that usually begins within months of initiating therapy and resolves within 1 to 4 weeks after discontinuing. Deteriorating renal function can occur in patients whose glomerular function depends on event arteriolar vasoconstriction by angiotensin II. A benign increase in serum creatinine may occur at the beginning of therapy, but medication should only be discontinued if there is a progressive or significant elevation of BUN/creatinine.
Angioedema is asymmetric swelling of subcutaneous tissue without itching or urticaria involving the face, mouth, and upper airway. ACE inhibitor-induced angioedema can occur anytime during therapy but most commonly occurs within the first 3 months of treatment. This adverse reaction is secondary to elevated bradykinin levels by inhibiting ACE, causing vasodilatation and extravasation of plasma into the submucosal tissue, leading to angioedema. The most crucial step in management is to discontinue the ACE inhibitors and note ACE inhibitors under the patient's allergies. Immediate symptomatic treatment and airway protection may be necessary.
Drug Interactions
Diuretics: When initiating lisinopril therapy, diuretics may further reduce blood pressure and lead to hypotension. When coadministered with thiazide-type diuretics or potassium-sparing diuretics (eg, amiloride, spironolactone, triamterene),lisinopril may increase the risk of hyperkalemia. Therefore, if coadministration of these diuretics is necessary, the patient’s serum potassium should be monitored frequently.
Antidiabetics: Concomitant administration of antidiabetic medicines (eg, insulins, oral hypoglycemic agents) and lisinopril can increase the risk of hypoglycemia.
NSAIDs:Non-steroidal anti-inflammatory agents can reduce the antihypertensive effect of ACE inhibitors. However, coadministration of NSAIDs and lisinopril in patients who are older, volume-depleted (including those on diuretic therapy), or renally impaired may worsen renal function and possibly cause acute renal injury (AKI). This is usually reversible, but renal function should be monitored periodically in patients receiving lisinopril and NSAID therapy.
Renin-angiotensin system blockers: Dual blockade of the renin-angiotensin with ACE inhibitors, angiotensin receptor blockers, or aliskiren is associated with higher risks of hyperkalemia, hypotension, and changes in renal function, including AKI, compared to monotherapy. Clinicians should closely monitor blood pressure, electrolytes, and renal function in these patients.
Lithium: Lithium toxicity has been reported in patients taking ACE inhibitors and lithium and is usually reversible. Clinicians should monitor serum lithium levels in these patients.
Rapamycin inhibitors: Patients using concomitant mechanistic targets of rapamycin inhibitors (eg, sirolimus, temsirolimus, everolimus) are at increased risk for angioedema.
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Contraindications
Lisinopril is contraindicated for patients with hyperkalemia, a history of angioedema, renal failure with prior lisinopril use, bilateral renal artery stenosis, concomitant use with aliskiren in patients with diabetes mellitus, and for patients receiving a neprilysin inhibitor or within 36 hours of taking one.
Box Warning
When pregnancy is detected, discontinue lisinopril as soon as possible.
Drugs that act directly on the renin-angiotensin system can cause injury and death to the developing fetus.
Warnings/Precautions
Fetal toxicity
Angioedema (concurrent mTOR inhibitor (eg, temsirolimus, sirolimus, everolimus) or neprilysin inhibitor administration)
Impaired renal function
Hypotension
Hyperkalemia
Hepatic failure
Risk of allergic reactions due to tartrazine in lisinopril doses of 20 mg, 30 mg, and 40 mg
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Monitoring
Caution is necessary when prescribing lisinopril for patients with high-potassium diets or who are taking other agents that might exacerbate hypotension and hyperkalemia, such as antihypertensive agents or aldosterone antagonists. Although cholestatic jaundice is a rare reaction associated with using the ACE inhibitor captopril, monitoring liver function during lisinopril use may be appropriate. Lisinopril should be discontinued immediately if elevated liver enzymes are detected.First-dose hypotension is an uncommon adverse effect of ACE inhibitors that clinicians should consider when prescribing lisinopril; a low starting dose is recommended to reduce the risk of this phenomenon.Clinicians should monitor serum potassium, blood pressure, and blood urea nitrogen/serum creatinine in patients taking lisinopril for up to 3 weeks after initiation.
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Toxicity
As lisinopril metabolism depends on renal excretion, overdose management consists of general supportive care. However, gastric emptying strategies, intravenous fluids, vasopressors, and hemodialysis may also be considered when appropriate. Maintaining optimal blood pressure using fluids is critical for hypotensive patients.Some reports suggest using angiotensin II administration as an alternative supportive treatment for the treatment of ACE inhibitor overdose.There is no antidote available for lisinopril.
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Enhancing Healthcare Team Outcomes
Lisinopril has been available for 3 decades and is a relatively safe medication for hypertension. Primary care clinicians, emergency department physicians, internists, and cardiologists often prescribe it. However, the drug requires monitoring. Potassium levels and renal function need periodic monitoring, which clinicians and nursing staff can oversee. Patients should understand how to avoid high-potassium diets, an area where clinicians, nurses, and pharmacists can provide counsel. Women need to be aware of the potential adverse effects of pregnancy while taking lisinopril, making interprofessional counsel from clinicians, nurses, and pharmacists vital. Pharmacists should also check for possible interactions and answer any patient questions. Even though lisinopril is a common and well-tolerated drug, an interprofessional team must optimize safety and therapeutic outcomes while minimizing or preventing adverse events.
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Review Questions
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Disclosure:Edgardo Olvera Lopez declares no relevant financial relationships with ineligible companies.
Disclosure:Mayur Parmar declares no relevant financial relationships with ineligible companies.
Disclosure:Venkata Satish Pendela declares no relevant financial relationships with ineligible companies.
Disclosure:Jamie Terrell declares no relevant financial relationships with ineligible companies.
Continuing Education Activity
Indications
Mechanism of Action
Administration
Adverse Effects
Contraindications
Monitoring
Toxicity
Enhancing Healthcare Team Outcomes
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK482230 PMID: 29489196
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Continuing Education Activity
Indications
Mechanism of Action
Administration
Adverse Effects
Contraindications
Monitoring
Toxicity
Enhancing Healthcare Team Outcomes
Review Questions
References
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Olvera Lopez E, Parmar M, Pendela VS, et al. Lisinopril. [Updated 2024 Oct 5]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from:
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Research Article
xml version="1.0" encoding="UTF-8"?Genetic and Epigenetic Alterations of the NF2 Gene in Sporadic Vestibular Schwannomas
Contributed equally to this work with:
Jong Dae Lee,
Tae Jun Kwon
Affiliation
Department of Otorhinolaryngology-Head and Neck Surgery, Soonchunhyang University College of Medicine, Bucheon, Korea
Contributed equally to this work with:
Jong Dae Lee,
Tae Jun Kwon
Affiliation
Department of Biology, College of Natural Sciences, Kyungpook National University, Daegu, Korea
E-mail: kimuk@knu.ac.kr
Affiliation
Department of Biology, College of Natural Sciences, Kyungpook National University, Daegu, Korea
Affiliation
Department of Otorhinolaryngology, Yonsei University College of Medicine, Seoul, Korea
xml version="1.0" encoding="UTF-8"?Genetic and Epigenetic Alterations of the NF2 Gene in Sporadic Vestibular Schwannomas
Figures
Abstract
Background
Mutations in the neurofibromatosis type 2 (NF2) tumor-suppressor gene have been identified in not only NF2-related tumors but also sporadic vestibular schwannomas (VS). This study investigated the genetic and epigenetic alterations in tumors and blood from 30 Korean patients with sporadic VS and correlated these alterations with tumor behavior.
Methodology/Principal Findings
NF2 gene mutations were detected using PCR and direct DNA sequencing and three highly polymorphic microsatellite DNA markers were used to assess the loss of heterozygosity (LOH) from chromosome 22. Aberrant hypermethylation of the CpG island of the NF2 gene was also analyzed. The tumor size, the clinical growth index, and the proliferative activity assessed using the Ki-67 labeling index were evaluated. We found 18 mutations in 16 cases of 30 schwannomas (53%). The mutations included eight frameshift mutations, seven nonsense mutations, one in-frame deletion, one splicing donor site, and one missense mutation. Nine patients (30%) showed allelic loss. No patient had aberrant hypermethylation of the NF2 gene and correlation between NF2 genetic alterations and tumor behavior was not observed in this study.
Conclusions/Significance
The molecular genetic changes in sporadic VS identified here included mutations and allelic loss, but no aberrant hypermethylation of the NF2 gene was detected. In addition, no clear genotype/phenotype correlation was identified. Therefore, it is likely that other factors contribute to tumor formation and growth.
Citation: Lee JD, Kwon TJ, Kim U-K, Lee W-S (2012) Genetic and Epigenetic Alterations of the NF2 Gene in Sporadic Vestibular Schwannomas. PLoS ONE 7(1):
e30418.
Editor: Sumitra Deb, Virginia Commonwealth University, United States of America
Received: August 29, 2011; Accepted: December 15, 2011; Published: January 25, 2012
Copyright: © 2012 Lee 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.
Funding: This study was supported by a grant from the Korea Healthcare technology R&D Project, Ministry of Health and Welfare, Republic of Korea (A100367). The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
Competing interests: The authors have declared that no competing interests exist.
Introduction
Vestibular schwannomas (VS) are benign tumors of the neural sheath that originate on the vestibular nerves. They occur as either sporadic unilateral tumors or bilateral tumors that are a manifestation of neurofibromatosis type 2 (NF2). The loss of both wild-type copies of the NF2 tumor suppressor gene is common in the pathogenesis of both sporadic and NF2-related schwannomas , . Although mutations in the NF2 gene have been reported in 53∼70% of sporadic VS (Neff et al., 2006), genetic alterations alone cannot explain the gene inactivation.
Recently, the importance of DNA methylation in tumors has been acknowledged, and the hypermethylation of promoter-associated CpG islands is emerging as the primary mechanism of epigenetic inactivation of tumor-suppressor genes in cancer development , . Therefore, hypermethylation in the regulatory region of the NF2 gene has been suggested as a possible mechanism of gene inactivation .
Attempts have been made to correlate the clinical expression of the sporadic VS tumors with specific NF2 mutations , , . Given the heterogeneity of the clinical response to various types of mutations, no clear genotype/phenotype correlation has been established . However, a recent study reported a significant relationship between tumor behavior and genetic alteration in Chinese patients .
The present study analyzed the genetic alterations in sporadic VS, including mutations, loss of heterozygosity (LOH), and epigenetic alterations of the NF2 gene, and correlated these alterations with tumor behavior in Korean patients.
Materials and Methods
Patients
A total of 30 unrelated subjects with sporadic unilateral VS were recruited from the Department of Otorhinolaryngology - Head and Neck Surgery, Soonchunhyang University Bucheon Hospital and the Department of Otorhinolaryngology, Yonsei University Health System Hospital, Korea. The subjects included 10 males and 20 females ranging in age from 11 to 73 years (mean age, 48.3 years). Blood was collected from all patients, and tumor tissue was obtained at the time of surgery and stored at −80°C before DNA isolation. DNA was isolated from both peripheral blood leukocytes and tumor tissues. Peripheral blood leukocyte DNA was used to investigate NF2 germ-line mutations and tumor tissue DNA was examined for somatic mutations of NF2. Written informed consent was obtained from all patients, and the study was approved by the ethics committee of Soonchunhyang University Hospital.
Tumor behavior
Tumor behavior was assessed using tumor size (maximum diameter in centimeters), a clinical growth index, and proliferative activity according to the Ki-67 labeling index (LI). The clinical growth index was calculated by dividing the tumor size (maximum diameter) by the duration of the initial presenting symptom in years. To determine the Ki-67 LI, immunohistochemistry was performed on formalin-fixed paraffin sections of the VS using an anti-MIB-1 antibody followed by microscopy (Olympus, Tokyo, Japan); all nuclear-stained tumor cells were counted at 400× magnification. The mean LI was determined as the number of positively stained nuclei in three randomly chosen fields divided by the total number of cells in the selected areas. The results are expressed as the percentage of the total cell number.
Mutational analysis
For the 30 subjects, genomic DNA was isolated from peripheral blood using a FlexiGene DNA extraction kit (QIAGEN, Hilden, Germany) and from the tumor tissue using a Puregene Buccal Cell Core Kit B (QIAGEN). For the blood and tumor DNA samples, the entire coding region (exon 1–16) and exon-intron boundaries of the NF2 gene were amplified by polymerase chain reaction (PCR) using appropriate primer sets. PCR was performed in 25-µl reactions containing 0.2 mM of each dNTP, 10 µM forward and reverse primers, 10× Taq buffer, 0.25 U of Taq DNA polymerase (Solgent, Daejeon, Korea), and 25 ng of genomic DNA from the blood or tumor, respectively. The PCR conditions consisted of an initial denaturation at 95°C for 15 min, followed by 30 cycles of 95°C for 20 s, 57°C for 40 s, and 72°C for 1 min, with a final 5-min extension at 72°C. The PCR products were confirmed by electrophoresis in 2% agarose gels stained with ethidium bromide (EtBr). For each PCR product, 7 µl were treated with 0.1 U of shrimp alkaline phosphatase (USB, Cleveland, OH, USA) and 1 U of exonuclease I (USB) at 37°C for 1 h, followed by incubation at 80°C for 15 min for enzyme inactivation. The purified PCR products were sequenced in 10-µl reactions containing 5 µM primer, 0.2 µl of ABI Big Dye Terminator v3.1 Cycle Sequencing Kit reagent (Applied Biosystems, Foster City, CA, USA), and 1 µl of 5× buffer. The PCR conditions consisted of 2 min at 95°C, followed by 30 cycles of 95°C for 20 s, 55°C for 20 s, and 60°C for 4 min. The sequencing reaction products were ethanol precipitated, and the pellets were resuspended in 10 µl of HiDi-formamide loading dye. An ABI 3130XL DNA sequencer was used to resolve the products, and the data were analyzed using ABI Sequencing Analysis (v5.0) and SeqScape software (v2.5). The resultant sequence was compared with the NF2 sequence in GenBank (GenBank ID NM_016418).
Loss of heterozygosity assay
Loss of heterozygosity (LOH) was determined using flanking and intragenic microsatellite markers in the NF2 region of chromosome 22. Three microsatellite markers were chosen for the NF2 gene region from the NCBI database (www.ncbi.nlm.nih.gov): D22S275 is centromeric, D22S929 is intragenic within intron 1 of the NF2 gene, and D22S268 is telomeric to the NF2 gene. These three microsatellite markers are tightly linked to the NF2 gene (mean distance between D22S275 and D22S268 is 0.7 Mb). PCR was performed to genotype the samples, and fluorescently labeled (FAM) markers were used. Each 10-µl reaction contained 0.2 mM of each dNTP, 10 µM labeled forward and reverse primers, 10× Taq buffer, 0.25 U Taq DNA polymerase (Solgent), and 25 ng of blood genomic DNA or tumor DNA. The PCR conditions consisted of an initial 2 min at 95°C, 35 cycles of 95°C for 20 s, 57°C for 40 s, and 72°C for 30 s, and a final extension at 72°C for 5 min. Subsequently, each PCR product was diluted with distilled water. HiDi-formamide and the GeneScan™ 500 LIZ® size standard (Applied Biosystems) were added to the diluted PCR products, and the samples were loaded (30 cm capillary, POP 7 polymer) in an ABI 3130XL genetic analyzer. The genotypes of the microsatellite markers were analyzed using GeneMapper software (v3.7).
Methylation-specific PCR (MSP)
Abnormal DNA methylation, i.e., hypermethylation, in the promoter region, including the CpG island, of the NF2 gene was examined using bisulfate modification of tumor tissue genomic DNA from all the patients. The bisulfate modification process was performed using an EZ-96 DNA methylation kit (Zymo Research, Orange, CA, USA). Bisulfate converts unmethylated cytosines in DNA into uracil, while methylated cytosines remain unchanged; uracil is then converted into thymidine in the subsequent PCR amplification. To test for methylation, methylation-specific PCR (MSP) was performed using a methylation-specific primer pair (forward 5′-GAGTTATTTTAAAGGAGGCGGGAC-3′ and reverse 5′-GAAACCCCTAAACGACAAC-GAC-3′; 304-bp product), and unmethylation-specific PCR was performed using an unmethylation-specific primer pair (forward 5′-AGTTATTTTAAAGGAGGTGGGATGG-3′ and reverse 5′-AAACAAAACCCCTAAACAACAA-3′; 307-bp product). The primers were designed using the MethPrimer program available at . Normal genomic DNA was used as the negative control in the MSP, and normal genomic DNA that was methylated using CpG methylase M.Sss1 (New England Biolabs, Beverly, MA) was used as the positive control. The PCR conditions consisted of an initial 5 min at 95°C, 35 cycles of 95°C for 20 s, 57°C for 40 s, and 72°C for 1 min, and a final extension at 72°C for 5 min. The PCR products were confirmed by 2% agarose gel electrophoresis and EtBr staining.
Statistical analysis
Statistical analysis was performed using the SPSS 18.0 statistics software for Windows (SPSS, Chicago, IL). Student's t-test was used to compare the tumor size, clinical growth index, and Ki-67 LI between the patients with and without mutations. P-values less than 0.05 were considered significant.
Results
Mutational analysis of the NF2 gene
Mutational analysis of the entire coding region, i.e., exons 1 to 16, and the flanking intronic sequences of the NF2 gene was performed using direct sequencing using genomic DNA from the blood and tumors of 30 patients with VS. No germ-line mutations were found in blood DNA, whereas 18 mutations were found in tissue DNA. These mutations included frameshift, nonsense, in-frame, missense, and splice site mutations. Nine different mutations (five single-nucleotide deletions or insertions, two multiple-nucleotide deletions in the coding region, and two several nucleotide deletions in the coding region and flanking intron sequences) were identified. Six of the eight frameshift mutations resulted in the premature termination of translation and the other two frameshift mutations generated a truncated protein via exon skipping caused by the loss of a splice acceptor or donor site. One 78 bp deletion mutation led to an in-frame deletion. In addition, seven nonsense mutations were identified that caused the generation of a prematurely terminated protein due to an early stop codon. Among the patients, two (cases 4 and 24) had two different mutations each. One patient had a missense mutation of a thymine to guanine transition at nucleotide 136 in exon 2, causing an amino acid change from leucine to arginine. One patient had a splice site mutation at a donor site, and the splice site mutation led to a substitution of guanine for adenine (GT to AT) at the donor site in intron 3, thus causing an in-frame mutation by skipping exon 3. Consequently, all of the mutations in patients with VS were somatic mutations of the NF2 gene.
Loss of heterozygosity
Genomic DNA (blood and tumor DNA) from all 30 patients was investigated for LOH using three highly polymorphic microsatellite markers near the NF2 gene region: two flanking markers (D22S268 and D22S275) and one intragenic marker within intron 1 (D22S929). Comparative analysis of the LOH of genomic DNA from blood and tumor samples showed heterozygous blood DNA and homozygous tumor DNA (Fig. 1). Of the 30 patients with VS, 9 (30%) had LOH, and 8 of the 9 showed LOH with the flanking marker D22S268 and the intragenic marker D22S929. Every patient with LOH had the D22S275 flanking marker. Thus, 21 of the 30 patients (70%) did not show LOH in their tumor DNA (Table 1).
A comparison of lymphocyte DNA and tumor DNA using the flanking, D22S268 and D22S275, and intragenic, D22S929, microsatellite markers in the NF2 gene region. In case, three markers showed loss of heterozygosity by allele loss in the tumor DNA.
A comparison of lymphocyte DNA and tumor DNA using the flanking, D22S268 and D22S275, and intragenic, D22S929, microsatellite markers in the NF2 gene region. In case, three markers showed loss of heterozygosity by allele loss in the tumor DNA.
CpG island hypermethylation of the NF2 gene
Methylation and unmethylation primers were designed for the region approximately 350 bp to 650 bp upstream of the start codon, which includes the CpG island, using the program MethPrimer (Li and Dahiya, 2002). The methylation state of the promoter region was examined using MSP and unmethylation-specific PCR (USP). The promoter region of the bisulfite-treated DNA of all patients and the negative control (NC) was amplified using USP, and the PCR products were identified using electrophoresis. The bisulfite-treated DNA positive control (PC) was amplified using only MSP. The results demonstrated that the expression of the NF2 gene in all of the VS patients was not repressed by hypermethylation of the promoter region.
Genetic alterations and tumor behavior
The average tumor sizes in the patients with mutations or LOH and patients without mutations and LOH were 16.11 and 20.83 cm, respectively. The average clinical growth index (CGI) values were 13.22 and 12.92, and the average Ki-67 labeling index (LI) values were 2.21 and 2.54, respectively. There was no significant difference in tumor size (p = 0.105), CGI (p = 0.878), or Ki-67 LI (p = 0.372) between the patients with mutations or LOH and the patients without mutations and LOH. Thus, there was no correlation between the nature of the NF2 gene mutation, including LOH, and tumor behavior.
Discussion
This study investigated the genetic alterations in sporadic VS, including mutations, loss of heterozygosity (LOH), and epigenetic alterations of the NF2 gene, and compared these alterations with the tumor behavior. The NF2 gene is frequently mutated in NF2-related vestibular schwannomas , , and mutations in this gene have also been found in sporadic unilateral schwannomas , , . The rate of mutations is higher in unilateral schwannomas than in familial schwannomas. We found 18 genetic alterations (60%) in 30 Korean patients with sporadic VS, which was in agreement with the 53–70% detection rate reported in previous studies .
The majority of the mutations (88%) in our study were protein-truncating mutations such as frameshift or nonsense mutations. Among constitutional NF2 mutations, nonsense mutations are significantly more common than frameshift mutations in NF2 patients, whereas frameshift mutations are significantly more common than nonsense mutations among somatic NF2 mutations in sporadic vestibular schwannomas . We also found that frameshift mutations (53%) were more common than nonsense mutations (35%) . Donor splice site mutations and missense mutations were rare, which was consistent with other studies , . According to Knudson's two-hit hypothesis , functional inactivation of a specific tumor-suppressor gene requires two separate genetic hits in the same gene in the same cell. For a sporadic tumor, the probability of both mutations occurring in the same cell is extremely small . Two patients (cases 4 and 24) each had two different mutations that inactivated the NF2 gene. It is possible that LOH represents the second hit on the remaining allele. Nine patients (30%) in our study had chromosome 22 LOH, and seven patients had both LOH and mutations. In our study, nine patients (30%) had two hits, and another nine patients (30%) had only one hit (mutation or LOH). It has been shown that mitotic recombination plays a role in the occurrence of LOH, even though the rate of mitotic recombination is relatively low in sporadic VS compared to NF2-related schwannomas . In the other cases, we postulate that other mechanisms, such as different genetic abnormalities or epigenetic factors, were involved in the pathogenesis of the vestibular schwannomas.
To search for other genes associated with vestibular schwannoma tumorigenesis, Welling et al. used cDNA microarray analysis to evaluate gene expression profiles. They identified a number of dysregulated genes involved in cell signaling, cell division, and angiogenesis. Caye-Thomasen et al. examined the gene expression profile in vestibular schwannomas using a microarray chip and identified 78 dysregulated genes involved in the cell cycle, morphogenesis, development, adhesion, differentiation, death, extracellular matrix, and protein binding.
Methylation of the promoter-associated CpG island is a possible mechanism of tumor-suppressor gene inactivation in a variety of human tumors . Lomas et al. proposed that aberrant NF2 hypermethylation may participate in the development of a significant proportion of sporadic meningiomas. However, Kullar et al. recently reported that hypermethylation of the CpG island of the NF2 gene was rare in sporadic vestibular schwannomas. None of our patients showed repressed expression of the NF2 gene due to hypermethylation of the promoter region. Instead, methylation of other tumor-related genes may have played a role in the development of these vestibular schwannomas .
Several studies have examined whether genotype predicts the disease severity . Irving et al. found no correlation between the nature of the NF2 gene mutation and tumor behavior. Conversely, Bian et al. reported a significant relationship between tumor behavior and genetic alterations in Chinese patients. Consequently, we investigated whether there was a racial difference in the genotype-phenotype correlation in sporadic vestibular schwannomas. To quantify tumor behavior, the previously mentioned studies used the clinical growth index (CGI). However, the clinical growth index is far from ideal because of the variability of the clinical symptoms and the uncertainty of their duration. Therefore, we used tumor size and CGI as a measure of tumor behavior. In the present study, we found no difference in tumor behavior between Korean patients with and without NF2 genetic alterations, suggesting that there is no racial difference.
In summary, the present study found 16 cases of mutations and 9 cases of LOH in the NF2 genes of 30 Korean patients with sporadic VS. However, no aberrant hypermethylation of the NF2 gene was detected, and no clear genotype/phenotype correlation was identified in our study. Therefore, other factors are likely to contribute to tumor formation and growth.
Author Contributions
Conceived and designed the experiments: JDL U-KK W-SL. Performed the experiments: TJK JDL. Analyzed the data: TJK JDL. Contributed reagents/materials/analysis tools: TJK JDL U-KK W-SL. Wrote the paper: TJK JDL U-KK.
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Malays J Med Sci
. 2000 Jul;7(2):3–8.
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New Advances in the Diagnosis of Typhoid and Detection of Typhoid Carriers
Asma Ismail
Asma Ismail
1 Centre for Medical Innovations and Technology Development and Department of Medical Microbiology and Parasitology, School of Medical Sciences, Universiti Sains Malaysia, 16150 Kubang Kerian, Kelantan, Malaysia
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1 Centre for Medical Innovations and Technology Development and Department of Medical Microbiology and Parasitology, School of Medical Sciences, Universiti Sains Malaysia, 16150 Kubang Kerian, Kelantan, Malaysia
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Correspondence : Profesor Dr. Asma Ismail, PHD, Centre for Medical Innovations and Technology, Development and Department of Medical Microbiology and Parasitology, School of Medical Sciences, Universiti Sains Malaysia, 16150 Kubang Kerian, Kelantan, Malaysia
© Penerbit Universiti Sains Malaysia, 2000
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PMCID: PMC3438001 PMID: 22977383
Abstract
For effective management of typhoid, diagnosis of the disease must be done with speed and accuracy. Development of such a test would require antigens that are specific for typhoid diagnosis. Attempts at finding the specific antigen have been carried out throughout the years. The finding of such an antigen can lead to carrier detection as well. Candidate antigens have been used in the development of antigen or antibody detection tests with variation in sensitivity and specificity. Further characterization and understanding of the candidate antigens combined with use of innovative technologies will allow for the ideal test for typhoid and typhoid carriers to be within reach.
Keywords: New advances, typhoid diagnosis, typhoid carriers, typhoid antigent
Introduction
Typhoid fever remains a public health problem in most developing countries with an estimated incidence of 540 per 100,000 population (1). Since typhoid may mimic the symptoms of other fevers including dengue, malaria, hepatitis and scub typhus, in typhoid endemic regions, results obtained from the laboratory are important in confirming the clinical diagnosis of typhoid and will contribute to the effective management and treatment of typhoid cases. The conventional diagnosis for typhoid include the culture method and antibody detection tests. Although variations to the conventional techniques have improved both tests, the search for better and improved tests still prevails.
The continued high incidence of typhoid is due to the dissemination of the disease via typhoid carriers (2). Hence there is an urgent need to increase the chance of detecting the carriers so as to decrease the risk that they pose to the communities. In urban areas where sewage disposal is lacking or inadequate, public water supplies are contaminated and typhoid fever is common. The contamination of food by food handlers who are carriers, forms the second commonest route of infection. Since spread of the disease is via fecal-oral route, attempts at breaking the transmission cycle would contribute toward the effective control of the disease. Control of typhoid outbreaks include screening of food and water samples to trace the source of the aetiologic agent. The availability of diagnostic tests that are rapid, sensitive, specific, simple to perfom and cost effective to detect for the pathogen in contaminated food, water and healthy human carriers, would provide an effective tool in controlling and preventing typhoid.
In the quest of developing an accurate test for typhoid fever, there is a need to discover antigens specific for typhoid diagnosis. The finding of such an antigen can lead to diagnostics for carrier detection as well. Studies have been done throughout the years to search for the specific antigen. Attempts at utilizing the antigens toward the development of a suitable diagnostic test have been reported with variation in sensitivity and specificity. Our increasing understanding of the candidate antigens and the use of innovative technologies for laboratory procedures are addressed in this review. It is the intention of this review to elucidate the use and effectiveness of the various diagnostic tests available for the diagnosis of typhoid fever and for the detection of typhoid carriers.
Conventional methods of typhoid diagnosis
Current diagnosis for typhoid is still via the method of culture and antibody detection by means of the Widal test. Isolation of Salmonella typhi has remained as the gold standard, with culture the bone marrow aspirate or a combination of specimens from blood, stool or urine. However, it is well recognised that facilities for culture are not readily available or are limited in many areas. Although the culture method may show specificity, it however lacks sensitivity and speed. If positive, culture produces results within 2–7 days, but culture negative typhoid is well recognised (3). Culture is also less sensitive for diagnosis of infection among children compared to adults (4,5,6). The culture method despite its shortcomings in speed and sensitivity is still useful for antibiotic sensitivity testing.
The value of the Widal test, which uses the bacterial agglutination technique for the diagnosis of typhoid and paratyphoid fevers, has been assessed by several investigators. In endemic areas where culture facilities are lacking or limited, the Widal test remains among the few tests available to differentiate enteric infection from other illnesses due to bacteria, viruses or animal parasites (7). However, it is also recognised that agglutination tests have serious shortcomings (8). Discrepancies in results between laboratories or even within the same laboratory have been reported especially when preparations of the antigens had come from different sources (9,10). There is also evidence that among patients who have been proven as typhoid cases, detection of antibody against the O and H antigens has not been demonstrated by the Widal test (11). On the other hand, antibodies against Salmonella typhi have been detected among nontyphoid Salmonella infections (12) and sometimes even in diseases not caused by Salmonella (13). For meaningful interpretation of the test, demonstration of a 4 fold rise in antibody titers between acute and convalescent sera, at least 10–14 days later, is essential. In the clinical settings, it is common practice to make an interpretation based on a single serum specimen which may not reflect the diagnostic value of the test. More often even when paired sera are obtained, a decrease in titer is commonly observed when comparing the convalescent titer to the acute titer. This could be due to the fact that most patients attended the hospital during the convalescent phase, after initial pretreatment by the general practioners failed.
When interpreting the Widal test it is of utmost importance that the test be interpreted against the background normal titer of the population in question. It is not uncommon to find what is considered positive in a non-endemic area may be considered normal in an endemic area. The interpretation of the tests may also vary among the endemic areas.
Despite problems of accurate diagnosis associated with the Widal test, studies have shown that the test may be useful among febrile paediatric patients in endemic areas (14).
Advances in typhoid diagnosis
An ideal diagnostic test for typhoid and typhoid carriers should be rapid, specific as well as sensitive. The development of a rapid and specific test combined with sensitive diagnosis would provide for prompt, effective management and control of typhoid fever. The existing conventional tests lack speed, sensitivity and specificity. To overcome the limitations of the existing tests, new specific antigens and new diagnostic techniques have been employed. Some of the antigenic candidates include outer membrane proteins (15), lipopolysaccharides (16) and heat shock proteins (17). The need for an alternative, low cost test for typhoid has also spurred the development of other serological assays including counterimmunoelectrophoresis (18), ELISA (19), RIA (20) and the haemagglutination assay (21). Coagglutination tests have also been used for the detection of antigens in urine and serum (22,23). and DNA probes have been suggested for the detection of S.typhi in blood (24). However, none of the tests have so far obtained widespread acceptance in microbiological laboratories. Since typhoid fever is common in developing and underdeveloped countries, the race toward development of the ultimate ideal test still continues. This is because the development of such a test will have a huge economic significance as well as impact on public health management for all endemic countries in the region.
Outer membrane proteins (OMP) due to their location have been primed as important candidates to elicit host immune response (16,25). Although several possible antigenic candidates have been elucidated from studies on the OMPs, only the 50 kDa protein has undergone a full scale multinational clinical trial in order to evaluate its diagnostic value (26,27,28,29). The 50 kD outer membrane protein was determined to be antigenic as well as specific for Salmonella typhi since it only reacted immunologically with typhoid sera (30). Further evaluation of the antigen using the dot enzyme immunosorbent assay (EIA) method revealed that the 50 kD antigen could detect for the presence of specific IgM and IgG in sera from patients with acute typhoid (31,32,33,34).
Evaluation of the tests in clinical settings, showed that the dot EIA test (TYPHIDOT) offers simplicity, speed ( 1–3 hours), specificity (75%), economy, early diagnosis, sensitivity (95%) and with high negative and postive predictive values (31 ). When interpreting the test, detection of IgM would reveal acute typhoid (early phase of infection) while detection of both IgG and IgM would also suggest acute typhoid (middle-phase of infection). In highly endemic areas where the rate of typhoid transmission is high, the detection of specific IgG will increase. Since IgG could persist for more than 2 years after typhoid infection (34) the detection of specific IgG could not differentiate between acute and convalescent cases. Furthermore, cases of false positive results due to previous infection may also occur. On the other hand, IgG positive may also occur in the event of current re-infection. In cases of re-infection, there will be a secondary immune response with a significant “boosting” effect of IgG over IgM such that the latter could not be detected hence “masking the effect” of IgM (35). One possible strategy to resolve the problems mentioned is to detect for the presence of IgM by making sure that its presence is “unmasked” (35,36). To increase diagnostic accuracy in these situations, a modification to the original TYPHIDOT test was done by inactivating total IgG in the serum sample. Studies with the modified test called TYPHIDOT-M have shown that inactivation of IgG would remove competitive binding and allow accessibility of the antigen to the specific IgM, when present. The detection of specific IgM (within 3 hours) would suggest acute typhoid infection. Evaluations on the TYPHIDOT and TYPHIDOT - M tests in clinical settings showed that both tests performed better than the Widal test and even the culture method ( 35,36 ).
In the laboratory diagnosis for typhoid fever, the method used as the gold standard should approach 100% in terms of its sensitivity, specificity, positive and negative predictive values. Evaluation studies have shown that TYPHIDOT-M was superior to the culture method (35,36). Although culture remained as the gold standard, it could not compete with TYPHIDOT-M in terms of sensitivity (>93%), negative predictive value as well as speed ( 35,36). TYPHIDOT-M could also be used to replace the Widal test when used in conjunction with the culture method for the rapid and accurate diagnosis of typhoid fever. The high negative predictive value of the test suggested the usefulness of TYPHIDOT-M in a highly endemic area.
Finding the typhoid carrier
The human population is a reservoir as well as natural host for several enteric pathogens. These asymptomatic carriers represent an important reservoir that helps to perpetuate the disease and is responsible for the outbreaks of enteric diseases including typhoid fever. Approximately, 2–5% of typhoid cases become chronic biliary carriers and hence perpetuate the endemicity of the disease (37). The chances of becoming a carrier increases with age and is evidently greater among women (38). The detection of these carriers thus is an important aspect of disease control. The current gold standard to detect for carriers is by means of stool culture. This is not only tedious and costly, it also has a low sensitivity (39). Multiple bacteriological examination of stools are also necessary to make a reliable diagnosis due to intermittent or light fecal excreters among carriers. There have been studies on carriers that showed positive fecal culture only after 196 negative culture results (40). Hence there is a need to have an alternative serological carrier detection system that is not only specific, sensitive and cost-effective but is also easy to use in the field.
When developing a serodiagnostic test it is important to determine the antibody that would be an indicator for carrier diagnosis. Studies with Vi antigens (41) have shown that IgG is the primary indicator for carriers and IgM does not play a role. IgA can be found in both the acute and carrier state. When further tested, IgA and secretory IgA were found most frequently in the sera of dysentery and typhoid carriers. No secretory IgA was detected among vaccinated individuals (42). When quantifying the content of immunoglobulins in different forms of typhoid fever, the carrier state showed high IgA and IgG content which began as early as the acute period (43). Other studies have also shown that IgA among carriers seemed to be elevated 2.4 times compared to non carriers (but with a previous typhoid history) while IgM is only elevated among acute typhoid cases (44 ). IgG was found to be high among typhoid and typhoid carriers. The high IgA content among carriers may reflect prolonged immunological stimulation since IgA when formed does not last long in the body (in cases of acute typhoid). It has been suggested that the continuous presence of IgA among typhoid carriers may be due to S.typhi being the primary occupant of the biliary system during chronic infection (41). Hence IgA detection among healthy individuals may also indicate typhoid carrier state.
Among the antigens used in the serological screening for carriers ( 45,46) none equaled the Vi antigen in terms of widespread acceptance as an indicator of typhoid carriers (47). Various techniques have been used in the development of a diagnostic test which uses the Vi antigen from Citrobacter freundii. Passive heamagglutination assay has been used and was found to show high specificity and sensitivity especially when used in highly endemic areas (47,48). However, the sera needed to be preabsorbed with sheep erythrocytes before being used and this may not be convenient for screening of large populations. Counterimmunoelectrophoresis has also been used and had high sensitivity and specificity when compared to the heamagglutination test (49). Converting the test to the ELISA format has also been tried without much success since the Vi antigen showed poor binding to the microtiter plates (50) except when tyraminated (41). The ELISA test using tyraminated Vi antigen was recommended for carrier diagnosis with IgG as the indicator at equal to or greater than the cut-off titer of 1:200. The test has a sensitivity of 86% and a specificity of 95% (41).
Conclusion
The anti-Vi test is still used for the lack of a better test for carrier diagnosis. The ideal carrier detection test should be easily used and interpreted in the field rather than in the laboratory to allow for immediate diagnosis. The development of a test using the immunochromatography or dipstick method may be more useful for convenient carrier detection in the field. Development of multi-test for simultaneous typhoid and typhoid carrier diagnosis will be able to have a greater impact on the control and management of typhoid fever. Multi-tests developed for the detection of the organism in environmental samples would also enhance typhoid control since it would allow for tracing of the source of contamination.
While carrier detection is important for public health, reports have also shown a close relationship between biliary disease and chronic carriers (51). Recent developments have shown that the risk of gallbladder carcinoma has increased among typhoid carriers (52,53,54). Hence a test to detect for typhoid carriers that is cheap, sensitive, specific and user friendly for field work would promote not only effective management but also reduce gallbladder carcinoma and dysfunction.
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Articles from The Malaysian Journal of Medical Sciences : MJMS are provided here courtesy of School of Medical Sciences, Universiti Sains Malaysia
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人口学
作者：吴忠观（主编）出版社：重庆大学出版社·重庆 馆藏单位：浙江大学 出版时间：1994-02 ISBN：7-5624-0862-9 资源类型：当代图书 标签： 添加标签主题：人口学(学科: 高等教育 学科: 教材)试读章节阅读
图书集合
人口学 重庆大学出版社·重庆 2005-07
人口学 重庆大学出版社·重庆 1994-02
人口学 重庆大学出版社·重庆 1994-02
目录
第一章 绪论 17
第一节人口学的产生和发展 17
一、古代的人口思想 17
二、近代人口学的产生 19
三、马尔萨斯人口论 20
四、马克思主义的人口理论 22
五、现代人口学的学科体系 24
第二节人口学的基本范畴 25
一、人口和人口属性 25
二、人口变动 26
三、人口规律 28
四、人口问题 29
第三节人口学的研究对象 30
一、人口学的科学定义 30
二、人口学的实用价值 32
第四节人口学的研究方法 34
第二章 两种生产的原理 40
第一节两种生产的涵义 40
一、两种生产的概念与内容 40
二、两种生产的特点 43
三、两种生产理论是马克思主义人口理论的基石 45
第二节两种生产的辩证关系 46
一、物质资料生产对人类自身生产的决定作用 46
二、人类自身生产对物质资料生产的反作用 47
三、两种生产的矛盾运动 48
四、两种生产在社会历史中的作用 48
第三节两种生产相适应的规律 49
第三章 人口再生产 55
第一节人口再生产的涵义 55
一、人口再生产的概念 55
二、宏观与微观，广义与狭义人口再生产 56
第二节生育、生育率及其决定因素 58
一、生育、生育率及出生率 58
二、影响生育率和出生率的因素 61
三、生育率模式的转变 68
第三节死亡、死亡率及其决定因素 71
一、死亡及死亡率 71
二、影响死亡率的因素 74
三、死亡率模式的转变 81
第四节人口再生产类型及转变 83
第四章 人口质量 92
第一节人口质量的涵义 92
一、人口质量的涵义 92
二、人口数量与人口质量 94
三、人口质量的衡量指标 96
第二节优生与人口质量 98
一、优生学的涵义及其体系 98
二、实行优生的意义 99
三、实行优生的措施 100
第三节社会经济发展与人口质量 107
第五章 人口结构 116
第一节人口结构的涵义及分类 116
一、人口结构的涵义 116
二、人口结构的分类 118
第二节人口的自然结构 119
一、人口的性别结构 119
二、人口的年龄结构 123
第三节人口的社会结构 129
一、人口的阶级结构 129
二、人口的产业结构和职业结构 132
三、人口的文化结构 136
四、人口的民族结构 139
第四节人口的地域结构 144
第六章 人口分布与迁移 154
第一节人口分布 154
一、人口分布的涵义 154
二、世界人口分布的特征 155
三、人口密度与人口经济密度 157
四、影响人口分布的因素 158
五、我国人口分布状况 160
第二节人口迁移 161
一、人口迁移的涵义与分类 161
二、世界人口迁移发展状况 162
三、影响人口迁移的因素 163
四、人口迁移的作用 165
五、我国的人口迁移与流动 166
第三节人口城市化 170
第七章 婚姻、家庭与人口再生产 178
第一节婚姻与人口再生产 178
一、婚姻的涵义与性质 178
二、婚姻制度的演变 179
三、婚姻状况与人口再生产 181
四、婚龄与人口再生产 183
第二节家庭与人口再生产 184
一、家庭的性质和职能 184
二、家庭规模、结构与人口再生产 186
三、家庭生命周期 188
四、家庭经济与人口再生产 189
第三节生育观 190
第八章 人口与自然环境 195
第一节人口与自然环境的一般关系 195
一、自然环境的涵义及生态系统 195
二、自然环境对人口的意义 196
三、人口对自然环境的利用和改造 197
四、生态系统 198
第二节人口与自然资源 203
一、自然资源的涵义及分类 203
二、人口增长对自然资源的压力 203
三、人口与自然资源关系上的几种观点 207
第三节人口与环境污染 209
一、环境污染的涵义及分类 209
二、人口增长与环境污染 210
三、环境污染的防治 214
第四节人口容量 216
第九章 人口政策 224
第一节人口政策的性质及种类 224
一、人口政策的涵义和性质 224
二、广义人口政策与狭义人口政策 225
三、世界各国人口政策的比较 226
第二节制定人口政策的主要依据 227
一、人口与经济发展的关系 228
二、人口与自然环境的结合状况 229
三、民族和军事的需要 230
四、社会的心理因素 231
第三节人口政策的措施体系 232
一、宣传教育措施 232
二、行政措施 233
三、经济措施 233
四、法律措施 235
五、技术措施 235
第四节我国人口政策 236
第十章 人口统计与计划 244
第二节人口统计方法 244
一、人口统计资料的收集和整理 244
二、人口普查 250
第二节人口统计指标体系 256
一、人口静态指标 256
二、人口动态指标 261
三、人口计划生育统计 265
第三节人口计划与人口预测 267
第十一章 世界人口的回顾与展望山 276
第十二章 当代西方人口理论评述 327
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188415 | https://www.youtube.com/watch?v=dKC7r2BNjTg | How to Estimate Square Root
Let's Do Math
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Posted: 12 Feb 2017
Don't be dependent on your calculator.
Develop your level of numeracy and learn to gauge a square root by using your existing multiplication knowledge.
Review of square numbers and square roots
Example problem: estimate the square root of 34. Explain your reasoning.
Bonus idea: apply this method to estimating the sidelength of a square.
Key points reiterated at end.
Video has Closed Captions by me (not YouTube).
If this helps you, please click LIKE and help me too. Thanks!
299 comments
Transcript:
how to estimate the square root of a number so far we know this about square numbers and square roots a number multiplied by itself can also be written like this a number squared when we work it out we get a square number as our answer and if you've got a square number like this it came from a square root so what is the square root of 49 what number multiplied by itself gives me 49 7 itself or 7^ SAR is 49 the square root of 49 is 7 I solved this square root problem by using my multiplication knowledge that's great if we can spot a number as a square number in our table's knowledge but when you can't spot a square number that's when we estimate the square root like in this problem estimate the square root of 34 explain your reasoning let's check it out using a multiplication chart here are the square numbers highlighted in blue notice that 34 is not one of the blue square numbers on our chart it's not on the chart at all that's why we have to estimate because we can't jump straight to it with our table's knowledge you can't get to 34 by multiplying a whole number by itself so here's what you do start off by finding the square numbers on either side of our number 34 they are 25 and 36 which numbers give us those squares we get 25 from 5 5 so the square root of 25 is 5 and of 36 the square Ro T is 6 straight away we know that the square root of 34 is somewhere between 5 and 6 when you do this for the first time in your math book it's a great idea to write a note to help you understand now and for revision later on my number is between the square numbers 25 and 36 the square root of 25 is 5 the square root of 36 is six so the square root of my number must be between five and six wait what is between five and six not a whole number between any two consecutive numbers is the decimal Zone we're looking for a decimal number it's going to be 5 point something but 5 point what the square OT of 34 might have a string of decimal places but estimating it to one decimal place is fine let's use a number line to help as we think it through here's my number and here are the square numbers on either side of it 34 is closer to 36 than it is to 25 that tells us that the square root value must be closer to six than to five and that means it has to be somewhere between 5.5 and 5.9 now I could just guess a number from any of these values but I can be more accurate in my estimate if I do a trial run like this 5.5 is halfway between 5 and six that gives us a square number halfway between 25 and 36 using the number line we can see that this would be about 30.5 it's too low the square root of 5.5 is too small okay 34 is very close to to 36 so here's my estimate the square < TK of 34 is approximately equal 2 5.8 that's between five and six and very close to Six this math sign means is approximately equal to and it's great for use in an estimate statement and I got to my estimate using only multiplication knowledge and reasoning I didn't touch a calculator when you have to estimate a square root for yourself never use a calculator instead use logic like guided this will develop your understanding of square roots let's review to estimate the square root of a number you're going to use your multiplication knowledge first find the square numbers on either side of your number to find the two whole numbers you're working between you can't do the estimate without this step a ruler might help you a number line can help you visualize the problem you can write in sentence form too and if you really want to impress your teacher you'll use the approximately equal to sign here's a bonus idea for you you know how we use the area of a square to find the missing side length well sometimes you're going to get a square whose area is not a square number guess what you can use to estimate the side length if you said this method you've got it remember think it through logically like I showed you and show your thinking now it's over to you try out some problems like this and build your math muscle |
188416 | https://ntrs.nasa.gov/api/citations/19980017535/downloads/19980017535.pdf | L. Danielle Koch Lewis Research Center, Cleveland, Ohio Design and Performance Calculations of a Propeller for Very High Altitude Flight NASA/TM—1998-206637 February 1998 The NASA STI Program Office . . . in Profile Since its founding, NASA has been dedicated to the advancement of aeronautics and space science. The NASA Scientific and Technical Information (STI) Program Office plays a key part in helping NASA maintain this important role.
The NASA STI Program Office is operated by Langley Research Center, the Lead Center for NASA’s scientific and technical information. The NASA STI Program Office provides access to the NASA STI Database, the largest collection of aeronautical and space science STI in the world.
The Program Office is also NASA’s institutional mechanism for disseminating the results of its research and development activities. These results are published by NASA in the NASA STI Report Series, which includes the following report types: • TECHNICAL PUBLICATION. Reports of completed research or a major significant phase of research that present the results of NASA programs and include extensive data or theoretical analysis. Includes compilations of significant scientific and technical data and information deemed to be of continuing reference value. NASA’s counterpart of peer-reviewed formal professional papers but has less stringent limitations on manuscript length and extent of graphic presentations.
• TECHNICAL MEMORANDUM. Scientific and technical findings that are preliminary or of specialized interest, e.g., quick release reports, working papers, and bibliographies that contain minimal annotation. Does not contain extensive analysis.
• CONTRACTOR REPORT. Scientific and technical findings by NASA-sponsored contractors and grantees.
• CONFERENCE PUBLICATION. Collected papers from scientific and technical conferences, symposia, seminars, or other meetings sponsored or cosponsored by NASA.
• SPECIAL PUBLICATION. Scientific, technical, or historical information from NASA programs, projects, and missions, often concerned with subjects having substantial public interest.
• TECHNICAL TRANSLATION. English-language translations of foreign scientific and technical material pertinent to NASA’s mission.
Specialized services that complement the STI Program Office’s diverse offerings include creating custom thesauri, building customized data bases, organizing and publishing research results . . . even providing videos.
For more information about the NASA STI Program Office, see the following: • Access the NASA STI Program Home Page at • E-mail your question via the Internet to help@sti.nasa.gov • Fax your question to the NASA Access Help Desk at (301) 621-0134 • Telephone the NASA Access Help Desk at (301) 621-0390 • Write to: NASA Access Help Desk NASA Center for AeroSpace Information 800 Elkridge Landing Road Linthicum Heights, MD 21090-2934 L. Danielle Koch Lewis Research Center, Cleveland, Ohio Design and Performance Calculations of a Propeller for Very High Altitude Flight NASA/TM—1998-206637 February 1998 National Aeronautics and Space Administration Lewis Research Center Available from NASA Center for Aerospace Information 800 Elkridge Landing Road Linthicum Heights, MD 21090-2934 Price Code: A07 National Technical Information Service 5287 Port Royal Road Springfield, VA 22100 Price Code: A07 DESIGN AND PERFORMANCE CALCULATIONS OF A PROPELLER FOR VERY HIGH ALTITUDE FLIGHT by LISA DANIELLE KOCH Submitted in partial fulfillment of the requirements for the degree of Masters of Science in Engineering Thesis Advisor: Dr. Eli Reshotko Department of Mechanical and Aerospace Engineering CASE WESTERN RESERVE UNIVERSITY DESIGN AND PERFORMANCE CALCULATIONS OF A PROPELLER FOR VERH HIGH ALTITUDE FLIGHT by LISA DANIELLE KOCH ABSTRACT Reported here is a design study of a propeller for a vehicle capable of subsonic flight in Earth’s stratosphere. All propellers presented were required to absorb 63.4 kW (85 hp) at 25.9 km (85,000 ft) while aircraft cruise velocity was maintained at Mach 0.40. To produce the final design, classic momentum and blade-element theories were combined with two and three-dimensional results from the Advanced Ducted Propfan Analysis Code (ADPAC), a numerical Navier-Stokes analysis code.
The Eppler 387 airfoil was used for each of the constant section propeller designs compared. Experimental data from the Langley Low-Turbulence Pressure Tunnel was used in the strip theory design and analysis programs written. The experimental data was also used to validate ADPAC at a Reynolds numbers of 60,000 and a Mach number of 0.20. Experimental and calculated surface pressure coefficients are compared for a range of angles of attack.
Since low Reynolds number transonic experimental data was unavailable, ADPAC was used to generate two-dimensional section performance predictions for Reynolds numbers of 60,000 and 100,000 and Mach numbers ranging from 0.45 to 0.75. Surface pressure coefficients are presented for selected angles of attack, in addition to the variation of lift and drag coefficients at each flow condition.
A three-dimensional model of the final design was made which ADPAC used to calculated propeller performance. ADPAC performance predictions were compared with strip-theory calculations at design point. Propeller efficiency predicted by ADPAC was within 1.5% of that calculated by strip theory methods, although ADPAC predictions of thrust, power, and torque coefficients were approximately 5% lower than the strip theory results. Simplifying assumptions made in the strip theory account for the differences seen.
ACKNOWLEDGEMENTS How fortunate I am to have so many blessings to count! For their patience and guidance, I would like to thank Dr. Eli Reshotko and Dr. Christopher Miller. It takes special skill to be a teacher, and I am sincerely grateful to have been able to learn by your examples.
For their unfailing enthusiasm, I would like to thank Dave Bents and Tony Colozza of the ERAST team. It has been a long and bumpy road, and your optimism has really made a difference.
For making financial support available, I would like to thank NASA, Jeff Haas, Wayne Thomas, and Chuck Mehalic. While slightly beyond the boundaries of my normal assigned duties, I believe that I am a better engineer because of the experience gained through this thesis work.
CONTENTS ABSTRACT ............................................................................................................ ii DEDICATION ........................................................................................................ iv ACKNOWLEDGEMENTS ..................................................................................... v NOMENCLATURE ...............................................................................................vii CHAPTER 1—Introduction...................................................................................... 1 CHAPTER 2—Selection of Airfoil and Airfoil Performance Data ............................ 6 CHAPTER 3—Grid Study and ADPAC Validataion ..................................…..........22 CHAPTER 4—Propeller Design Using Experimental Airfoil Data ...........................42 CHAPTER 5—ADPAC Two-Dimensional Aerodynamic Predictions .............…..... 52 CHAPTER 6—Propeller Design and Analysis Using Two-Dimensional ADPAC Predictions............................................ 86 CHAPTER 7—ADPAC Three-Dimensional Performance Calculations .......….........97 CHAPTER 8—Conclusion ...................................................................................114 REFERENCES .....................................................................................................116 NOMENCLATURE a = speed of sound c = chord Cd = drag coefficient = D/(qS) Cl = lift coefficient = L/(qS) Cp = pressure coefficient=( ) ( ) ( ) ( ) 2 2 1 2 1 1 2 2 2 1 γ γ γ γ γ M M M ∞ ∞ − + − + − − CP = power coefficient = P/(ρn3d5) CQ = torque coefficient = Q/(ρn2d5) CT = thrust coefficient = T/(ρn2d4) d = diameter D = drag e = specific stored energy J = advance ratio = V/(nd) L = lift M = Mach number = V/a n = rotational speed P = power q = dynamic pressure = 0.5ρV2 Q = torque r = radius Rtip = tip radius S = wing area T = thrust u = x component of velocity v = y component of velocity V = freestream velocity w = z component of velocity x = chordwise distance y = distance from airfoil surface y+ = dimensionless distance = (y/ν)(τwall/ρ)0.5 α = angle of attack β = twist angle η = efficiency = (CTJ)/CP µ = dynamic viscosity ν = kinematic viscosity ρ = density τwall = shear stress at the wall 1 CHAPTER 1--INTRODUCTION Background Concern for the environment and determination to overcome a new challenge can make very high altitude subsonic flight possible. Driven by the needs of the atmospheric research community, a remotely piloted vehicle capable of flying subsonically in the stratosphere is being developed by a consortium of federal, industrial, and academic partners under NASA’s Environmental Research Aircraft and Sensor Technology (ERAST) program.
In-situ measurements at altitudes between 24.4 km (80,000 ft) and 30.5 km (100,000 ft) are needed to further our understanding of the dynamics and chemistry of Earth’s atmosphere. These measurements would augment laboratory research, data from samples of the lower stratosphere taken by the ER-2 aircraft, and measurements from satellites and balloons. A better fundamental understanding of the atmosphere can help us to make more responsible decisions in the way we choose to live, work, and travel.
Undeniably, development of a suitable propulsion system is the most formidable challenge. Currently, there are no existing propulsion systems capable of meeting either the program’s near term altitude goal of 25.3 km (83,000 ft) or ultimate goal of 30.5 km (100,000 ft). Studies summarized by the ERAST program’s Leadership Team ( Ref. 1) suggest that the near term goal may be met by either a modified gas turbine power plant or a turbocharged reciprocating combustion engine.
2 Non-airbreathing or hybrid systems are the most likely candidates for propelling an aircraft to the ultimate altitude goal.
For many of the conceptual aircraft being considered, power produced is transferred to thrust by a propeller. Presented here is an aerodynamic design study of a propeller to meet the near term ERAST vehicle requirements using the most readily available computational tools, design and analysis methods, and experimental data.
Strip-theory design and analysis methods are used with two and three-dimensional results from the Advanced Ducted Propfan Analysis Code-Version 7 (ADPAC), a numerical Navier-Stokes analysis code, to develop a propeller design.
3 Propeller Requirements Design is an iterative process. It begins with a carefully chosen set of requirements that are based on the best results of any conceptual or experimental studies done. The propeller requirements in Table 1 have been derived from the Near-Term ERAST mission requirements shown in Table 2 and the expected propulsion system performance (Ref. 1).
Table 1. Propeller Requirements Altitude 25.9 km (85,000 ft) Power 63.4 kW (85 hp) Maximum Relative Mach Number 0.80 Cruise Mach Number 0.40 Table 2. ERAST Mission Requirements Mission Near-Term Goal (1998 - 2000) Long-Term Goal (2000 +) Mission altitude 25.3 km (83,000 ft) 30.5 km (100,000 ft) Operational Radius 1000 km (539 nmi) 4000 km (2160 nmi) Payload weight 150 kg (330 lbm) 225 kg (496 lbm) Payload accommodations Access to undisturbed free stream Airspeed range 0.40 < M < 0.85 Operational Constraints Can operate in moderate turbulence Operation in ambient air temperatures to -100° C Crosswind Capability Takeoff and landing in moderate crosswinds (minimum 15 knots) Deployment To remote base of operations at airfields worldwide 4 Overview of the Design and Analysis Methods Used Having a firm set of requirements, the designer must then decide upon a method to meet them. The steps taken to arrive at the final design presented in this thesis are briefly described below. Each step will be discussed in depth in the chapters to follow.
Step 1. The Eppler 387 airfoil was chosen for the constant section propeller blade.
While this airfoil was considerably thicker than those typically used for propeller sections, it was selected because it was known to have good performance at low Reynolds numbers. Experimental lift and drag data that could be used in the strip-theory programs, as well as the surface static pressure data that could be used to validate ADPAC was available for this airfoil.
Step 2. A grid study was conducted to determine an acceptable computational mesh density for the two-dimensional ADPAC performance calculations of the Eppler 387 airfoil. Having identified an appropriate mesh, ADPAC was validated by comparing calculated surface pressure coefficients to experimental pressure coefficients for a range of angles of attack.
Step 3. A strip-theory design program was written based on the procedures described by Adkins and Liebeck (Ref. 2). The low Reynolds number, low Mach number 5 experimental data from the Eppler 387 were incorporated into the program and a series of propellers were designed.
Step 4. The ADPAC code was used to calculate two-dimensional airfoil section performance for higher Mach numbers since no experimental data were found for this regime. The Reynolds number-Mach number combinations were identified by the results of Step 3.
Step 5. The results from Step 4 were incorporated into the strip-theory design and analysis codes. Another set of propellers was designed, examined and compared to arrive at the final geometry. Off design performance calculations were made for the final propeller design with the strip-theory analysis program.
Step 6. A three-dimensional computational mesh for the final propeller design was made, and ADPAC was used to make a three-dimensional performance prediction.
The ADPAC result was compared to the strip-theory result at the design point.
6 CHAPTER 2 Selection of Airfoil and Airfoil Performance Data The propeller designer’s most critical decisions lie in the selection of blade airfoil sections and airfoil performance data. Accuracy of a propeller design or analysis conducted using strip theory methods will be compromised if any variation of airfoil performance with Reynolds number and Mach number is not taken into account.
While the designer may be aided by modern numerical design and analysis programs, results cannot be used with any confidence until calculated performance is in good agreement with experimental data.
With the renewed interest in high altitude remotely piloted vehicles, sailplanes, and wind energy conversion, much research has been done to gain a better understanding of the behavior of steady and unsteady, two and three dimensional subsonic flows at low Reynolds numbers. Low Reynolds numbers are those below 106, as defined by Mueller in Reference 3, since the unpredictiblity of the boundary layer and the effects of laminar separation and reattachment play an important role in this regime. Results of this research has led to the development of many computational tools to design and analyze airfoils in this regime. Even so, behavior of the laminar boundary layer is still not completely understood and measurement and modelling of the boundary layer over airfoils at low Reynolds numbers still presents a challenge (Ref. 3).
For subsonic flows, the laminar boundary layer over an airfoil at low Reynolds number has been observed to behave in three different patterns as reported in 7 Reference 3. The laminar boundary layer may either transition naturally to a turbulent boundary layer, may fully separate from the airfoil surface, or may separate and reattach to the airfoil surface forming a laminar separation bubble.
Performance is best if the laminar boundary layer naturally transitions to a turbulent boundary layer before reaching the adverse pressure gradient. With increased energy, the turbulent boundary layer is able to withstand the adverse pressure gradient and the flow will be able to stay attached to the airfoil surface much longer yielding good lift and drag characteristics.
Full separation results in a severe performance degradation. The airfoil will stall at high angles of attack when the laminar boundary layer completely separates from the airfoil surface near the leading edge. Full separation can also occur at lower angles of attack if the laminar boundary layer is unable to overcome an adverse pressure gradient.
A separation bubble is formed when the laminar boundary layer, unable to overcome an adverse pressure gradient, separates but then reattaches to the airfoil surface after transitioning to turbulent. Typically, the laminar shear layer will begin transitioning immediately downstream of the separation point. The separated turbulent shear layer will then grow quickly entraining flow from the free stream until it reattaches to the airfoil surface. Within the bubble is a region of slow moving reversed flow with the center of the vortex lying near the reattachment point. The airfoil surface pressure remains nearly constant across the bubble region and increases rapidly near the point where the flow reattaches.
8 The length of the separation bubble depends upon how rapidly the laminar shear layer is able to transition. Generally, separation bubbles will lengthen as chord Reynolds number is decreased. While not always clearly seen, increasing free stream turbulence levels, airfoil surface roughness, and employing boundary layer tripping devices have usually been effective in shortening the laminar separation bubble and improving airfoil performance (Ref. 3). Increasing the adverse pressure gradient has also been seen to reduce the length of the separation bubble by reducing the time needed for transition (Ref. 4).
Preliminary studies showed that at an altitude of 25.9 km (85,000 ft) propeller blade Reynolds numbers could vary from 50,000 to 200,000 while relative free-stream Mach numbers could range from 0.40 to the design limit of 0.80. No experimental data was found for this transonic low Reynolds number regime. Experimental results were found for an Eppler 387 airfoil that had been tested at the Langley Low-Turbulence Pressure Tunnel (LTPT) at Reynolds numbers of 60,000 to 460,000 and a Mach number range of 0.03 to 0.13 (Ref. 5). Because of the high quality of the experimental data, and the availability of airfoil surface pressure, lift, and drag coefficients at each angle of attack, the Eppler 387 airfoil was chosen over other airfoils with known good performance at low Reynolds numbers for the constant section propeller. This section is much thicker than the transonic airfoils normally chosen for propellers tip sections. The simplification of one airfoil for the entire blade was made for academic purposes only, simplifying programming and modelling.
9 The variation of the lift and drag coefficients for the Eppler 387 measured at the Langley LTPT for Reynolds numbers of 60,000, 100,000 and 200,000 are shown in Figures 1 and 2. At a Reynolds number of 60,000, laminar separation was not always observed to be followed by turbulent reattachment. This phenomenon can be seen in Figure 1, as the airfoil stalls around α = 3.00° and the boundary layer did not reattach until an angle of attack of 7.50°. In fact, both phenomena, separation with and without reattachment, were observed at an angle of attack of 4.00°. The measurement techniques used were not able to resolve the unsteady nature of the flow at this Reynolds number. This behavior may result from short separation bubbles bursting, although further experimentation would be needed to prove this. McGhee, Walker, and Millard observed no hysteresis loops for this airfoil in the Langley LTPT in a set of experiments designed to study hysteresis effects for Reynolds numbers from 60,000 to 300,000.
Figures 3 through 5 show the variation of pressure coefficients on the airfoil surface for Reynolds numbers ranging from 60,000 to 200,000 for an angle of attack of 8.00°. From these figures and others reported in Reference 5, surface pressure measurements and oil flow visualization techniques showed that the length of the separation bubble decreased as Reynolds number was increased. Oil flow visualization indicated that the boundary layer naturally transitioned for a Reynolds number of 200,000 and α = 8.00°.
10 The Eppler 387 has been tested in several other facilities and the data from the Langley Low-Turbulence Pressure Tunnel have been used to validate numerous airfoil design and analysis codes. Comparisons of these results shed light on the many challenges still faced. Reference 5 presents experimental results from tests of two dimensional models of the Eppler 387 in the Low Turbulence Wind Tunnel at Delft and the Model Wind Tunnel at Stuttgart. As reported in Reference 5 and seen again in Figures 6 through 11 , observations at Langley for a Reynolds number of 60,000 were confirmed in tests of an Eppler 387 section in the Low Turbulence Wind Tunnel at Delft, but not by tests in the Model Wind Tunnel at Stuttgart. Reasons for these discrepancies are still unknown but have been associated with differences in tunnel turbulence levels, model quality, model mounting configurations, and force measurement methods.
The Eppler-Somers code (Ref. 5) and Drela’s XFOIL and ISES codes (Ref. 6, 7) are among the design and analysis codes that have been validated with the Eppler 387 data taken at Langley. The XFOIL and ISES codes use an inviscid/viscous interaction technique while the Eppler-Somers code couples complex mapping, potential flow and boundary layer techniques together to solve for the flow field around an airfoil. General agreement between the calculated and experimental results was good for each of these codes, although agreement degraded as Reynolds number decreased. This degradation is inherent to the technique used since as Reynolds number decreases and the boundary layer thickens the interaction between the inviscid and viscous region gets stronger and the boundary layer approximations become less 11 accurate (Ref. 8). At Reynolds numbers above 100,000 these codes are practical design tools because they can solve for an airfoil’s performance quickly.
12 5 0 5 10 15 20 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re = 60,000 M = 0.05 Re = 100,000 M = 0.08 Re = 200,000 M = 0.06 Figure 1: Variation of Lift Coefficient with Reynolds Number as Measured at the Langley Low-Turbulence Pressure Tunnel α, degrees 13 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Re = 60,000 M = 0.05 Re = 100,000 M = 0.08 Re = 200,000 M = 0.06 Figure 2: Variation of Drag Coefficient with Reynolds Number as Measured at the Langley Low-Turbulence Pressure Tunnel Drag Coefficient, Cd 14 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 Figure 3: Pressure Coefficient versus Non-Dimensional Chordwise Position, Re = 60,000 and α = 8.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.08 Figure 4: Pressure Coefficient versus Non-Dimensional Chordwise Position, Re = 100,000 and α = 8.00° X/C X/C Laminar Separation Oil Flow Visualization Turbulent Reattachment Oil Flow Visualization 15 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.06 Figure 5: Pressure Coefficient versus Non-Dimensional Chordwise Position, Re = 200,000 and α = 8.00° X/C Natural Transition Oil Flow Visualization 16 5 0 5 10 15 20 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 Langley Data, Tunnel Turbulence Level = 0.16% Delft Data, Tunnel Turbulence Level = 0.03% Stuttgart Data, Tunnel Turbulence Level = 0.08% Figure 6: Comparison of Measured Lift Coefficients versus Angle of Attack from Tests at Langley, Delft, and Stuttgart at Re = 60,000 α, degrees 17 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 Langley Data, Tunnel Turbulence Intensity = 0.16% Delft Data, Tunnel Turbulence Intensity = 0.03% Stuttgart Data, Tunnel Turbulence Intensity = 0.08% Figure 7: Comparison of Measured Lift Coefficients versus Drag Coefficients Tests at Langley, Delft, and Stuttgart at Re = 60,000 Drag Coefficient, Cd 18 5 0 5 10 15 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Langley Data, Tunnel Turbulence Level = 0.06% Delft Data, Tunnel Turbulence Level = 0.03% Stuttgart Data, Tunnel Turbulence Level = 0.08% Figure 8: Comparison of Measured Lift Coefficients versus Angle of Attack from Tests at Langley, Delft, and Stuttgart at Re = 100,000 α, degrees 19 0.01 0.025 0.04 0.055 0.07 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Langley Data, Tunnel Turbulence Level = 0.06% Delft Data, Tunnel Turbulence Level = 0.03% Stuttgart Data, Tunnel Turbulence Level = 0.08% Figure 9: Comparison of Measured Lift Coefficients versus Drag Coefficients Tests at Langley, Delft, and Stuttgart at Re = 100,000 Drag Coefficient, Cd 20 5 0 5 10 15 20 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Langley Data, Tunnel Turbulence Level = 0.06% Delft Data, Tunnel Turbulence Level = 0.03% Stuttgart Data, Tunnel Turbulence Level = 0.08% Figure 10: Comparison of Measured Lift Coefficients versus Angle of Attack from Tests at Langley, Delft, and Stuttgart at Re = 200,000 α, degrees 21 0 0.01 0.02 0.03 0.04 0.05 0.06 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Langley Data, Tunnel Turbulence Level = 0.06% Delft Data, Tunnel Turbulence Level = 0.03% Stuttgart Data, Tunnel Turbulence Level = 0.08% Figure 11: Comparison of Measured Lift Coefficients versus Drag Coefficients Tests at Langley, Delft, and Stuttgart at Re = 200,000 Drag Coefficient, Cd 22 CHAPTER 3 Grid Study and ADPAC Validation ADPAC is a three-dimensional time-marching Euler/Navier-Stokes analysis code that was originally developed to enable researchers to study the effects of compressor casing and endwall treatments (Ref. 9). The code is flexible enough to allow it to be used for analysis applications other than compressors. ADPAC gains much of its flexibility from the use of a multiple-block grid system. This feature is helpful when studying complicated geometries where it may not be possible to create a single structured grid of the flowfield. The multiple-block grid system allows one to create different grids for different areas of the flowfield. Special commands are provided to allow the different blocks to communicate with each other.
While the code had been validated for several turbomachinery and non-turbomachinery applications, ADPAC was unvalidated in the regime of interest for the ERAST propeller. The two-dimensional Eppler 387 experimental data from the Langley LTPT were used for this validation.
The first task at hand was to conduct a grid study. The purpose of the grid study is to identify the minimum computational mesh density for which a solution is independent of the number of grid points. For subsonic low Reynolds number flows, the grid must be sufficiently dense to resolve the boundary layer and any separation bubbles formed. Several ‘C’ grids of the flow field around the Eppler 387 geometry were created. Measured coordinates from the Langley pressure model of the airfoil were used. The “Level 4” grid was the most dense with 58,277 grid points. The 23 “Level 3” grid had 23,409 points, and the “Level 2” grid had 5,945 points. Each grid extended ten chordlengths upstream, downstream, above, and below the blade. The minimum acceptable number of points is desired to reduce computational time.
Increasing the number of grid points past that of the Level 4 mesh was considered prohibitive. Views of the entire computational domain for the three two-dimensional grids can be found in Figures 12 through 14. Figures 15 through 17 show the portion of each mesh close to the airfoil surface and display the packing of grid points in the boundary layer region.
Two other files are required by ADPAC in order to run a calculation, an input file and a boundary data file. The input file contains parameters that allow one to scale the non-dimensional grid. In the boundary data file, one can specify the angle of attack and how the boundaries of the grid are to be treated. A combination of parameters within the input and boundary data files set the flow conditions. For all the two-dimensional airfoil performance calculations, the freestream Mach number was fixed on the horizontal straight sections of the outer boundary of the computational grid. Total temperature and total pressure were fixed on the inlet curved section of the outer boundary, and static pressure was fixed at the straight vertical exit plane. A no-slip condition was imposed on the blade surface, forcing the velocity at the blade to be zero. Because ADPAC is a compressible code, a freestream Mach number of 0.20 was set. The Mach numbers for the experimental data were generally below 0.10.
Each grid was used to calculate the flow at Reynolds numbers of 60,000, 100,000 and 200,000 while angle of attack was nominally 8.00°. Angle of attack was 24 set to match that given by the experiment. Figures 18 through 20 show the calculated airfoil surface pressure coefficients plotted against the non-dimensionalized chordwise position and compared to the experimental data. Good agreement between the computations with the Level 4 grid and the experimental data was seen even at a Reynolds number of 60,000 which would have the longest separation bubble of all three cases.
Figure 21 shows a comparison of the calculated value of the lift coefficient as a function of the inverse number of mesh points for each Reynolds number. As Reynolds number increases, more grid lines must be packed towards the airfoil surface to resolve the thinner boundary layer. The dimensionless distance of the grid line away from the wall, y+, is defined by Schlichting in Reference 10 as: y y wall + = τ ρ ν where y is the dimensional distance of the grid line away from the airfoil surface, τwall is the shear stress at the wall, ρ is the fluid density, and ν is the fluid kinematic viscosity. The values of y+ for the first grid line of the Level 4 mesh at the quarter chord point are 0.0624, 0.0915, and 0.1537 for Reynolds numbers of 60,000, 100,000, and 200,000, respectively.
Accompanying each solution is a set of convergence plots. Figures 22 through 24 contain two convergence plots for each solution: the Root Mean Square (RMS) Error, and the Number of Separated points at each iteration. The Root Mean Square 25 Error was defined to be the sum of the squares of the residuals of all the cells in the mesh, the residual being the sum of the changes of the five conservative variables, ρ, ρu, ρv, ρw, and ρe. The ADPAC definition of a separated point was a cell whose value for vx was negative. Generally, for the low Mach number cases, the calculation was run until the number of separated points seemed to be constant and the residuals were reduced by at least three orders of magnitude.
The Level 4 grid was then used for a series of calculations at a Reynolds number of 60,000 and angles of attack ranging from -2.94° to 12.00°. The Level 4 grid was chosen because of the good agreement between the surface pressures calculated by ADPAC and the experimental data at a Reynolds number of 60,000 (Fig.
18). Lift coefficients were calculated for each angle of attack and can be seen compared to the Langley results in Figure 25. Pressure coefficient distributions are presented in Figures 26 through 31 for each of the colored points in Figure 25.
Figures 32 through 37 are the corresponding convergence histories for each of the selected ADPAC calculations. Since the viscous drag result was unavailable from ADPAC, it was estimated by calculating skin friction drag on both sides of a flat plate drag (Ref. 10): Cd = 1328 .
Re The total of the viscous plus pressure drag coefficients were then plotted against the Langley data and are shown in Figure 38.
26 There was generally good agreement between the ADPAC calculated values and the experimental values. Examination of the plots shows that ADPAC, like many other numerical analysis codes, was unable to predict the laminar stall that was measured at Langley at angles of attack from 3.00° to 7.50°. The marked increase in measured drag in this region was also unpredicted by this version of ADPAC. For angles of attack less than 8.01° there is not good agreement between the calculated and experimental data near the trailing edge with ADPAC predicting recompression farther upstream than was seen in experiments in the Langley LTPT (Fig. 26 - 29).
Agreement between the ADPAC prediction and the experimental results at the trailing edge improves at angles of attack of 8.01° and greater (Fig. 30 - 31).
27 Figure 12: Entire Level 2 Computational Mesh of the Eppler 387 Airfoil Figure 13: Entire Level 3 Computational Mesh for the Eppler 387 Airfoil 28 Figure 14: Entire Level 4 Computational Mesh for the Eppler 387 Airfoil Figure 15: Portion of Level 2 Mesh Close to the Airfoil Surface 29 Figure 16: Portion of Level 3 Mesh Close to the Airfoil Surface Figure 17: Portion of Level 4 Computational Mesh Close to the Airfoil Surface 30 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 Level 4 Mesh, M = 0.20 Level 3 Mesh, M = 0.20 Level 2 Mesh, M = 0.20 Figure 18: Pressure Coefficient vs. Non-dimensional Chordwise Position and Different Computational Mesh Densities, Re = 60,000 and α = 8.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.08 Level 4 Mesh, M = 0.20 Level 3 Mesh, M = 0.20 Level 2 Mesh, M = 0.20 Figure 19: Pressure Coefficient vs. Non-dimensional Chordwise Position and Different Computational Mesh Densities, Re = 100,000 and α = 8.00° X/C X/C 31 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.06 Level 4 Mesh, M = 0.20 Level 3 Mesh, M = 0.20 Level 2 Mesh, M = 0.20 Figure 20: Pressure Coefficient vs. Non-dimensional Chordwise Position and Different Computational Mesh Densities, Re = 200,000 and α= 8.00° 0 0.5 1 1.5 2 Re = 60,000 Re = 100,000 Re = 200,000 0.001 .
1 10 5 Figure 21: Calculated Lift Coefficient vs. Inverse of Number of Computational Mesh Points at α = 8.01° 1/(Number of Mesh Points) X/C 32 0 500 1000 1500 2000 2500 3000 3500 4000 10 8 6 4 2 Level 4 Convergence Level 3 Convergence Level 2 Convergence 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Level 4 Convergence Level 3 Convergence Level 2 Convergence Figure 23: Comparison of the Convergence Histories, Re = 100,000 M = 0.2 0 500 1000 1500 2000 2500 3000 3500 4000 10 8 6 4 2 Level 4 Convergence Level 3 Convergence Level 2 Convergence 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Level 4 Convergence Level 3 Convergence Level 2 Convergence Figure 22: Comparison of the Convergence Histories, Re = 60,000 M = 0.2 Iteration Number Iteration Number Iteration Number Iteration Number 33 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 10 8 6 4 2 Level 4 Convergence Level 3 Convergence Level 2 Convergence 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 1000 2000 3000 Level 4 Convergence Level 3 Convergence Level 2 Convergence Figure 24: Comparison of the Convergence Histories, Re = 200,000 M = 0.2 Iteration Number Iteration Number 34 5 0 5 10 15 0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 25: Comparison of Measured and Calculated Lift Coefficient versus Angle of Attack for Re = 60,000 α, degrees 35 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 26: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 27: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.99° X/C X/C 36 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 28: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 6.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 29: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 7.00° X/C X/C 37 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 30: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 8.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 31: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 9.00° X/C X/C 38 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1 104 10 8 6 4 2 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1 104 0 1000 2000 3000 Figure 32: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 4.00° 0 1000 2000 3000 4000 5000 6000 7000 8 6 4 2 0 1000 2000 3000 4000 5000 6000 7000 0 1000 2000 3000 Figure 33: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 4.99° Iteration Number Iteration Number Iteration Number Iteration Number 39 0 1000 2000 3000 4000 5000 6000 7000 8000 10 8 6 4 2 0 1000 2000 3000 4000 5000 6000 7000 8000 0 2000 4000 Figure 34: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 6.01° 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10 8 6 4 2 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 2000 4000 Figure 35: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 7.00° Iteration Number Iteration Number Iteration Number Iteration Number 40 0 500 1000 1500 2000 2500 3000 3500 4000 10 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 36: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 8.01° 0 1000 2000 3000 4000 5000 6000 7000 8000 10 8 6 4 2 0 1000 2000 3000 4000 5000 6000 7000 8000 0 2000 4000 6000 Figure 37: ADPAC Convergence History, Re = 60,000 M = 0.20 α = 9.00° Iteration Number Iteration Number Iteration Number Iteration Number 41 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 Langley Data, M = 0.05 ADPAC Calculation, M = 0.20 Figure 38: Comparison of Measured and Calculated Drag Coefficient versus Lift Coefficient for Re = 60,000 Drag Coefficient, Cd 42 CHAPTER 4 Propeller Design Using Experimental Airfoil Data Fundamentals of propeller theory were established by Glauert as early as 1926 (Ref. 11). Primarily because of the absence of computers, solutions of Glauert’s analysis theory could only be obtained after making a number of simplifying assumptions. While the basic theory has remained the same, Adkins and Liebeck have recently removed most of the assumptions, establishing iterative design and analysis procedures which, with the aid of a computer, can solve for the geometry or performance of a propeller quickly.
Glauert used a combination of momentum and blade element theory to model the propeller. In the momentum theory, the flow upstream and downstream of the propeller is considered to be a potential flow, that is, the fluid is assumed to be inviscid, incompressible, and irrotational. The propeller is thought of to have a large number of blades so that it could be represented as an ‘actuator disc.’ Further, Glauert assumed that the axial velocity passing through the actuator disc is continuous, and the pressure over the surface of the disc is constant although it increases discontinuously after passing through the disc. Glauert used a blade element theory to get more detailed information on the performance of the propeller blades. In this theory, flow past any blade airfoil section is assumed to be two-dimensional and the lift at each section results from the circulation of flow around the blade. Trailing vortices are shed from the blade and 43 pass downstream in a helical vortex sheet, and interference from these vortices cause the rise in axial and radial velocities through the propeller. Forces on the blade elements can be resolved, and once integrated over the length of the blade, ultimately yield propeller thrust, power, and efficiency.
Strip-theories other than Glauert’s momentum-blade element theory exist, differing only in the way in which the induced velocities are found (Ref.12). It was Adkins and Liebeck (Ref. 2), however, who published algorithms for iterative design and analysis procedures in which many of the simplifying assumptions were eliminated.
Specifically, Adkins and Liebeck’s procedures eliminated the small angle assumption, and the lightly loaded assumption in the Prandtl approximation for momentum loss due to radial flow. Their procedures continue to neglect contraction of the wake. If the vortex sheet is assumed to form a rigid helical surface, the Betz condition for minimum energy loss will be met. A design will be optimized when viscous as well as momentum losses are minimized. To do this, the designer should specify that each section operate at an angle of attack corresponding to the maximum lift-to-drag ratio.
Adkins and Liebeck give eleven steps describing the iterative design procedure in Reference 2. Briefly, the parameters specified at the beginning of a design are: power, hub and tip radii, rotational speed, flight velocity, number of blades, number of radial stations along the blade, and either a lift coefficient or chord length distribution along the blade. The program iterates to find blade twist angles, chord or lift coefficient distributions (depending on which was specified in the input), radial and axial interference factors, Reynolds number, and relative Mach number.
44 A Reynolds number distribution was specified for the initial propeller designs (Fig. 39). This ensured that the two-dimensional airfoil experimental data available to the design program would be representative of the blade sections. The program used simple conditional statements to apply the experimental data from the Langley tests shown in Figures 1 and 2. For chord Reynolds number equal to or exceeding 100,000, the experimental data for Reynolds number of 100,000 was used, while the experimental data for Reynolds number of 60,000 was used if the chord Reynolds number was below 100,000.
Several designs were produced using the specified Reynolds number distribution. The propeller diameter and number of blades were varied until the maximum lift coefficient along the blade did not exceed 80% of the maximum experimental lift coefficient for the section. For a two-bladed propeller, these criteria were met when the diameter was increased to 6.8 m (22.3 ft). A 4.6 m (15.1 ft) three-bladed propeller and a 3.5 m (11.5 ft) four-bladed propeller also met these requirements. Of the three designs, the three-bladed propeller was the most feasible.
While there are several benefits to a two-bladed design, the extremely large diameter required raises manufacturing questions that would be avoided in the smaller three-bladed design. The three-bladed design yielded better efficiency than the even smaller four-bladed design because of the decreased disk loading. Efficiencies of the three-bladed propeller and the four-bladed propeller were 85.3% and 81.5%, respectively.
Comparisons of the propeller geometries can be found in Figures 40 through 42.
45 Examination of the blade twist distributions for all three designs (Fig. 40) showed a ‘hook’ in the curve as the blade was twisted through the stalled portion of the lift curve for the experimental data at the Reynolds number of 60,000 (Fig. 1).
Since viscous losses were not minimized in these designs, the three-bladed design was optimized by relaxing the requirement that section lift coefficient be less than 80% of the maximum and specifying a constant lift coefficient corresponding to the maximum lift-to-drag ratio point in the experimental data set (Fig. 43 - 44). Blade twist, chord, lift coefficient, and chord Reynolds number distributions can be found in Figures 45 through 47. While this eliminated the ‘hook’ in the twist distribution (Fig. 40), it was clear that the lift coefficients near the hub would have to be decreased in order to increase the chordlengths of the inboard sections. The exercise of refining the hub sections was deferred to the next design trials.
The design studies using the Langley experimental data were useful in several ways. These studies showed that the specified Reynolds number distribution did yield a reasonable propeller geometry. With refinement, chordlengths of the inboard sections could be improved and the hook in the twist distribution curve could be eliminated. Knowing the Reynolds number and relative Mach number distribution over the blade would help to account for compressibility effects that were so far neglected in the present designs. As will be shown in the next chapter, ADPAC would be used to make performance predictions for the Eppler 387 airfoil operating at these 46 transonic low Reynolds number conditions. Finally, the design studies were useful in identifying feasible values for the propeller diameter and number of blades.
47 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 6 104 8 104 1 105 1.2 105 1.4 105 1.6 105 Figure 39: Design Reynolds Number vs. Non-dimensional Radial Position 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 30 40 50 60 70 80 90 Blade Twist Angle, 2 Bladed Propeller, Rtip = 3.4 m Blade Twist Angle, 3 Bladed Propeller, Rtip = 2.3 m Blade Twist Angle, 4 Bladed Propeller, Rtip = 1.75 m Figure 40: Blade Twist Angle (degrees) vs. Non-dimensional Radial Position, Designed Using Langley 2-D Experimental Data Only r/Rtip r/Rtip 48 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.4 0.6 0.8 1 Chord, 2 Bladed Propeller, Rtip = 3.4 m Chord, 3 Bladed Propeller, Rtip = 2.3 m Chord, 4 Bladed Propeller, Rtip = 1.75 m Figure 41: Chord (meters) vs. Non-dimensional Radial Position, Designed Using Langley 2-D Experimental Data Only 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 1.2 Lift Coefficient, 2 Bladed Propeller, Rtip = 3.4 m Lift Coefficient, 3 Bladed Propeller, Rtip = 2.3 m Lift Coefficient, 4 Bladed Propeller, Rtip = 1.75 m Figure 42: Lift Coefficient vs. Non-dimensional Radial Position, Designed Using Langley 2-D Experimental Data Only r/Rtip r/Rtip 49 4 2 0 2 4 6 8 10 12 14 0 5 10 15 20 25 30 35 40 45 50 55 60 Re = 60,000 M = 0.05 Re = 100,000 M = 0.08 Figure 43: Lift-to-Drag Ratio versus Angle of Attack for the Langley LTPT Experimental Data α, degrees 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.12 1.14 1.16 1.18 1.2 Figure 44: Lift Coefficient vs. Non-dimensional Radial Position, Three Bladed Design, Tip Radius = 2.3 m, Langley 2-D Experimental Data Only 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 40 50 60 70 80 90 Figure 45: Blade Twist Angle (degrees) vs. Non-dimensional Radial Position, Three Bladed Design, Tip Radius = 2.3 m, Langley 2-D Data Only r/Rtip r/Rtip 51 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 4 1 10 5 1.510 5 100000 60000 Figure 47: Reynolds Number vs. Non-dimensional Radial Position, Three Bladed Design, Tip Radius = 2.30 m, Langley 2-D Data Only 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.4 0.6 0.8 1 Figure 46: Chord (meters) vs. Non-dimensional Radial Position, Three Bladed Design, Tip Radius = 2.3 m, Langley 2-D Experimental Data Only r/Rtip r/Rtip 52 CHAPTER 5 ADPAC Two-Dimensional Aerodynamic Predictions ADPAC two-dimensional results were combined with the strip theory design program in an effort to account for compressibility effects so far neglected. To do so, the blade was divided into four segments. The specified propeller Reynolds number distribution and the resulting Mach number distribution were used to identify an average value of the Reynolds number and the Mach number for each segment (Figure 48). These values are listed in Table 3.
Table 3. Reynolds Number-Mach Number Combinations Segment Reynolds Number Mach Number 1 60,000 0.45 2 100,000 0.55 3 100,000 0.65 4 60,000 0.75 The Level 4 computational mesh of the Eppler 387 (Figures 14 and 15) was used to generate section performance predictions for a range of angles of attack for each of the Reynolds number-Mach number combinations in Table 3. Surface pressure coefficients were compared to Langley experimental data that had been corrected for the elevated relative Mach number using the Prandtl-Glauert compressibility correction (Ref. 13) Cp Cp M corrected = − ∞ 0 2 1 53 where Cp0 is the value of the incompressible pressure coefficient and M8 is the free stream Mach number. The pressure coefficients calculated with ADPAC were in turn integrated to yield lift and pressure drag coefficients. The total drag coefficient was found by adding the viscous drag component once again estimated by drag on both sides of a flat plate given by Schlichting in Reference 10: Cd = 1328 .
Re Lift and drag curves are found in Figures 49 through 52. Plots of the surface pressure coefficients and corresponding ADPAC convergence history plots for angles of attack of 4°, 5°, 6°, 7°, 8°, and 9° are given for Segment 1 in Figures 53 through 64, Segment 2 in Figures 65 through 76, Segment 3 in Figures 77 through 88, and Segment 4 in Figures 89 through 100. Where appropriate, the number of supersonic points at each iteration are included in the convergence history plots.
An examination of the Segment 1 results at Re = 60,000 and M = 0.45 shows that agreement between the ADPAC calculated values and the corrected Langley data is generally good. The calculated suction side pressure recoveries are sharper than the corrected data would suggest for low angles of attack. Just as was seen in the code validation, ADPAC did not predict a laminar stall at angles of attack between 5.00° and 6.49°. For angles of attack between 7.51° and 12.00°, there appears to be periodic shedding of the separated shear layer, similar to that reported by Pauley, et.
al. in Reference 4. Figure 101 shows the pressure distribution over the airfoil at an angle of attack at 8.01° and the streamlines shown in Figure 102 for the same 54 condition indicate multiple separation and reattachment points near the trailing edge of the suction side. These results, as well as all of the other ADPAC results presented here, are not time-accurate, but represent the steady state flow solution achieved with local time stepping. The local time stepping technique advances each cell in time by an increment equal to the maximum allowable time step for that cell. Generally, larger cells away from a boundary layer will have a larger time step than smaller cells closer to a solid surface.
Examination of the results in Segment 2 for a Reynolds number of 100,000 and a Mach number of 0.55 shows that there was also generally good agreement for angles of attack ranging from -2.88° to 4.00°. Shock waves near the leading edge are seen in the ADPAC predictions for angles of attack greater than 6.00°. Full stall is not predicted at 14.04°. The Prandtl-Glauert correction is inadequate for these conditions and an experiment would be needed to validate these predictions.
As Mach number is increased from 0.55 in Segment 2 to 0.65 in Segment 3, more differences are seen between the corrected experimental data and the ADPAC predictions. Shock waves are seen for all angles of attack above 3.00°, and like in Segment 2, full stall of the airfoil at 14.04° is not predicted by ADPAC.
Representative of the tip sections, Segment 4 with a Reynolds number of 60,000 and a Mach number of 0.75 shows gross differences between the corrected Langley data and the ADPAC predictions as would be expected from the inaccuracy of the Prandtl-Glauert correction at these conditions. ADPAC solutions indicate that the 55 airfoil is unable to produce a strong leading edge suction as a result of the shock waves seen at every angle of attack.
56 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 Relative Mach Number Reynolds Number/100,000 1 0.89 0.25 Figure 48: Segmented Reynolds Number and Mach Number Distributions 0.55 Segment 1 Segment 2 Segment 3 Segment 4 57 4 2 0 2 4 6 8 10 12 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 Re = 60,000 M = 0.2 Least Squares Polynomial Fit of Re = 60,000 M = 0.2 Re = 60,000 M = 0.45 Least Squares Polynomial Fit of Re = 60,000 M = 0.45 Re = 60,000 M = 0.75 Least Squares Polynomial Fit of Re = 60,000 M = 0.75 Figure 49: Comparison of ADPAC Predicted Lift Coefficients at Re = 60,000 and M = 0.20, 0.45, 0.75 α, angle of attack, degrees 58 4 2 0 2 4 6 8 10 12 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Re = 60,000 M = 0.2 Least Squares Polynomial Fit of Re = 60,000 M = 0.2 Re = 60,000 M = 0.45 Least Squares Polynomial Fit of Re = 60,000 M = 0.45 Re = 60,000 M = 0.75 Least Squares Polynomial Fit of Re = 60,000 M = 0.75 Figure 50: Comparison of ADPAC Predicted Drag Coefficients vs Angle of Attack at Re = 60,000 and M = 0.20, 0.45, 0.75 α, angle of attack, degrees 59 4 2 0 2 4 6 8 10 12 14 16 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 Re = 100,000 M = 0.55 Least Squares Polynomial Fit of Re = 100,000 M = 0.55 Re = 100,000 M = 0.65 Least Squares Polynomial Fit of Re = 100,000 M = 0.65 Figure 51: Comparison of ADPAC Predicted Lift Coefficients at Re = 100,000 and M = 0.55, 0.65 α, angle of attack, degrees 60 4 2 0 2 4 6 8 10 12 14 16 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Re = 100,000 M = 0.55 Least Squares Polynomial Fit of Re = 100,000 M = 0.55 Re = 100,000 M = 0.65 Least Squares Polynomial Fit of Re = 100,000 M = 0.65 Figure 52: Comparison of ADPAC Predicted Drag Coefficients vs Angle of Attack at Re = 100,000 and M = 0.55, 0.65 α, angle of attack, degrees 61 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 53: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 54: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.99° X/C X/C 62 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 55: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 6.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 56: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 7.00° X/C X/C 63 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 57 : Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 8.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.45 Figure 58: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 9.00° X/C X/C 64 0 500 1000 1500 2000 2500 3000 3500 4000 4500 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 1000 2000 3000 Figure 59: ADPAC Convergence History, Re = 60,000 M = 0.45 α =4.00° 0 500 1000 1500 2000 2500 3000 3500 4000 4500 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 2000 4000 Figure 60: ADPAC Convergence History, Re = 60,000 M = 0.45 α =4.99° Iteration Number Iteration Number Iteration Number Iteration Number 65 0 500 1000 1500 2000 2500 3000 3500 4000 4500 7 6 5 4 3 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 2000 4000 Figure 62: ADPAC Convergence History, Re = 60,000 M = 0.45 α =7.00° 0 200 400 600 800 1000 1200 1400 8 6 4 2 0 200 400 600 800 1000 1200 1400 0 2000 4000 Figure 61: ADPAC Convergence History, Re = 60,000 M = 0.45 α =6.01° Iteration Number Iteration Number Iteration Number Iteration Number 66 0 500 1000 1500 2000 2500 3000 3500 4000 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 63: ADPAC Convergence History, Re = 60,000 M = 0.45 α = 8.01° 0 500 1000 1500 2000 2500 3000 3500 4000 4500 7 6 5 4 3 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 2000 4000 6000 Figure 64: ADPAC Convergence History, Re = 60,000 M = 0.45 α = 9.00° Iteration Number Iteration Number Iteration Number Iteration Number 67 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 65: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 4.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 66: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 5.01° X/C X/C 68 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 67: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 6.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 68: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 7.00° X/C X/C 69 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 69: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 8.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.55 Figure 70: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 9.00° X/C X/C 70 0 500 1000 1500 2000 2500 3000 3500 4000 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 0 1000 2000 3000 Figure 71: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 4.00° 0 500 1000 1500 2000 2500 3000 3500 4000 8 6 4 2 0 500 1000 1500 2000 2500 3000 3500 4000 0 1000 2000 3000 Figure 72: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 5.01° Iteration Number Iteration Number Iteration Number Iteration Number 71 8 6 4 2 4000 1 0 200 400 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 73: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 6.00° 8 6 4 2 4000 1 0 500 1000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 74: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 7.00° Iteration Number Iteration Number 72 8 6 4 2 4000 1 0 500 1000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 75: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 8.00° 8 6 4 2 4000 1 0 1000 2000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 76: ADPAC Convergence History, Re = 100,000 M = 0.55 α = 9.00° Iteration Number Iteration Number 73 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 77: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 4.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 78: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 5.01° X/C X/C 74 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 79: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 6.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 80: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 7.00° X/C X/C 75 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 81: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 8.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.65 Figure 82: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 100,000 α = 9.00° X/C X/C 76 8 6 4 2 2500 1 0 500 1000 2500 1 500 1000 1500 2000 2500 0 2000 4000 Figure 83: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 4.00° 8 6 4 2 4000 1 0 1000 2000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 84: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 5.01° Iteration Number Iteration Number 77 8 6 4 2 1200 1 0 1000 2000 1200 1 200 400 600 800 1000 1200 0 2000 4000 Figure 85: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 6.00° 8 6 4 2 4000 1 0 1000 2000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 86: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 7.00° Iteration Number Iteration Number 78 8 6 4 2 4000 1 0 1000 2000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 87: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 8.00° 8 6 4 2 4000 1 0 1000 2000 4000 1 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 Figure 88: ADPAC Convergence History, Re = 100,000 M = 0.65 α = 9.00° Iteration Number Iteration Number 79 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 89: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 90: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 4.99° X/C X/C 80 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 92: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 7.00° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 91: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 6.01° X/C X/C 81 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 93: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 8.01° 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 1 0 1 2 3 4 5 Corrected Langley Data ADPAC Calculation, M = 0.75 Figure 94: Pressure Coefficient vs. Non-dimensional Chordwise Position, Re = 60,000 α = 9.00° X/C X/C 82 9 7 5 3 5200 4200 0 1000 2000 5200 4200 4200 4400 4600 4800 5000 5200 0 2000 4000 Figure 95: ADPAC Convergence History, Re = 60,000 M = 0.75 α = 4.00° 8 6 4 2 5500 1000 0 1000 2000 5500 1000 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 0 2000 4000 Figure 96: ADPAC Convergence History, Re = 60,000 M = 0.75 α = 4.99° Iteration Number Iteration Number 83 8 5 2 4100 2000 0 1000 2000 4100 2000 2000 2500 3000 3500 4000 0 2500 5000 Figure 97: ADPAC Convergence History, Re = 60,000 M = 0.75 α = 6.01° 8 6 4 2 4000 5 0 1000 2000 4000 5 0 500 1000 1500 2000 2500 3000 3500 4000 0 2500 5000 Figure 98: ADPAC Convergence History, Re = 60,000 M = 0.75 α = 7.00° Iteration Number Iteration Number 84 8 6 4 2 4000 0 0 2000 4000 4000 0 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 6000 Figure 99: 1ADPAC Convergence History, Re = 60,000 M = 0.75 α = 8.01° 8 6 4 2 4000 0 0 1000 2000 4000 0 0 500 1000 1500 2000 2500 3000 3500 4000 0 2000 4000 6000 Figure 100: ADPAC Convergence History, Re = 60,000 M = 0.75 α = 9.00° Iteration Number Iteration Number 85 Figure 101: Pressure Distribution for Re = 60,000 M = 0.45 α= 8.01° Figure 102: Streamlines Over the Suction Surface of the Airfoil, Re = 60,000 M = 0.45, α= 8.01° 86 CHAPTER 6 Propeller Design and Analysis Using Two-Dimensional ADPAC Predictions The designs in Chapter 4 using only the low speed Langley data neglected the effects of compressibility. Even the Prandtl-Glauert compressibility corrections are inadequate for the tip sections. To account for the compressibility effects, the resulting lift and drag curves for each of the conditions listed in Table 3 were incorporated into the strip theory design and analysis programs (Figures 49 through 52). Again, simple conditional statements were used to apply these predicted values along the length of the blade.
The first design exercise was to see what effect elevated Mach numbers would have on the propellers designed using only the Langley experimental data. Keeping the number of blades, diameter, and Reynolds number distribution the same as was used for the initial designs, new designs were made. The new twist, chord, and lift coefficient distributions can be found in Figures 103 through 105.
Examination of the twist distribution still show a ‘hook’ near the tip. In these designs, though, this ‘hook’ results from the degradation in performance as Mach number is increased, rather than from the laminar stall found in the Langley data at low angles of attack. Efficiencies of the three and four-bladed propellers using the ADPAC predictions were 80.3% and 76.4%. These values are roughly 5% less than the efficiencies predicted for the low speed designs using the low speed Langley data alone.
87 To optimize the design, a lift coefficient distribution asymptotically reaching the average maximum lift-to-drag ratio point along the blade was specified (Fig. 106).
This was done to increase the chordlengths of the inboard sections, as was seen to be necessary in Chapter 4. The resulting propeller twist, chord, Reynolds number, and relative Mach number distributions can be found in figures 107 through 110.
Efficiency for this propeller at design point was 85.1%. This propeller was considered to be the final design.
The strip-theory analysis program was used to generate off-design propeller performance predictions for the final design. Input to the analysis program includes the propeller geometry, advance ratio, pitch angle, cruise Mach number, and altitude.
The program iterates to find the induced velocities and induced angle of attack. From this information, the lift coefficient distribution along the blade can be found and propeller thrust, power, and torque coefficients as well as efficiency can be calculated.
Among the many tests of programming integrity, performance calculated by the design program should and did exactly match that calculated by the analysis program at design point.
The off-design performance maps were created by changing the advance ratio and pitch angle of the propeller. Variations of the propeller efficiency, thrust, power, and torque coefficients for a range of pitch angles are shown in figures 111 through 114. At design point, the value of the advance ratio is 1.814 and the blade twist angle at the 75% radial position is 42.52°.
88 Examination of the off-design performance curves for the propellers indicates that there may be some merit to operating the propeller at a slightly higher pitch angle than that at the design point. While maintaining the design advance ratio of 1.814, increasing the pitch angle so that the angle at the 75% radial position is 45°, a 25% increase in thrust can be realized for a penalty of 5% in efficiency. Operating at this point may be desirable if sufficient engine power is available. Increasing the pitch setting past this point may not be recommended since the blade begins to stall when the angle at the 75% radial station reaches 50° as shown in Figure 115. To decrease the advance ratio while maintaining a constant cruise velocity may not yield the predicted increase in thrust since as tip speeds increase, shock waves will affect a larger portion of the blade degrading performance. A comparison of the design point conditions and those for this high thrust case can be found in Table 4. Actual performance for both cases will vary from the predicted values as the blades twist in operation. A structural analysis would need to be conducted to understand the propeller’s performance more fully.
89 Table 4. Comparison of the Design Point and Maximum Thrust Point Design Point Pitch Angle at 75%R = 42.52° Maximum Thrust Point Pitch Angle at 75%R = 45.00° Efficiency 0.8509 0.8107 Thrust Coefficient 0.1411 0.1780 Power Coefficient 0.3007 0.3983 Torque Coefficient 0.0479 0.0634 Thrust 450.9 N (101.3 lbs) 569.0 N (127.9 lbs) Power 63.4 kW (85 hp) 83.9 kW (112.6 hp) Torque 703.7 N-m (519.0 ft-lbs) 932.1 N-m (687.5 ft-lbs) Advance Ratio 1.814 1.814 90 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 30 40 50 60 70 80 Blade Twist Angle, 2 Bladed Propeller, Rtip = 3.4 m Blade Twis Angle, 3 Bladed Propeller, Rtip = 2.3 m Blade Twist Angle, 4 Bladed Propeller, Rtip = 1.75 m Figure 103: Blade Twist Angle (degrees) vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.4 0.6 0.8 1 Chord, 2 Bladed Propeller, Rtip = 3.4 m Chord, 3 Bladed Propeller, Rtip = 2.3 m Chord, 4 Bladed Propeller, Rtip = 1.75 m Figure 104: Chord (meters) vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only r/Rtip r/Rtip 91 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.2 0.4 0.6 0.8 1 Lift Coefficient, 2 Bladed Propeller, Rtip = 3.4 m Lift Coefficient, 3 Bladed Propeller, Rtip = 2.3 m Lift Coefficient, 4 Bladed Propeller, Rtip = 1.75 m Figure 105:Lift Coefficients vs Non-dimensional Radial Position 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 106: Lift Coefficient vs. Non-dimensional Radial Position, ADPAC Predicted Values Only r/Rtip r/Rtip 92 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 30 40 50 60 70 80 Figure 107: Blade Twist Angle (degrees) vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 Figure 108: Chord (meters) vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only r/Rtip r/Rtip 93 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 4 1 10 5 1.510 5 2 10 5 100000 60000 Figure 109: Reynolds Number vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.4 0.5 0.6 0.7 0.8 Figure 110: Relative Mach Number vs. Non-dimensional Radial Position, Designed Using ADPAC Predicted Values Only r/Rtip r/Rtip 94 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0.5 0.6 0.7 0.8 0.9 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees--Design Point Beta at 75% R = 40.00 degrees Figure 111: Propeller Efficiency versus Advance Ratio for a Range of Pitch Angles 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.05 0.1 0.15 0.2 0.25 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees - Design Point Beta at 75% R = 40.00 degrees Figure 112: Thrust Coefficient versus Advance Ratio for a Range of Pitch Angles Advance Ratio, J Advance Ratio, J Advance Ratio, J 95 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.1 0.2 0.3 0.4 0.5 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees--Design Point Beta at 75% R = 40.00 degrees Figure 113: Power Coefficient versus Advance Ratio for a Range of Pitch Angles 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.02 0.04 0.06 0.08 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees - Design Point Beta at 75% R = 40.00 degrees Figure 114: Torque Coefficient versus Advance Ratio for a Range of Pitch Angles Advance Ratio, J Advance Ratio, J 96 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees -- Design Point Beta at 75% R = 40.00 degrees Beta at 75% R = 50.00 degrees Figure 115: Lift Coefficient Distributions for a Range of Pitch Angles Radius, meters 97 CHAPTER 7 ADPAC Three-Dimensional Performance Calculations A three-dimensional model of the propeller was made from the twist and chord distributions (Fig. 107-108) calculated using Adkins’ and Liebeck’s strip theory design methods (Ref. 5). Views of the propeller’s solid surfaces can be seen in Figures 116 and 117. The computational mesh of the flowfield around the propeller extends one radius upstream of the propeller, one radius downstream, and one radius radially outward from the blade tip. The entire mesh contains 4,157,138 grid points. The hub surface was modelled as a simple cylinder. The blade suction and pressure surfaces were established as “no-slip” surfaces while the cylindrical hub was established as a “slip” surface. Total temperature and total pressure were fixed on the vertical inlet plane, and freestream static pressure was fixed at the vertical exit plane. The freestream Mach number was fixed on the cylindrical outer boundary of the computational grid.
A program called TCGRID written by Roderick V. Chima (Ref. 14) was used to create a fine “C” grid around one propeller blade as well as a somewhat coarser “H” grid farther upstream of the blade. Since TCGRID was originally intended to generate meshes for turbomachinery blading, it was a challenge to create an appropriate mesh for such a large blade with this program. The two grid blocks made with TCGRID were modified so that the final mesh consisted of ten separate grid blocks. As shown in Figure 118, the fine “C” grid block was broken into four separate blocks (Blocks 1-4), and the coarse inlet block is shown in yellow as Block 5. The fine mesh was 98 retained for Block 1 closest to the blade, while the mesh was coarsened by eliminating every other grid line from the original fine mesh to yield Blocks 2 through 4. The coarse mesh was broken at the corners of the “C” grid to avoid possible problems within ADPAC concerning stretching ratios, or the relative size of neighboring computational cells. Figure 119 is a view forward looking aft of the computational domain surrounding one of the three propeller blades. Figures 120 through 122 show the portion of the computational mesh near the inlet block region, near the blade hub, and near the exit region, respectively. Separate FORTRAN programs were written to extend the mesh radially outwards creating Blocks 6 through 10 (Fig. 123).
There are many features which are desired of a computational mesh—some for physical reasons and some to comply with the ADPAC program format. Physically, there again should be enough grid points packed close to the blade surface so that the boundary layer can be resolved. The grid should also extend well into the free stream ahead of the blade, behind the blade, and radially outward. It is also recommended that the gridlines follow the trailing edge angle of the blade sections so that separation and any vortices that may be shed can be seen. The challenge when creating a mesh is meeting all these requirements while keeping the number of grid points to a minimum to reduce computational time. Practically, the upstream and downstream extents of the mesh were sacrificed in order to add grid points closer to the blade surface. Even so, the three-dimensional mesh is much coarser than the two-dimensional meshes seen earlier. Values for y+ of the first grid line above the blade surface at approximately the 99 quarter chord location were 2.266 for the section at r/Rtip = 0.406, 2.692 for the section at r/Rtip = 0.666, and 3.671 at r/Rtip = 0.905.
For a mesh to be suitable for an ADPAC three-dimensional calculation, stretching ratios for the cells should be around 1.3. TCGRID did not contain input parameters to vary stretching ratios directly, so this requirement was met by adding grid points until the stretching ratios near the tip were within acceptable bounds.
ADPAC uses a multigrid method to speed convergence. This technique generates intermediate solutions by coarsening the mesh by eliminating grid points. Three levels of multigrid are recommended and can be achieved if the number of cells in each direction for each block are divisible by four. Coordinates for the leading edge and trailing edge of the blade must also meet this criterion if the multigrid technique is to be used. If an “H” grid is used for an inlet block, the upstream edge of the “C” grid must be square and must not overlap the inlet block cells. Finally, coordinates of all grid blocks must be ordered to form a left-handed coordinate system.
ADPAC was used to calculate the steady state viscous flow over the rotating propeller blade at the design conditions. The convergence history plots are shown in Figure 124. After over 2500 iterations, the solution was considered to be converged because the RMS error had decreased by three orders of magnitude and the number of separated points was constant. Figure 125 shows three blade sections near the hub, midspan, and tip of the blade where the pressure and Mach number contours are shown in Figures 126 through 131. The ADPAC solution at the design point indicated no separation bubbles that had been seen in the low Reynolds number low Mach 100 number two-dimensional cases studied earlier. Full separation near the trailing edge was seen, along with a shock wave along approximately one quarter of the blade near the tip. The area of supersonic flow is more clearly seen in Figures 132 and 133.
Efficiency, as well as thrust, power, and torque coefficients were calculated and compared to results obtained from strip theory calculations. The comparisons can be in Table 5 or graphically in Figures 134 through 137.
Table 5. Comparison of Strip Theory and ADPAC Results at Design Point Strip Theory Result at the Design Point ADPAC Result at the Design Point Efficiency 0.8509 0.8624 Thrust Coefficient 0.1411 0.1355 Power Coefficient 0.3007 0.2850 Torque Coefficient 0.04785 0.04536 Thrust 450.9 N (101.3 lbs) 433.0 N (97.3 lbs) Power 63.4 kW (85 hp) 60.1 kW (80.6 hp) Torque 703.7 N-m (519.0 ft-lbs) 667.0 N-m (491.9 ft-lbs) Examination of the results show that the ADPAC calculation of efficiency was within 1.5% of the strip theory value, but that the ADPAC values for the thrust, 101 power, and torque coefficients were approximately 5% lower. This difference can be attributed to several factors. First, because of the simplistic application of the two-dimensional low Reynolds number transonic predictions, the strip theory design did not account for the shock from r/Rtip of approximately 0.75 to 0.90. Secondly, the three-dimensional mesh was considerably less dense than the two-dimensional meshes and may be below the value for which the solution is independent of the number of grid points. Finally, the tip vortices as shown in Figure 138, were not taken into account in the strip theory design and analysis programs. The tip vortex from one blade and the blade wake can also be seen in the Mach number distribution at the exit of the computational domain shown in Figure 139.
102 Figure 116: The Three-Bladed Propeller and Extended Hub Surface, Side View Figure 117: Front View of Three-Bladed Propeller 103 Figure 118: Computational Mesh Blocks and Solid Surfaces, Axial View Figure 119: Forward Looking Aft View of Propeller Blade and Computational Mesh Blocks Block 1 Block 2 Block 3 Block 4 Block 5 104 Figure 120: Magnified View of Inlet Block Region Figure 121: Magnified View of Blade Region 105 Figure 122: Magnified View of Trailing Edge Region Figure 123: View of Computational Mesh Upstream and Above the Blade Tip— Forward Looking Aft 106 0 500 1000 1500 2000 2500 3000 8 6 4 0 500 1000 1500 2000 2500 3000 0 5000 1 104 1.5 104 Figure 124: Convergence History for the Three-Dimensional ADPAC Calculation Figure 125: Hub, Midspan, and Tip Sections Iteration r/Rtip = 0.406 r/Rtip = 0.666 r/Rtip = 0.905 107 Figure 126: Pressure Ratio Contours for r/Rtip = 0.406 Figure 127: Mach Number Contours for r/Rtip = 0.0406 108 Figure 128: Pressure Ratio Contours for r/Rtip = 0.666 Figure 129: Mach Number Contours for r/Rtip = 0.666 109 Figure 130: Pressure Ratio Contours for r/Rtip = 0.905 Figure 131: Mach Number Contours for r/Rip = 0.905 110 Figure 132: Suction Surface Pressure Distribution Figure 133: Mach Number Distribution Over the Suction Side of the Blade 111 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0.5 0.6 0.7 0.8 0.9 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees--Design Point Beta at 75% R = 40.00 degrees ADPAC 3-D Calculation at the Design Point Figure 134: Comparison of Strip Theory and ADPAC Calculations of Efficiency 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.05 0.1 0.15 0.2 0.25 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees - Design Point Beta at 75% R = 40.00 degrees ADPAC 3-D Calculation at the Design Point Figure 135: Comparison of Strip Theory and ADPAC Calculations of Thrust Coefficient Advance Ratio, J Advance Ratio, J 112 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.1 0.2 0.3 0.4 0.5 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees--Design Point Beta at 75% R = 40.00 degrees ADPAC 3-D Calculation at the Design Point Figure 136: Comparison of Strip Theory and ADPAC Calculations of Power Coefficient 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 0 0.02 0.04 0.06 0.08 Beta at 75% R = 45.00 degrees Beta at 75% R = 42.52 degrees - Design Point Beta at 75% R = 40.00 degrees ADPAC 3-D Calculation at the Design Point Figure 137: Comparison of Strip Theory and ADPAC Calculations of Torque Coefficient Advance Ratio, J Advance Ratio, J 113 Figure 138: Tip Vortex Streamlines Figure 139: Pressure Distribution at the Exit Plane of the Computational Domain y 114 CHAPTER 8--CONCLUSION Comparison of the strip theory design and analysis results with those from ADPAC, a numerical Navier-Stokes analysis code, sheds light on the strengths and weaknesses of both techniques. Propeller efficiency predicted by ADPAC was within 1.5% of that calculated with the strip theory methods at the design point while ADPAC predictions of thrust, power, and torque coefficients were approximately 5% lower than the strip theory results.
The main advantage to the strip theory design and analysis methods is speed.
Since design is an iterative process, the designer must be able to change conditions easily and determine the propeller’s geometry quickly and Adkins’ and Liebeck’s procedures prove to be a useful tool. Results from strip theory analyses can be obtained quickly as well, a necessity when calculating performance at many off-design conditions in order to create a propeller map. Despite the simplifying assumptions made, reasonable accuracy was achieved with Adkins’ and Liebeck’s strip theory methods.
The main advantages of using ADPAC was the elimination of simplifying assumptions required to make the performance calculation and the visualization of results. Shocks and tip vortices ignored in the strip theory analysis were clearly seen in viewing the ADPAC results. However, the improvement in the analysis required a substantial amount of time to learn how to run the code, to create appropriate meshes, and to post-process the output files. This time is in addition to that required to actually run the calculation. The three-dimensional mesh consumed a great amount of 115 time since approximately 22 hours were needed to complete 100 iterations on a dedicated workstation with an R8000 processor.
In conclusion, the fusion of two dimensional strip theory design and analysis methods with ADPAC, a three-dimensional Navier-Stokes yielded a good first-generation propeller design for a vehicle capable of subsonic flight in Earth’s stratosphere. Lift and drag coefficients for the Eppler 387 at low Reynolds number transonic conditions were generated with ADPAC since experimental results were unavailable for this regime. Examination of the three-dimensional results showed that the Eppler 387 was not a suitable airfoil for at least the tip sections of the propeller because of the shock waves seen on the suction surface of the relatively thick outboard sections. Improvements can be made to this design by combining two-dimensional airfoil ADPAC results for other airfoils into the strip theory design and analysis methods (if experimental data is unavailable), making a full three-dimensional prediction only after several design iterations are made.
116 REFERENCES ERAST Leadership Team. 1996. A Review of Remotely Piloted Aircraft (RPA) Technology Required for High Altitude Civil Sciences Missions. Washington, D.C.: NASA, Office of Aeronautics (Code R), Office of Mission to Planet Earth (Code Y). Photocopied.
Adkins, C.N. and R.H. Liebeck. 1983. Design of Optimum Propellers. New York: American Instintue of Aeronautics and Astronautics. AIAA-83-0190.
Mueller, T. J. 1985. Low Reynolds Number Vehicles. Neuilly-Sur-Seine, France: Advisory Group for Aerospace Research and Development. NTIS, AGARDograph No. 288.
Pauley, L.L. and P. Moin and W.C. Reynolds. 1989. The Instability of Two-Dimensional Laminar Separation. In Low Reynolds Number Aerodynamics: Proceedings of the Conference in Notre Dame, Indiana. June 5-7, 1989 by Springer-Verlag, 82-92. New York: Springer-Verlag.
McGhee, R.J. and B.J. Walker and B.F. Millard. 1988. Experimental Results for the Eppler 387 Airfoil at Low Reynolds Numbers in the Langley Low-Turbulence Pressure Tunnel. Washington, D.C. NASA TM-4062.
Drela, M.1989. XFOIL: An Analysis and Desing System for Low Reynolds Number Airfoils. In Low Reynolds Number Aerodynamics: Proceedings of the Conference in Notre Dame, Indiana, June 5-7. 1989. By Springer-Verlag.1-12.
New York: Springer-Verlag Drela, M. 1992. Transonic Low-Reyonlds Number Airfoils. Journal of Aircraft.
Vol. 29 No. 6 (Nov.-Dec.): 1106-1113.
Coiro, D.P. and C.deNicola. 1989. Prediction of Aerodynamic Performance of Airfoils in Low Reynolds Number Flows. In Low Reynolds Number Aerodynamics: Proceedings of the Conference in Notre Dame, Indiana. June 5-7, 1989. By Springer-Verlag. 13-23, New York: Springer-Verlag.
Hall, E.J. and D.A. Top and R.A. Delaney. Task 7-ADPAC Users Manual.Cleveland: NASA Lewis Research Center, NASA CR-195472.
Schlichting, H. 1960. Boundary Layer Theory. New York: McGraw-Hill.
Glauert, H. 1926. The Elements of Aerofoil and Airscrew Theory, Cambridge: 117 The University Press.
McCormick, B.W. 1979. Aerodynamics, Aeronautics, and Flight Mechanics, New York: John Wiley and Sons.
Anderson, J.D. 1989. Introduction to Flight. New York: McGraw-Hill Book Company.
Chima, R.V. 1990. TCGRID. Cleveland: NASA Lewis Research Center.
This publication is available from the NASA Center for AeroSpace Information, (301) 621–0390.
REPORT DOCUMENTATION PAGE 2. REPORT DATE 19. SECURITY CLASSIFICATION OF ABSTRACT 18. SECURITY CLASSIFICATION OF THIS PAGE Public reporting burden for this collection of information is estimated to average 1 hour per response, including the time for reviewing instructions, searching existing data sources, gathering and maintaining the data needed, and completing and reviewing the collection of information. Send comments regarding this burden estimate or any other aspect of this collection of information, including suggestions for reducing this burden, to Washington Headquarters Services, Directorate for Information Operations and Reports, 1215 Jefferson Davis Highway, Suite 1204, Arlington, VA 22202-4302, and to the Office of Management and Budget, Paperwork Reduction Project (0704-0188), Washington, DC 20503.
NSN 7540-01-280-5500 Standard Form 298 (Rev. 2-89) Prescribed by ANSI Std. Z39-18 298-102 Form Approved OMB No. 0704-0188 12b. DISTRIBUTION CODE 8. PERFORMING ORGANIZATION REPORT NUMBER 5. FUNDING NUMBERS 3. REPORT TYPE AND DATES COVERED 4. TITLE AND SUBTITLE 6. AUTHOR(S) 7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES) 11. SUPPLEMENTARY NOTES 12a. DISTRIBUTION/AVAILABILITY STATEMENT 13. ABSTRACT (Maximum 200 words) 14. SUBJECT TERMS 17. SECURITY CLASSIFICATION OF REPORT 16. PRICE CODE 15. NUMBER OF PAGES 20. LIMITATION OF ABSTRACT Unclassified Unclassified Technical Memorandum Unclassified National Aeronautics and Space Administration Lewis Research Center Cleveland, Ohio 44135–3191 1. AGENCY USE ONLY (Leave blank) 10. SPONSORING/MONITORING AGENCY REPORT NUMBER 9. SPONSORING/MONITORING AGENCY NAME(S) AND ADDRESS(ES) National Aeronautics and Space Administration Washington, DC 20546–0001 February 1998 NASA TM—1998-206637 E–11102 WU–529–10–13–00 130 A07 Design and Performance Calculations of a Propeller for Very High Altitude Flight L. Danielle Koch Propeller; High-altitude; Low Reynolds number aerodynamics; ADPAC Unclassified -Unlimited Subject Categories: 02 and 07 Distribution: Nonstandard Reported here is a design study of a propeller for a vehicle capable of subsonic flight in Earth’s stratosphere. All propellers presented were required to absorb 63.4 kW (85 hp) at 25.9 km (85,000 ft) while aircraft cruise velocity was maintained at Mach 0.40. To produce the final design, classic momentum and blade-element theories were combined with two and three-dimensional results from the Advanced Ducted Propfan Analysis Code (ADPAC), a numerical Navier-Stokes analysis code. The Eppler 387 airfoil was used for each of the constant section propeller designs compared. Experimental data from the Langley Low-Turbulence Pressure Tunnel was used in the strip theory design and analysis programs written. The experimental data was also used to validate ADPAC at a Reynolds numbers of 60,000 and a Mach number of 0.20. Experimental and calculated surface pressure coefficients are compared for a range of angles of attack. Since low Reynolds number transonic experimental data was unavailable, ADPAC was used to generate two-dimensional section performance predictions for Reynolds numbers of 60,000 and 100,000 and Mach numbers ranging from 0.45 to 0.75. Surface pressure coefficients are presented for selected angles of attack, in addition to the variation of lift and drag coefficients at each flow condition. A three-dimensional model of the final design was made which ADPAC used to calculated propeller performance. ADPAC performance predictions were compared with strip-theory calculations at design point. Propeller efficiency predicted by ADPAC was within 1.5% of that calculated by strip theory methods, although ADPAC predictions of thrust, power, and torque coefficients were approximately 5% lower than the strip theory results. Simplifying assumptions made in the strip theory account for the differences seen.
This report was submitted as a thesis in partial fulfillment of the requirements for the degree of Masters of Science in Engineering to Case Western Reserve University, Cleveland, Ohio, January 1998. Responsible person, L. Danielle Koch, organization code 7565, (216) 433–5656. |
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188418 | https://maxim.ece.illinois.edu/teaching/fall21/notes/week5.pdf | ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu 1 Introduction Last time, we introduced the basic definition of Hilbert spaces, proved the projection theorem as well as several other properties of Hilbert spaces. This week, we focus on the structural and geometric properties of Hilbert spaces. We study several applications and optimization problems formulated in Hilbert spaces.
2 Minimum Norm Problems 2.1 The Orthogonality Principle Let H be the Hilbert space and let S ⊂H be a closed linear subspace. We claim that the minimum norm problem min z∈S ∥x −z∥ (1) has a unique solution y ∈S that satisfies x −y ∈S⊥. This is known as the orthogonality principle.
Theorem 2.1 (Orthogonality Principle) y ∈S is the unique minimizer of (1) if and only if x −y ∈S⊥.
Proof: ( = ⇒) Let y ∈S be the unique minimizer of (1) and suppose there exist some s ∈S such that ⟨x −y, s⟩= δ ̸= 0. Without loss of generality, let ∥s∥2 = 1 ∥x −(y + δs)∥2 = ∥x −y∥2 −2⟨x −y, δs⟩+ ∥δs∥2 = ∥x −y∥2 −2δ2 + δ2 = ∥x −y∥2 −δ2 < ∥x −y∥2.
We show that if δ ̸= 0, there exist some y′ = y + δs such that ∥x −(y + δs)∥< ∥x −y∥. This contradicts the assumption that y is the minimizer of (1). Therefore, δ = 0 ∀s ∈S.
( ⇐ = ) Let y ∈S such that x −y ∈S⊥, for any z ∈S, z ̸= y we have ∥x −z∥2 =∥x −y + y −z∥2 =∥x −y∥2 + 2⟨x −y, y −z⟩+∥y −z∥2 =∥x −y∥2 +∥y −z∥2 >∥x −y∥2 .
Therefore y is the unique minimizer of (1).
2.2 Minimum Norm on a Hyperplane The orthogonality principle is fundamental for solving minimum norm problems in Hilbert spaces.
In this section we see that it can be used to find the point on a hyperplane that has the smallest norm.
1 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu Definition 2.1 (Hyperplanes in Hilbert Spaces) Let S = {z ∈H | ⟨h, z⟩= c} for h ∈H{0} and c ∈R. If c = 0, then S is a close subspace of H because all vectors in S are orthogonal to h. If c ̸= 0 , then S is a closed hyperplane.
Consider the problem of finding the closest point to the origin on a hyperplane S = {z ∈H | ⟨h, z⟩= c}. We can formulate this as a constrained minimization problem of the form min z∈H ∥z∥ s.t.
⟨h, z⟩= c.
(2) Pick any fixed z0 ∈H such that ⟨h, z0⟩= c, we can rewrite the hyperplane as the set S = {ξ + z0 | ξ ∈ker(l)}, where l(x) = ⟨h, x⟩. We can turn (2) into a minimization problem over ker(l): min z∈S ∥z∥= min ξ∈ker(l) ∥z0 + ξ∥.
Now using the orthogonality principle, we see that the problem above has a unique minimizer ξ∗ that satisfies ξ∗+ z0 ∈ker(l)⊥. Setting z∗= ξ∗+ z0 gives a solution to the original problem.
2.3 Minimum Norm Solution to Linear Systems Suppose L: H →K is a bounded linear operator with dim(K) < ∞.
The adjoint operator L∗: K →H is defined as the unique linear operator that satisfies ⟨k, Lh⟩K = ⟨L∗k, h⟩H for all k ∈H and k ∈K. We consider the situation where the linear system Lx = y0 is under-determined, meaning that it has more than one solution. In this case we want to find the minimum norm solution. In other words, we want to solve the problem min ∥x∥H s.t.
Lx = y0 (3) We have the following result.
Theorem 2.2 (Pseudoinverse) Assume LL∗: K →K is invertible. Then x0 = L∗(LL∗)−1y0 is the unique minimizer of (3). The operator L∗(LL∗)−1 is called the pseudo-inverse of L.
Remark 2.1 We note that the definition of the pseudo-inverse above in specific to the case where LL∗is invertible. Similarly, if L∗L is invertible, then the pseudo-inverse is defined as (L∗L)−1L∗ instead.
Proof: To prove that x0 is the unique minimizer, we first show that x0 is feasible: Lx0 = LL∗(LL∗)−1y0 = y0.
2 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu Next, we show that ⟨x1 −x0, x0⟩= 0 for any x1 satisfies Lx1 = y0. Let y1 = (LL∗)−1y0 we have ⟨x1 −x0, x0⟩= ⟨x1, x0⟩−⟨x0, x0⟩ = ⟨x1, L∗y1⟩−⟨x0, L∗y1⟩ = ⟨Lx1, y1⟩−⟨Lx0, y1⟩ = ⟨y0, y1⟩−⟨y0, y1⟩ = 0.
Finally, we show that for any x1 such that Lx1 = y0, we have ∥x1∥≥∥x0∥.
∥x1∥2 = ∥x1 −x0 + x0∥2 = ∥x0∥2 + 2⟨x1 −x0, x0⟩+ ∥x1 −x0∥2 = ∥x0∥2 + ∥x1 −x0∥2 ≥∥x0∥2.
The operator L∗(LL∗)−1 is called the pseudoinverse of L.
3 Approximation Problems Let {h1, . . . , hn} be a linearly independent set in a real Hilbert space H, and S = span{h1, . . . , hn} be its linear span. The approximation problem for y ∈S is min y∈S ∥x −y∥= min α1,...,αn∈R x − n X i=1 αihi .
(4) By Theorem 2.1, x −x∗∈S⊥if and only if x∗= Pn i=1 α∗ i hi is the minimizer of (4). We can use this property to find α∗ 1 . . . , α∗ n. Since x − n X i=1 α∗ i hi, hj + = 0 j = 1, . . . , n, which is equivalent to ⟨x, hj⟩= n X i=1 α∗ i ⟨hi, hj⟩ j = 1, . . . , n, (5) the problem reduces to finding the Gram matrix G(h1, . . . , hn) := ⟨h1, h1⟩ ⟨h1, h2⟩ . . .
⟨h1, hn⟩ ⟨h2, h1⟩ ⟨h2, h2⟩ . . .
.
.
.
.
.
.
.
.
.
...
.
.
.
⟨hn, h1⟩ . . .
. . .
⟨hn, hn⟩ .
3 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu The determinant of Gram matrix det G(h1, . . . , hn) = g(h1, . . . , hn) is referred to as the Gramian.
Note that g must be non-zero for this system to have a solution. Next, we can apply Gram–Schmidt process to orthonormalize {h1, . . . , hn} to obtain {ϕ1, . . . , ϕn}. Then ⟨x, ϕj⟩= n X i=1 α∗ i ⟨ϕi, ϕj⟩= α∗ j ∀j.
Therefore, we have x∗= n X i=1 ⟨ϕi, x⟩ϕi.
Theorem 3.1 The Gramian g(h1, . . . , hn) ̸= 0 if and only if h1, . . . , hn are linearly independent.
Proof: Let H = [h1, . . . , hn], we have G(h1, . . . , hn) = HT H. If g(h1, . . . , hn) = 0, then there exist some x ̸= 0 such that HT Hx = 0. We have HT Hx = 0 = ⇒H⊤(Hx) = 0 = ⇒ h⊤ 1 .
.
.
h⊤ n h x1h1 · · · xnhn i = 0 = ⇒ h⊤ 1 (x1h1 + . . . + xnhn) .
.
.
h⊤ n (x1h1 + · · · + xnhn) = 0.
For a vector x1h1+· · ·+xnhn to be orthogonal to all n basis vector in span{h1, . . . , hn}, we must have x1h1 + · · · + xnhn = 0, which mean h1, . . . , hn are linear dependent if and only if g(h1, . . . , hn) = 0.
By contrapositive, g(h1, . . . , hn) ̸= 0 if and only if h1, . . . , hn are linear independent.
4 Approximation in Convex Hulls Definition 4.1 For a set G ⊂H, the convex hull of G is the set of convex combinations: conv(G) = J X j=1 λjgj λj ≥0, J X j=1 λj = 1, gj ∈G .
Example 4.1 (Neural Nets) Let G = {x 7→ασ(⟨w, x⟩+ b) | α, b ∈R, w ∈Rd}, where σ: R →R is a nonlinear map referred to as the activation function. Common examples include the hyperbolic tangent σ(u) = tanh u = eu−e−u eu+e−u or the threshold function σ(u) = 1{u≥0}. The convex hull conv(G) consist of 1-layer neural nets with activation function σ.
4 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu The set conv(G) is a rather rich object consisting not only of arbitrary convex combinations of the elements of G, but also of all norm limits of such convex combinations. Hence, given f ∈F = conv(G) and an integer n ≥1, it is of interest to approximate f by a finite convex combination ˆ fn of the form ˆ fn = n X j=1 λjgj, g1, . . . , gn ∈G, and we want a method that works for general Hilbert spaces. This is made possible by the following lemma: Lemma 4.1 (Maurey’s Empirical Method) Let G be a subset of some Hilbert Space H. As-sume ∥g∥≤M < ∞for all g ∈G. Then for all f ∈F = conv(G), there exists g1, . . . , gn ∈G such that f −1 n n X j=1 gj 2 ≤c n, where c > M2 −∥f∥2.
Proof: Since f ∈F, we know that f = R G g P(dg) for some probability measure P on G. Let g1, . . . , gn be n i.i.d. samples from P, and define ˆ fn := (1/n) Pn j=1 gj. Then we have E[gi] = f, E[⟨f −gi, f −gj⟩] = EP ∥f −g∥2 · 1{i = j}.
Hence, EP f −ˆ fn 2 = 1 n2 EP n X j=1 (f −gj) 2 = 1 nEP ∥f −g∥2 = 1 n EP ∥f∥2 −2⟨f, EP [g]⟩+ EP ∥g∥2 = 1 n EP ∥g∥2 −∥f∥2 ≤1 n(M2 −∥f∥2) ≤c n.
Note that, for any random variable X, E[X] ≤c implies P(X ≤c) > 0.
Thus, there exist g1, . . . , gn ∈G such that f −1 n n X j=1 gj ≤ r c n.
5 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu 5 Greedy Algorithm for Maurey’s Empirical Method In Lemma 4.1 we gave a non-constructive proof that any f ∈F = conv(G) can be well-approximated by a convex combination of some g1, . . . , gn ∈G. In this section we present a simple greedy algorithm that explicitly constructs the desired approximation.
Algorithm 1 Greedy Algorithm Choose any g0 ∈G and set f0 = g0.
for n = 1, . . . do Compute gn ∈G and αn ∈(0, 1] as the solution to the following minimization problem ∥(1 −αn)fn−1 + αngn −f∥= min α∈[0,1] min g∈G ∥(1 −α)fn−1 + αg −f∥ Set fn = (1 −αn) + αngn.
end for Theorem 5.1 ([Jon92]) Let fn be the sequence of functions generated by Algorithm 1. Then we have ∥fn −f∥≤c/√n for some constant c > 0.
Proof: Let fn be defined as in Algorithm 1, and set en = ∥fn −f∥. Then e2 n = ∥fn −f∥2 = (1 −αn) fn−1 + αngn −f 2 = (1 −αn) (fn−1 −f) + αn (gn −f) 2 = (1 −αn)2 e2 n−1 + 2αn (1 −αn) ⟨fn−1 −f, gn −f⟩+ α2 n∥gn −f∥2 = inf 0≤α≤1 inf g∈G{(1 −α)2 e2 n−1 + 2α (1 −α) ⟨fn−1 −f, g −f⟩+ α2∥g −f∥2} (6) We claim that there exists some g ∈G such that the cross term ⟨fn−1 −f, g −f⟩can be arbitrarily small. For any δ > 0, since f is in the closure of the convex hull of G, we may find φ1, . . . , φm ∈G such that fn−1 −f, m X i=1 αiφi −f + < δ, where αi ≥0 and Pm i=1 αi = 1. We can rewrite the inequality above as Pm i=1 αi⟨fn−1 −f, φi −f⟩< δ.
It follows that there exists at least one index j such that ⟨fn−1 −f, φj −f⟩< δ. Since this is true for arbitrarily small δ > 0, setting g = φj on the last line of (6) yields e2 n ≤ inf 0≤α≤1{(1 −α)2 e2 n−1 + α2∥φj −f∥2} ≤ inf 0≤α≤1{(1 −α)2 e2 n−1 + α2M2}, 6 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu where M := M + ∥f∥. Note that the second inequality follows from the fact that ∥φj −f∥≤ ∥φi∥+ ∥f∥≤M + ∥f∥, since φj ∈G. Now we substitute a particular choice of α = e2 n−1 e2 n−1+M 2 , to get e2 n ≤ M2 e2 n−1 + M2 2 e2 n−1 + e2 n−1 e2 n−1 + M2 2 M2 = M2 1 + M2 e2 n−1 −1 Taking the inverse on both sides we get 1 e2 n ≥ 1 M2 + 1 e2 n−1 ≥ 2 M2 + 1 e2 n−2 ≥· · · ≥ n M2 + 1 e2 0 Taking the inverse again we have e2 n ≤ 1 (n/M2) + (1/e2 0) ≤c2 n for some constant c > 0. This completes the proof.
6 Applications in Control Theory 6.1 Controllability First we consider the problem of controllability. Given some state vector x(t) ∈Rn and control u(t) ∈Rm, we want to solve the following problem: min Z t1 t0 ∥u(t)∥2 dt s.t.
˙ x(t) = A(t)x(t) + B(t)u(t) x(t0) = x0, x(t1) = x1, (7) where A : [t0, t1] →Rn×n and B : [t0, t1] →Rn×m are some given matrix-valued functions. Here we treat x0, x1, t0, t1 as data of the problem. Furthermore, we assume that u ∈L2([t0, t1], Rm).
6.1.1 Drift-free Case First we consider the drift-free case A(t) = 0. Let z(t) denote the state-vector in this case. The optimization problem then becomes min Z t1 t0 ∥u(t)∥2 dt s.t.
˙ z(t) = G(t)u(t) z(t0) = z0, z(t1) = z1.
(8) 7 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu Furthermore, we assume that Z t1 t0 ∥G(t)∥2 dt < ∞ where ∥G(t)∥= sup |ξ|=1 |G(t)ξ| |ξ| .
Now the fundamental theorem of calculus yields z(t1) = z(t0) + Z t1 t0 G(s)u(s) ds, t0 ≤t ≤t1.
Define a linear operator L: L2([t0, t1], Rm) →Rn where L(u) := Z t1 t0 G(t)u(t) dt.
It is easy to see that L is linear. To see that it is bounded, note that ∥Lu∥= Z t1 t0 G(t)u(t) dt ≤ Z t1 t0 G(t)u(t) dt ≤ Z t1 t0 G(t) u(t) dt ≤ sZ t1 t0 G(t) 2 dt · sZ t1 t0 u(t) 2 dt = sZ t1 t0 G(t) 2 dt · ∥u∥L2.
Now we can rewrite problem (8) as min ∥u∥H s.t.
Lu = z1 −z0 (9) where H = L2([t0, t1], Rm).
Define next the operator L∗: Rn →L2([t0, t1], Rm) as L∗ξ(t) = G(t)⊤ξ(t). It is easy to see that ⟨ξ, Lu⟩= ⟨Lu∗ξ, u⟩, ∀ξ ∈Rn so that L∗is indeed the adjoint of L. Applying theorem 2.2 we see that the optimal solution is given by u(t) = L∗(LL∗)−1(z1 −z0), assuming that (LL∗)−1 is invertible. Note that here L∗(LL∗)−1(ξ) = G(t)⊤ Z t1 t0 G(t)G(t)⊤dt !−1 ξ.
We can summarize our result in the following theorem: 8 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu Theorem 6.1 If the matrix R t1 t0 G(t)G(t)⊤dt is non-singular, then u(t) = G(t)⊤ Z t1 t0 G(t)G(t)⊤dt !−1 (z1 −z0) is optimal for problem 8.
6.1.2 General Case Now we are ready to solve problem 7. We first recall from basic ODE theory that the solution of ˙ x(t) = A(t)x(t) + B(t)u(t) is given by x(t) = Φ(t, t0)x(t0) + Z t1 t0 Φ(t, s)B(s)u(s) ds, (10) where Φ(t, s) is the fundamental solution such that d dtΦ(t, t0) = A(t)Φ(t, t0), Φ(t0, t0) = In.
Note that the fundamental solution satisfies the so-called flow property, which states that Φ(t, s)Φ(s, r) = Φ(t, r) for any t < s < r.
Set z(t) = Φ(t0, t)x(t), then from (10) we see that z(t) = x(t0) + Z t t0 Φ(t0, s)B(s)u(s) ds.
Taking the derivative on both sides we get ˙ z(t) = G(t)u(t), where G(t) = Φ(t0, t)B(t).
By definition, we have z(t0) = Φ(t0, t0)x(t0) = x0 and z(t1) = Φ(t0, t1)x(t1) = Φ(t0, t1)x1. Now we can apply theorem 2.2 to see that the optimal control is given by u(t) = G(t)⊤[W(t0, t1)]−1[Φ(t0, t1)x(t1) −x0] = B(t)⊤Φ(t0, t)⊤[W(t0, t1)]−1[Φ(t0, t1)x(t1) −x0] where W(t0, t1) is called the controllability Gramian, defined as W(t0, t1) := Z t1 t0 Φ(t0, t)B(t)B(t)⊤Φ(t0, t)⊤dt.
9 ECE 580: 9/21/21, 9/23/21 Scribe: Gavin Zhang, Hong-Ming Chiu, Peng Xu 6.2 Observability Now consider the observability problem, where we observe y(t) = C(t)x(t) and x(t) satisfies ˙ x(t) = A(t)x(t). Here, y(t) ∈Rp, and C : [t0, t1 →Rp×n is a given matrix-valued function. Our goal is to recover x(t0) given y(t), t0 ≤t ≤t1.
Using the same notation as the previous section, we have y(t) = C(t)Φ(t1, t0)x(t0). Now define a linear operator M : Rn →L2([t0, t1], Rp) defined by Mx(t) = C(t)Φ(t, t0)y(t).
It is easy to see that x(t) is observable if and only if M is invertible. Let M∗be the adjoint of M, then we can compute the observability Gramian M∗M as M∗M = Z t1 t0 φ(t, t0)⊤C(t)⊤C(t)Φ(t1, t0) dt.
Since ker(M) = ker(M∗M), we see that x(t) is observable if and only if M∗M is invertible.
References [Jon92] Lee K Jones. A simple lemma on greedy approximation in hilbert space and convergence rates for projection pursuit regression and neural network training. The annals of Statistics, pages 608–613, 1992.
10 |
188419 | https://www.mathway.com/popular-problems/Algebra/965736 | Find the Center x^2-y^2-4x-4y-4=0 | Mathway
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Find the Center x^2-y^2-4x-4y-4=0
x 2−y 2−4 x−4 y−4=0 x 2-y 2-4 x-4 y-4=0
Step 1
Find the standard form of the hyperbola.
Tap for more steps...
Step 1.1
Add 4 4 to both sides of the equation.
x 2−y 2−4 x−4 y=4 x 2-y 2-4 x-4 y=4
Step 1.2
Complete the square for x 2−4 x x 2-4 x.
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Step 1.2.1
Use the form a x 2+b x+c a x 2+b x+c, to find the values of a a, b b, and c c.
a=1 a=1
b=−4 b=-4
c=0 c=0
Step 1.2.2
Consider the vertex form of a parabola.
a(x+d)2+e a(x+d)2+e
Step 1.2.3
Find the value of d d using the formula d=b 2 a d=b 2 a.
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Step 1.2.3.1
Substitute the values of a a and b b into the formula d=b 2 a d=b 2 a.
d=−4 2⋅1 d=-4 2⋅1
Step 1.2.3.2
Cancel the common factor of −4-4 and 2 2.
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Step 1.2.3.2.1
Factor 2 2 out of −4-4.
d=2⋅−2 2⋅1 d=2⋅-2 2⋅1
Step 1.2.3.2.2
Cancel the common factors.
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Step 1.2.3.2.2.1
Factor 2 2 out of 2⋅1 2⋅1.
d=2⋅−2 2(1)d=2⋅-2 2(1)
Step 1.2.3.2.2.2
Cancel the common factor.
d=2⋅−2 2⋅1 d=2⋅-2 2⋅1
Step 1.2.3.2.2.3
Rewrite the expression.
d=−2 1 d=-2 1
Step 1.2.3.2.2.4
Divide−2-2 by 1 1.
d=−2 d=-2
d=−2 d=-2
d=−2 d=-2
d=−2 d=-2
Step 1.2.4
Find the value of e e using the formula e=c−b 2 4 a e=c-b 2 4 a.
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Step 1.2.4.1
Substitute the values of c c, b b and a a into the formula e=c−b 2 4 a e=c-b 2 4 a.
e=0−(−4)2 4⋅1 e=0-(-4)2 4⋅1
Step 1.2.4.2
Simplify the right side.
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Step 1.2.4.2.1
Simplify each term.
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Step 1.2.4.2.1.1
Cancel the common factor of (−4)2(-4)2 and 4 4.
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Step 1.2.4.2.1.1.1
Rewrite −4-4 as −1(4)-1(4).
e=0−(−1(4))2 4⋅1 e=0-(-1(4))2 4⋅1
Step 1.2.4.2.1.1.2
Apply the product rule to −1(4)-1(4).
e=0−(−1)2⋅4 2 4⋅1 e=0-(-1)2⋅4 2 4⋅1
Step 1.2.4.2.1.1.3
Raise −1-1 to the power of 2 2.
e=0−1⋅4 2 4⋅1 e=0-1⋅4 2 4⋅1
Step 1.2.4.2.1.1.4
Multiply 4 2 4 2 by 1 1.
e=0−4 2 4⋅1 e=0-4 2 4⋅1
Step 1.2.4.2.1.1.5
Factor 4 4 out of 4 2 4 2.
e=0−4⋅4 4⋅1 e=0-4⋅4 4⋅1
Step 1.2.4.2.1.1.6
Cancel the common factors.
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Step 1.2.4.2.1.1.6.1
Factor 4 4 out of 4⋅1 4⋅1.
e=0−4⋅4 4(1)e=0-4⋅4 4(1)
Step 1.2.4.2.1.1.6.2
Cancel the common factor.
e=0−4⋅4 4⋅1 e=0-4⋅4 4⋅1
Step 1.2.4.2.1.1.6.3
Rewrite the expression.
e=0−4 1 e=0-4 1
Step 1.2.4.2.1.1.6.4
Divide 4 4 by 1 1.
e=0−1⋅4 e=0-1⋅4
e=0−1⋅4 e=0-1⋅4
e=0−1⋅4 e=0-1⋅4
Step 1.2.4.2.1.2
Multiply−1-1 by 4 4.
e=0−4 e=0-4
e=0−4 e=0-4
Step 1.2.4.2.2
Subtract 4 4 from 0 0.
e=−4 e=-4
e=−4 e=-4
e=−4 e=-4
Step 1.2.5
Substitute the values of a a, d d, and e e into the vertex form (x−2)2−4(x-2)2-4.
(x−2)2−4(x-2)2-4
(x−2)2−4(x-2)2-4
Step 1.3
Substitute (x−2)2−4(x-2)2-4 for x 2−4 x x 2-4 x in the equation x 2−y 2−4 x−4 y=4 x 2-y 2-4 x-4 y=4.
(x−2)2−4−y 2−4 y=4(x-2)2-4-y 2-4 y=4
Step 1.4
Move −4-4 to the right side of the equation by adding 4 4 to both sides.
(x−2)2−y 2−4 y=4+4(x-2)2-y 2-4 y=4+4
Step 1.5
Complete the square for −y 2−4 y-y 2-4 y.
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Step 1.5.1
Use the form a x 2+b x+c a x 2+b x+c, to find the values of a a, b b, and c c.
a=−1 a=-1
b=−4 b=-4
c=0 c=0
Step 1.5.2
Consider the vertex form of a parabola.
a(x+d)2+e a(x+d)2+e
Step 1.5.3
Find the value of d d using the formula d=b 2 a d=b 2 a.
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Step 1.5.3.1
Substitute the values of a a and b b into the formula d=b 2 a d=b 2 a.
d=−4 2⋅−1 d=-4 2⋅-1
Step 1.5.3.2
Simplify the right side.
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Step 1.5.3.2.1
Cancel the common factor of −4-4 and 2 2.
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Step 1.5.3.2.1.1
Factor 2 2 out of −4-4.
d=2⋅−2 2⋅−1 d=2⋅-2 2⋅-1
Step 1.5.3.2.1.2
Move the negative one from the denominator of −2−1-2-1.
d=−1⋅−2 d=-1⋅-2
d=−1⋅−2 d=-1⋅-2
Step 1.5.3.2.2
Rewrite −1⋅−2-1⋅-2 as −−2--2.
d=−−2 d=--2
Step 1.5.3.2.3
Multiply−1-1 by −2-2.
d=2 d=2
d=2 d=2
d=2 d=2
Step 1.5.4
Find the value of e e using the formula e=c−b 2 4 a e=c-b 2 4 a.
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Step 1.5.4.1
Substitute the values of c c, b b and a a into the formula e=c−b 2 4 a e=c-b 2 4 a.
e=0−(−4)2 4⋅−1 e=0-(-4)2 4⋅-1
Step 1.5.4.2
Simplify the right side.
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Step 1.5.4.2.1
Simplify each term.
Tap for more steps...
Step 1.5.4.2.1.1
Cancel the common factor of (−4)2(-4)2 and 4 4.
Tap for more steps...
Step 1.5.4.2.1.1.1
Rewrite −4-4 as −1(4)-1(4).
e=0−(−1(4))2 4⋅−1 e=0-(-1(4))2 4⋅-1
Step 1.5.4.2.1.1.2
Apply the product rule to −1(4)-1(4).
e=0−(−1)2⋅4 2 4⋅−1 e=0-(-1)2⋅4 2 4⋅-1
Step 1.5.4.2.1.1.3
Raise −1-1 to the power of 2 2.
e=0−1⋅4 2 4⋅−1 e=0-1⋅4 2 4⋅-1
Step 1.5.4.2.1.1.4
Multiply 4 2 4 2 by 1 1.
e=0−4 2 4⋅−1 e=0-4 2 4⋅-1
Step 1.5.4.2.1.1.5
Factor 4 4 out of 4 2 4 2.
e=0−4⋅4 4⋅−1 e=0-4⋅4 4⋅-1
Step 1.5.4.2.1.1.6
Move the negative one from the denominator of 4−1 4-1.
e=0−(−1⋅4)e=0-(-1⋅4)
e=0−(−1⋅4)e=0-(-1⋅4)
Step 1.5.4.2.1.2
Multiply−(−1⋅4)-(-1⋅4).
Tap for more steps...
Step 1.5.4.2.1.2.1
Multiply−1-1 by 4 4.
e=0−−4 e=0--4
Step 1.5.4.2.1.2.2
Multiply−1-1 by −4-4.
e=0+4 e=0+4
e=0+4 e=0+4
e=0+4 e=0+4
Step 1.5.4.2.2
Add 0 0 and 4 4.
e=4 e=4
e=4 e=4
e=4 e=4
Step 1.5.5
Substitute the values of a a, d d, and e e into the vertex form −(y+2)2+4-(y+2)2+4.
−(y+2)2+4-(y+2)2+4
−(y+2)2+4-(y+2)2+4
Step 1.6
Substitute −(y+2)2+4-(y+2)2+4 for −y 2−4 y-y 2-4 y in the equation x 2−y 2−4 x−4 y=4 x 2-y 2-4 x-4 y=4.
(x−2)2−(y+2)2+4=4+4(x-2)2-(y+2)2+4=4+4
Step 1.7
Move 4 4 to the right side of the equation by adding 4 4 to both sides.
(x−2)2−(y+2)2=4+4−4(x-2)2-(y+2)2=4+4-4
Step 1.8
Simplify 4+4−4 4+4-4.
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Step 1.8.1
Add 4 4 and 4 4.
(x−2)2−(y+2)2=8−4(x-2)2-(y+2)2=8-4
Step 1.8.2
Subtract 4 4 from 8 8.
(x−2)2−(y+2)2=4(x-2)2-(y+2)2=4
(x−2)2−(y+2)2=4(x-2)2-(y+2)2=4
Step 1.9
Divide each term by 4 4 to make the right side equal to one.
(x−2)2 4−(y+2)2 4=4 4(x-2)2 4-(y+2)2 4=4 4
Step 1.10
Simplify each term in the equation in order to set the right side equal to 1 1. The standard form of an ellipse or hyperbola requires the right side of the equation be 1 1.
(x−2)2 4−(y+2)2 4=1(x-2)2 4-(y+2)2 4=1
(x−2)2 4−(y+2)2 4=1(x-2)2 4-(y+2)2 4=1
Step 2
This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.
(x−h)2 a 2−(y−k)2 b 2=1(x-h)2 a 2-(y-k)2 b 2=1
Step 3
Match the values in this hyperbola to those of the standard form. The variable h h represents the x-offset from the origin, k k represents the y-offset from origin, a a.
a=2 a=2
b=2 b=2
k=−2 k=-2
h=2 h=2
Step 4
The center of a hyperbola follows the form of (h,k)(h,k). Substitute in the values of h h and k k.
(2,−2)(2,-2)
Step 5
x 2−y 2−4 x−4 y−4=0 x 2-y 2-4 x-4 y-4=0
and,,passing
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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188420 | https://talkpal.ai/vocabulary/%E5%A4%9A-duo-vs-%E5%A4%9A%E5%B0%91-duoshao-decoding-quantity-adjectives-in-chinese/ | 多 (duō) vs. 多少 (duōshǎo) - Decoding Quantity Adjectives in Chinese - Talkpal
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多 (duō) vs. 多少 (duōshǎo) – Decoding Quantity Adjectives in Chinese
When learning Mandarin Chinese, two terms that often cause confusion among learners are 多 (duō) and 多少 (duōshǎo). Both are used to express quantities, but their usage and contexts differ significantly. Understanding these differences is crucial for anyone looking to achieve fluency in Chinese. This article will explore these two terms, their meanings, usage, and provide practical examples to help you master their applications.
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Understanding 多 (duō)
多 (duō) directly translates to “many” or “much” in English. It is used to describe an abundance or a large quantity of something. However, unlike its English counterparts, 多 is used in specific grammatical structures in Mandarin.
When using 多 to describe nouns, it is typically placed after the noun and followed by a measure word if the noun is countable. For instance:
他有很多书。
(Tā yǒu hěn duō shū.)
“He has many books.”
Here, 很多 (hěn duō) is used to express a large quantity, and the structure remains the same whether discussing books, cars, or any other countable noun.
多 is also frequently used in questions to express the extent or degree of an action or quality. It is placed before an adjective or another adverb in this context. For example:
你怎么这么多吃？
(Nǐ zěnme zhème duō chī?)
“Why do you eat so much?”
In this sentence, 多 is used to intensify the verb “to eat” (吃 chī), asking about the extent of the action.
Understanding 多少 (duōshǎo)
On the other hand, 多少 (duōshǎo) translates to “how many” or “how much” and is used exclusively in interrogative sentences to inquire about the quantity or amount of something. Unlike 多, 多少 is placed directly before the noun it modifies, without needing a measure word in between. For example:
这个班有多少学生？
(Zhège bān yǒu duōshǎo xuéshēng?)
“How many students are in this class?”
Here, 多少 is used to ask for the specific number of students, emphasizing the need for a numeric answer.
It is important to note that 多少 can also be used to inquire about non-countable quantities, such as time, money, or weight, thus making it a versatile question word. For instance:
这本书多少钱？
(Zhè běn shū duōshǎo qián?)
“How much is this book?”
Comparing 多 and 多少 in Practical Contexts
While both 多 and 多少 deal with quantities, their applications in sentences are quite distinct. To further clarify, let’s compare them in practical contexts.
If you want to express that you have a lot of something and emphasize the large quantity, you would use 多. For example:
我有很多朋友。
(Wǒ yǒu hěn duō péngyǒu.)
“I have many friends.”
However, if you were asked to specify exactly how many friends you have, the question would be phrased with 多少:
你有多少朋友？
(Nǐ yǒu duōshǎo péngyǒu?)
“How many friends do you have?”
Understanding when to use 多 and 多少 is key to mastering their applications. 多 is used for stating facts or making declarations about quantity, while 多少 is used for seeking specific numeric information.
Tips for Practicing 多 and 多少
To effectively learn when to use 多 and 多少, here are a few tips:
– Practice forming questions with 多少 to become comfortable with asking about quantities.
– When using 多, try to pair it with 很 (hěn) as in 很多, to strengthen your declarations about large quantities.
– Listen to native speakers or watch Chinese media to hear how and in what contexts these terms are used.
By understanding and practicing the usage of 多 and 多少, you will enhance your ability to communicate more precisely in Chinese, making your learning journey both successful and enjoyable.
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188421 | https://afmmath.files.wordpress.com/2015/04/sequences-and-series-word-problems.pdf | Adapted from Name_______Date___Period_A#D-4_ Arithmetic and Geometric Sequences and Series: Applications For each of the problems below: A. Identify whether the pattern is arithmetic or geometric. B. Determine if you need to calculate a term in a sequence or the value of a series. C. Solve the problem. 1. An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats. How many seats are in the theatre? 2. Domestic bees make their honeycomb by starting with a single hexagonal cell, then forming ring after ring of hexagonal cells around the initial cell, as shown. The numbers of cells in successive rings form an arithmetic sequence. a. Write a rule for the number of cells in the nth ring. b. What is the total number of cells in the honeycomb after the 9th ring is formed? (Hint: Do not forget to count the initial cell.) 3. Suppose you go to work for a company that pays one penny on the first day, 2 cents on the second day, 4 cents on the third day and so on. If the daily wage keeps doubling, what will your total income be for working 31 days? 4. The first year a toy manufacturer introduces a new toy, its sales total $495,000. The company expects its sales to drop 10% each succeeding year. Find the total expected sales in the first 6 years. Adapted from 5. The deer population in an area is increasing. This year, the population was 1.025 times last year’s population of 2537. a. Assuming that the population increases at the same rate for the next few years, write an explicit formula for the sequence where n is the number of years since the population was 2537. b. Find the expected deer population for the tenth year. 6. Logs are stacked in a pile with 24 logs on the bottom row and 15 on the top row. There are 10 rows in all with each row having one more log than the one above it. How many logs are in the stack? 7. A ball is dropped from a height of 16 feet. Each time it drops, it rebounds 80% of the height from which it is falling. Find the total distance traveled in 15 bounces. 8. A company is offering a job with a salary of $30,000 for the first year and a 5% raise each year after that. If the 5% raise continues every year, find the amount of money you would earn in a 40-year career. |
188422 | https://spinningnumbers.org/a/sign-convention.html | Published Time: 2025-03-27T14:48:36-07:00
Sign convention for passive components | Spinning Numbers
Sign convention for passive components
Spinning Numbers- [x]
AboutExpand
Sign convention for passive components
The sign convention for passive components is a widely understood way to assign polarity to voltages and currents. It defines what we mean by positive and negative voltage and current.
When you label the voltage and current of a circuit element, the convention says you should point the current arrow into the terminal with the positive voltage polarity.
Contents
Symbolic labels for current and voltage
A resistor experiment
Label resistors to make Ohm’s Law easy
Sign convention for passive components
Example 1
Example 1x - the other sign convention
Example 2
Exceptions
Power
Summary
Where we’re headed
The sign convention for passive components is an arbitrary but widely accepted rule that says,
Point the current arrow into the positive voltage terminal of an element.
This article is based on conventional current direction, not electron current direction.
Veterans: Some military electronic training programs (for example the U.S. Navy NEETS program from the 1960’s) use the opposite sign convention, defining current to flow in the direction of electron motion. We don’t use that convention here at Spinning Numbers. We follow the SI convention for current direction.
Symbolic labels for current and voltage
Here’s a resistor,
For now it is just sitting here all by itself. The only thing we know about is its resistance, R=100 Ω\text R = 100 \,\Omega R=1 0 0 Ω. The other thing we know is Ohm’s Law, v=i R v = i\,\text R v=i R. We don’t know specific values for i i i or v v v because it’s not in a circuit, yet.
One of the first things we might do is add symbolic labels on our resistor to represent voltage and current. That will let us talk about them and include them in equations.
There are two possible directions you might point the current arrow, and two possible orientations for the voltage +++ and −-− polarity signs.
If you mix and match all the choices, there are 4 4 4 possible ways to label the resistor,
Please check that I drew all the variations correctly.
Notice there are really only two different versions. See how a.a.a. and d.d.d. are the same thing, just mirrors of each other? The current arrow points into the +++ voltage polarity.
b.b.b. and c.c.c. are also twins because the current arrow points into the −-− voltage polarity.
So there are just two ways to put symbolic labels on the resistor,
BUT, one way is a better choice than the other. When possible, you should point the current arrow into the +++ voltage sign. Why? Because it means we use the normal version of Ohm’s Law, v=i R v = i\,\text R v=i R. If we use the other labeling method (arrow pointing into the negative sign) we have to remember to include a minus sign in Ohm’s Law, v=−i R v = -i\,\text R v=−i R.
A resistor experiment
Let’s apply a current to our resistor. Let the current be i=10 mA i = 10\,\text{mA}i=1 0 mA just for discussion.
Ohm’s Law is v=i R v = i \,\text R v=i R. We know the value of R\text R R is always positive.
Are there negative resistors?
Every resistor in this course has a positive resistance. There are some exotic devices called tunnel diodes that seem like they have negative resistance. Tunnel diodes are quite rare. As Borg electrical engineers say, “Resistance is positive.”
Suppose we connect a real battery to a real resistor and touch voltmeter probes to the resistor. The red voltmeter probe defines which resistor terminal we consider to be the +++ polarity. The black probe defines the −-− voltage polarity.
The following diagram shows two versions of the experiment. The red probe touches the upper terminal of the resistor in both versions. That means the +++ voltage polarity is the upper terminal in both cases. What’s different is the current direction. Some external circuit causes a 10 mA 10\,\text{mA}1 0 mA current to flow in the resistor,
a.\text{a.}a. The current flows down from the top.
b.\text{b.}b. The current flows up from the bottom.
a.\text{a.}a. The meter reads +1.0 V+1.0\,\text V+1.0 V, so the red probe is 1.0 V 1.0\,\text V 1.0 V above the black probe. v=+1 V\goldD v = +1\,\text V v=+1 V.
b.\text{b.}b. The meter reads −1.0 V-1.0\,\text V−1.0 V, so the red probe is 1.0 V 1.0\,\text V 1.0 V below the black probe. v=−1 V\goldD v = -1\,\text V v=−1 V.
The current is reversed between a.\text{a.}a. and b.\text{b.}b., so the voltage shown on the meter changes from +++ to −-−. That makes sense.
Voltmeter diagram b.\text{b.}b. can be a real mind-bender. The negative sign on the voltmeter display is telling us the black probe is at a higher voltage than the red probe.
Our voltmeter diagrams represent the two alternative conventions for labeling resistors. On the left the current arrow goes into the +++ sign of the voltage polarity. On the right the current arrow goes into the −-− sign. The voltmeters show the same magnitude voltage, but the one on the right has a minus sign.
Label resistors to make Ohm’s Law easy
Let’s see if Ohm’s Law agrees with what our meters say.
First, point the current arrow into the +++ sign,
Let R=100 Ω\text R = 100 \,\Omega R=1 0 0 Ω and i=+10 mA i = +10\,\text{mA}i=+1 0 mA.
Find v v v using Ohm’s Law.
Plug the values into Ohm’s Law: v=i R=+10 mA×100 Ω=+1 V v = i\,\text R = +10\,\text{mA} \times 100\,\Omega = +1\,\text V v=i R=+1 0 mA×1 0 0 Ω=+1 V.
This is great. This is what the voltmeter said. Theory matches experiment!
Now do it again with the current arrow pointing into the −-− sign.
We use the exact same problem statement. Let’s see what happens if we blindly apply the usual version of Ohm’s Law.
Let R=100 Ω\text R = 100 \Omega R=1 0 0 Ω and i=+10 mA i = +10\,\text{mA}i=+1 0 mA.
Find v v v using Ohm’s Law.
v=i R=+10 mA⋅100 Ω=+1 V v = i\,\text R = +10\,\text{mA} \cdot 100\,\Omega = +1\,\text V v=i R=+1 0 mA⋅1 0 0 Ω=+1 V
But that isn’t what the voltmeter says! The voltmeter reads −1 V-1\,\text V−1 V.
This labeling convention forces us to learn what to do when the current arrow points in this direction. We adapt Ohm’s Law by including a minus sign, v=−i R v = -i\,\text R v=−i R.
Any time a current arrow goes into the negative side of a resistor we have to use this version of Ohm’s Law,
v=−i R=−10 mA×100 Ω=−1 V v = -i\,\text R = -10\,\text{mA} \times 100\,\Omega = -1\,\text V v=−i R=−1 0 mA×1 0 0 Ω=−1 V
Now the answer comes out matching the voltmeter.
Here’s the problem. That little minus sign is a source of a lot of silly errors in circuit analysis. So what do engineers do? We try not to label components this way. We discipline ourselves to point the current arrow into the plus sign when possible. A whole lot of potential errors simply vanish.
Point the current arrow into the positive voltage polarity,
The fancy name for this idea is the sign convention for passive components.
Sign convention for passive components
We apply the convention to all passive components like this,
The labeling convention helps you get the right answer when analyzing a circuit.
What is that voltage arrow?
The images above show voltage using two notations: with +++ and −-− signs, and also with an orange voltage arrow. The voltage arrow points from −-− to +++. The polarity signs and the arrow are redundant, they mean exactly the same thing. You can use either or both in your schematics.
Example 1
The voltage polarity for this 250 Ω 250\,\Omega 2 5 0 Ω resistor has been assigned with +++ at the top. This polarity direction was an arbitrary choice. Something (not shown) in the surrounding circuit causes 2 V 2\,\text V 2 V to appear across the resistor.
Now we add the current arrow using the sign convention for passive components,
We point the current arrow into the positive terminal. This was not an arbitrary choice. The sign convention for passive components tells us to point the current arrow into the +++ sign.
What is current i i i?
To find the current, apply Ohm’s Law,
i=v R i = \dfrac{v}{\text R}i=R v
i=+2 V 250 Ω i = \dfrac{+2\,\text V}{250\,\Omega}i=2 5 0 Ω+2 V
i=+8 mA i = +8 \,\text{mA}i=+8 mA
The voltage polarity tells us the top of the resistor is 2 V 2\,\text V 2 V above the bottom of the resistor. Ohm’s Law tells us the current is +8 mA+8 \,\text{mA}+8 mA. The +++ sign on current means current is flowing in the direction of the arrow, from top to bottom. (Conventional current, not electron current.)
Example 1x - the other sign convention
What would happen if we labeled the resistor with the other sign convention? The diagram below shows the same resistor with the same voltage polarity, but the current arrow points out of the positive terminal, so the sign convention for passive components is not being used.
Apply Ohm’s Law, exactly the same as Example 1,
i=+2 V 250 Ω=+8 mA i = \dfrac{+2\,\text V}{250\,\Omega} = +8 \,\text{mA}i=2 5 0 Ω+2 V=+8 mA
This is telling us the current is +8 mA+8 \,\text{mA}+8 mA. The +++ sign means it is flowing in the direction of the arrow. What? That can’t be. In a real resistor the current flows the other way. We got the wrong answer. Oh wait! To get the right answer we have to remember to include a −-− sign in Ohm’s Law.
i=−i R=−+2 V 250 Ω=−8 mA i = -i\,\text R = -\dfrac{+2\,\text V}{250\,\Omega} = -8 \,\text{mA}i=−i R=−2 5 0 Ω+2 V=−8 mA
Lesson: You make fewer errors if you use the sign convention for passive components.
Example 2
This 10 k Ω 10\,\text k\Omega 1 0 k Ω resistor has been labeled with the sign convention for passive components, just like Example 1: The voltage polarity has +++ at the top and the blue current arrow points into the positive sign. This time, the current is specified instead of the voltage. The value of the current is −20 μ A-20 \,\mu\text A−2 0 μ A. This may look a little odd, to show −20 μ A-20 \,\mu\text A−2 0 μ A current flowing in the direction of the arrow, but let’s see what happens.
Find voltage v v v.
We use Ohm’s Law to solve for the unknown voltage. Since we’ve been careful to use the sign convention all we have to do is plug in the actual values shown on the schematic,
v=i R v = i\,\text R v=i R
v=−20 μ A⋅10 k Ω v = -20 \,\mu\text A \cdot 10\,\text k\Omega v=−2 0 μ A⋅1 0 k Ω
v=−20×1 0−6⋅10×1 0+3=−200×1 0−3 v = -20 \times 10^{-6} \cdot 10 \times 10^{+3} = -200 \times 10^{-3}v=−2 0×1 0−6⋅1 0×1 0+3=−2 0 0×1 0−3
v=−0.2 V v = -0.2\,\text V v=−0.2 V
The voltage came out with a minus sign, meaning the terminal with the +++ voltage polarity is 0.2 V 0.2 \,\text V 0.2 V below the terminal with the −-− sign. We used the sign convention, and we let the math produce the right sign, even with the negative current.
Exceptions
You will run into cases every now and then where you can’t or don’t want to use the sign convention for passive components. In those cases, the current arrow will be pointing into the negative terminal of the element. When this happens, you don’t need to freak out, but your spidey sense should be tingling. You deal with this the same way we did in Example 1x where we included a −-− sign in Ohm’s Law.
This situation came up when I wrote the formal derivation of the RC natural response.
spidey sense
"The so-called ‘spidey sense’ or ‘spider sense’ generally refers to an extraordinary ability to sense imminent danger, attributed to the comic-book superhero Spider Man."
Power
The power in a resistor is,
P=i v P = i \, v P=i v
Power is energy transferred over a period of time, measured in joules/second.
The sign convention has an impact on how we think about power. Power can be generated or it can be dissipated. When we use the sign convention, power dissipation ends up with a positive sign, and power generation ends up with a negative sign. Let’s find the power dissipated by the 250 Ω 250\,\Omega 2 5 0 Ω resistor,
First find the current,
i=v R=2 V 250 Ω=8 mA i = \dfrac{v}{\text R} = \dfrac{2\text V}{250 \,\Omega} = 8 \,\text{mA}i=R v=2 5 0 Ω 2 V=8 mA
Then find the power,
P resistor=i v=8 mA⋅2 V=+16 mW P_\text{resistor} = i\,v = 8\,\text{mA} \cdot 2\,\text V = +16\,\text{mW}P resistor=i v=8 mA⋅2 V=+1 6 mW
Power dissipation has a positive sign.
What happens if we apply the passive sign convention to the voltage source?
We know the voltage source will provide 8 mA 8\,\text{mA}8 mA coming out of its top terminal (Ohm’s Law for the resistor tells us that). With the current arrow pointing in the direction shown, the current is i=−8 mA i = -8\,\text{mA}i=−8 mA.
Something interesting happens when we compute the power of the voltage source.
P voltage source=i v=−8 mA⋅2 V=−16 mW P_\text{voltage source} = i\,v = -8\,\text{mA} \cdot 2\,\text V = -16\,\text{mW}P voltage source=i v=−8 mA⋅2 V=−1 6 mW
The voltage source is a power generator. Power generation has a negative sign.
The negative sign is a side effects of using the sign convention for passive components on power-generating elements like the voltage source.
Is there such a thing as negative power?
Power is never actually negative. The minus sign comes from using the sign convention for passive components. If you are talking to someone about power it is clearer to use the words dissipate and generate rather than numeric signs +++ and −-−.
If you are an engineer in the power generation industry you might not want to go around bragging that you built a −100 Megawatt-100 \,\text{Megawatt}−1 0 0 Megawatt solar panel facility, so you are forgiven if you don't mention the −-− sign.
What is negative power good for?
The idea of negative power is not a bad thing. If you are creating a power budget for a complicated system, you compute all the positive power dissipated by passive elements and balance that against all the negative power from power-generating elements. Everything should add up to zero.
Summary
The sign convention for passive components says,
The current arrow points into positive voltage terminal of the element.
With this sign convention we directly apply Ohm’s Law (v=i R)(v = i\,\text R)(v=i R) to resistors.
If you ever see the sign convention being violated, it should grab your attention and remind you to include a minus sign in Ohm’s Law.
When you use the sign convention on passive elements (R,L,C)(\text R, \text L, \text C)(R,L,C), power P=i v P = i \, v P=i v has a positive sign. Positive power is associated with power dissipation.
If you apply the passive sign convention to a power-generating element, the power comes out with a negative sign. Negative power is associated with power generation.
Questions
dot_, 03 September 2021
i wish there’s a like button on this page. i would give 1000 likes. you are my saviour.
↪︎ Reply to dot_
kris Chedumbarum, 14 September 2020
Easy to understand
↪︎ Reply to kris Chedumbarum
Jonah, 02 April 2020
In the second power example, is the current not flowing into the positive terminal and so should therefore be positive? Since current flows into the positive terminal and out the negative terminal?
↪︎ Reply to Jonah
Willy McAllister, 02 April 2020
The second power example is where I apply the sign convention for passive components to an “active” (not passive) circuit element, the voltage source. We are abusing the passive sign convention somewhat when we do this, but let’s see what happens. The current arrow points in to the + terminal of the voltage source. The current arrow defines which direction we consider as positive current. That doesn’t mean current actually flows that way, but if it did, its value would have a + sign. In fact, we know the current flows the other way, out of the + battery terminal and into the top of the resistor. So with the current arrow pointing into the battery, the value of the current is a negative value, i = -8mA. If you do a power calculation on the voltage source you multiply a positive voltage times a negative current, ending up with a negative power for an object that generates power. That is an outcome of using the passive sign convention.
Fernando, 13 July 2017
“An electric potential (also called the electric field potential or the electrostatic potential) is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration.” cf.
This definition of potential seems to me explain this passive element convention through physics, isn’t it? Positive charges go from higher potential to lower potential.
Is this correct?
Thank you!
↪︎ Reply to Fernando
Willy McAllister, 14 July 2017
This definition applies to the term “voltage”. “Volt” is the honorary name of electric potential difference.
The Passive Sign Convention is more of a bookkeeping rule. If we want to have only one version of Ohm’s Law, with no minus signs in it, we need to have a rule that the voltage and current polarities have to be arranged in the proper way. It would be a drag to have two versions of Ohm’s Law so we could let everyone draw polarity any which way. Sometimes you would use v = iR, and other times v = -iR, and there would be tons of errors.
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188423 | https://www.glowm.com/resources/glowm/cd/pages/drugs/f022.html | flumazenil
flumazenil Romazicon Pharmacologic classification: benzodiazepine antagonist Therapeutic classification: antidote Pregnancy risk category C Available forms Available by prescription only Injection: 0.1 mg/ml in 5-ml and 10-ml multiple-dose vials Indications and dosages Complete or partial reversal of sedative effects of benzodiazepines after anesthesia or short diagnostic procedures (conscious sedation).Adults: Initially, 0.2 mg I.V. over 15 seconds. If patient doesn’t reach desired level of consciousness after 45 seconds, repeat dose. Repeat at 1-minute intervals until a cumulative dose of 1 mg has been given (initial dose plus four additional doses). Most patients respond after 0.6 to 1 mg of drug. If resedation occurs, dose may be repeated after 20 minutes, but no more than 1 mg should be given at one time, and patient shouldn’t receive more than 3 mg/hour. Management of suspected benzodiazepine overdose.Adults: Initially, 0.2 mg I.V. over 30 seconds. 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Use cautiously in alcohol-dependent or psychiatric patients, in those at high risk for seizures, and in those with head injuries, signs of seizures, or recent high intake of benzodiazepines, such as patients in the intensive care unit. Interactions Drug-drug.Antidepressants, drugs that can cause seizures or arrhythmias: May cause seizures or arrhythmias after flumazenil removes the effects of the benzodiazepine overdose. Don’t use flumazenil in mixed overdose, especially when seizures (from any cause) are likely to occur. Drug-food.Any food: Ingestion of food during I.V. flumazenil infusion increases drug clearance by 50%. Be aware of this interaction. Adverse reactions CNS:dizziness, abnormal or blurred vision, headache,seizures, agitation, emotional lability, tremor, insomnia. CV:arrhythmias, cutaneous vasodilation, palpitations. GI:nausea, vomiting. Respiratory:dyspnea, hyperventilation. Skin:diaphoresis, pain at injection site. Effects on lab test results None reported. Overdose and treatment In clinical trials, large doses of flumazenil were administered I.V. to volunteers in the absence of a benzodiazepine agonist. No serious adverse reactions, clinical signs or symptoms, or altered laboratory tests were noted. In patients with benzodiazepine overdose, large doses of flumazenil may produce agitation or anxiety, hyperesthesia, increased muscle tone, or seizures. Seizures may be treated with barbiturates, phenytoin, or benzodiazepines. Special considerations • Because flumazenil has a duration of action shorter than that of benzodiazepines, monitor patient carefully and administer additional drug as needed. Duration and degree of effect depend on plasma levels of the sedating benzodiazepine and the dose of flumazenil. • Monitor patient for resedation after reversal of benzodiazepine effect. Usually, serious resedation is unlikely in patient who fails to show signs of resedation 2 hours after a 1-mg dose of flumazenil. • Flumazenil can be given by direct injection or diluted with a compatible solution. Breast-feeding patients • It isn’t known whether drug appears in breast milk. Use cautiously in breast-feeding women. Pediatric patients • Because no clinical data exist regarding risks, benefits, or dosage range in children, manufacturer doesn’t recommend its use in this age-group. Patient education • Because of risk of resedation, advise patient to avoid hazardous activities (such as driving a car), alcohol, CNS depressants, and OTC drugs within 24 hours of the procedure. Reactions may be common, uncommon, life-threatening, or COMMON AND LIFE THREATENING. ◆ Canada only ◇ Unlabeled clinical use |
188424 | https://www.youtube.com/watch?v=jB6NSwfcR20 | Khan Academy Tutorial: scientific notation
West Explains Best
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hi this is Mr West and today we're doing a Khan Academy tutorial on scientific notation this was requested by the chatha family thank you so much for requesting it I do try to get to all of my requests I can't get to all of them but I get to as many as I can so thank you so much leave a comment if you would like your own video main so the basics of scientific scientific notation is this that if we have 10 times 10 times 10 so this is the first part I'm looking at you're going to get a thousand okay notice how there's three zeros there so one two three three zeros now we could also represent this in scientific notation and it's just a shorter way to write things when uh we're talking about really big things or really small things we could have a lot of decimals so a quicker way to do it is with scientific notation notice how this 10 to the third takes up less space than this 10 times 10 times 10 or even that one thousand it takes up less space okay so uh it's the same thing 10 to the third power three zeros okay we get a thousand I'm going to show another thing in just second two okay so let's move on to the next one here we have 0.1 or 0.10 times 0.10 times 0.10 so instead of ten we have 10 hundredths or point one okay and that equals 0.001 so notice how the pattern it's kind of the similar here was one thousand here this is one thousandth okay so we have the tens hundreds and that's the thousandths place so that's one thousandth instead of one thousand okay and then a quicker way to write that is ten to the negative three if we multiply um 0.10 times itself three times we're gonna get the same thing point zero zero one now notice how numbers get smaller when there's a negative exponent and then they get bigger when there's a positive exponent I wanted to Circle this in green so as we multiply by 10 to some sort of positive power we're going to move the decimal to the right and if we multiply it by a negative power okay so 10 to a negative power we're going to move the decimal to the left and that's kind of like the guiding principle right here so if you understand that okay that's going to be one of the most important parts okay so let's look at this example notice how this became a number that's smaller than one okay so we already know that we're going to have some sort of 10 to the negative power so we're going to do here is it's got to be a number when we do scientific notation it's got to be a number between 0 and 10 as we express it so we wouldn't want to say like 82.35 for example okay we're going to want to express this as 8.235 and it needs to be exact so we need to include all those digits but we want this number to be between 0 and 10. not 82.35 or 823.5 whatever okay and it can't be 0.8235 so it really has got to be between one sorry I didn't mean uh zero I meant between one and ten all right so from there between 1 and 10 not 0 and 10 between 1 and 10 we're going to want to okay so we're doing 8.235 how do we go from eight point two three five into this okay we already know we're going to be multiplying by 10 to some negative number what we're going to do now is we're going to put we're going to place it where we want it between the number and one and ten so I'm placing this one uh decimal right here and I'm counting over how many places I moved over one two three four okay another way to do it is just go this way we go one two three four to get it to eight point two three five so we're going to say this is going to be 10 to the negative 4 times 8.235 okay so since we're just counting over how many places we move that decimal over to the left or to the right as I showed in that second example so I'm going to write that out I'm going to have actually I'm going to go 8.235 times make sure you don't use the ax you use the times and then we go 10 I need a power here to the negative 4 okay because I moved it over four times and we'll check it and let's move on okay so we're done with that example we got a hang of it so now we're going in standard form all right so with standard form this is a little bit easier we have 9.647 raised to a positive power so we're going to move this decimal to the right seven times one two three four fill in a zero five fill into zero six fill in zero and then seven fill in a zero and technically we could put the decimal there but it does make a difference so we're going to write in 9647 with four zeros behind it and that's going to be our answer nine six four seven four zeros nine six four seven with one two three four four zeros check it and let's move on nine ten thousandths okay we better figure this out so better know your decimal places here so we have the ones place we have the tens place we have the hundreds thousands ten thousands so we have nine ten thousands okay so again this was thousands right here so that's ten thousands right there and then hundred thousands etcetera so this is the number so if we were to move this over okay one two actually should be red right because we're going a negative number we go one two three four to get it between nine point zero okay to get between a number in one and ten so we have nine times ten and we moved over how many times one two three four times to the negative four so nine times ten to the negative fourth nine times ten to the negative fourth power check it alright next question uh this one's not bad okay it's very close but notice it can't be this is the reason why we can't have it between zero and ten I don't know why I said that earlier it's between one and ten okay so how do we get this between one and ten well we can just move the decimal place over one time okay actually should be in red right move it over one time and it's going to be 5.401 okay but because we moved it over uh and it's a number smaller than one we're gonna do times ten to the negative first power so if we did times 10 to the negative First it would be the same thing 5.401 5.401 times 10 to the negative first power okay here we go three more questions wow it's the same one Express this number in standard form I think we just had this uh verbatim so we're just going to go one decimal over let me write it out just to be thorough so we're going to go 1 over to the left okay make it a smaller number anytime it's negative exponent we're making it smaller so it's just going to be 0.5401 and that's our final answer let's move it over one place 763 thousands okay so all right zero point and then notice that thousands has to be our last number hundreds and then seven this is the number okay 0.763 we write this number and then this last digit needs to be in the thousands place and that's that so 763 thousands looks like that 0.763 but we need to express in scientific notation so let's move the decimal over one time so it's a number between one and ten we can't move it over twice otherwise it'd be 76.3 that's no good we want 7.63 and to get there we only moved it over once okay and since this is a number that's smaller than one it's a smaller number we have to multiply it by 10 times uh excuse 10 to the negative first Power Times 7.63 okay so we're gonna I'll I don't know why I keep saying that reverse 7.63 times 10 to the what I say negative first Power yes just moved over once times 10 to the negative first power all right one more Express this number in scientific notation okay so what we're going to do here is it's a big number we want it to be 3.868 and how many times we moved over the decimal one two three should be green sorry let's go green one two three four five six seven eight nine nine places I'm pretty sure I counted that right looks good hopefully this isn't a mess up so I'm going to say 3.868 times 10 to the ninth power because it's a very big number not a small number and I think we should be good 10 3.868 3.868 times 10 to the ninth power hopefully I counted correctly and I did all right I hope you enjoyed this video make sure to leave a comment if you need help with this or if you want to make a suggestion and look forward to seeing you next time right here on West 16 Fest |
188425 | https://www.lovethegarden.com/uk-en/article/oedema | Oedema - treatment and control | Love The Garden
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Via-interconnect ball-grid-array (ViB) fabricated by die-middle PCB-like process | IEEE Conference Publication | IEEE Xplore
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Conferences>2015 10th International Micro...
Via-interconnect ball-grid-array (ViB) fabricated by die-middle PCB-like process
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Charlie Lu; H.K. Tien; J.C. Lin; Kidd Lee; Pie Shih
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Package Construction and Warpage Simulation by Fea
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This is the very first package made by PCB-like process with die-middle concept in the industry, and is named as die-middle ViB-W. ViB is a BGA package with via to electr...Show More
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Abstract:
This is the very first package made by PCB-like process with die-middle concept in the industry, and is named as die-middle ViB-W. ViB is a BGA package with via to electrically interconnect the chip I/O pads to package. The suffix W means the ViB is fabricated by wet PCB-like process which interprets lower cost when compared with the wafer Fab-like dry process. Die-middle process has its advantage for potentially high production yield. Thus die-middle ViB-W package possesses high yield/low cost feature which will make it become dominating in low-end to mid-range applications. In this paper, FEA warpage simulation was served to determine the package construction for a ViB-W with 15mm×15mm body size. Two dielectric materials were built into two groups of samples to compare the warpage and reliability performances. Both the groups with 15mm×15mm body size succeeded to meet the requirements of warpage and reliability. Warpage characterization by Shadow Moire was carried out. Warpage behavior is analyzed in details to understand the effect of reflow process and moisture soaking. Bigger body size ViB-W with 24mm×24mm and 29mm×29mm were built to explore the capability boundary of the package construction and BOM designed for 15mm×15mm.
Published in:2015 10th International Microsystems, Packaging, Assembly and Circuits Technology Conference (IMPACT)
Date of Conference: 21-23 October 2015
Date Added to IEEE Xplore: 28 December 2015
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DOI:10.1109/IMPACT.2015.7365199
Publisher: IEEE
Conference Location: Taipei, Taiwan
Contents
Introduction
Packaging technology has been evolving very fast and diversified since last decade. From application point of view, CSPs (Chip Scale Packages) and WLPs (Wafer Level Packages) has played dominating role in the low-end/mobile segment owing to their smaller form factor. For high-end application, 3D-IC has demonstrated its superior performance, yet it is still a long way for industry to adopt widely not only because of its very high cost but also mainly due to technical challenges, e.g. EDA (Electronic Design Automation) and design tool availability, KGD (Known Good Die), thermal management, and test-ability etc. , . 2.5D-IC is then invented as an alternative solution to dodge the difficulty of 3D-IC . However, 2.5D-IC is still too high cost to be adopted for most of the high-end to midrange application. Thus, lots of alternative technologies were proposed to further reduce the cost while trying to maintain the similar level of performance. One of them is ViB packages .
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188427 | https://emj.bmj.com/content/39/11/818 | Published Time: 2022-11-01
Management of patients presenting to the emergency department with sudden onset severe headache: systematic review of diagnostic accuracy studies | Emergency Medicine Journal
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Management of patients presenting to the emergency department with sudden onset severe headache: systematic review of diagnostic accuracy studies
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Systematic review
Management of patients presenting to the emergency department with sudden onset severe headache: systematic review of diagnostic accuracy studies
Walton1,
Robert Hodgson1,
Alison Eastwood1,
Melissa Harden1,
James Storey2,
Taj Hassan3,
Stuart Randall4,
Abu Hassan3,
John Williams5,
Ros Wade1
1 Centre for Reviews and Dissemination, University of York, York, UK
2 Department of Acute Internal Medicine, Leeds Teaching Hospitals NHS Trust, Leeds, UK
3 Department of Emergency Medicine, Leeds Teaching Hospitals NHS Trust, Leeds, UK
4 Department of Adult Neurology, Leeds Teaching Hospitals NHS Trust, Leeds, UK
5 Patient representative, UK, UK
Correspondence to Mr Matthew Walton, University of York, York, UK; matthew.walton@york.ac.uk
Abstract
Objective Advances in imaging technologies have precipitated uncertainty and inconsistency in the management of neurologically intact patients presenting to the Emergency Department (ED) with non-traumatic sudden onset severe headache with a clinical suspicion of subarachnoid haemorrhage (SAH). The objective of this systematic review was to evaluate diagnostic strategies in these patients.
Methods Studies assessing any decision rule or diagnostic test for evaluating neurologically intact adults with a severe headache, reaching maximum intensity within 1 hour, were eligible. Eighteen databases (including MEDLINE and Embase) were searched. Quality was assessed using QUADAS-2. Where appropriate, hierarchical bivariate meta-analysis was used to synthesise diagnostic accuracy results.
Results Thirty-seven studies were included. Eight studies assessing the Ottawa SAH clinical decision rule were pooled; sensitivity 99.5% (95% CI 90.8 to 100), specificity 24% (95% CI 15.5 to 34.4). Four studies assessing CT within 6 hours of headache onset were pooled; sensitivity 98.7% (95% CI 96.5 to 100), specificity 100% (95% CI 99.7 to 100). The sensitivity of CT beyond 6 hours was considerably lower (≤90%; 2 studies). Three studies assessing lumbar puncture (LP; spectrophotometric analysis) following negative CT were pooled; sensitivity 100% (95% CI 100 to 100), specificity 95% (95% CI 86.0 to 98.5).
Conclusion The Ottawa SAH Rule rules out further investigation in only a small proportion of patients. CT undertaken within 6 hours (with expertise of a neuroradiologist or radiologist who routinely interprets brain images) is highly accurate and likely to be sufficient to rule out SAH; CT beyond 6 hours is much less sensitive. The CT–LP pathway is highly sensitive for detecting SAH and some alternative diagnoses, although LP results in some false positive results.
emergency department
diagnosis
computed tomography
headache
Data availability statement
All data relevant to the study are included in the article or uploaded as supplementary information. Not applicable.
This is an open access article distributed in accordance with the Creative Commons Attribution 4.0 Unported (CC BY 4.0) license, which permits others to copy, redistribute, remix, transform and build upon this work for any purpose, provided the original work is properly cited, a link to the licence is given, and indication of whether changes were made. See:
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emergency department
diagnosis
computed tomography
headache
Key messages
What is already known on this subject
Guidelines typically recommend non-contrast CT head followed by lumbar puncture in patients who present with headache symptoms suspicious for subarachnoid haemorrhage.
More recently, studies have questioned the need for routine lumbar puncture after a normal CT head.
Additionally, a decision rule to direct imaging has been widely studied.
What this study adds
In this systematic review and meta-analysis, we found that the Ottawa subarachnoid haemorrhage clinical decision rule has low specificity, and could result in significant additional unnecessary testing.
CT head within 6 hours of headache onset, with images assessed by a neuroradiologist or radiologist who routinely interprets brain images, is highly accurate; around 658 CT-negative patients would have to undergo further investigation to identify a single case of subarachnoid haemorrhage.
CT head undertaken beyond 6 hours is much less sensitive, therefore additional testing is more likely to be beneficial.
In healthcare systems and settings in which neuroradiology expertise is unavailable, caution should be exercised when translating the diagnostic accuracy of CT head in the literature to clinical decision making.
How this study might affect research, practice or policy
CT head within 6 hours of headache onset and with access to neuroradiology expertise is likely to be sufficient to rule out subarachnoid haemorrhage.
The diagnostic accuracy of CT head may be contingent on time since symptom onset, which must be accounted for in practice, and investigated in future research.
Risk tolerance of the patient and physician for the potential consequences of investigation and missed diagnoses will continue to inform practice.
Introduction
Non-traumatic acute headache accounts for around 2% of adult Emergency Department (ED) attendances.1 Sudden onset severe headaches may be caused by a primary headache disorder or may be secondary to a more serious underlying pathology, such as subarachnoid haemorrhage (SAH). Diagnosis of SAH is particularly challenging in alert, neurologically intact patients presenting with acute severe headache. Clinical features separating these patients from higher volume complaints with a similar presentation (eg, migraine) are often unreliable indicators of who requires further investigation.2
Advances in imaging technologies have precipitated uncertainty and inconsistency in the optimal management of neurologically intact patients presenting to the ED with non-traumatic sudden onset severe headache.3 4 Given increasing evidence on the potentially low therapeutic value of lumbar puncture (LP) following CT of the head, and its associated adverse effects,3 5–7 updated evidence-based guidance is needed. We therefore undertook a systematic review of evidence on diagnostic strategies for neurologically intact adult patients presenting to hospital with non-traumatic sudden onset severe headache, reaching maximum intensity within 1 hour.
Methods
The review protocol is registered on PROSPERO (CRD42020173265). This paper conforms to the recommendations of the Preferred Reporting Items for a Systematic Review and Meta-Analysis of Diagnostic Test Accuracy Studies statement.8
Search strategy and selection criteria
Eighteen databases (including MEDLINE and Embase) were systematically searched in February 2020. Further details of the search strategy are presented in online supplemental file 1. To meet inclusion criteria, studies had to assess any care pathway for ruling out SAH (including clinical decision rules and specific diagnostic tests, such as CT or LP) in neurologically intact adult patients presenting to hospital with a sudden onset severe headache (reaching maximum intensity within 1 hour), with a clinical suspicion of SAH. Studies of patients who had suffered a head injury (ie, traumatic headache) were excluded. Any primary study design (other than single case study) was eligible for inclusion. Outcomes of interest included diagnostic accuracy, quality of life and adverse events. Two researchers (MW and RW) independently screened the titles and abstracts of all retrieved records and subsequently all full text publications for inclusion. Disagreements at each stage of the study selection process were resolved through discussion. Authors of potentially relevant conference abstracts were contacted for additional information. Relevant foreign language studies were translated and included in the review.
Supplemental material
[emermed-2021-211900supp001.pdf]
Data extraction and quality assessment
Data were extracted on study methods, patient, intervention and reference standard characteristics, outcome measures, adverse events and results (presented in online supplemental file 2). Data extraction and quality assessment were undertaken by one researcher and independently checked by a second. The majority of studies were assessed for quality using the Quality Assessment of Diagnostic Accuracy Studies 2 (QUADAS-2) tool.9 The QUADAS-2 tool was not appropriate for studies where a reference standard test was not used, therefore, a quality assessment tool was developed by RW specifically for the review, piloted and refined before use (see online supplemental file 3 for details).
Supplemental material
[emermed-2021-211900supp002.pdf]
Supplemental material
[emermed-2021-211900supp003.pdf]
Data analysis
Where sufficient information was reported, diagnostic accuracy data were extracted into 2×2 tables to calculate sensitivity, specificity, false positive and false negative rates. Where equivalent diagnostic strategies or tools were used in three or more studies, the hierarchical bivariate model described by Reitsma et al10 was fitted, along with an extension described by Simmonds and Higgins11 to meta-analyse sensitivity and specificity while accounting for correlation between the two, and within-person correlation between test results. Meta-analyses used standard random-effects DerSimonian-Laird methods. Subgroups were analysed separately to account for underlying differences in diagnostic strategies. The diagnostic accuracy of CT conducted <6 hours from headache onset was analysed separately, as CT accuracy is known to drop rapidly outside of this time frame.12 The accuracy of different methods of cerebrospinal fluid (CSF) analysis was also assessed. Where results could not be pooled, they were synthesised narratively along with reported adverse event data.
Public and patient involvement
A patient collaborator with experience of presenting to an ED with a sudden onset severe headache was involved throughout the project. Three additional patients were recruited to an advisory group. The patients provided input during protocol development and interpretation of review findings.
Results
The search strategy identified 15 750 records; 37 cohort/before and after studies were eligible for inclusion (figure 1 and table 1). More detailed study characteristics and results are presented in online supplemental file 2.
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Figure 1
Flow diagram of the study selection process. Any study which recruited patients before the year 2000 was considered to have used outdated CT technology.
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Table 1
Studies included in the systematic review
Twelve studies had a low risk of bias for all domains, the other 25 were at risk of bias. Twenty-eight studies were assessed using the QUADAS-2 tool; results are summarised in figure 2.9 Nine studies did not use a reference standard test, therefore, QUADAS-2 was inappropriate; a quality assessment tool developed specifically for the review was used instead. Quality assessment results are presented in the online supplemental file 3.
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Figure 2
Quality Assessment of Diagnostic Accuracy Studies 2 (QUADAS-2) results.
Clinical decision rules
Thirteen studies assessed the clinical decision rules developed by Perry et al for screening patients according to the presence of clinical characteristics associated with a high risk of SAH.13–25 The predecessors of the Ottawa SAH Rule (sometimes termed the ‘Canadian clinical decision rules 1, 2 and 3’) were evaluated in six studies. Results of these studies can be found in online supplemental file 2. Rule 1 was refined to develop the final Ottawa SAH Rule, which states that alert patients with new severe atraumatic headache, reaching maximum intensity within 1 hour, require investigation if one of the following are present: age ≥40 years, neck pain/stiffness, witnessed loss of consciousness, onset during exertion, thunderclap headache or limited neck flexion.21
A summary of the diagnostic performance of the Ottawa SAH Rule in the individual studies and pooled results generated from the bivariate meta-analysis are presented in table 2. Perry et al (2017) is excluded,22 due to patient overlap with the larger Perry et al (2020) study.23 The overall SAH prevalence in the studies ranged from 1.6%24 to 10%14 with a population-weighted mean prevalence of 5.0%. The Ottawa SAH Rule is highly sensitive, but specificity was low; strict application of the rule would result in 76% of SAH-negative patients undergoing further investigation with no additional benefit. There was considerable heterogeneity in false positive rates (FPR), potentially due to study population differences or inconsistent application of the rule. No studies assessed the accuracy of the Ottawa SAH Rule in patient subgroups by time to headache peak.
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Table 2
Diagnostic performance of Ottawa SAH Rule
Pathway of CT followed by LP
The pathway of non-contrast CT followed by LP was assessed in six studies.7 26–30 Only one reported complete diagnostic data, so meta-analysis was not performed. Overall, the pathway was highly sensitive, but specificity was low in some studies owing to the high FPR for LP. Importantly, this pathway also identified other significant pathologies, such as intracerebral haemorrhage, brain tumour and meningitis. More detailed results for this pathway can be found in online supplemental file 2.
Computed tomography
The diagnostic accuracy of CT was assessed in nine studies,7 12 20 23 26 30–33 although three studies had significant patient overlap,12 20 33 therefore, only the results for the largest of the three are presented.12
CT undertaken within 6 hours of headache onset
Four studies of CT <6 hours from headache onset were included in bivariate meta-analysis (table 3).12 23 30 32 In all four studies, CT scans were assessed by neuroradiologists or radiologists who routinely interpret head CT images. Perry et al (2020) classed two incidental aneurysms with traumatic tap on subsequent LP as SAH, and thus as false negatives. This is inconsistent with the other included studies and with our interpretation of what constitutes a false negative. Therefore, these two patients were reclassified as true negatives.
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Table 3
Diagnostic performance of CT (<6 hours from headache onset)
The recruitment of patients from SAH patient databases in Backes et al32 meant that SAH patients were over-represented in the study population (41.5%). SAH prevalence ranged from 9.2%23 to 12.7%12 in the other three studies, with a population-weighted average prevalence of 10.8%. Assuming that these patients are representative of those presenting to EDs in practice, the pre-test probability of SAH in patients with headache who undergo CT within 6 hours is 10.8%. Using the pooled estimate of diagnostic accuracy, the post-test probability of having suffered a SAH after a negative <6 hour CT result is 0.15%. Assuming a hypothetical follow-up test (eg, LP) has 100% accuracy, this means that 658 (95% CI 250 to 1749) patients would have to undergo further investigation to identify a single case of SAH.
One additional study assessed the diagnostic accuracy of CT <6 hours, but was excluded from the meta-analysis as it did not report sufficient diagnostic accuracy data to construct a 2×2 table to calculate sensitivity and specificity.26 In this study, 760 patients had a negative CT (assessed by a staff radiologist) and subsequently underwent LP; 7% of CSF samples were initially considered positive for SAH, but subarachnoid blood was identified in only one patient on review by two neuroradiologists and a neurologist. The negative predictive value for detection of blood on CT by staff radiologists was 99.9% (95% CI 99.3 to 100).
CT undertaken at any time interval from headache onset
Three studies of CT undertaken at any time interval from headache onset were included in bivariate meta-analysis (table 4).7 12 32 In all three studies, CT scans were assessed by neuroradiologists or radiologists who routinely interpret head CT images. The prevalence of SAH in patients undergoing CT at any time since headache onset was lower than in those who underwent CT within 6 hours. Prevalence was 2.7% in the study by Cooper et al7 and 7.7% in the study by Perry et al.12 As noted above, SAH patients were over-represented in the Backes et al study population (35.2%).32
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Table 4
Diagnostic performance of CT (at any time)
The pooled sensitivity of CT at any time since headache onset was 94.1% (95% CI 91.0 to 96.2). This result includes patients who had CT <6 hours, as well as CT >6 hours, from symptom onset. Results from Perry et al12 and Backes et al32 suggest CT scans performed >6 hours after symptom onset have significantly poorer performance, reporting sensitivities of 85.7% (95% CI 78.3 to 90.9) and 90.0% (95% CI 76.3 to 97.2), respectively. The bimodal nature of the diagnostic performance of CT means that the ‘CT at any time’ statistics are misleading, as the timing of CT has a significant impact on the pre-test and post-test probabilities of SAH.
One additional CT study compared interpretation by emergency physicians (images viewed on standard resolution desktop screens) with the reference standard of neuroradiologists’ readings (images viewed using dedicated high definition screens).31 The sensitivity of CT interpreted by emergency physicians was 84% (95% CI 63.9 to 95.5) and specificity was 95% (95% CI 90.9 to 97.2). However, this study was considered to have a high risk of bias due to the difference in hardware used between the two specialties for examining CT images.
Lumbar puncture
The diagnostic accuracy of LP in patients judged to be SAH-negative using CT was assessed in 11 studies.7 30 34–42 The method of assessing CSF for xanthochromia varied, with Canadian and American studies predominantly using visual inspection and UK and European studies predominantly using spectrophotometry. LP was not always undertaken ≥12 hours from symptom onset. The standard UK NHS practice is to take the CSF sample ≥12 hours from symptom onset to allow xanthochromia to develop, with samples analysed using spectrophotometry.43
Spectrophotometric CSF analysis
Three studies reported diagnostic accuracy data for spectrophotometric CSF analysis following negative CT (table 5).7 36 40 Samples were analysed for presence of bilirubin using the UK National External Quality Assessment Service protocol/assay.43 The prevalence of SAH in these studies was only 0.65%, likely due to prescreening with CT. The FPR (and subsequent rate of angiography) was particularly high in Perry et al (2006), perhaps due to reported limitations in the spectrophotometric equipment used by the authors. The FPR in the more recent studies was substantially lower and likely better represents the diagnostic accuracy of CSF spectrophotometry in current practice.
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Table 5
Diagnostic performance of spectrophotometric CSF inspection (UK National External Quality Assessment Service)
Three further studies assessed CSF spectrophotometry in patients who underwent LP after negative CT, but reporting was insufficient for meta-analysis.34 38 42 Horstman et al included 30 patients with a negative CT result for whom bilirubin was detected in the CSF; aneurysms were identified in 13 patients; however, all cases presented 4–14 days after symptom onset.38 Brunell et al included 453 patients, 400 (88%) of whom presented with thunderclap headache; 14 (3%) patients had a pathological diagnosis based on LP, most commonly aseptic meningitis, and 5 (1.1%) had SAH.34 Four of the five SAH patients had non-aneurysmal SAH which did not require surgical intervention and the other SAH patient had reduced consciousness, therefore did not strictly meet the inclusion criteria for this review.34 Sansom et al included 60 CT-negative patients with thunderclap headache; all samples were negative for xanthochromia but 8/60 CSF examinations were abnormal for other CSF parameters (protein, glucose, cells, microscopy), with cerebral infarction confirmed in two of these patients on subsequent investigation.42
Visual CSF inspection
Five studies examined the diagnostic accuracy of visible xanthochromia in CT-negative patients with further investigation and follow-up used as a reference standard.35 36 39–41 Three studies included sufficient information to calculate diagnostic accuracy (table 6). Sensitivity varied widely (50%–93%), due to the low prevalence of SAH (2%). The pooled false negative rate of 15% for visual inspection was higher than that for spectrophotometric analysis (0%).
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Table 6
Diagnostic performance of visual CSF inspection across identified studies
Migdal et al assessed 245 patients with ‘low risk clinical features’, which aligned with the population in this review, but identified no cases of SAH. However, 13/245 (5.3%) patients had LP-related complications that resulted in a return visit to the ED or hospitalisation.39 Perry et al examined the diagnostic accuracy of visible xanthochromia in ‘abnormal’ CSF samples drawn from 1739 (mostly) CT-negative patients; there were 15 (0.9%) patients classed as having aneurysmal SAH, 7 of whom had visible xanthochromia in their CSF.41
Red blood cell-based CSF analysis thresholds
Two studies explored methods to distinguish SAH from ‘traumatic tap’, where blood enters the CSF sample due to the LP procedure itself. Perry et al found that the presence of fewer than 2000×10 6/L red blood cells (RBCs) with no xanthochromia excluded a diagnosis of aneurysmal SAH (sensitivity 100% (95% CI 74.7 to 100), specificity 91.2% (95% CI 88.6 to 93.3)) in patients who had previously undergone CT.41 Heiser et al assessed the same RBC cut-off, reporting 81.6% sensitivity (95% CI 68.0 to 91.2) and 97.3% specificity (95% CI 95.7 to 98.4); the incidence of traumatic LP was 24.4%.37 These results are not directly comparable to those reported by Perry et al,41 as this population was not prescreened with CT.
Finally, Valle Alonso et al assessed 74 patients who underwent LP (method of analysis not specified) following negative CT <6 hours.30 LP was positive in one patient and inconclusive in two; further imaging ruled out bleeding in all three patients. Seven patients experienced postpuncture headache, two of whom were admitted for pain control.
CT angiography
Two small studies assessed CTA after normal CT/LP; no cases of SAH were identified, although other vascular abnormalities (including incidental aneurysms, cerebral venous thrombosis and reversible vasoconstriction syndrome) were identified.44 45
History and examination
Three studies explored the use of historical and emergent clinical factors as predictors of SAH.2 46 47 Two studies investigated the adequacy of assessment for SAH and one study assessed neurological examination for neck stiffness as a predictor of SAH. Using physicians’ clinical suspicion had a sensitivity 93% and specificity of 49%.46 Presence of individual clinical factors (age >65 years, temperature >38°C, systolic BP >160 mm Hg, neck stiffness) were poor predictors of secondary headache (sensitivity 37.8%, specificity 82.1%).2 Presence of neck stiffness was more strongly predictive of SAH in patients who had other high-risk clinical characteristics (eg, age ≥40 years, vomiting, transient loss of consciousness).47 Recording of history in medical records was poor.2 46 47
Discussion
In summary, the Ottawa SAH Rule does little to aid clinical decision making for patients with sudden onset severe headache. The FPR was high, such that 76% of SAH-negative patients would undergo further investigation with CT and/or LP with no diagnostic value with regard to SAH, resulting in greater healthcare resource use and higher rates of adverse events related to LP and CT radiation exposure. Evidence on use of the rule in patient subgroups by time to headache peak is lacking but could be informative for clinical practice given the importance of headache incipiency.
LP (with spectrophotometric CSF analysis) following negative CT was highly sensitive, although there was a 4.8% FPR. Spectrophotometry-based CSF analysis appeared to have a higher sensitivity but lower specificity than visual inspection for xanthochromia. Two studies reported rates of LP-related complications resulting in a return to the ED or hospitalisation (5%–10%). In view of the reduced sensitivity of CT >6 hours after headache onset, LP may be beneficial in these patients where a clinical suspicion of SAH remains. The CT–LP pathway also identified other significant pathologies, such as intracerebral haemorrhage, brain tumour and meningitis, meaning that its value could extend beyond the identification of SAH.
Non-contrast CT <6 hours from headache onset, with CT scans assessed by a neuroradiologist or radiologist who routinely interprets head CT images, is highly accurate for identifying SAH, and results in a very low post-test probability of SAH. This means that very large numbers of patients (estimated at 658) would have to undergo further testing to yield an additional case of SAH.
However, the relatively high rate of false positive LP results (4.8% using spectrophotometry) is likely to lead to yet more testing downstream with the potential for diagnosing incidental aneurysms, leading to difficult decisions about invasive procedures. A 2016 survey of UK clinicians reported a higher risk tolerance for missed SAH diagnoses among emergency clinicians than neurospecialists, with the former accepting over 2.5 times the risk of a missed SAH (2.8% vs 1.1%; p=0.03), and the latter more likely to advocate routine LP following a negative CT result (74% vs 39%; p=0.01).4 Emergency clinicians were also more inclined to omit LP if CT had been conducted within 6 hours of headache onset (35% vs 3%; p=0.002).
Draft guidelines by the National Institute for Health and Care Excellence (publication delayed due to COVID-19) recommend that when there is no evidence of SAH on CT images taken <6 hours from symptom onset, LP should not be routinely offered, and alternative diagnoses should instead be considered.48 However, we consider that in smaller centres without access to specialist neuroradiology expertise, or radiologists who routinely interpret head CTs, the accuracy of early CT may be reduced; studies included in our meta-analyses benefited from neuroradiology expertise. Introduction of universal access to expert interpretation of CT images could improve SAH-related patient outcomes through optimised targeting of further investigations while increasing efficiency of resource allocation. This may be achieved through widened neuro-specific training and teleradiology using other centres with relevant expertise. While interpretation of CT images using diagnostic deep learning algorithms (artificial intelligence) has the potential to improve consistency across centres, this has yet to be reliably demonstrated in high-quality studies.49
The prevalence of SAH was higher in patients who received CT <6 hours from headache onset than in the wider population of patients presenting to the ED with sudden onset severe headache (10.8% vs 7.0%). It is unclear whether this difference in pre-test probability can be assumed to exist at the point of patient assessment in the ED. Instead, triage based on severity of symptoms may have reduced wait time for CT, equally, symptom severity associated with true SAH could drive earlier presentation.
A limitation of this review was the substantial heterogeneity in the study methods and population characteristics of the included studies. The evidence base included too few patients, given the rarity of SAH events, missed diagnoses and alternative non-SAH pathologies. This led to heterogeneity in the results of some meta-analyses, and potentially meant uncertainty was underestimated in others.
There was a lack of research evidence on the small subgroup of patients who present to hospital several days after headache onset. Diagnosis of SAH in such patients is particularly challenging and there is a lack of guidance and consistency in how these patients are assessed.
Conclusions
The Ottawa SAH Rule rules out further investigation in only a small proportion of patients; its introduction into practice could result in substantially increased rates of unnecessary investigation. Assuming the availability of neuroradiology expertise, early head CT (<6 hours) appears to be sufficient to rule out SAH in patients with sudden onset severe headache in the vast majority of patients. CT undertaken >6 hours from headache onset is much less sensitive, therefore, LP is more likely to be beneficial, where a clinical suspicion of SAH remains. Risk tolerance of the patient and the physician, the expertise of the CT reader and consequences of additional investigations must all be considered.
Data availability statement
All data relevant to the study are included in the article or uploaded as supplementary information. Not applicable.
Ethics statements
Patient consent for publication
Not applicable.
Ethics approval
Not applicable.
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Supplementary materials
Supplementary Data
This web only file has been produced by the BMJ Publishing Group from an electronic file supplied by the author(s) and has not been edited for content.
Data supplement 1
Data supplement 2
Data supplement 3
Footnotes
Handling editor Richard Body
Twitter @tajekbhassan
Contributors MW was involved in all aspects of the systematic review process including study selection, data extraction, validity assessment, synthesis of the included studies and drafting the protocol and manuscript. MW is responsible for the overall content as guarantor. RH was involved in reviewing economic studies and commented on the protocol and manuscript. AE provided expertise in evidence synthesis and project management, input at all stages of the project and commented on the protocol and manuscript. MH was responsible for devising the search strategies, carrying out the literature searches, maintaining the literature database and writing the sections of the protocol and final report relating to the searches. JS provided clinical expertise, input at all stages of the project and commented on the protocol and manuscript. TH provided clinical expertise, input at all stages of the project and commented on the protocol and manuscript. MSR provided clinical expertise, input at all stages of the project and commented on the protocol and manuscript. AH provided clinical expertise, input at all stages of the project and commented on the protocol and manuscript. John Williams provided expertise as a patient collaborator. He provided input at all stages of the project and commented on the protocol and manuscript. RW was the principal investigator who led the application for funding and took overall managerial responsibility for the project. She was responsible for the day-to-day management of the systematic review, was involved in study selection, data extraction, validity assessment, synthesis of the included studies and drafting the protocol and manuscript.
Competing interests All authors have completed the ICMJE uniform disclosure form at www.icmje.org/coi_disclosure.pdf and declare: all authors had financial support from the National Institute for Health Research (NIHR) Research for Patient Benefit (RfPB) Programme for the submitted work; no financial relationships with any organisations that might have an interest in the submitted work in the previous 3 years; no other relationships or activities that could appear to have influenced the submitted work.
Patient and public involvement Patients and/or the public were involved in the design, or conduct, or reporting, or dissemination plans of this research. Refer to the 'Methods' section for further details.
Provenance and peer review Not commissioned; externally peer reviewed.
Supplemental material This content has been supplied by the author(s). It has not been vetted by BMJ Publishing Group Limited (BMJ) and may not have been peer-reviewed. Any opinions or recommendations discussed are solely those of the author(s) and are not endorsed by BMJ. BMJ disclaims all liability and responsibility arising from any reliance placed on the content. Where the content includes any translated material, BMJ does not warrant the accuracy and reliability of the translations (including but not limited to local regulations, clinical guidelines, terminology, drug names and drug dosages), and is not responsible for any error and/or omissions arising from translation and adaptation or otherwise.
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188428 | https://www.youtube.com/watch?v=8cyYqzlvSmg | Chapter 2 - Force Vectors
STATICS THE EASY WAY
41000 subscribers
10382 likes
Description
836537 views
Posted: 10 Sep 2015
Chapter 2:
4 Problems for Vector Decomposition. Determining magnitudes of forces using methods such as the law of cosine and law of sine.
415 comments
Transcript:
hello guys how are you doing H today we're going to be solving some problems I'm going to do some practice problems so you can go and review them H at home and basically it's going to be about chapter two we're going to be working with vectors and just basically Vector the composition the first problem that I want to work with you is this one and it says if the magnitude of the resultant force is to be 500 Newton directed along the positive y AIS determine the magnitude of the force F and his Direction angle or this angle here not the direction angle this angle here so the first thing that you have to do as always is to learn how to pick up the clues from the problems and the first clue that we have is the magnitude of the resultant Force but we also know the direction of the resultant Force because it say that it's going to be directed along the positive y AIS meaning the resultant force is going to be acting in this axis and it's positive so that means that it's going to be pointing out this problem basically the way it's stated H you can do it in two ways the first way you can do it by using trigonometry and using the law of H signs and cosiness and this is the first approach that we are going to do so basically if we're going to do the parallelogram law H what we have to do is is build a parallelogram here this is going to be parallel to the force and at the same time we're going to have the other one parall to this one in this direction that should coincide with the N of the force if this is made out of a scale but I know for sure that the resultant Force has to be here so this is going to be my resultant one from here to here and I know the value because they are telling me here the value the value for this one for this resultant force is 500 500 Newton I'm not going to put the units but you know it's Newtons in this case so once again remember what we have to look for what what we're looking for here is the value of this Force F and this angle in this position what else do we know regarding this problem well we know this angle here which is 15° is a given with the horizontal axis and if this is 15° and this is 90 because X and Y are orthogonal it doesn't say it but H they are orthogonal usually when when the angles the axis are refer as X Y is because they are orthogonal if they were in orthogonal they will use the different type of letters like u v or something like that so the total angle if this is 90° here and this is 15° the total angle between these two vectors or these two distances is going to be 90 + 15 so it's going to be 105° this angle here remember this angle is 90° and this is 15 90 + 15 is going to be 105 and the value of the force F the magnitude is going to be this side of this triangle Force F this is going to be Force F Well if you remember the law of cosiness law of cosine say that if you have a triangle you can say that c² is equal to a² + b² - 2 a B time cosine of the angle between A and B well in this particular example that we are f following this will be or this will be a and this will be B and this is the angle between both of them and what I'm looking for is f so I can say that f is going to be equal I'm going to take just the square root on both sides here and here to eliminate the C Square so f is going to be the square root of a but a is 700 700 squar plus b but B is this other Vector that we have here the other four is 500 so it's going to be 500 squar minus 2 700 500 cosine of the angle between A and B cosine of 105 once you solve this F is going to be equal to 95978 and remember to use the units Newton first part of the problem done this is the first part of the problem magnitude now the second part of the problem it says find its direction this direction here this angle I can find that angle in different ways but the easiest way or not the easiest way the way I'm going to show you right now is finding this angle through this other angle here I'm going to call this angle Alpha and of course this angle Theta is going to be 90° minus Alpha and we're going to apply the other law which is the law of signs and remember the law of signs what it says is a if you have a triangle I'm going to put it here if you have a triangle like this a and this angle is a this side is B and this angle is B and then this is the angle C and this is the side C in the same way that you have c² = a² b² - 2 a a cosine Alpha well in this case Alpha will be this angle here seene the law for the law of cosin the law of sign establish that a / by the S of a has to be equal to B / by the S of B has to be equal to C / by the S of C so the side divided by the sign of the opposite Ang the sign of the opposite angle has to be equal it's the same ratio for H every side of the triangle so if I'm going to apply the law of signs for this one I know this value here because this distance from here to here is going to be 700 how do I know that because it's 700 here and I know this one because this one is 500 and I know also this one because this one here I just calculated like 9 959 also I know this angle here this angle here is 105 what else do I know if the this angle here is 105 that means that this angle here is also 105 because these are two parallel lines cut by a straight line or you can say alternate internate triangle the two parallel lines cut by one straight line so this is 105 this is Alpha if I apply the law of signs I can say perfectly that the sign of alpha of this angle divid by 700 which is basically the same formulation I'm just flipping it both sides has to be equal to the sign of 105 divided by this side and this side I just calculated and it's 95978 from here we can solve for Alpha and you can say that Alpha is going to be equal to the inverse or 700 95978 ultip by S of 105 and Alpha is going to come to be 44. 79° but this is not the answer because Alpha will be this angle here the angle Theta is the one that we are looking for and that angle is going to be 90° minus Alpha because the alpha and Zeta are complementary so 90° minus 44. 79 Dees and Theta is going to be equal to 45.2 De that's the first way of approaching this problem as you can see it's really easy problem it's a very easy problem H but I don't know why a lot of people struggle with the law of signs and cosiness I don't know if the the physics background that you need or geometry trigonometry background is not sufficient uh from some students but if this is the case and you don't even remember what we're discussing here I I I'm going to ask you please to go and and review that part okay let's check the second approach which is a a more like a vector approach in this problem what we're going to do H we're going to use the components of the forces and we're going to use also the principles of equilibrium and basically if this is the same problem and we know the resultant force is going to be directed and it's going to be 500 and I like this approach more because don't don't get me wrong it's not that I don't like math or trigonometry but I like more physics I like to see what is happening I like to see what happens in the real world so I know that if this is 500 Newton and is acting vertical going up is because the horizontal Force resultant force in X is zero there's nothing X the resultant force is only vertical so then I can come and DEC compose this in the corresponding components the horizontal component of this 700 Force which is this one is going to be 700 multip by cosine of 15 and also the horizont Al component of this Force F which will be acting here is going to be F cosine of the angle and I know the resultant force in X is zero or I can say the summation of these two is zero so this is going to be then F cosine Theta positive because it's going in the positive direction of x minus remember negative it's acting in the negative Direction pointing to the left 700 time cosine of 15 Dees and that's going to be equal to0 or I can say also that F cosine Theta = 700 cosine 15° and that's I'm going to call that equation one now I'm going to do the same thing H with the vertical components the vertical component of this Force here f acting in this direction I can say that the summation I didn't copy this here before but summation of forces in X is going to be equal zero and now I'm going to say that the summation of forces in y is going to be equal to the resultant Force which is 500 why because the problem states that is telling you that 500 directed along the positive y AIS so if I know that then I'm going to do it this component here of F this one this one once again is going to be F Time s of the angle and this other component of this other Force here is going to be F which is 700 multip by S of 15 and this is 500 500 so what I have is f s of the angle minus 700 s 15 it's really important that you keep in in mind and keep track of the signs negative because it's acting to the opposite side of the positive side in this case downwards that has to be equal to 5 100 which is the resultant force and from here I can solve and say f sign of the angle is going to be equal to 500 + 700 s of 15° and I'm going to call this my equation number two this is a number sin 15 700 that's a number uh what I'm going to do now is I'm going to divide equation two term by term by equation one and then I'm going to get f sign of the angle equal remember this is a number 500 + 700 sin 15 divid by F cosine Theta which is the second equation and the second term is 700 cosine 50 degrees once you solve this this and this cancel out and then you have S / cosine is tangent of the angle and that's going to be equal to 1.0 074 and before you say anything I know the number of decimal places and significant figures here doesn't match the other one but I don't care I'm just showing a problem if you want to run out the result at the end you will give the appropriate amount of significant figures and then I can say that Alpha is equal to the inverse tangent of 1.74 not Alpha I mean Theta and Theta equal 45.2 Dees which is the same result that we got before here the other procedure now the other thing that we have to do is calculate the force and by for doing that I just have to plug this either into equation two or into equation one equation one is a little bit easier not that is that much easier but I'm going to plug in one plug in one and then you're going to have that F cine of 45.2 has to be equal 700 cosine 15 which is equation one and then f is going to be equal to 95978 Newton there you go two approaches for solving the same problem now I'm going to try to uh do another problem in the same recording I hope that is not too long the other I was going to do it several several recordings for several problems but I'm going to try I'm going to do a a new recording name a a new problem the same recording let's try to solve now H this other problem this is a problem that usually H has a lot of difficulties and the student always have a issues trying to solve these type of problems why because students are used to say Define the projection of the vectors projection and one thing is the projection and the other thing is are the components what happened is that the projection matches the components if the axis are perpendicular for example let me explain my point this is y and this is X and these two axes are perpendicular if I have a vector like that and I want to find the components of that Vector there's no problem with that easy why because the components of this Vector if this is Vector I don't know F the force F then this Force f is going to have a projection which is no another thing that this is like the shadow that that H Vector projects over the axis and like that this is the projection Vector this is the projection of the force F over the x- axis and this is the projection of the force F over the y axis you see it's like the shadow of that vector and of course if this is angle is Alpha FX is going to be equal to F cosine Alpha and this one which is the same side size of FY then is going to be f s of the am and this is what you are used to do now what happened in a situation like this what happened in a situation where you have one axis like that and the other axis like that and then you say oh wait a second this axis is U and this axis is V and then you have a force or any other Vector which is going to be acting somewhere here well this is the tricky part if I want the projection I'm going to look for the shadow of the is over this axis and if this angle is Alpha let's say the projection is going to be this distance and guess what if this is f the projection of f over the axis U is going to be F cosine Alpha the projection but this is not the component in order to find the components I have to go and I have to go parallel to the axis parallel to the axis so if I go parallel to this axis here and I and I look for this line and this line is parallel to this one and in this other case I'm going to make another line parallel to this other axis here also then the projections of that Vector I mean the components of that Vector are going to be this and this you see this is the component this is Fu and this is FB and as you can see f u is very different than the projection because the projection I told you was this one this is the projection this is f you so if you say F cosine of alpha what you're calculating is this and not the component that's why a lot of people has a lot lot of trouble doing this type of situation but I don't know why is the the trouble because you apply law of ss or law or law of cosiness which by the way the law of sign is exactly or cosine what you do in this case difference is the angle here is 90° that's the only difference and when you say F cosine or f sign in this particular case this is a particularity of this case but this is the general case so what you do is you say once again this side ID by S of alpha has to be equal to this side ID by sign this side divid by sign of this angle and has to be equal to this side divid by sign of this Angle now if I said let's say that I'm in this case or you're going to say but but you're missing Here sign of the other angle because you have the this one here or you have this one here yes but the other angle is 90° and when I say f look at this case this is 90° right so if I apply the law of signs here I'm going to say f / by S of 90 has to be equal to FY divided by S of alpha so f y soling for FY I pass this to this side is going to be f s of alpha ID by S of 9 by sign of 9 is one so basically I have this which is what you did before but that's a particular case of the law of signs when you have a right triangle this is not what happens in this case this case as you can see you have two axes u and v and those axes are not perpendicular so because the axis the axis u and v are not not perpendicular then we have to follow the law of signs remember the law of signs once again says that a / s of a has to be equal to B / by S of B has to be equal to C / s of C where a b and c Capital are the sides and a B and C Small caps are the Angles and then we come to this H problem and we have to try to solve it first read the problem the problem is asking us to resolve resolve not project resolve the force F1 into components acting along the u and v axis only fub1 not F2 not resultant not nothing else just fub1 well if I want to find the components of F1 with respect to these two axes what I have to do is come parallel to the axis I have a line here which is the axis and I have to put the other line parallel to the a the U axis which is here right that like that and my components are going to be this one up to here and this one up to here right there now I have to find H the rest of my angles in order for me to solve this problem so how I'm going to do that well I know for sure these angles how do I know this angle once again and I have two parallel lines intersected by a straight line This angle and this angle are the same so this angle is 30° here that's the first thing that that I have there what else do I know ER let's think about it what else do you know here well if you look closely to the to the picture I have ER looking at the big picture here I have the U axis here and the v- axis here u and v and this angle here is 70° and 70° is the same angle here because they're H vertical angles opposed by the vertex 70 so this is 70 that means that if this is 70 this here has to be 40 and for supplementary angles you know the the not supplementary angles the summation of all the internal angles in a triangle has to be 180 that means that this internal angle here has to be 110 110° now I know the three angles and I know this side and this side is 300 so I can say immediately that applying this law 300 divided by the sign of the opposite this is 300 00 and this is the opposite s of 110 has to be equal to this side which in this case we're going to say this is F1 u fub1 u divided by H the sign of the opposite angle which is 40 and from here I pass this to the other side and I can calculate fub1 U and F1 U is equal to 25.2 Newton and it's positive because it's acting in the positive direction of the U Axis or I can say also that 300 ided by the sign of 110 has to be equal now I'm going to the other one which is this one this side is F1 V because it's acting in v fub1 v divided by the sign of the opposite but the sign of the opposite is 30 and then I can calculate F1 V and that's H finding the components of the other one this one F2 but F2 is the same I I just move my Axis here to the bottom one and then my components are going to be this one and this one the problem is not asking for that but let's say that what if the problem is asking me also for F2 and you and f2v well if that's the case I know this angle is 45 I know this side can I calculate the rest yes this angle here is 70 here why how do I know it's 70 well I know this is 70 because this one is 110 and this one is 110 110 + 70 is going to be 180 for supplementary angles so this is 70 here you can have something like that something like that that and something like that that's what you have this angle here is 70 this angle here I know it because it's a given 45° and because this is 70 and this is 45 this is 115 uh to 180 this has to be 65° and I know this value here which is 500 so this side of the triangle this value is going to be a F2 right yes F2 U and this is going to be F2 V I do the same thing I say F2 U / by the S of 45 has to be equal to H 500 divided by the S of 70 and even more I can say it also that it's going to be equal to f2v / by S of 65 and then I select whatever part I if I want to calculate this I select this part if I want to calculate the other one I select this part and then I get that F2 U is equal to 37624 Newton and F2 V equal to 48224 Newton now pay attention to this if you look closely even though this is going down because this other thing that you are used to do you are you are you are using your mind to say everything that goes down is negative no sir it's not like that this is not negative because it's coming in the positive direction of the v-axis wherever you have the letter that's the positive direction so this component is positive but look at this one the component of U is negative and guess what the law of signs is not going to give you that negative that negative you have to add it this is the one if I want to calculate the resultant Force which the problem is not asking me then I have to add everything that goes in U together so I'm going to have 25.2 minus 37624 remember this goes in the U Direction plus whatever goes in the v Direction h + 48224 and this goes in the v Direction whatever this operation is I didn't calculate it you can do it now once again pay attention this is going to be negative and this is going to be positive why because this is bigger than this so if these are your axes once again this is U and this is V then and the U component is going to be negative and the V component is going to be positive and it's going to follow the axis all the time that means that ER your resultant force is going to be somewhere here how are you going to calculate that resultant for for I don't know you have this angle here because you have it you have this value you have this value you apply the low of sign again and calculate the resultant do not apply Pythagoras which is the square root of this square plus this Square because Pythagoras once again applies only to right angles and this is not a right triangle I mean right triangles and this is not the right triangle so don't do it like that you can calculate R if you want to by the law of cosiness because you know these Ang here and then you can say that R is equal to this square plus this Square minus 2 this time this cosine of that angle and then you can calculate the uh resultant R in that way I hope this problem helps you uh clarify that situation let me see I think I have I have another problem that I can do we have time yes I think we have time let me see another problem let's try to do okay this problem is nice let's do this problem now this is a nice problem and also allows us [Music] to I don't know help you at least to read the problem and pick up the clues that are intrinsic in the problem let's read the problem and one of the things that you have to learn is how to read problems it says if the magnitude of the resultant force acting on the ibolt is 600 Newton magnitude of the resultant Force magnitude of the resultant force is 600 Newton now is the other part what is also important because the part A lot of people just skip it and don't read this clockwise part here it's Direction measure clockwise clockwise meaning this direction from the positive axis xaxis is 30° that means that my resultant force is going to be acting in this direction this is going to be the resultant Force this is going to be 600 Newtons and this angle from here to here because it say clockwise otherwise if it doesn't say clockwise it say each direction is 30° it has to be counterclockwise should be counterclockwise because uh per default H the direction is measured counterclockwise from the positive direction of the x-axis but in this particular problem it's telling you the problem is telling you that is measured clockwise from the positive a xaxis so these angle Le 30° right there and this is 600 now we have to find the magnitude of fub1 and the angle how are we going to do this problem basically this problem is solved in the same way that we solved the other the other problem we're going to decompose every Force H in in components then we're going to add X at y the summation of the forces in X has to be equal to the component X of the resultant force and the summation in y has to be equal to the component in y of the resultant Force that's basically what we're going to do so let's start with the first uh problem with the first force and then the first force is going to be let's go summation of forces in x equals z let's do that not equal zero equal H the resultant force in X this is the Force One the force one has this horizontal component the value for this horizontal component is going to be now the axes are perpendicular so then it's going to be the f1x which is this one is going to be fub1 multiply by cosine of the angle why cosine because it's adjacent so fub1 cosine of the angle to the right so it's positive because this is going to the right now plus the horizontal component of this one is this one what is this value I don't know I need this angle here how much is that angle over there that angle over there either either ER you say that oh wait a second this is 60 and this is 30 so you can use this one or you can use this one also if you use this one then you're going to say s of 30 but if you use H if you use the other one which is H this one here you're going to say cosine 60 is is the same thing let's use this this Force this component here because I don't want to mess up this this drawing more so this component is going to be F2 sin 30 and it's positive also plus F2 sin 30 which is the same cosine 60 remember here is the opposite but here is the adjon to 60 uh what else now I have this one you you know this little triangle here the component the horizontal component of F3 is going to be H this one and this little triangle what is telling you is the ratio between geometry and force so basically what is telling you is that this Force this horizontal component is going to be 3 / 5 3 / 5 multip 450 and this is negative why because it's pointing to the left to the negative Direction and that's important minus 450 3 / 5 and this F2 remember I didn't put it but this FS2 is 500 because it's given also the summation of forces in X is this one and that has to be equal to the horizontal component of the resultant Force and the horizontal component of the resultant force is 600 let me put it here this is the resultant Force this is the horizontal component of the resultant Force so it's going to be 600 cosine 30 oh or I can say that fub1 cosine of the angle and I solve all these numbers because those are numbers I take this pass it to other side negative I take this pass it to the other side positive and what we have there now is that is going to be equal to 539 62 Newton and I call this equation one and then I do the same thing and I say summation of forces in y has to be equal to the resultant component in y and summation of forces in y is this direction here so the Y component of fub1 is going to be this side of the triangle which is going to be F1 sign of the angle so I'm going to have fub1 sign of that angle this one and now I have the vertical component of this one of F2 is this one and this one is no other thing that first at all is negative why negative because it's pointing down negative negative what 500 multip by cosine 30 and the vertical component of this one is also this side and this side is four this time you see four / 5 and it's negative 4 / 5 because that's what the real triangle is telling us 4 / 5 ultip by 450 and that has to be equal to the summation of forces in y summation of forces in y I'm going to copy it in this side because I got R out of space here but the summation of forces in y the resultant force in y is this this one over here and this this one over here is negative First at all why negative same reason it's acting downwards right so negative remember this is the component -600 sin 30 it's going to be equal to that from here I get this pass it to this side positive this pass it to that side positive and I get that F1 sign of the angle has to be equal to 4 93.1 Newton second equation and you can solve it whatever way you want to but it's easier for me to divide 2 by one and if I do that then I'm going to have fub1 sin Theta = 493 01 and the other equation is going to be F1 cosine Theta equal 5 3962 and this and this cancel out and then if you go here you have that tangent of the angle is going to be equal to this divided by this which is 0.91 and then you take the inverse tangent of this and the angle is going to be 42. 42° and this is the value of H the angle once we have the angle we can plug it into here or plug it into here and we can calculate fub1 solving for here let's say that we plug it into this one then fub1 is going to be equal to 49301 this is going to pass to the other side dividing s of 42 42 and then fub1 is 73092 Newton there you go another problem that you can go and check um practice on your own and I hope that these problems helped you a little bit with the material that's my intention at least that's what I'm doing it last problem for today I didn't think I was going to do so many problems but I'm going to do it let's check this problem and this problem is interesting because H there are several ways that you can H represent the vector you can represent a vector using the magnitude and the cosine directors or the unit Vector you can represent a vector by the cosine directors and you can represent a vector or you can decompose a vector using the projection angles and you have to be able to identify which one is which in this particular case we have two now let's start with F1 F1 is this one here well let's start by reading the problem the problem says that and the bracket is subject to two forces shown Express the each force in cartisian Vector form and then determine the resultant Force find the magnitude and coordinate Direction angles of the resultant Force that's basically what we have to do okay let's start let's start with the angle F1 this is a rope here and this is a rope here acting on that bracket and both of them are going to be a applying a tensile force in this direction and we have to calculate the result and in order to do that the easiest way all the time is find the cartisian coordinates and even more now that we are in 3D for the cartisian components Express the forces in cartisian notation add whatever goes in I with whatever goes in I I'll add all the JS together and add all the K's together and that's going to be the Force One once we have the force find the magnitude how do we find the magnitude the square root of the summation of the squares of each one of the components and how do we find the direction by finding the cosine directors and I explained that in class how to do it but anyway let's do it again so F1 is this Force here the first thing that we know about F1 H is that if you look this angle here and this angle here are projection angles how do I know there are projection angles because they are orthogonal you see so the first thing that I'm going to do is I'm going to project this angle this force over the plane XY so the projection of fub1 let's say over the XY plane it will be this side and that side is going to be fub1 multiply by cosine of this angle and this angle is 35 how do I know is 35 is given so the projection of the force fub1 over the XY plane is 250 multiply by cosine of 35° perfect I have this and now I can take this to the Y AIS and to the x axis because I have this other triangle here so this Force I know is this one if I I want to find the X component which is this one this is the X component remember fub1 X is going to be equal to the projection fub1 over the plane XY multiply by sine of 25 this is the angle this is 25 now over the plane and this is the opposite so it's going to be S of 25° and fub1 Y which is the projection over this axis here Y is going to be cosine 25 so it's going to be fub1 XY cosine 25 but if the projection over the plane XY is this one so if I want to calculate f1x I just plug this into here and that's going to be 250 cosine 35 sin 25 and F1 Y is going to be 250 cosine 35 cosine 25° this value is 8655 remember H our forces are in Newton and this value here is er 8C 85 60 Newton also wait a second why two components only if we are XYZ oh look this is the Z component of the force Direct that's the easiest one to see why because this angle here is vertical angle like that and it's 35° so this one is going to be fub1 fub1 which is 250 multiply by S of 35° attention here the the Z component is pointing downwards negative it's pointing in the opposite direction to the letter negative the other two were pointing in the same direction of the letters so be careful with this one and that's 14339 Newton what we got so far is F1 as a vector and F1 as a vector is 8655 I + 18560 J minus 143 39 K and that's our Force F1 in cartisian representation or cartisian notation now if we go to F2 F2 I'm going to keep working working here if we go for FS2 now look at the angles that we have for F2 for F2 we have angles with respect to the axis look at these angles these angles are like inclin with respect to the axis and for these you have the cosine director angles or the director angles alpha beta and gamma and you have to remember the definition of alpha beta and gamma Alpha is the angle between the x- axis and the force which in this case is clearly stated as 120 so Alpha is 120 beta is the angle between the Y AIS and the force and that is 45° now look at this 45 and 60 oh wait a second 45 and 60 that shouldn't be 90° here no it shouldn't be because those are not in the same plane these 45 and these 60 are inclin angles so that's one of the indications also that you are dealing with cosine director angles and not with a projection angles now gamma is 60 how do I know because it's given but even if I don't know one of them remember that I said in class and I stated that a you have a relation direct relation between the three angles those are not independent alpha beta and gamma they are not independent they always have to be a always always it has to it has to happen that cosine Square Alpha plus cosine Square beta plus cosine Square gamma has to be equal one so in case I don't have one of them I can calculate them by using this expression and try with these angle I say try with this and this and calculate gamma and you're going to see that and is is true what I'm saying okay so FS2 as a vector I'm going to use the cosine directors and they are called cosine directors because the cosine of this the cosine of this and the cosine of these are going to give me the unit vector and I multiply the unit vector by the magnitude and the magnitude is 400 so 400 is the magnitude multiply by the unit Vector which is going to be cosine 120 I plus cosine 45 J plus cosine 60 K this is my cosine director remember this is Newton also so FS2 is going to be equal to 200 I + 282 84 J plus 200 K that's F2 that's my F2 now what do I have to do add this with this so I'm going to put F1 here remember this is as a vector an arrow on top of the uh of the vector notation so F1 was 86. 55 I the other one and J is 185 60 J and the other one is 14339 K now I have these two remember this is Newton this are Newton and I'm going to add them in order to calculate the resultant force my resultant force is going to be I together so this is negative 11345 in I and this is plus here and even if you don't have the J please put zero here in case that you didn't have it and put it on on top of the J because I know you're going to mess and you're going to add this with this if you are in a rush in a test or something so be organized when you solve the problem now when you do this plus this this is going to be 46844 J plus 6 56.61 K this is going to be my resultant Force as a vector but uh actually yeah resultant Force f r i should be calling this f r not R and now they are asking me for the magnitude and coordinate Direction angles of the resultant Force the magnitude is really easy because the magnitude of the resultant force is just the square root of 11345 squar + 46844 squ + 56.61 squ and then you're going to say oh where is X how come you didn't put this negative so what I don't care about the negative remember the negative sign is going to be a square so it's going to be positive no matter what so then when I do that the magnitude of the resultant force is going to be equal to 4 48530 Newton that's my value for the magnitude of the resultant force and the other part that is asking the problem is the coordinate Direction angles well the coordinate Direction Les if you remember from class cosine Alpha is going to be equal to what it's going to be equal to the component of I of the force divided by the magnitude of the resultant Force that's cosine Alpha so that mean the alpha is going to be the inverse of this Alpha is going to be the inverse cosine of this divided by this but remember this is negative now you have to put the negative gam H beta is going to be H also this 4 68 44 divided by the magnitude of the force and remember this is not the value beta this is a inverse cosine of this and Gamma is going to be the inverse cosine of H this value 56.61 divid by the magnitude of that Force which give us Alpha equal 103 52° beta equal 15.15 de and Gamma equal 83. 30° and our results for the problem are magnitude first the force as Vector then the magnitude of the force and then the angle the director angles for the force well you have four problems h i solv several I solve two of them in different ways ER I hope that you enjoy them I hope hope that they clarify a lot practice practice practice do not procrastinate and you will be successful in this class I'm going to prepare another one ER with chapter three probably and I'm going to try to keep doing this for every chapter but at least for this chapter you have some problems that you can practice see you next class |
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Hepatocellular adenoma: when and how to treat? Update of current evidence
Maarten G Thomeer
Mirelle Broker
Joanne Verheij
Michael Doukas
Turkan Terkivatan
Diederick Bijdevaate
Robert A De Man
Adriaan Moelker
Jan N IJzermans
Email: m.thomeer@erasmusmc.nl
Issue date 2016 Nov.
This article is distributed under the terms of the Creative Commons Attribution-NonCommercial 3.0 License ( which permits non-commercial use, reproduction and distribution of the work without further permission provided the original work is attributed as specified on the SAGE and Open Access page(
Abstract
Hepatocellular adenoma (HCA) is a rare, benign liver tumor. Discovery of this tumor is usually as an incidental finding, correlated with the use of oral contraceptives, or pregnancy. Treatment options have focused on conservative management for the straightforward, smaller lesions (<5 cm), with resection preferred for larger lesions (>5 cm) that pose a greater risk of hemorrhage or malignant progression. In recent years, a new molecular subclassification of HCA has been proposed, associated with characteristic morphological features and loss or increased expression of immunohistochemical markers. This subclassification could possibly provide considerable benefits in terms of patient stratification, and the selection of treatment options. In this review we discuss the decision-making processes and associated risk analyses that should be made based on lesion size, and subtype. The usefulness of this subclassification system in terms of the procedures instigated as part of the diagnostic work-up of a suspected HCA will be outlined, and suitable treatment schemes proposed.
Keywords: adenoma, immunohistochemistry, liver, MRI
Introduction
Hepatocellular adenomas (HCAs) are an uncommon, solid, benign tumor of the liver, with an estimated incidence of 3–4 per 100,000 women [Bioulac-Sage et al. 2010]; this frequency is based on population research including women using oral contraceptives (OCs) [Baum et al. 1973]. A causal role for hormone activity in HCA growth is suggested by data linking adenoma regression to the cessation of OC use, and growth associated with pregnancy [Cobey and Salem, 2004].
Typically, HCAs are treated conservatively, with patients advised to avoid oral contraception. The risks of growth and rupture of HCAs during pregnancy has to be underlined, especially in larger HCAs. Tumor progression, suggested by internal bleeding and malignant transformation, necessitates a more aggressive therapeutic approach, with lesions larger than 5 cm considered as the primary risk factor [Marrero et al. 2014]. The introduction of a new subclassification system for HCA has been suggested to help clinicians to stratify patients according to imaging criteria, expression of associated immunohistochemical markers or molecular findings. These data may influence the treatment selected [Marrero et al. 2014] since certain subtypes of HCA pose a greater risk of progression to hepatocellular carcinoma (HCC) than others. For example, a subtype of HCA defined by the reduced expression of liver fatty acid-binding protein (LFABP) ordinarily indicates a subtype with a less aggressive course and a tendency towards a benign phenotype.
Based on the recent literature, we describe the impact of this newly instigated HCA subclassification, and discuss whether this knowledge, combined with imaging data, improves our risk analyses for patients with HCA. Furthermore, we outline the different therapeutic options indicated by each HCA subtype.
The Bordeaux classification of HCA
In recent years, four distinct subtypes of HCA have been recognized: inflammatory HCA (40–50%, IHCA), HNF1A-mutated HCA (30–40%, H-HCA), β-catenin activated HCA (10–15% b-HCA), and unclassified HCA (10–25%, UHCA) [Nault et al. 2013]. In these different subtypes, several genetic mutations are identified, causing (benign) proliferation of hepatocytes and in some HCA, malignant transformation [Pilati et al. 2014].
Patients presenting with IHCA demonstrate both serum, and lesional indicators of an active inflammatory response. In these lesions, increased expression is seen of the markers serum amyloid A and C-reactive protein, both classic indicators of the acute phase response [Bioulac-Sage et al. 2007a]. Patients within this HCA category frequently demonstrate increased body weight, and a high alcohol intake [Bioulac-Sage et al. 2007b, 2009; Paradis et al. 2007]. In approximately 10–20% of these lesions, a β-catenin mutation is found [van Aalten et al. 2011b].
A second subtype of HCA, H-HCA, is characterized by a downregulation of the LFABP; this phenotype, which is not apparent in the other HCA subtypes, rarely leads to malignant progression [Zucman-Rossi et al. 2006].
Subtype b-HCA is typified by activating mutations of β-catenin that resist phosphorylation-mediated down-regulation by the GSKB/APC/AXIN complex [Nault et al. 2013]. Particularly the exon 3 mutation of β-catenin plays a significant role in malignant progression in contrast to exon 7/8 mutations [Pilati et al. 2014]. The result is an accumulation of nuclear β-catenin which, combined with deletion of APC, favors progression to HCC [Nault et al. 2013]. The comparatively small number of β-catenin positive nuclei can lead to this phenotype being overlooked in small biopsies [van Aalten et al. 2011b]. The b-HCA subtype also demonstrates an overexpression of glutamate-ammonia ligase (GLUL) encoding glutamine synthase (GS), which can be used as a sensitive diagnostic biomarker for this subtype [van Aalten et al. 2011b].
The final subtype, UHCA, is not yet defined by any specific genetic mutation, but is instead characterized by various histologic criteria that are unusual in the other subtypes; the underlying pathogenesis of this subtype remains unclear [Blanc et al. 2015].
Magnetic resonance imaging of the different subtypes of HCA
The primary differential diagnosis for HCA is focal nodular hyperplasia (FNH). If in doubt, a biopsy should be taken, especially for larger lesions, as the clinical management will differ for either pathology. In most cases these diagnoses can be differentiated according to signal intensity and dynamic vascular patterns after intravenous aspecific gadolinium injection [(conventional magnetic resonance imaging (MRI)] [van Aalten et al. 2011a]. Different patterns can be used for confident diagnosis as proposed by Thomeer and colleagues [Thomeer et al. 2014a].
In more challenging cases, specific hepatobiliary contrast agents can be used. Currently there are two agents available, gadobenate dimeglumine, and gadoxetate disodium [Grazioli et al. 2013; Thomeer et al. 2014a; McInnes et al. 2015]. If the lesion turns hypointense to the surrounding liver in the hepatobiliary phase, FNH can be excluded in most cases. If the lesion becomes iso- to hyperintense the differential diagnosis is FNH or in exceptional cases HCC. However, it should be noted that IHCA can also be isointense in the hepatobiliary phase [Agarwal et al. 2014; Thomeer et al. 2014b]. This might be explained by the presence of internal bile duct proliferation, previously thought to be only visible in FNH. This diagnostic pitfall can be visualized when using gadobenate dimeglumine [Thomeer et al. 2014b], or gadoxetate disodium [Agarwal et al. 2014]. A recent systematic review about the value of gadoxetate disodium has shown that apart from this pitfall, adequate differentiation is possible in most cases [McInnes et al. 2015]. It was reported that the hepatobiliary phase has a sensitivity of 91–100% and a specificity of 87–100% for differentiating HCA from FNH. In the largest study this was only seen in 13% of cases [Bieze et al. 2012].
In conclusion, in the vast majority lesions can easily be differentiated based on a combination of typical findings on conventional MRI and features on hepatobiliary phase MRI.
Some typical MRI features allow us to discriminate different subtypes of HCA (Table 1): IHCA can be hyperintense on T2-weighted images, with persistent enhancement on delayed imaging in the venous phase [Laumonier et al. 2008]. Ronot and colleagues validated this feature as being highly specific for IHCA, with a sensitivity of 82% [28/34, confidence interval (CI) 65–93%], and an optimal specificity of 100% (12/12, CI 75–100%) [Ronot et al. 2011]. Another diagnostic indicator for IHCA is the atoll-sign [van Aalten et al. 2011a], a hyperintense rim on T2-weighted images at the periphery of the lesion (resembling an atoll) that is enhanced late in the venous phase. This feature is present in 27% of IHCAs in this study [van Aalten et al. 2011a].
Table 1.
Typical MRI findings according to subtypes of HCA.
Subtype | Most typical MRI signs
IHCA | Hyperintense on T2-weighted images, with persistent enhancement in the venous phase;atoll sign
H-HCA | Diffuse and homogenous fat deposition (Figure 1)
b-HCA | (Vaguely demarcated scar)
UHCA | No typical sign
b-HCA, β-catenin activated HCA; HCA, hepatocellular adenoma; H-HCA, HNF1A-mutated HCA; IHCA, inflammatory HCA; MRI, magnetic resonance imaging; UHCA, unclassified HCA.
Whilst H-HCAs are typically characterized by a large amount of aberrant fat which can be readily appreciated by out-of-phase imaging, or on a fat-saturated T1-weighted image [Laumonier et al. 2008; van Aalten et al. 2011a], van Aalten and colleagues failed to detect fat by MRI for as many as 22% of cases (2/9) [van Aalten et al. 2011a]. Ronot and colleagues validated the diagnostic feature of diffuse and homogeneous signal dropout on out-of-phase T1 weighted imaging, with a reported sensitivity of 90% (10/11, CI 58–99), and specificity of 88% (32/36, CI 73–96) [Ronot et al. 2011]. The main drawback of this marker is that diffuse intralesional steatosis may also be present in up to 11% (4/34) of IHCAs [Ronot et al. 2011], although, according to the authors, this does not represent a major pitfall as fat is usually distributed heterogeneously within IHCAs (Figure 1).
Figure 1.
In- and out-of-phase MRI of a typical case of histochemistry proved H-HCA which was resected. Note the diffuse and homogenous signal drop-off on the out-of-phase image (a) versus the in-phase image (b). This correlates with diffuse intralesional fat identified by histology. MRI differentiation between H-HCA and IHCA would not be possible when the signal drop is more heterogeneous.
HCA, hepatocellular adenoma; H-HCA, HNF1A-mutated HCA; IHCA, inflammatory HCA; MRI, magnetic resonance imaging.
The MRI features of b-HCA are not well defined, principally because these lesions are comparatively rare. Van Aalten and colleagues reported poorly delimited, high-signal intensity areas, to be typical of this subtype (5/7, 71%), but additional investigations are warranted [van Aalten et al. 2011a]. Table 2 shows the various MRI features reported in the literature for b-HCA, although, where reported, these features are inconsistent. Despite significant numbers of false negatives, the specificity of these MRI features is very high, leading us to conclude that if any one of these signs are present, a diagnosis of the corresponding MRI subtype can be made with some certainty. Larger datasets will be needed to determine the true value of MRI in HCA imaging for all subtypes; currently, this technique is of most use in evaluating prognosis.
Table 2.
Recently published b-HCAs with their typical characteristics defined by MRI. Note the low prevalence in the literature of MRI data, with inconsistent findings.
Authors | Year of publication | Number of β-catenin HCA | MRI findings
[Van Aaltenet al.2011a] | 2012 | 7 | Vaguely defined scar on T2W sequences (3 cases)
[Laumonieret al.2008] | 2008 | 5 | Marked hyperintensity on T2W sequences and persistent delayed enhancement (3 cases)Isointensity on T2W sequences, with strong arterial enhancement and delayed washout (2 cases)
[Yonedaet al.2012] | 2012 | 1 | Vaguely defined scar on T2W sequences
b-HCA, β-catenin positive HCA; HCA, hepatocellular adenoma; MRI, magnetic resonance imaging; T2W, T2-weighted.
Reviewing the known complications
Intralesional bleeding
On reviewing the recent literature, van Aalten and colleagues detected evidence of hemorrhage in 27.2% of all patients (315/1160) with one or more HCAs, giving a 15.8% chance of hemorrhage for every HCA (118/748) [van Aalten et al. 2012]. Acute rupture and intraperitoneal bleeding were reported in 17.5% of patients. A size for the smallest HCAs showing hemorrhage was reported for 13 of the 28 articles reviewed; hemorrhage generally arose in the larger lesions (>5 cm), although smaller lesions could also bleed (Table 3), albeit at much lower rates. These data should be interpreted with caution, as only the resected cases were included. The actual chance of bleeding in the different subtypes is likely to be significantly lower. The risk of bleeding was inconsistent across the subtypes of HCA: IHCA showed a higher propensity for macroscopic hemorrhage (30%), than H-HCA (8%) [Dokmak et al. 2009] which can presumably be attributed to the larger number of venous structures, or telangiectatic changes in this subtype.
Table 3.
Summary of the findings of an earlier review of 12 articles in which the percentage of hemorrhaged HCAs, and minimal lesion sizes, were reported. Hemorrhage occurred mostly in larger HCAs (>5 cm; minimally 42.2%), but smaller lesions also showed some bleeding (range 8.3–11.5%).
Series | Patients with hemorrhaged HCA | Size of smallest HCA (cm) | Percentage <5 cm of total (%)
[Reddyet al.2001] | 3 of 25 | 4 | –
[Hunget al.2001] | 4 of 25 | 4.2 | –
[Tosoet al.2005] | 10 of 25 | 1.7 | –
[Choet al.2008] | 12 of 41 | 1 | 8.3 (1/12)
[Bioulac-Sageet al.2009] | 23 of 128 | <5 | –
[Edmondsonet al.1976;Dokmaket al.2009] | 26 of 122 | <5 | 11.5 (3/26)
[Edmondsonet al.1976] | 10 of 42 | >5 | 0
[Leeseet al.1988] | 2 of 24 | 5 | 0
[Aultet al.1996] | 4 of 12 | 6 | 0
[Clossetet al.2000] | 7 of 16 | 7 | 0
[Deneveet al.2009] | 31 of 124 | >5 | 0
[Chunget al.1995] | – | 5 | 0
HCA, hepatocellular adenoma.
Although there may be a difference in prevalence of internal bleeding, all subtypes bear this intrinsic risk [Laumonier et al. 2008; Dokmak et al. 2009; Ronot et al. 2011; van Aalten et al. 2011b], which diminishes the utility of subtype classification in terms of the clinical management of this risk. Furthermore, more data are needed to prove any correlate between reduced bleeding and the H-HCA subtype.
Bieze and colleagues described a series of 45 patients with 195 lesions. In this cohort, there was a tendency to an enhanced risk of bleeding when the lesion was located in the left lateral liver (11/32 versus 31/163 in other regions), and showed exophytic growth (16/24 versus 9/82) [Bieze et al. 2014] (Figure 1). The latter phenomenon is probably due to the subcapsular location, with no intrinsic capsule, and minimal surrounding liver with which to prevent rupture of the hematoma into the abdominal cavity. However, no other data are available to support this theory, and preventive treatment in these cases does not appear to be warranted.
As for the clinical application of these findings, there is no evidence that supports the use of subtype classification in the stratification and management of individual patients. Moreover, size still remains the most important marker to predict those at risk for larger bleeding in follow up.
Malignant transformation
Malignant transformation of HCA to HCC is rarely reported, but is an accepted risk, particularly when the diameter of the adenoma exceeds 5 cm (Figure 2) [Stoot et al. 2010; Grazioli et al. 2013]. In a systematic review, Stoot and colleagues reported an overall frequency of malignant transformation of 4.2% for HCAs [Stoot et al. 2010] (67 of 1635 HCAs, CI 0–100%). Only three cases showed malignant transformation for tumors <5 cm in diameter, which represented 4.4% of the total number of HCCs arising from HCAs (3 out of 67). As suggested for the internal bleeding data, these reports should be interpreted with caution.
Figure 2.
A 32-year-old female using oral contraceptives with a 11 cm lesion in the liver. Based on MRI this lesion was compatible with a HCA. However, both on T2-weighting (Figure 1(a)) as on the images after contrast injection the lesion appeared heterogeneous with a focus of diffusion restriction (typically a low ADC value, (b)). Diffusion restriction is thought to be a typical sign of malignancy in liver lesions. Based on the findings above and because the lesion was larger than 5 cm, the lesion was resected 3 months later. On gross pathology there was a focal nidus (Figure 1(c), arrow, concordant with the nidus on MRI) which appeared to be an HCC in a HCA (‘nodule-into-nodule’). On histology (H–E × 25, (d)) at the interface HCA/HCC, the upper part of the tumor showed proliferation of hepatocytes without obvious cytological anomalies, intermingled with thin/isolated vessels (down side of dotted line), favoring an HCA. ‘Nodule-into-nodule’ consists of clearer cells with mild atypia (above dotted-line, (d)), disorganized or decreased reticulin fibers (e) and obvious positivity for Glypican-3 (f), favoring an HCC (well differentiated). (g) GS-staining pattern at the periphery of the HCA. Glypican-3, Serum Amyloid A and CRP were negative in the HCA. β-catenin staining showed only membranous expression. Based on the above we interpreted this HCA as a b-HCA.
CRP, C-reactive protein; HCA, hepatocellular adenoma; HCC, hepatocellular carcinoma; GS staining, glutamine synthetase immunostaining indicative of b-HCA; MRI, magnetic resonance imaging.
Of the four HCA subtypes, (exon 3) b-HCAs are known to trigger a potent mitogenic signaling pathway that is prominent in HCC [Zucman-Rossi et al. 2007; Chu and Moon, 2013; Pilati et al. 2014], which suggests a positive correlate between the two. Zucman-Rossi and colleagues reported an incidence of HCCs, or borderline malignant tumors in b-HCAs, of up to 46%; this malignant progression was seldom seen in other subtypes [Zucman-Rossi et al. 2007], and was over-represented for male patents (5 cases, 38%; p = 0.02) [Hussain et al. 2006]. Since the β-catenin pathway can also be activated in IHCA [van Aalten et al. 2011b], both the b-HCA and IHCA subtypes may necessitate more aggressive treatment than either the H-HCA or UHCA, although the clinical relevance of this determination has yet to be broadly accepted. In follow up, malignant progression of HCA to HCC has only rarely been demonstrated, with questionable quality of the imaging data for those rare, reported cases. Therefore it is presently difficult to prove that HCC is a transition from HCA, although the presence of β-catenin has been suggested as a criterion for the selection of HCA, or well-defined HCC, for resection [Bioulac-Sage et al. 2013]. Interestingly, Figure 2 shows a lesion with a typical nodule-in-nodule appearance which suggests a form of transition from HCA to HCC. Another problem is that corroboration of the pathology is seldom available, due to the fact that biopsies of HCA are rarely performed, with diagnoses generally made with MRI [Hussain et al. 2006]. A final diagnosis of b-HCA based solely on MRI findings would be helpful, but the MRI findings published to date for this subtype are sparse (Table 2). Finally, it should be mentioned that HCA shows a higher risk of malignant transformation in men [Farges et al. 2011]. In these cases, the possibility of hepatitis, an underlying glycogen storage disease (Figure 3), or sex steroid hormone abuse, should all be considered as all predispose to HCC [Yoneda et al. 2012]. A more aggressive treatment is advised in such cases, even for lesions <5 cm.
Figure 3.
A 50-year-old male with multiple hypervascular lesions. These lesions were diagnosed as HCA or HCC based on imaging and clinical (glycogen storage disease) findings. (a) An axial MR image, with T1 weighting, after contrast injection in the arterial phase. In segment 5, a small lesion with a cystic central portion (large arrow) was biopsied, and subsequently diagnosed as HCC following positive GS staining with negative β-catenin staining. Posteriorly, a second, smaller lesion was visualized (<1 cm, small arrow). Histologic sample of a lesion with diffuse GS-positivity (b). Axial MR image with T1-weighing after contrast injection in the arterial phase (c). In this image, taken 3 years later, the second lesion has grown (now 3 cm, small arrow). The large arrow shows the resection/ablated part of the liver (from lesion 1). A new hypervascular lesion (curved arrow) was also detected outside the liver, which proved to be a trajectory metastasis. These lesions (large arrow, curved arrow) were successfully ablated. This patient is currently being followed at regular, short intervals, and is on the waiting list for a liver transplantation.
GS staining, glutamine synthetase immunostaining indicative of b-HCA, even with a negative β-catenin staining; HCA, hepatocellular adenoma; HCC, hepatocellular carcinoma; MR, magnetic resonance.
Finally, according to the recent literature, H-HCA almost never degenerate into HCC, although some very rare cases have been reported [Stueck et al. 2015]. The low risk of H-HCA degeneration may help to simplify the management of liver adenomas as will be discussed later.
As for clinical application, mainly b-HCA and IHCA are prone to malignant degeneration, and mostly if >5 cm. In these instances, invasive treatment is recommended.
Pregnancy
Women with HCA who are pregnant, or wish to become pregnant (Figure 4), should be closely monitored for HCA size (with ultrasound or MRI) during their pregnancy, due to the tendency of the lesion to grow, especially during the third trimester when high levels of estrogens are present [Cobey and Salem, 2004]. Hormone-induced growth, and possible rupture, may result in potentially lethal complications for the mother and unborn child. Treatment of HCA during pregnancy may be indicated when the lesion shows signs of growth or bleeding, however specific figures for the risk of HCA complication during pregnancy are not yet available.
Figure 4.
A 25-year-old female with a MRI diagnosis of single HCA, probably inflammatory subtype. As the patient wished to become pregnant, despite growth of her HCA, a decision to treat with TAE was taken. Coronal MR image with T1-weighting of the upper abdomen (a). A hypervascular HCA is indicated (small arrow), adjacent to the gallbladder (long arrow). Ablation was contraindicated due to the close proximity of the gallbladder. Angiogram (b) before TAE showing an arterial tumor ‘blush’ in the HCA (short arrow), with the gallbladder perfused by the same local hepatic artery division (long arrow). This finding contraindicated TAE due to the possibility of gallbladder necrosis following infarction. Instead, a decision to operate was made, with resection of the gallbladder, and subsequent intraoperative RFA of the HCA. Axial postoperative CT image after contrast injection in the venous phase (c). The gallbladder was resected, in combination with intraoperative RFA (arrow).
CT, computerized tomography; HCA, hepatocellular adenoma; RFA, radiofrequency ablation; TAE, transcatheter arterial embolization.
Whether some subtypes are more prone to complications during pregnancy is not known, mainly because the majority is diagnosed non-invasively.
The choice of follow up, surgery, radiofrequency ablation (RFA) or transcatheter arterial embolization (TAE) for the treatment of HCAs in pregnancy is often a matter of debate. Surgery of lesions located at the periphery of the liver can be performed safely within the first or second trimester, and will usually be indicated by the size of the lesion, and its change in size. Radiation exposure or exposure to iodinated contrast media during RFA or TAE may be contraindicated during the early phase of pregnancy, with the treatment of smaller lesions not being indicated. Given the increased risk of hemorrhage in larger HCAs (>5 cm), or when a previous pregnancy was complicated by either minor or major bleeding, we currently advocate a preemptive treatment strategy before pregnancy, as proposed by Broker and colleagues [Broker et al. 2012]. Whenever a HCA is discovered during pregnancy, the second trimester is the optimal moment for invasive treatment, if indicated, as anesthesia is well tolerated at this stage, and the fetus is not yet so large as to interfere with liver surgery [Parangi et al. 2007].
Liver adenomatosis
Hepatic adenomatosis (HCAs more than 10) is regarded by some authors as a different entity [Barthelmes and Tait, 2005; Frulio et al. 2014]. There seems not to be a strong association with estrogen or anabolic steroid use [Chiche et al. 2000; Grazioli et al. 2000]. However, there is a strong association with glycogen storage disease [Chiche et al. 2000; Frulio et al. 2014]. Mostly, these adenomas are of the H-HCA and IHCA subtypes [Frulio et al. 2014]. The nodules in hepatocellular adenomatosis are often of the same subtypes. Although one might assume that multiple HCAs increase the propensity for lesional bleeding, previous data have shown no significant difference in macroscopic bleeding between single and multiple HCAs (p < 0.155) [Dokmak et al. 2009]. According to literature there seems to be no indication to suggest that the risk of malignant transformation is increased in hepatic adenomatosis compared with solitary HCAs [Barthelmes and Tait, 2005]. However, hepatic adenomatosis are more often found in glycogen storage disease and in man [Chiche et al. 2000], and as such at risk for increased malignant potential. Presently, there is no systematic review available which evaluates the malignant potential of hepatic adenomatosis. As for clinical management, and there are no data suggesting that hepatic adenomatosis should be treated differently from solitary HCAs.
Biopsy in the management of HCA
Since the introduction of the HCA subclassification system, several authors have attempted to further refine the diagnostic work-up using additional techniques, including immunohistochemistry [Zucman-Rossi et al. 2006]. The primary motivation for the introduction of additional biopsies was the prospect of identifying HCAs with greater malignant potential (such as exon-3 β-catenin mutated HCAs). There seem to be no unique MRI features with which to assign a b-HCA subtype risk, which offers one argument for the expansion of the use of diagnostic biopsy in order to arrive at a correct diagnosis.
However, at present there is no consensus regarding the diagnostic work-up of HCA [Nault et al. 2013; Marrero et al. 2014]. Nault and colleagues regard histologic analysis as the backbone of HCA diagnosis, with the detection or exclusion of b-HCA as the main input [Nault et al. 2013]. They argue that biopsy should be offered in all cases of HCA <5 cm with no typical MRI sign of H-HCA. Lesions >5 cm do not require biopsy since they are all preferably resected. In our opinion, and in accordance with recent American College of Gastroenterology guidelines for liver lesions, the diagnostic workup of suspected HCA should be based primarily on MRI findings, with biopsy in cases where the lesion cannot be clearly differentiated from FNH [Marrero et al. 2014]. Other indications for biopsy are an atypical presentation of the HCA on imaging, with the main differential diagnosis being HCC in a noncirrhotic liver [Marrero et al. 2014].
The biopsy of all HCAs (with the exclusion of typical H-HCAs based on MRI) found by imaging would be impractical. Most patients with HCA are young, with minor symptoms on malignant progression; invasive procedures should preferably be avoided. While the risk of bleeding complications is very low when using an 18G core needle biopsy (0.6%) [Haage et al. 1999; Kadri Aribas et al. 2010; Aribas et al., 2012], the risks are not negligible, and deaths due to bleeding complications have been reported [Stattaus et al. 2007]. In our practice a biopsy has not been performed to date, except where the diagnosis of a specific adenoma subtype was expected to alter clinical management.
Unquestionably, a biopsy for further characterization may add important information in well-defined cases. For example, a biopsy with the additional help of immunostaining could facilitate better discrimination between HCA and FNH, as shown in a large retrospective study in France where the investigators compared biopsies against a final diagnosis based on surgical resection [Bioulac-Sage et al. 2012]. A total of 239 needle biopsies were compared with the final diagnosis made on resection. A difference in sensitivity of 74.3% with immunostaining versus 58.6% achieved with routine analyses without GS or other molecular features was reported [Bioulac-Sage et al. 2012]. These data suggest that immunostaining should be made available in centers that routinely treat HCAs.
What is the best approach in cases where differentiation between HCA and HCC is not evident based on MRI? In cases where there is a major suspicion of malignancy (e.g. HCC in noncirrhotic liver) based on a combination of clinical findings, size of the lesion, increased serum α-fetoprotein, and MRI findings (such as heterogeneous presentation with heterogeneous enhancement, washout, and true capsule) (Figure 2), resection without prior biopsy can be recommended. Although biopsy of each suspect lesion would undoubtedly help in detecting HCC, this approach may be impractical due to the significant level of false-negative findings, the chances of seeding (Figure 3), and the enhanced risk of bleeding, which is particularly relevant when multiple biopsies are taken. Furthermore, in cases with a typical presentation, a biopsy will not influence the decision to remove the lesion. Therefore, we suggest to biopsy in selected cases only. Interestingly, the HCC literature documents a similar debate on whether it is acceptable to diagnose HCC in a cirrhotic liver based solely on MRI findings, or whether the use of routine biopsies should be advocated for all suspected lesions in patients with liver cirrhosis [Sherman and Bruix, 2015; Torbenson and Schirmacher, 2015]. Even in high grade dysplastic nodes, follow up by imaging is still preferred above biopsies.
As for daily practice, we recommend biopsy only in very selected cases where HCA cannot be differentiated from FNH with any imaging modality. The clinical repercussion of a wrong diagnosis of either HCA or FNH can have a large influence on a patient’s future in terms of treatment and prognosis. When there are signs of malignancy patients should preferentially be forwarded to an experienced referral center for further evaluation. One should be aware that in some cases MRI or biopsy will be unable to differentiate between HCA and well-differentiated HCA.
Treatment options for HCA
Historically, HCAs were treated with a wait-and-see policy, with surgical intervention preferred for larger (>5 cm) growing tumors. Current management options for HCAs may also include RFA, and TAE, mainly due to the advantages of these minimally-invasive techniques. In the following section we will discuss routine, as well as less commonly used HCA treatment options, with a focus on minimally-invasive, image-guided, treatment options.
Conservative treatment
When HCAs are <5 cm, or regress (to <5 cm) following cessation of OCs, with no further growth detected, a wait-and-see policy is warranted. Although no widely accepted approach has yet been published, we prefer to schedule a patient for follow up, including MRI, or ultrasound in a yearly follow up until menopause.
Surgery
Surgery has long been considered the treatment of choice because complete surgical removal of the lesion can be achieved in a controlled and relatively safe manner. Elective surgical resection is considered for all lesions >5 cm in diameter. With a mortality of 1.1% (review by Lin and colleagues, n = 170), surgery is a relatively safe procedure. In one review, 93% of patients with ruptured, or nonruptured HCAs, were primarily treated with surgery, with complications that included two deaths, one biloma, one bile leakage, one infection, and one case of sepsis [Lin et al. 2011]. In another, single-center retrospective analysis of 41 cases, no perioperative mortality was found, and only minor complications arose. These included pleural effusion requiring drainage (n = 2), pneumonia (n = 1), and wound infection (n = 1) [Cho et al. 2008].
In the latter study, nine cases were operated on laparoscopically, a technique that is increasingly popular, where appropriate. De’Angelis and colleagues described 62 HCA patients who underwent either an open procedure or laparoscopy [De’Angelis et al. 2014]. They found no difference in postoperative morbidity and zero mortality, with no longterm complications or recurrence of HCA. However, patients with smaller lesions were preferentially treated with laparoscopy (68 versus 9). These authors concluded that laparoscopic liver resections may be limited by lesion size and location, and that the technique requires advanced surgical skills. Given the precision required, robotic surgery may prove to be useful in the future, and could reduce complications; we await further evaluation of its efficacy [Jackson et al. 2015].
In rare circumstances, the treatment of HCA may also involve liver transplantation, a procedure described in a case report by Venarecci and colleagues [Vennarecci et al. 2013]. Obesity, steatosis, and diabetes, are frequent co-morbidities in patients with HCAs, particularly the inflammatory subtype. These factors, and especially obesity, make surgery less attractive. For those patients who are poor candidates for surgery (centrally-located lesions, multiple adenomas, or morbid obesity), RFA and TAE may instead be offered.
Radiofrequency ablation
RFA is a minimally invasive technique used in the treatment of HCC, other liver lesions such as colorectal metastases [Solbiati et al. 2001; Cabibbo et al. 2013], and HCAs [van Aalten et al. 2010; van Vledder et al. 2011]. Laparoscopic RFA, or perioperative RFA, may also be considered when the anatomical location [i.e. close proximity to the bowel or gallbladder (Figure 3)] leads to an increased risk using a percutaneous approach. The use of RFA in the treatment of HCAs has only been described in small case series.
Van Vledder and colleagues described one case series including 45 HCAs, in 18 patients, that were ablated in 32 RFA sessions (open, n = 4; percutaneous, n = 28) [Van Vledder et al. 2011]. A total of 26 of 45 HCAs were successfully treated in one RFA session, with no visible residual disease. A further nine HCAs were totally ablated following a second RFA session. There were 3 HCAs that required 3 or more RFA sessions, with all but 7 of the 45 HCAs totally ablated after 3 or more sessions. The treated HCAs had a median size of 3.0 cm (ranging from 0.8 to 7.3 cm). Only minor complications were attributable to the RFA procedure; none of which required additional intervention (class A according to the Society of Interventional Radiology scoring system for complications). A single class D major complication was reported; a cerebrovascular accident during open surgery combined with RFA. Though severe, this complication was linked to anesthesiological and hemodynamic changes during laparotomy, rather than the RFA procedure. In conclusion, RFA can be effectively and safely used in the treatment of HCA, although multiple sessions may be required for larger lesions.
In a review of HCA cases reported between 1998–2008, Lin and colleagues identified 356 HCA patients in reports from China, Europe, North America, and South-East Asia [Lin et al. 2011]. Only 14 (3.9%) of these cases were treated with RFA, and no severe complications were reported. However, no results in terms of efficacy were provided. In 2008, Rhim and colleagues. assessed the therapeutic efficacy and safety of RFA for HCAs [Rhim et al. 2008], and reported their initial experience in 10 patients with 12 HCAs. Tumor sizes ranged from 1.5 to 4.5 cm. As no complications were reported after RFA, and no progression or recurrence was noted, RFA was considered a safe and effective treatment option. A minimal ablative margin of 5 mm is recommended during the radical treatment of lesions when using thermal ablation of HCCs. It is presently unclear if a similar margin should be applied to HCAs, as these lesions are assumed to be benign. No data are yet available regarding the ideal ablative margin during thermal ablation of HCAs. In our opinion, volume reduction is more important than an ablation margin, as the former correlates strongly with a risk of bleeding, and malignant transformation. Follow-up imaging after both RFA, and TAE, is routinely performed in our institution by MRI, 6 months after treatment.
As HCAs requiring treatment are generally large (>5 cm), a promising alternative for RFA may be microwave ablation (MWA). Based on preliminary data, MWA was shown to produce larger ablation zones, in less time, in patients treated for HCC and colorectal metastases. MWA delivers high frequency microwaves (0.9–2.4 GHz) into tumor tissue, which causes fast spinning of molecules and thus destroys tissue. No data are available concerning efficacy and complications after MWA. MWA has specific advantages over RFA, such as larger ablation zones, higher treatment temperatures, and less susceptibility to local cooling by adjacent large blood vessels (heatsink). Although larger zones of ablation can probably be achieved by using single-electrode needles and MWA to treat HCAs, to the best of our knowledge, no data currently exists to substantiate this idea.
Transcatheter arterial embolization
As HCAs are hypervascular arterial lesions, bleeding may be treated by selective transcatheter arterial embolization (TAE) in cases where patients present with hemodynamic instability. Embolization of HCAs is a safe but relatively challenging procedure due to multiple small feeding vessels [van Aalten et al. 2010]. Nonetheless, in cases of spontaneous rupture and bleeding, TAE should be considered as a first line treatment as it is highly successful and minimally invasive in an acute setting. Although high success rates have been described for TAE, there is only a sparse literature comparing TAE with either surgery, or conservative management. In one study Karkar and colleagues described 52 patients with 100 HCAs, of which 37% were treated with TAE [Karkar et al. 2013]. In most of these cases TAE was performed in a (semi) elective setting, with rupture and hemorrhage as indications in 20%, and suspected malignancy in 56%. Success rates of up to 92% were claimed for TAE (32), and of the 37 HCAs embolized, only 3 required secondary interventions (8.1%). All other embolized lesions were treated successfully; some disappeared (5/34), most decreased in size (22/34), or remained stable (7/34). Recurrence rates were also low. It is worth noting that the HCAs embolized were relatively small, with a median diameter of 2.6 cm. However, we feel that resection is indicated if malignancy is suspected and no contraindications for surgery exist. In a report by Erdogan and colleagues, six HCAs were primarily embolized [Erdogan et al. 2007], four because of bleeding, and two electively, 1 year after bleeding. No complications were reported, and all HCAs ceased bleeding. A total of two of the lesions were resected after embolization, two regressed visibly on follow-up imaging, and two HCAs were only seen after resorption of hematoma on follow-up imaging. These last two patients were managed with a wait-and-see policy. In a retrospective study by Dheodar and colleagues, 17 embolizations were successfully performed in eight patients [Deodhar et al. 2011], with five patients undergoing more than one embolization. The mean size of the treated HCAs was 3.6 cm, and regression was noted in all embolized HCAs after embolization. As noted by Karkar and colleagues, TAE may also be used in an elective setting where no acute intervention is needed [Karkar et al. 2013]. This approach is of clinical interest and deserves further consideration.
Proposed management strategy
One of the major discussions on the management strategy of hepatocellular adenomas involves the clinical application of these recent findings in the dynamic field of adenoma subtyping. How should we take into account these new insights into daily practice? In our view, more data are needed to implement this subclassification in the diagnosis and treatment of adenomas, balancing the risk of an invasive liver biopsy with the additional benefits in terms of individualized therapy and prognostic stratification. A major effort should be made by expert centers involved in the diagnosis and treatment of hepatocellular adenomas to work on this collaboratively, preferably in research setting, to gather more data on the potential benefits for an individual patient.
Based on our review of the current literature, we propose a management strategy applicable to most cases in which there is a suspicion of HCA (Figure 5). This decision tree may not be appropriate for all patients; for some, a more customized approach may be required. In standard situations, mainly when a lesion is >5 cm, OCs should be stopped and MRI performed after 6 months. If the lesion has contracted to <5 cm, clear signs of an H-HCA should be ruled in or out (Figure 1). In scenarios where H-HCAs are subsequently identified, therapy can be less aggressive as inherent malignant progression is very low. Follow up is then advised, initially every 6 months, and if the lesion shows no further alteration, follow up can be stopped or simply repeated yearly until menopause [Marrero et al. 2014]. Since typical H-HCAs are easily identified using out-of-phase MRI sequences, intravenous contrast can be obviated at follow up. A second option is to apply sonography during follow up which is cheaper and less inconvenient for patients. For small lesions (<5 cm) that are categorized as IHCA, therapy should ordinarily not be altered (in standard cases). However, some may opt for a biopsy in order to exclude β-catenin mutation. This could also be the case if a subclassification cannot be made with MRI. For larger lesions (>5cm) with a β-catenin mutation or if the patient has an aggravating status such as male sex, steroid use, glycogen storage disease, or underlying viral hepatitis, intervention may be the first alternative. Treatment can be primarily surgical, and, in selected cases, RFA or TAE may be used. Depending on the underlying risks (obesity, diabetes, centrally located tumor), the best option would be to evaluate these patients in an expert referral center.
Figure 5.
The management decision tree used in our tertiary academic medical center. This decision tree may not be appropriate for all patients; for some, a customized approach should be considered.
One option is to biopsy those lesions where a subtyping diagnosis by imaging is impossible to achieve, or those lesions with typical signs of IHCA. Currently, this option is considered impractical given the large number of biopsies involved.
Follow up is advised initially, at 6-monthly intervals. Thereafter, if the lesion shows no further alteration, follow up can be stopped, or repeated yearly until menopause. If the lesion is a typical H-HCA, follow up can be less stringent, possibly involving sonography, or MRI without contrast.
Referral to an expert center is advised for the evaluation of any indication requiring intervention. This decision should be taken with consideration of contraindications (obesity, diabetes, centrally located tumor, ASA classification). Treatment can be primarily surgical, and in selected cases, RFA or local embolization.
b-HCA, β-catenin activated HCA; GSD, glycogen storage disease; HCA, hepatocellular adenoma; H-HCA, HNF1A-mutated HCA; IHCA, inflammatory HCA; M, months; RFA, radiofrequency ablation.
Conclusion
MRI is the preferred tool in the management of HCA, its current principle use being size evaluation (cutoff 5 cm), identification of signs of malignancy, and exclusion of H-HCAs, recognized for their benign course and permitting a conservative approach. Until now, there is no reliable MRI characteristic to diagnose non-invasively b-HCA, being the most important lesion to diagnose as it may have the highest malignant potential.
Conservative management remains the strategy of choice for uncomplicated small HCAs, and surgery may be indicated if imaging shows heterogeneous signal and growing smaller lesions suspected of being highly-differentiated HCC. Further prospective cohort studies are warranted to support the choices made between these treatment strategies and to determine the role of biopsy in the subclassification of HCAs. In cases where a HCA requires treatment, and surgical resection of smaller lesions (<3 cm) carries an unacceptable risk, RFA or TAE may be considered.
Footnotes
Funding: This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.
Conflict of interest statement: The authors declare that there is no conflict of interest.
Contributor Information
Maarten G. Thomeer, Department of Radiology, Erasmus Medical Center Rotterdam, P.O Box 2040, 3015 CE Rotterdam, The Netherlands.
Mirelle Broker, Department of Surgery, Erasmus MC University Medical Center, Rotterdam, The Netherlands.
Joanne Verheij, Department of Pathology, Amsterdam University Medical Center, Amsterdam, The Netherlands.
Michael Doukas, Department of pathology, Erasmus MC University Medical Center, Rotterdam, The Netherlands.
Turkan Terkivatan, Department of Surgery, Erasmus MC University Medical Center, Rotterdam, The Netherlands.
Diederick Bijdevaate, Department of Radiology, Erasmus MC University Medical Center, Rotterdam, The Netherlands.
Robert A. De Man, Department of Gastroenterology and Hepatology, Erasmus MC University Medical Center, Rotterdam, The Netherlands
Adriaan Moelker, Department of Radiology, Erasmus MC University Medical Center, Rotterdam, The Netherlands.
Jan N. IJzermans, Department of Surgery, Erasmus MC University Medical Center, Rotterdam, The Netherlands
References
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188430 | https://www.youtube.com/watch?v=9ORnGlJ3BK4 | Atelectasis Chest X Ray Signs
Dr M Free Medical Courses
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Posted: 5 Dec 2023
🔍 Explore the intricate details of Atelectasis through the lens of Chest X-ray findings in our latest video! 🫁 In this comprehensive guide, we break down the key aspects of Atelectasis, a condition where the lung or a Lobe of it collapses, leading to potential respiratory challenges.
Atelectasis
Lobar Collapse
Lung Collapse
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Video Chapters
0:00 Introduction
0:09 Definition and signs of volume loss
2:22 Causes of atelectasis
3:40 Right upper lobe collapse
6:18 Middle lobe collapse
8:10 Right lower lobe collapse
12:11 Left lower lobe collapse
16:17 Left upper lobe collapse
18:50 Lung collapse
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Transcript:
Introduction here we will explain the chest x-ray findings in all the different types of aasis first let's define aasis so aasis Definition and signs of volume loss is failure of all or part of the lung to expand due to loss of air in the Alvi in that part and aasis is on types so it is either l or collapse means that it's a collapse of a particular lobe of the lung and we know that the right lung has three Lopes the upper middle and lower and the left lung has two Lopes the upper and lower so either one of these if collapsed we call it low bar collapse but if all of the lung collapse then we call it lung collapse which is collapse of the whole lung and we must differentiate the atlases which is the collapse from the consolidation when there is consolidation of a certain lung Globe this means that the air in that lobe is replaced by fluid in comparison with lober collapse at which the air in that lobe will be gone and there is nothing replacing it and because of the fact that the air in a certain lob of the lung or a whole lung will be gone and there is nothing replacing it this means that we lost volume from the chest and we will see signs of volume loss on a just x-ray of flowar or lung collapse signs of volume loss mainly include displacement of the mediastinum and trachea towards the collapsed lung and there is also elevation of the hemid diaphragm and there is compensatory hyber inflation of adjacent loes or opposite lung all of this happen because to replace the volume that is lost because of the low bar or lung collapse and we have many causes of Causes of atelectasis atrasis for example consolidation so the Consolidated lob the lob that is filled with fluid might collapse after a while aasis might occur also because of bronchial obstruction so when the when there is a bronchial obstruction there is no air reaching the a certain lobe or a certain lung and this might lead to collapse of that lobe or that lung for example tumor inside the broni or compressing from an outside region also mucus blogging might uh block the broncus and inhaled foreign bodies and sometimes iatrogenics for example tube inserted too far this might lead to bronchial obstruction aasis may also occur because of surfactant problems for example in acute respiratory distress syndrome also inflammatory etiologies and lung fibrosis now that is a very simplified drawing of the frontal and lateral chest x-rays that we will explain the lower and lung collapse on and let's start Right upper lobe collapse with the right upper lob collapse so we know that in the right lung we have the horizontal Fisher and the oque fissure and those fissures would split the lung into three loes we have the upper lobe the middle lob and the lower lob so in the right upper lob collapse the right upper L collapses upward and medially the horizontal Fisher gets pulled up by the collapsed slope the horizontal fissure that you can see here would go up when the right upper L collapses and it would reach into this area here and the right upper l would appear as a thickened Barat tral tissue and this is the right upper lobe in this picture on the lateral x-ray we have the horizontal fissure here and the obique fissure here so the right upper L is right here and when the right upper L collapses the horizontal Fisher moves upward as we mentioned and the right upper L would appear as increas in density in this area right here and here we have a chest x-ray of a 16-month old child with pneumonia and we can see that the right upper L is collapsing in this picture so the horizontal Fisher is normally right here but in this case it moved up to be right here and the right upper lob would appear as increas in density in the right patrial region we can also see signs of volume loss on this Chase x-ray we can see that the trachea is deviated to the right this is the tracha here and the mediastinum is also shifted so normally there is 2/3 of the heart are to the left and 1/3 is to the right but in this picture we can see that there is less than 2/3 to the left and more than than 1/3 to the right meaning the medium is shifted to the right and there's also elevation of the right H diaphragm here and this x-ray have other findings that you can appreciate for example there is consolidation of the right middle lobe Middle lobe collapse here now let's move on to talk about the middle lob collapse or the right middle love collapse we can call it just the middle love collapse because we only have one middle and in Middle Lo collapse the collapse middle lobe sits next to the right heart border and appear as increase in density leading to loss of the right heart border on the chest x-rays which is the silhouette sign so the collapsed right middle lob sits next to the right heart border and obscure it because Anatomy wise the right right middle lobe sits adjacent to the right heart border and when there is increase in density in that right middle lobe for any reason for example a middle lob collapse this lead to the loss of the right heart border because now the middle lobe and the right heart border is of the same density and we cannot perceive the interface between them anymore and with any collapse we can always appreciate the sign of volume loss like the mediastinal tracheal shift or elevation of the hemid diaphragm or compensatory hyper inflation so we mentioned the horizontal fure in the right upper L collapse and we said that it would move upward when the right upper L collapses but the opposite occur when the middle L collapses because the horizontal Fisher moves downward for the middle L to sits next to the heart on the lateral x-ray the horizontal fissure moves downward and there is increase in density over this area right here now let's talk Right lower lobe collapse about the right lower lobe collapse so the right lower lobe collapses inferiorly and medially and appears as a triangular hyper density referred to as the sail side so the right lower l appears as increasing density in the middle and lower lung zones in this area right here and when there is r lower L collapse there is loss of the definition of the right Hemy diaphragm so we cannot see the right hemid diaphragm anymore because the right lower lobe sits next to the hemid diaphragm and when there is increase in density in the right lower lobe we fail to perceive re the interface between the right hem diaphragm and the right lower lobe because now they are of the same density so we lose the definition of the right H diaphragm but we can still see the right heart border so the right heart border would be seen in the right lower lobe collapse so we can still see the right heart border but we lose the diaphragm and that is the opposite of what happened in the right middle Lo collapse because in the right middle Lo collapse we cannot see the right heart border but we can still see the right hem diaphragm and on the lateral chest x-ray the right lower lobe collapses inferiorly and medially and appears as increasing density in the posterior inferior aspect of the X-ray so here we have a case of a right middle and lower lobe collapse together so we can see the sale sign right here we cannot see the diaphragm so it means that we have right lower L collapse and we also cannot see the right heart border means we have the right middle love collapse and because we don't see the diaphragm and the middle Lo together means we have a right middle and lower lob collapse on this picture we can also see that the right H diaphragm has a tent in it so it goes like that it's like tenting right hand diaphragm that is because there is a ligament that connects the diaphragm with theum a ligament here and when there's volume loss the ligament would get pulled and it would pull the diaphragm part with it leaving attenting like projection and this is called the jxa frenic beak sign and it is caused by a volume loss that occurred due to the right middle and lower lobe collapse also in the right lower lob collapse the right tyum gets pulled down by the collapsed lobe and we also see signs of volume loss like mediastinal and tracheal shift so on this example we can see that the trachea here is shifted to the right and we can also see that the medus dyum is also shifted to the right so normally there is 2/3 of the heart to the left and 1/3 to the right but in this example there is less than 2/3 of the heart to the left and less than 1/3 to the right meaning there is mediastinal shift to the right and we can see also that the rest of the lung looks darker than normal that is because of the compensatory hyperinflation of the lungs because of the volume loss and the right diaphragm is elevated in comparison with the left Left lower lobe collapse now let's talk about the left lower lobe collapse so in the left lower lobe collapse the left lower lobe collapses inferiorly and medially and appear as a triangular hyper dense area be behind the heart it is called the retrocardiac sail sign so the left lower lob collapses like a triangle as we can see here so it appears as a triangle behind the heart and because it collapses behind the heart it is referred to as the retrocardiac sail sign in comparison with just the sail sign for the right lower lobe collapse and in left lower lob collapse there is lot of definition of the left Hemi diaphragm because the left lower lobe in close contact with the left hemid diaphragm so when the left lower lobe become hyperdense it would match the density of the left H diaphragm leading to the loss of the interface between them so we cannot see the edge of the diaphragm anymore and this is called the silhouette sign and we also o lose the interface between the left H diaphragm and the descending aorta so we cannot see the descending aorta anymore but the left heart border stay visible and not obse secured so the left hard border is staying visible in the left lower lob collapse same with the right lower L collapse in which the right hard border stay visible same applies to the left the left level of collapse the left heart B stay visible but the hemid diaphragm is lost in both of them and on the lateral chest x-ray we have the applier running like that and the left lower lob collapses inferiorly and medially and appears as increased density in this area right here so here we have a frontal just x-ray of a left lower low collapse and a right lower lob collapse so on the left we can see the edge of the collapsed lobe this is The Edge it comes like that and run through this path and we can see the left heart border this is the left heart border it is not compromised it's still visible so as you can see we cannot see the the descending aorta and we can also not see a part of the left H diaphragm so those are both obscured by the collapsed loope also on the left lower L collapse the left highum gets pulled down by the collapsed lob and there is signs of volume loss like mediastinal shift or tracheal deviation there's also elevation of the Hamid diam and compensatory hyperinflation the left lower L collapse is a tricky to detect because the left lower L collapses behind the heart and normally you can see some pulmonary vessels through the heart if they are not seen then put the lower love collapse in your differential diagnosis so in this example this is a normal chest x-ray if you focus here you can see some pulmonary vessels on a zoomed in picture you can see some pulmonary vessels running through through this area so this means that there is no left lower L collapse in comparison with the previous frontal chest x-ray that we seen we cannot see that there is pulmonary vessels running behind the heart meaning there's some type of abnormality in that area now Left upper lobe collapse let's talk about the left Abal lope collapse so in the left aope collapse the left aope collapses Inward and upward producing increase in density in most of the left lung field so we can see that there is increase in density over most of the left lung field also in left upper low collapse there is lots of definition of the left heart border and the left medum so we cannot see the left heart border anymore and the left H diaphragm is not obscured so we can see the left H diaphragm in left uppop collapse and on the lateral Chase x-ray we can see that the left Upp lope collapses over most of the area of the anterior lung field so it collapses over all of this area so here we have an x-ray of a left upper L collapse we can see that there is increase in density over all of the left lung field and we can see that the left heart border is lost and also we can see the diaphragm so the diaphragm is seen But the left heart border is not seen and also on the left of low collapse the left tyum gets pulled upward by the collapsed lobe and oftenly we can see aent outline over the aortic arch called the lle sign so we can see aent outline over the Arctic Arch here in most of the cases of the left Upp low collapse is called the left CLE sign but on this case it's not seen because we have a mass here and we can also appreciate some signs of volume loss so on this x-ray we can see that the trachea is deviated to the left you can see it very well here and the Heart is also shifted so more than 2/3 to the left meaning there is a left shifting of the midus stum and there is also elevation of the left hemid diaphragm here and on the lateral chest x-ray we can see that the left upper lob collapses over most of the anterior chest so it appears as increasing density over most of the anterior chest area so after we finished our talk about Lung collapse the low bar collapse now let's talk about the L collapse so in lung collapse there is collapse of the entire lung and it is shown as increas in density all over the length field of the affected L and there's also signs of volume loss like medial shift and tral deviation and elevation of the hemid diaphragm and compensatory hyperinflation so here we have an x-ray of a right lung collapse we can see that there is increase in density over the whole right lung and there is tracheal deviation also and the left lung looks darker than normal meaning there is compensatory hyperinflation of the left lung and with that we reach the end of this video now this video is a part of a collection of another videos that is called the just x-ray Master Class it will appear on your monitor right now if you want to check it out you can also support this video by liking it and commenting your ideas and questions |
188431 | https://www.cdc.gov/vaccines/hcp/imz-schedules/child-adolescent-notes.html | Child Immunization Schedule Notes | Vaccines & Immunizations | CDC
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Child Immunization Schedule Notes
Recommendations for Ages 18 Years or Younger, United States, 2025
July 2, 2025
Purpose
Guide health care providers in determining recommended vaccine types, dosing frequencies and intervals, and considerations for special situations.
How to use the schedule
Vaccines and Other Immunizing Agents in the Child and Adolescent Immunization Schedule
To make vaccination recommendations, healthcare providers should:
Determine recommended vaccine by age (Table 1 – By Age)
Determine recommended interval for catch-up vaccination (Table 2 – Catch-up)
Assess need for additional recommended vaccines by medical condition or other indication (Table 3 – By Medical Indication)
Review vaccine types, frequencies, intervals, and considerations for special situations (Notes)
Review contraindications and precautions for vaccine types (Appendix)
Review new or updated ACIP guidance (Addendum)
Notes
For vaccination recommendations for persons ages 19 years or older, see the Recommended Adult Immunization Schedule, 2025.
Additional information
For calculating intervals between doses, 4 weeks = 28 days. Intervals of ≥4 months are determined by calendar months.
Within a number range (e.g., 12–18), a dash (–) should be read as "through."
Vaccine doses administered ≤4 days before the minimum age or interval are considered valid. Doses of any vaccine administered ≥5 days earlier than the minimum age or minimum interval should not be counted as valid and should be repeated as age appropriate. The repeat dose should be spaced after the invalid dose by the recommended minimum interval.For further details, see Table 3-2, Recommended and minimum ages and intervals between vaccine doses, in the Timing and Spacing of Immunobiologics section of the General Best Practices for Immunization.
Information on travel vaccination requirements and recommendations is available at www.cdc.gov/travel/.
For vaccination of persons with immunodeficiencies, see Table 8-1, Vaccination of persons with primary and secondary immunodeficiencies, in the Altered Immunocompetence section of the General Best Practices for Immunization, and Immunization in Special Clinical Circumstances (In: Kimberlin DW, Barnett ED, Lynfield Ruth, Sawyer MH, eds. Red Book: 2021–2024 Report of the Committee on Infectious Diseases. 32nd ed. Itasca, IL: American Academy of Pediatrics; 2021:72–86).
For information about vaccination in the setting of a vaccine-preventable disease outbreak, contact your state or local health department.
The National Vaccine Injury Compensation Program (VICP) is a no-fault alternative to the traditional legal system for resolving vaccine injury claims. All vaccines included in the child and adolescent vaccine schedule are covered by VICP except dengue, PPSV23, RSV, Mpox, and COVID-19 vaccines. Mpox and COVID-19 vaccines are covered by the Countermeasures Injury Compensation Program (CICP). For more information, see www.hrsa.gov/vaccinecompensation or www.hrsa.gov/cicp.
Expand All
COVID-19 vaccination
(minimum age: 6 months [Moderna and Pfizer-BioNTech COVID-19 vaccines], 12 years [Novavax COVID-19 Vaccine])
Shared clinical decision-making
Ages 6 month–17 years who are NOT moderately or severely immunocompromised.
Shared clinical decision-making vaccinations are individually based and informed by a decision process between the health care provider and the patient or parent/guardian. Where the parent presents with a desire for their child to be vaccinated, children 6 months and older may receive COVID-19 vaccination, informed by the clinical judgment of a healthcare provider and personal preference and circumstances.
Age 6 months–4 years
All vaccine doses should be from the same manufacturer.
Unvaccinated:
2 doses 2024–25 Moderna at 0, 4–8 weeks
3 doses 2024–25 Pfizer-BioNTech at 0, 3–8, and at least 8 weeks after dose 2
Incomplete initial vaccination series before 2024–25 vaccine with:
1 dose Moderna:complete initial series with 1 dose 2024–25 Moderna 4–8 weeks after most recent dose
1 dose Pfizer-BioNTech:complete initial series with 2 doses 2024–25 Pfizer-BioNTech 8 weeks apart (administer dose 1 3–8 weeks after most recent dose).
2 doses Pfizer-BioNTech:complete initial series with 1 dose 2024–25 Pfizer-BioNTech at least 8 weeks after the most recent dose.
Completed initial vaccination series before 2024–25 vaccine with:
2 or more doses Moderna:1 dose 2024–25 Moderna at least 8 weeks after the most recent dose.
3 or more doses Pfizer-BioNTech:1 dose 2024–25 Pfizer-BioNTech at least 8 weeks after the most recent dose.
Age 5–11 years
Unvaccinated:1 dose 2024–25 Moderna or Pfizer-BioNTech
Previously vaccinated before 2024–25 vaccine with 1 or more doses Moderna or Pfizer-BioNTech:1 dose 2024–25 Moderna or Pfizer-BioNTech at least 8 weeks after the most recent dose.
Age 12–17 years
Unvaccinated:
1 dose 2024–25 Moderna or Pfizer-BioNTech
2 doses 2024–25 Novavax at 0, 3–8 weeks
Previously vaccinated before 2024–25 vaccine with:
1 or more doses Moderna or Pfizer-BioNTech: 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech at least 8 weeks after the most recent dose.
1 dose Novavax: 1 dose 2024–25 Novavax 3–8 weeks after most recent dose. If more than 8 weeks after most recent dose, administer 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech.
2 or more doses Novavax: 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech at least 8 weeks after the most recent dose.
Routine vaccination
Age 18 years who are NOT moderately or severely immunocompromised
Unvaccinated:
1 dose 2024–25 Moderna or Pfizer-BioNTech
2 doses 2024–25 Novavax at 0, 3–8 weeks
Previously vaccinated before 2024–25 vaccine with:
1 or more doses Moderna or Pfizer-BioNTech:1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech at least 8 weeks after the most recent dose.
1 dose Novavax: 1 dose 2024–25 Novavax 3–8 weeks after most recent dose. If more than 8 weeks after most recent dose, administer 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech.
2 or more doses Novavax: 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech at least 8 weeks after the most recent dose.
Special situations
Persons who ARE moderately or severely immunocompromised.
Age 6 months–4 yearsUse vaccine from the same manufacturer for all doses (initial vaccination series and additional doses).
Unvaccinated:
4 doses (3-dose initial series 2024–25 Modernaat 0, 4 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna 6 months later [minimum interval 2 months]).May administer additional doses.
4 doses (3-dose initial series 2024–25 Pfizer-BioNTechat 0, 3 weeks, and at least 8 weeks after dose 2, followed by 1 dose 2024–25 Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses.
Incomplete initial 3-dose vaccination series before 2024–25 vaccine:
Previous vaccination with Moderna
1 dose Moderna:complete initial series with 2 doses 2024–25 Moderna at least 4 weeks apart (administer dose 1 4 weeks after most recent dose), followed by 1 dose 2024–25 Moderna 6 months later (minimum interval 2 months). May administer additional doses of Moderna.
2 doses Moderna:complete initial series with 1 dose 2024–25 Moderna at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna 6 months later (minimum interval 2 months). May administer additional doses of Moderna.
Previous vaccination with Pfizer-BioNTech
1 dose Pfizer-BioNTech:complete initial series with 2 doses 2024–25 Pfizer-BioNTech at least 8 weeks apart (administer dose 1 3 weeks after most recent dose), followed by 1 dose 2024–25 Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Pfizer-BioNTech.
2 doses Pfizer-BioNTech:complete initial series with 1 dose 2024–25 Pfizer-BioNTech at least 8 weeks after most recent dose, followed by 1 dose 2024–25 Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Pfizer-BioNTech.
Completed initial 3-dose vaccination series before 2024–25 vaccine with:
3 or more doses Moderna:2 doses 2024–25 Moderna 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Moderna.
3 or more doses Pfizer-BioNTech:2 doses 2024–25 Pfizer-BioNTech 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Pfizer-BioNTech.
Age 5–11 yearsUse vaccine from the same manufacturer for all doses in the initial vaccination series.
Unvaccinated:
4 doses (3-dose initial series 2024–25 Modernaat 0, 4 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses.
4 doses (3-dose initial series 2024–25 Pfizer-BioNTechat 0, 3 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses.
Incomplete initial 3-dose vaccination series before 2024–25 vaccine:
Previous vaccination with Moderna
1 dose Moderna:complete initial series with 2 doses 2024–25 Moderna at least 4 weeks apart (administer dose 1 4 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Pfizer-BioNTech.
2 doses Moderna:complete initial series with 1 dose 2024–25 Moderna at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Pfizer-BioNTech.
Previous vaccination with Pfizer-BioNTech
1 dose Pfizer-BioNTech:complete initial series with 2 doses 2024–25 Pfizer-BioNTech at least 4 weeks apart (administer dose 1 3 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Pfizer-BioNTech.
2 doses Pfizer-BioNTech:complete initial series with 1 dose 2024–25 Pfizer-BioNTech at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Pfizer-BioNTech.
Completed initial 3-dose vaccination series before 2024–25 vaccine with:
3 or more doses Moderna or 3 or more doses Pfizer-
BioNTech: 2 doses 2024–25 Moderna or Pfizer-BioNTech
6 months apart (minimum interval 2 months). Administer
dose 1 at least 8 weeks after the most recent dose. May
administer additional doses of Moderna or Pfizer-BioNTech.
Age 12–17 yearsUse vaccine from the same manufacturer for all doses in the initial vaccination series.
Unvaccinated:
4 doses (3-dose initial series Modernaat 0, 4 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
4 doses (3-dose initial series Pfizer-BioNTechat 0, 3 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
3 doses (2-dose initial series Novavaxat 0, 3 weeks, followed by 1 dose Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Incomplete initial vaccination series before 2024–25 vaccine:
Previous vaccination with Moderna
1 dose Moderna:complete initial series with 2 doses 2024–25 Moderna at least 4 weeks apart (administer dose 1 4 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 doses Moderna:complete initial series with 1 dose 2024–25 Moderna at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Previous vaccination with Pfizer-BioNTech
1 dose Pfizer-BioNTech:complete initial series with 2 doses 2024–25 Pfizer-BioNTech at least 4 weeks apart (administer dose 1 3 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 doses Pfizer-BioNTech:complete initial series with 1 dose 2024–25 Pfizer-BioNTech at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Previous vaccination with Novavax
1 dose Novavax:complete initial series with 1 dose 2024–25 Novavax at least 3 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Completed initial 3-dose vaccination series before 2024–25 vaccine with:
3 or more doses Moderna or 3 or more doses Pfizer-BioNTech:2 doses 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 or more doses Novavax:2 doses 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Age 18 years who ARE moderately or severely immunocompromised
Use vaccine from the same manufacturer for all doses in the initial vaccination series.
Unvaccinated:
4 doses (3-dose initial series Modernaat 0, 4 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
4 doses (3-dose initial series Pfizer-BioNTechat 0, 3 weeks, and at least 4 weeks after dose 2, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
3 doses (2-dose initial series Novavaxat 0, 3 weeks, followed by 1 dose Moderna or Novavax or Pfizer-BioNTech 6 months later [minimum interval 2 months]). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Incomplete initial vaccination series before 2024–25 vaccine:
Previous vaccination with Moderna
1 dose Moderna:complete initial series with 2 doses 2024–25 Moderna at least 4 weeks apart (administer dose 1 4 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 doses Moderna:complete initial series with 1 dose 2024–25 Moderna at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Previous vaccination with Pfizer-BioNTech
1 dose Pfizer-BioNTech:complete initial series with 2 doses 2024–25 Pfizer-BioNTech at least 4 weeks apart (administer dose 1 3 weeks after most recent dose), followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 doses Pfizer-BioNTech:complete initial series with 1 dose 2024–25 Pfizer-BioNTech at least 4 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Previous vaccination with Novavax
1 dose Novavax:complete initial series with 1 dose 2024–25 Novavax at least 3 weeks after most recent dose, followed by 1 dose 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months later (minimum interval 2 months). May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Completed initial 3-dose vaccination series before 2024–25 vaccine with:
3 or more doses Moderna or 3 or more doses Pfizer-BioNTech:2 doses 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
2 or more doses Novavax:2 doses 2024–25 Moderna or Novavax or Pfizer-BioNTech 6 months apart (minimum interval 2 months). Administer dose 1 at least 8 weeks after the most recent dose. May administer additional doses of Moderna or Novavax or Pfizer-BioNTech.
Additional doses of 2024–25 COVID-19 vaccine for moderately or severely immunocompromised: based on shared clinical decision-making and administered at least 2 months after the most recent dose. For description of moderate and severe immunocompromising conditions and treatment, see
Unvaccinated persons have never received any COVID-19 vaccine doses. There is no preferential recommendation for the use of one COVID-19 vaccine over another when more than one recommended age-appropriate vaccine is available. Administer an age-appropriate COVID-19 vaccine product for each dose.
For information about transition from age 4 years to age 5 years or age 11 years to age 12 years during COVID-19 vaccination series, see Tables 1 and 2 at
For information about interchangeability of COVID-19 vaccines, see www.cdc.gov/vaccines/covid-19/clinical-considerations/interim-considerations-us.html#Interchangeability.
Current COVID-19 schedule and dosage formulation available at www.cdc.gov/covidschedule. For more information on Emergency Use Authorization (EUA) indications for COVID-19 vaccines, see
Contraindications and Precautions
For contraindications and precautions to COVID-19 vaccination, see COVID-19 Appendix
Dengue vaccination
(minimum age: 9 years)
Routine vaccination
Age 9–16 years living in areas with endemic dengue AND have laboratory confirmation of previous dengue infection
3-dose series administered at 0, 6, and 12 months
Endemic areas include Puerto Rico, American Samoa, US Virgin Islands, Federated States of Micronesia, Republic of Marshall Islands, and the Republic of Palau. For updated guidance on dengue endemic areas and pre-vaccination laboratory testing see www.cdc.gov/mmwr/volumes/70/rr/rr7006a1.htm and www.cdc.gov/dengue/hcp/vaccine/eligibility.html.
Dengue vaccine should not be administered to children traveling to or visiting endemic dengue areas.
Contraindications and Precautions
For contraindications and precautions to dengue vaccination, see Dengue Appendix
Diphtheria, tetanus, and pertussis (DTaP) vaccination
(minimum age: 6 weeks [4 years for Kinrix or Quadracel])
Routine vaccination
5-dose series (3-dose primary series at age 2, 4, and 6 months, followed by booster doses at ages 15–18 months and 4–6 years)
Prospectively: Dose 4 may be administered as early as age 12 months if at least 6 months have elapsed since dose 3.
Retrospectively: A 4th dose that was inadvertently administered as early as age 12 months may be counted if at least 4 months have elapsed since dose 3.
Catch-up vaccination
Dose 5 is not necessary if dose 4 was administered at age 4 years or older and at least 6 months after dose 3.
For other catch-up guidance, see Table 2.
Special situations
Children younger than age 7 years with a contraindication specific to the pertussis component of DTaP: may administer Td for all recommended remaining doses in place of DTaP. Encephalopathy within 7 days of vaccination when not attributable to another identifiable cause is the only contraindication specific to the pertussis component of DTaP. For additional information, see www.cdc.gov/pertussis/hcp/vaccine-recommendations/td-offlabel.html.
Wound management in children younger than age 7 years with history of 3 or more doses of tetanus-toxoid-containing vaccine: For all wounds except clean and minor wounds, administer DTaP if more than 5 years since last dose of tetanus-toxoid-containing vaccine. For detailed information, see www.cdc.gov/mmwr/volumes/67/rr/rr6702a1.htm.
Contraindications and Precautions
For contraindications and precautions to Diphtheria, tetanus, pertussis (DTaP) vaccination, see DTaP Appendix
Haemophilus influenzae type b vaccination
(minimum age: 6 weeks)
Routine vaccination
ActHIB, Hiberix, Pentacel, or Vaxelis: 4-dose series (3-dose primary series at age 2, 4, and 6 months, followed by a booster dose at age 12–15 months)
Vaxelis is not recommended for use as a booster dose. A different Hib-containing vaccine should be used for the booster dose.
PedvaxHIB: 3-dose series (2-dose primary series at age 2 and 4 months, followed by a booster dose at age 12–15 months)
American Indian and Alaska Native infants: Vaxelis and PedvaxHIB preferred over other Hib vaccines for the primary series.
Catch-up vaccination
Dose 1 at age 7–11 months:Administer dose 2 at least 4 weeks later and dose 3 (final dose) at age 12–15 months or 8 weeks after dose 2 (whichever is later).
Dose 1 at age 12–14 months:Administer dose 2 (final dose) at least 8 weeks after dose 1.
Dose 1 before age 12 months and dose 2 before age 15 months:Administer dose 3 (final dose) at least 8 weeks after dose 2.
2 doses of PedvaxHIBbefore age 12 months:Administer dose 3 (final dose) at age 12–59 months and at least 8 weeks after dose 2.
1 dose administered at age 15 months or older:No further doses needed
Unvaccinated at age 15–59 months:Administer 1 dose.
Previously unvaccinated children age 60 months or olderwho are not considered high risk:Catch-up vaccination not required.
For other catch-up guidance, see Table 2. Vaxelis can be used for catch-up vaccination in children younger than age 5 years. Follow the catch-up schedule even if Vaxelis is used for one or more doses. For detailed information on use of Vaxelis see www.cdc.gov/mmwr/volumes/69/wr/mm6905a5.htm.
Special situations
Chemotherapy or radiation treatment:Age 12–59 months
Unvaccinated or only 1 dose before age 12 months: 2 doses, 8 weeks apart
2 or more doses before age 12 months: 1 dose at least 8 weeks after previous dose
Doses administered within 14 days of starting therapy or during therapy should be repeated at least 3 months after therapy completion.
Hematopoietic stem cell transplant (HSCT):
3-dose series 4 weeks apart starting 6 to 12 months after successful transplant regardless of Hib vaccination history
Anatomic or functional asplenia (including sickle cell disease):Age 12–59 months
Unvaccinated or only 1 dose before age 12 months: 2 doses, 8 weeks apart
2 or more doses before age 12 months: 1 dose at least 8 weeks after previous dose
Unvaccinated persons age 5 years or older
1 dose
Elective splenectomy:Unvaccinated persons age 15 months or older
1 dose (preferably at least 14 days before procedure)
HIV infection:Age 12–59 months
Unvaccinated or only 1 dose before age 12 months: 2 doses, 8 weeks apart
2 or more doses before age 12 months: 1 dose at least 8 weeks after previous dose
Unvaccinated persons age 5–18 years
1 dose
Immunoglobulin deficiency, early component complement deficiency, or early component complement inhibitor use:Age 12–59 months
Unvaccinated or only 1 dose before age 12 months: 2 doses, 8 weeks apart
2 or more doses before age 12 months: 1 dose at least 8 weeks after previous dose
Unvaccinated = Less than routine series (through age 14 months) or no doses (age 15 months or older)
Contraindications and Precautions
For contraindications and precautions to Haemophilus influenzae type b (Hib) vaccination, see Hib Appendix
Hepatitis A vaccination
(minimum age: 12 months for routine vaccination)
Routine vaccination
2-dose series (minimum interval: 6 months) at age 12–23 months
Catch-up vaccination
Unvaccinated persons through age 18 years should complete a 2-dose series (minimum interval: 6 months).
Persons who previously received 1 dose at age 12 months or older should receive dose 2 at least 6 months after dose 1.
Adolescents age 18 years or older may receive HepA-HepB (Twinrix) as a 3-dose series (0, 1, and 6 months) or 4-dose series (3 doses at 0, 7, and 21–30 days, followed by a booster dose at 12 months).
International travel
Persons traveling to or working in countries with high or intermediate endemic hepatitis A
(www.cdc.gov/travel/)
Infants age 6–11 months: 1 dose before departure; revaccinate with 2 doses (separated by at least 6 months) between age 12–23 months.
Unvaccinated age 12 months or older: Administer dose 1 as soon as travel is considered.
Contraindications and Precautions
For contraindications and precautions to Hepatitis A (HepA) vaccination, see HepA Appendix
Hepatitis B vaccination
(minimum age: birth)
Routine vaccination
Mother is HBsAg-negative
3-dose series at age 0, 1–2, 6–18 months (use monovalent HepB vaccine for doses administered before age 6 weeks)
Birth weight ≥2,000 grams: 1 dose within 24 hours of birth if medically stable
Birth weight <2,000 grams: 1 dose at chronological age 1 month or hospital discharge (whichever is earlier and even if weight is still <2,000 grams).
Infants who did not receive a birth dose should begin the series as soon as possible (see Table 2for minimum intervals).
Administration of 4 doses is permitted when a combination vaccine containing HepB is used after the birth dose.
Minimum intervals (see Table 2):when 4 doses are administered, substitute “dose 4” for “dose 3” in these calculations
Final (3rd or 4th) dose:age 6–18 months(minimum age 24 weeks)
Mother is HBsAg-positive
Birth dose (monovalent HepB vaccine only): administer HepB vaccine and hepatitis B immune globulin (HBIG) (in separate limbs) within 12 hours of birth, regardless of birth weight.
Birth weight <2000 grams: administer 3 additional doses of HepB vaccine beginning at age 1 month (total of 4 doses)
Final (3rd or 4th) dose: administer at age 6 months (minimum age 24 weeks)
Test for HBsAg and anti-HBs at age 9–12 months. If HepB series is delayed, test 1–2 months after final dose. Do not test before age 9 months.
Mother is HBsAg-unknownIf other evidence suggestive of maternal hepatitis B infection exists (e.g., presence of HBV DNA, HBeAg-positive, or mother known to have chronic hepatitis B infection), manage infant as if mother is HBsAg-positive
Birth dose (monovalent HepB vaccine only):
Birth weight ≥2,000 grams: administer HepB vaccine within 12 hours of birth. Determine mother’s HBsAg status as soon as possible. If mother is determined to be HBsAg-positive, administer HBIG as soon as possible (in separate limb), but no later than 7 days of age.
Birth weight <2,000 grams: administer HepB vaccine and HBIG (in separate limbs) within 12 hours of birth. Administer 3 additional doses of HepB vaccine beginning at age 1 month (total of 4 doses)
Final (3rd or 4th) dose: administer at age 6 months (minimum age 24 weeks)
If mother is determined to be HBsAg-positive or if status remains unknown, test for HBsAg and anti-HBs at age 9–12 months. If HepB series is delayed, test 1–2 months after final dose. Do not test before age 9 months.
Catch-up vaccination
Unvaccinated persons should complete a 3-dose series at 0, 1–2, 6 months. See Table 2 for minimum intervals
Adolescents age 11–15 years may use an alternative 2-dose schedule with at least 4 months between doses (adult formulation Recombivax HB only).
Adolescents age 18 years may receive:
Heplisav-B: 2-dose series at least 4 weeks apart
PreHevbrio: 3-dose series at 0, 1, and 6 months
HepA-HepB (Twinrix): 3-dose series (0, 1, and 6 months) or 4-dose series (3 doses at 0, 7, and 21–30 days, followed by a booster dose at 12 months).
Note: PreHevbrio is not recommended in pregnancy due to lack of safety data in pregnant women.
Special situations
Revaccination is generally not recommended for persons with a normal immune status who were vaccinated as infants, children, adolescents, or adults.
Post-vaccination serology testing and revaccination (if anti-HBs<10mlU/mL) is recommended for certain populations, including:
Infants born to HBsAg-positive mothers
Persons who are predialysis or on maintenance dialysis
Other immunocompromised persons
For detailed revaccination recommendations, see www.cdc.gov/mmwr/volumes/67/rr/rr6701a1.htm.
Contraindications and Precautions
For contraindications and precautions to Hepatitis B (HepB) vaccination, see HepB Appendix
Human papillomavirus vaccination
(minimum age: 9 years)
Routine and catch-up vaccination
HPV vaccination routinely recommended at age 11–12 years (can start at age 9 years) and catch-up HPV vaccination recommended for all persons through age 18 years if not adequately vaccinated
2- or 3-dose series depending on age at initial vaccination:
Age 9 –14 years at initial vaccination: 2-dose series at 0, 6–12 months (minimum interval: 5 months; repeat dose if administered too soon)
Age 15 years or older at initial vaccination: 3-dose series at 0, 1–2 months, 6 months (minimum intervals: dose 1 to dose 2 = 4 weeks; dose 2 to dose 3 = 12 weeks; dose 1 to dose 3 = 5 months; repeat dose if administered too soon)
No additional dose recommended when any HPV vaccine series of any valency has been completed using recommended dosing intervals.
Special situations
Immunocompromising conditions, including HIV infection: 3-dose series, even for those who initiate vaccination at age 9 through 14 years.
History of sexual abuse or assault: Start at age 9 years.
Pregnancy: Pregnancy testing not needed before vaccination; HPV vaccination not recommended until after pregnancy; no intervention needed if vaccinated while pregnant
Contraindications and Precautions
For contraindications and precautions to Human papillomavirus (HPV) vaccination, see HPV Appendix
Influenza vaccination
(minimum age: 6 months [IIV3], 2 years [LAIV3],18 years [recombinant influenza vaccine, RIV3])
Routine vaccination
Use any influenza vaccine appropriate for age and health status annually:
Age 6 months–8 years who have received fewer than 2 influenza vaccine doses before July 1, 2024, or whose influenza vaccination history is unknown: 2 doses, separated by at least 4 weeks. Administer dose 2 even if the child turns 9 years between receipt of dose 1 and dose 2.
Age 6 months–8 years who have received at least 2 influenza vaccine doses before July 1, 2024: 1 dose.
Age 9 years or older: 1 dose
Age 18 years solid organ transplant recipients receiving immunosuppressive medications: high-dose inactivated (HD-IIV3) and adjuvanted inactivated (aIIV3) influenza vaccines are acceptable options. No preference over other age-appropriate IIV3 or RIV3.
For the 2024–25 season, see www.cdc.gov/mmwr/volumes/73/rr/rr7305a1.htm.
For the 2025–26 season, see the 2025–26 ACIP influenza vaccine recommendations.
Note:Persons with an egg allergy can receive any influenza vaccine (egg-based and non-egg-based) appropriate for age and health status.
Special situations
Close contacts (e.g., household contacts) of severely immunosuppressed persons who require a protected environment: should not receive LAIV3. If LAIV3 is given, they should avoid contact with, or caring for such immunosuppressed persons for 7 days after vaccination.
Note:Persons with an egg allergy can receive any influenza vaccine (egg-based or non-egg-based) appropriate for age and health status.
Contraindications and Precautions
For contraindications and precautions to Influenza vaccination, see IIV4 Appendix, LAIV4 Appendix, ccIIV4 Appendix, and RIV4 Appendix.
Measles, mumps, and rubella vaccination
(minimum age: 12 months for routine vaccination)
Routine vaccination
2-dose series at age 12–15 months, age 4–6 years
MMR or MMRV may be administered
Note: For dose 1 in children age 12–47 months, it is recommended to administer MMR and varicella vaccines separately. MMRV may be used if parents or caregivers express a preference.
Catch-up vaccination
Unvaccinated children and adolescents: 2-dose series at least 4 weeks apart
The maximum age for use of MMRV is 12 years.
Note:If MMRV is used, the minimum interval between MMRV doses is 3 months
Special situations
International travel
Infants age 6–11 months: 1 dose before departure; revaccinate with 2-dose series at age 12–15 months (12 months for children in high-risk areas) and dose 2 as early as 4 weeks later.
Children age 12 months or older:
Unvaccinated: 2-dose series (separated by at least 4 weeks) before departure
Previously received 1 dose: administer dose 2 at least 4 weeks after dose 1
In mumps outbreak settings, for information about additional doses of MMR (including 3rd dose of MMR), see www.cdc.gov/mmwr/volumes/67/wr/mm6701a7.htm
Note: If MMRV is used, the minimum interval between MMRV doses is 3 months
Contraindications and Precautions
For contraindications and precautions to Measles, mumps, rubella (MMR), see MMR Appendix
Meningococcal serogroup A,C,W,Y vaccination
(minimum age: 2 months [MenACWY-CRM, Menveo], 2 years [MenACWY-TT, MenQuadfi]), 10 years [MenACWY-TT/MenB-FHbp, Penbraya])
Routine vaccination
2-dose series at age 11–12 years; 16 years
Note: MenACWY vaccines may be administered simultaneously with MenB vaccines if indicated, but at a different anatomic site, if feasible.
Catch-up vaccination
Age 13–15 years: 1 dose now and booster at age 16–18 years (minimum interval: 8 weeks)
Age 16–18 years: 1 dose
Note: MenACWY vaccines may be administered simultaneously with MenB vaccines if indicated, but at a different anatomic site, if feasible.
Special situations
Anatomic or functional asplenia (including sickle cell disease), HIV infection, persistent complement component deficiency, complement inhibitor (e.g., eculizumab, ravulizumab) use:
Menveo
Dose 1 at age 2 months: 4-dose series (additional 3 doses at age 4, 6, and 12 months)
Dose 1 at age 3–6 months: 3- or 4- dose series (dose 2 [and dose 3 if applicable] at least 8 weeks after previous dose until a dose is received at age 7 months or older, followed by an additional dose at least 12 weeks later and after age 12 months)
Dose 1 at age 7–23 months: 2-dose series (dose 2 at least 12 weeks after dose 1 and after age 12 months)
Dose 1 at age 24 months or older: 2-dose series at least 8 weeks apart
MenQuadfi
Dose 1 at age 24 months or older: 2-dose series at least 8 weeks apart
Travel to countries with hyperendemic or epidemic meningococcal disease, including countries in the African meningitis belt or during the Hajj
(www.cdc.gov/travel/):
Children younger than age 24 months:
Menveo(age 2–23 months)
Dose 1 at age 2 months: 4-dose series (additional 3 doses at age 4, 6, and 12 months)
Dose 1 at age 3–6 months: 3- or 4- dose series (dose 2 [and dose 3 if applicable] at least 8 weeks after previous dose until a dose is received at age 7 months or older, followed by an additional dose at least 12 weeks later and after age 12 months)
Dose 1 at age 7–23 months: 2-dose series (dose 2 at least 12 weeks after dose 1 and after age 12 months)
Children age 2 years or older: 1 dose Menveo or MenQuadfi
First-year college students who live in residential housing (if not previously vaccinated at age 16 years or older) or military recruits: 1 dose Menveo or MenQuadfi
Adolescent vaccination of children who received MenACWY prior to age 10 years:
Children for whom boosters are recommended because of an ongoing increased risk of meningococcal disease (e.g., those with complement component deficiency, HIV, or asplenia): Follow the booster schedule for persons at increased risk.
Children for whom boosters are not recommended (e.g., a healthy child who received a single dose for travel to a country where meningococcal disease is endemic): Administer MenACWY according to the recommended adolescent schedule with dose 1 at age 11–12 years and dose 2 at age 16 years.
Menveo has two formulations: lyophilized and liquid. The liquid formulation should not be used before age 10 years. See www.cdc.gov/vaccines/vpd/mening/downloads/menveo-single-vial-presentation.pdf.
Note:For MenACWYbooster dose recommendations for groups listed under “Special situations” and in an outbreak setting and additional meningococcal vaccination information, see www.cdc.gov/mmwr/volumes/69/rr/rr6909a1.htm.
Children age 10 years or older may receive a single dose of Penbraya as an alternative to separate administration of MenACWY and MenB when both vaccines would be given on the same clinic day, (see “Meningococcal serogroup B vaccination” section below for more information).
Contraindications and Precautions
For contraindications and precautions to Meningococcal ACWY (MenACWY)
[MenACWY-CRM (Menveo); MenACWY-D (Menactra); MenACWY-TT (MenQuadfi)], see Meningococcal ACWY (MenACWY) Appendix
For contraindications and precautions to Meningococcal ABCWY, (MenACWY-TT/MenB-FHbp) [Penbraya], see Meningococcal ABCWY Appendix
Meningococcal serogroup B vaccination
(minimum age: 10 years [MenB-4C, Bexsero; MenB-FHbp, Trumenba; MenACWY-TT/MenB-FHbp, Penbraya])
Shared Clinical Decision-Making
Adolescents not at increased risk age 16–23 years (preferred age 16–18 years) based on shared clinical decision-making.
Bexsero or Trumenba (use same brand for all doses): 2–dose series at least 6 months apart (if dose 2 is administered earlier than 6 months, administer dose 3 at least 4 months after dose 2)
To optimize rapid protection (e.g., for students starting college in less than 6 months), a 3-dose series (0, 1–2, 6 months) may be administered.
For additional information on shared clinical decision-making for MenB, see www.cdc.gov/vaccines/media/pdfs/2025/03/2024-isd-job-aid-scdm-menb-508-remediated.pdf
Note: MenB vaccines may be administered simultaneously with MenACWY vaccines if indicated, but at a different anatomic site, if feasible.
Special situations
Anatomic or functional asplenia (including sickle cell disease), persistent complement component deficiency, complement inhibitor (e.g., eculizumab, ravulizumab) use.
Bexsero or Trumenba (use same brand for all doses including booster doses) 3-dose series at 0, 1–2, 6 months (if dose 2 was administered at least 6 months after dose 1, dose 3 not needed; if dose 3 is administered earlier than 4 months after dose 2, a 4th dose should be administered at least 4 months after dose 3)
For MenB booster dose recommendations for groups listed under “Special situations” and in an outbreak setting and additional meningococcal vaccination information, see
www.cdc.gov/mmwr/volumes/69/rr/rr6909a1.htm.
Note: MenB vaccines may be administered simultaneously with MenACWY vaccines if indicated, but at a different anatomic site, if feasible.
Children age 10 years or older may receive a dose of Penbraya (MenACWY–TT/MenB–FHbp) as an alternative to separate administration of MenACWY and MenB when both vaccines would be given on the same clinic day. For age-eligible children not at increased risk, if Penbraya is used for dose 1 MenB, MenB-FHbp (Trumenba) should be administered for dose 2 MenB. For age-eligible children at increased risk of meningococcal disease, Penbraya may be used for additional MenACWY and MenB doses (including booster doses) if both would be given on the same clinic day and at least 6 months have elapsed since most recent Penbraya dose.
Contraindications and Precautions
For contraindications and precautions to Meningococcal B (MenB), MenB-4C [Bexsero], MenB-FHbp [Trumenba], see MenB Appendix
For contraindications and precautions to Meningococcal ABCWY, (MenACWY-TT/MenB-FHbp) [Penbraya], see Meningococcal ABCWY Appendix
Mpox vaccination
(minimum age: 18 years [Jynneos])
Special situations
Age 18 years and at risk for mpox infection: complete 2-dose series, 28 days apart.
Risk factors for mpox infection include:
Gay, bisexual, or other MSM, or a person who has sex with gay, bisexual, or other MSM who in the past 6 months have had one of the following:
A new diagnosis of at least 1 sexually transmitted disease
More than 1 sex partner
Sex at a commercial sex venue
Sex in association with a large public event in a geographicarea where mpox transmission is occurring
Persons who are sexual partners of the persons described above
Persons who anticipate experiencing any of the situations described above
Pregnancy:There is currently no ACIP recommendation for Jynneos use in pregnancy due to lack of safety data in pregnant women. Pregnant women with any risk factor described above may receive Jynneos
For detailed information, see:www.cdc.gov/mpox/hcp/vaccine-considerations/vaccination-overview.html
Contraindications and Precautions
For contraindications and precautions to Mpox vaccination, see Mpox Appendix
Pneumococcal vaccination
(minimum age: 6 weeks [PCV15], [PCV 20]; 2 years [PPSV23])
Routine vaccination with PCV
4-dose series at 2, 4, 6, 12–15 months
Catch-up vaccination with PCV
Healthy children ages 2–4 years with any incomplete PCV series: 1 dose PCV
For other catch-up guidance, see Table 2.
Note: For children without risk conditions, PCV20 is not indicated if they have received 4 doses of PCV13 or PCV15 or another age appropriate complete PCV series.
Special situations
Children and adolescents with cerebrospinal fluid leak; chronic heart disease; chronic kidney disease (excluding maintenance dialysis and nephrotic syndrome); chronic liver disease; chronic lung disease (including moderate persistent or severe persistent asthma); cochlear implant; or diabetes mellitus:
Age 2–5 years
Any incomplete PCV series with:
3 PCV doses: 1 dose PCV (at least 8 weeks after the most recent PCV dose)
Less than 3 PCV doses: 2 doses PCV (at least 8 weeks after the most recent dose and administered at least 8 weeks apart)
Completed recommended PCV series but have not received PPSV23
Previously received at least 1 dose of PCV20: no further PCV or PPSV23 doses needed
Not previously received PCV20: administer 1 dose PCV20 or 1 dose PPSV23 administer at least 8 weeks after the most recent PCV dose.
Age 6–18 years
Not previously received any dose of PCV13, PCV15, or PCV20: administer 1 dose of PCV15 or PCV20. If PCV15 is used and no previous receipt of PPSV23, administer 1 dose of PPSV23 at least 8 weeks after the PCV15 dose.
Received PCV before age 6 years but have not received PPSV23
Previously received at least 1 dose of PCV20: no further PCV or PPSV23 doses needed
Not previously received PCV20: 1 dose PCV20 or 1 dose PPSV23 administer at least 8 weeks after the most recent PCV dose.
Received PCV13 only at or after age 6 years: administer 1 dose PCV20 or 1 dose PPSV23 at least 8 weeks after the most recent PCV13 dose.
Received 1 dose PCV13 and 1 dose PPSV23 at or after age 6 years: no further doses of any PCV or PPSV23 indicated.
Children and adolescents on maintenance dialysis, or with immunocompromising conditions such as nephrotic syndrome; congenital or acquired asplenia or splenic dysfunction; congenital or acquired immunodeficiencies; diseases and conditions treated with immunosuppressive drugs or radiation therapy, including malignant neoplasms, leukemias, lymphomas, Hodgkin disease, and solid organ transplant; HIV infection; or sickle cell disease or other hemoglobinopathies:
Age 2–5 years
Any incomplete PCV series:
3 PCV doses: 1 dose PCV (at least 8 weeks after the most recent PCV dose)
Less than 3 PCV doses: 2 doses PCV (at least 8 weeks after the most recent dose and administered at least 8 weeks apart)
Completed recommended PCV series but have not received PPSV23
Previously received at least 1 dose of PCV20: no further PCV or PPSV23 doses needed
Not previously received PCV20: administer 1 dose PCV20 or 1 dose PPSV23 at least 8 weeks after the most recent PCV. If PPSV23 is used, administer 1 dose of PCV20 or dose 2 PPSV23 at least 5 years after dose 1 PPSV23.
Age 6–18 years
Not previously received any dose of PCV13, PCV15, or PCV20: administer 1 dose of PCV15 or 1 dose of PCV20. If PCV15 is used and no previous receipt of PPSV23, administer 1 dose of PPSV23 at least 8 weeks after the PCV15 dose.
Received PCV before age 6 years but have not received PPSV23
Previously received at least 1 dose of PCV20: no additional dose of PCV or PPSV23
Not previously received PCV20: administer 1 dose PCV20 or 1 dose PPSV23 at least 8 weeks after the most recent PCV dose. If PPSV23 is used, administer either PCV20 or dose 2 PPSV23 at least 5 years after dose 1 PPSV23.
Received PCV13 only at or after age 6 years: administer 1 dose PCV20 or 1 dose PPSV23 at least 8 weeks after the most recent PCV13 dose. If PPSV23 is used, administer 1 dose of PCV20 or dose 2 PPSV23 at least 5 years after dose 1 PPSV23.
Received 1 dose PCV13 and 1 dose PPSV23 at or after age 6 years: administer 1 dose PCV20 or 1 dose PPSV23 at least 8 weeks after the most recent PCV13 dose and at least 5 years after dose 1 PPSV23.
Pregnancy: no recommendation for PCV or PPSV23 due to limited data. Summary of existing data on pneumococcal vaccination during pregnancy can be found at www.cdc.gov/mmwr/volumes/72/rr/rr7203a1.htm
For guidance on determining which pneumococcal vaccines a patient needs and when, please refer to the mobile app, which can be downloaded here: www.cdc.gov/pneumococcal/hcp/vaccine-recommendations/app.html
Incomplete series = Not having received all doses in either the recommended series or an age-appropriate catch-up series. See Table 2 in ACIP pneumococcal recommendations at stacks.cdc.gov/view/cdc/133252
When both PCV15 and PPSV23 are indicated, administer all doses of PCV15 first. PCV15 and PPSV23 should not be administered during the same visit.
Contraindications and Precautions
For contraindications and precautions to Pneumococcal conjugate (PCV), see PCV Appendix and Pneumococcal polysaccharide (PPSV23), see PPSV23 Appendix
Poliovirus vaccination
(minimum age: 6 weeks)
Routine vaccination
4-dose series at ages 2, 4, 6–18 months, 4–6 years; administer the final dose on or after age 4 years and at least 6 months after the previous dose.
4 or more doses of IPV can be administered before age 4 years when a combination vaccine containing IPV is used. However, a dose is still recommended on or after age 4 years and at least 6 months after the previous dose.
Catch-up vaccination
In the first 6 months of life, use minimum ages and intervals only for travel to a polio-endemic region or during an outbreak.
Adolescents age 18 years known or suspected to be unvaccinated or incompletely vaccinated: administer remaining doses (1, 2, or 3 IPV doses) to complete a 3-dose primary series. Unless there are specific reasons to believe they were not vaccinated, most persons aged 18 years or older born and raised in the United States can assume they were vaccinated against polio as children.
Series containing oral poliovirus vaccine (OPV), either mixed OPV-IPV or OPV-only series:
Total number of doses needed to complete the series is the same as that recommended for the U.S. IPV schedule. See www.cdc.gov/mmwr/volumes/66/wr/mm6601a6.htm.
Only trivalent OPV (tOPV) counts toward the U.S. vaccination requirements.
Doses of OPV administered before April 1, 2016, should be counted (unless specifically noted as administered during a campaign).
Doses of OPV administered on or after April 1, 2016, should not be counted.
For guidance to assess doses documented as “OPV,” see www.cdc.gov/mmwr/volumes/66/wr/mm6606a7.htm.
For other catch-up guidance, see Table 2.
Special situations
Adolescents aged 18 years at increased risk of exposure topoliovirusandcompletedprimaryseries: may administer one lifetime IPV booster
Note:Complete primary series consist of at least 3 doses of IPV or trivalent oral poliovirus vaccine (tOPV) in any combination.
For detailed information, see: www.cdc.gov/vaccines/vpd/polio/hcp/recommendations.html
Contraindications and Precautions
For contraindications and precautions to Poliovirus vaccine, inactivated (IPV), see Appendix
Respiratory syncytial virus immunization
(minimum age: birth [Nirsevimab, RSV-mAb, Beyfortus])
Routine immunization
Infants born October – March in most of the continental United States
Mother did not receive RSV vaccine or mother’s RSV vaccination status is unknown or mother received RSV vaccine in previous pregnancy: administer 1 dose nirsevimab within 1 week of birth—ideally during the birth hospitalization
Mother received RSV vaccine less than 14 days prior to delivery: administer 1 dose nirsevimab within 1 week of birth—ideally during the birth hospitalization
Mother received RSV vaccine at least 14 days prior to delivery: nirsevimab not needed but can be considered in rare circumstances at the discretion of healthcare providers (seewww.cdc.gov/rsv/hcp/vaccine-clinical-guidance/infants-young-children.html)
Infants born April–September in most of the continental United States
Mother did not receive RSV vaccine or mother’s RSV vaccination status is unknown or mother received RSV vaccine in previous pregnancy: administer 1 dose nirsevimab shortly before start of RSV season
Mother received RSV vaccine less than 14 days prior to delivery: administer 1 dose nirsevimab shortly before start of RSV season
Mother received RSV vaccine at least 14 days prior to delivery: nirsevimab not needed but can be considered in rare circumstances at the discretion of healthcare providers (see www.cdc.gov/rsv/hcp/vaccine-clinical-guidance/infants-young-children.html)
Infants with prolonged birth hospitalization (e.g., for prematurity) discharged October through March should be immunized shortly before or promptly after discharge.
Note: While the timing of the onset and duration of RSV season may vary, administration of nirsevimab is recommended October through March in most of the continental United States (optimally October through November or within 1 week of birth). Providers in jurisdictions with RSV seasonality that differs from most of the continental United States (e.g., Alaska, jurisdiction with tropical climate) should follow guidance from public health authorities (e.g., CDC, health departments) or regional medical centers on timing of administration based on local RSV seasonality.
Note: Nirsevimab can be administered to children who are eligible to receive palivizumab. Children who have received nirsevimab should not receive palivizumab for the same RSV season.
Special situations
Ages 8–19 months with chronic lung disease of prematurity requiring medical support (e.g., chronic corticosteroid therapy, diuretic therapy, or supplemental oxygen) any time during the 6-month period before the start of the second RSV season; severe immunocompromise; cystic fibrosis with either weight for length <10th percentile or manifestation of severe lung disease (e.g., previous hospitalization for pulmonary exacerbation in the first year of life or abnormalities on chest imaging that persist when stable):
1 dose nirsevimab shortly before start of second RSV season
Ages 8–19 months who are American Indian or
Alaska Native: 1 dose nirsevimab shortly before start of second RSV season
Age-eligible and undergoing cardiac surgery with cardiopulmonary bypass: 1 additional dose of nirsevimab after surgery. See www.accessdata.fda.gov/drugsatfda_docs/label/2023/761328s000lbl.pdf
Note: While the timing of the onset and duration of RSV season may vary, administration of nirsevimab is recommended October through March in most of the continental United States (optimally October through November or within 1 week of birth). Providers in jurisdictions with RSV seasonality that differs from most of the continental United States (e.g., Alaska, jurisdiction with tropical climate) should follow guidance from public health authorities (e.g., CDC, health departments) or regional medical centers on timing of administration based on local RSV seasonality.
Note:Nirsevimab can be administered to children who are eligible to receive palivizumab. Children who have received nirsevimab should not receive palivizumab for the same RSV season.
For further guidance, see www.cdc.gov/mmwr/volumes/72/wr/mm7234a4.htm and www.cdc.gov/vaccines/vpd/rsv/hcp/child-faqs.html
Contraindications and Precautions
For contraindications and precautions to RSV monoclonal antibody (RSV-mAb), see RSV monoclonal antibody Appendix
Respiratory syncytial virus vaccination
(RSV [Abrysvo])
Routine vaccination
Pregnant at 32 weeks 0 days through 36 weeks and 6 days gestation from September through January in most of the continental United States: 1 dose Abrysvo. Administer RSV vaccine regardless of previous RSV infection.
Either maternal RSV vaccination with Abrysvo or infant immunization with nirsevimab (RSV monoclonal antibody) is recommended to prevent severe respiratory syncytial virus disease in infants.
All other pregnant women: RSV vaccine not recommended.
Subsequent pregnancies: additional doses not recommended. No data are available to inform whether additional doses are needed in subsequent pregnancies. Infants born to pregnant women who received RSV vaccine during a previous pregnancy should receive nirsevimab.
Note: Providers in jurisdictions with RSV seasonality that differs from most of the continental United States (e.g., Alaska, jurisdictions with tropical climate) should follow guidance from public health authorities (e.g., CDC, health departments) or regional medical centers on timing of administration based on local RSV seasonality.
Contraindications and Precautions
For contraindications and precautions to Respiratory syncytial virus vaccine (RSV), see RSV Appendix
Rotavirus vaccination
(minimum age: 6 weeks)
Routine vaccination
Rotarix:2-dose series at age 2 and 4 months
RotaTeq:3-dose series at age 2, 4, and 6 months
If any dose in the series is eitherRotaTeqor unknown, default to 3-dose series.
Catch-up vaccination
Do not start the series on or after age 15 weeks, 0 days.
The maximum age for the final dose is 8 months, 0 days.
For other catch-up guidance, see Table 2.
Contraindications and Precautions
For contraindications and precautions to Rotavirus (RV) [RV1 (Rotarix), RV5 (RotaTeq)], see Rotavirus Appendix
Tetanus, diphtheria, and pertussis (Tdap) vaccination
(minimum age: 11 years for routine vaccination, 7 years for catch-up vaccination)
Routine vaccination
Age 11–12 years: 1 dose Tdap (adolescent booster)
Pregnancy:1 dose Tdap during each pregnancy, preferably in early part of gestational weeks 27–36.
Note: Tdap may be administered regardless of the interval since the last tetanus- and diphtheria-toxoid-containing vaccine.
Catch-up vaccination
Age 13–18 years who have not received Tdap:1 dose Tdap (adolescent booster)
Age7–18yearsnotfullyvaccinatedwithDTaP: 1 dose Tdap as part of the catch-up series (preferably the first dose); if additional doses are needed, use Td or Tdap.
Tdap administered at age 7–10 years:
Age 7–9 years who receive Tdap should receive the adolescent Tdap booster dose at age 11–12 years.
Age 10 years who receive Tdap do not need the adolescent Tdap booster dose at age 11–12 years.
DTaP inadvertently administered on or after age 7 years:
Age 7–9 years: DTaP may count as part of catch-up. Administer adolescent Tdap booster dose at age 11–12 years.
Age 10–18 years: Count dose of DTaP as the adolescent Tdap booster dose.
For other catch-up guidance, see Table 2.
Fully vaccinated = 5 valid doses of DTaP OR 4 valid doses of DTaP if dose 4 was administered at age 4 years or older
Note: Tdap may be administered regardless of the interval since the last tetanus- and diphtheria-toxoid-containing vaccine.
Special situations
Wound managementin persons age 7 years or older with history of 3 or more doses of tetanus-toxoid-containing vaccine: For clean and minor wounds, administer Tdapor Td if more than 10 years since last dose of tetanus-toxoid-containing vaccine; for all other wounds, administer Tdap or Td if more than 5 years since last dose of tetanus-toxoid-containing vaccine. Tdap is preferred for persons age 11 years or older who have not previously received Tdap or whose Tdap history is unknown. If a tetanus-toxoid-containing vaccine is indicated for a pregnant adolescent, use Tdap.
Note: Tdap may be administered regardless of the interval since the last tetanus- and diphtheria-toxoid-containing vaccine.
For detailed information, see www.cdc.gov/mmwr/volumes/69/wr/mm6903a5.htm.
Contraindications and Precautions
For contraindications and precautions to Tetanus, diphtheria, and acellular pertussis (Tdap) and Tetanus, diphtheria (Td), see Tdap and Td Appendix
Varicella vaccination
(minimum age: 12 months)
Routine vaccination
2-dose series at age 12–15 months, 4–6 years
VAR or MMRV may be administered
Dose 2 may be administered as early as 3 months after dose 1 (a dose inadvertently administered after at least 4 weeks may be counted as valid)
Note:For dose 1 in children age 12–47 months, it is recommended to administer MMR and varicella vaccines separately. MMRV may be used if parents or caregivers express a preference.
Catch-up vaccination
Ensure persons age 7–18 years without evidence of immunity (see MMWR at www.cdc.gov/mmwr/pdf/rr/rr5604.pdf) have a 2-dose series:
Age 7–12 years: Routine interval: 3 months (a dose inadvertently administered after at least 4 weeks may be counted as valid)
Age 13 years and older: Routine interval: 4–8 weeks (minimum interval: 4 weeks)
The maximum age for use of MMRV is 12 years.
Contraindications and Precautions
For contraindications and precautions to Varicella (VAR), see VAR Appendix
On This Page
How to use the schedule
Notes
July 2, 2025
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188432 | https://letstranzact.com/blogs/marginal-rate-transformation | Home
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Marginal Rate of Transformation (MRT): Definition and Calculation
Marginal Rate of Transformation (MRT): Definition and Calculation
By Team TranZact | Published on Dec 5, 2023
The marginal rate of transformation (MRT) is an important concept shaping decision-making and resource distribution in any business. This concept of MRT highlights the complex nature of decision-making by showing the rate at which one good can be sacrificed for another in the hope of increasing the business’s performance and overall profits.
In this article, we will explain the marginal rate of transformation, with a clear definition and explore the different methods of its calculation.
What Is the Marginal Rate of Transformation (MRT)?
In this section, we will define marginal rate of transformation:
The Marginal Rate of Transformation (MRT) represents the quantity of one product that can be gained by sacrificing some quantity of another product. It measures the opportunity cost of producing an additional unit of output. This concept measures the difference between goods X and Y, showing how many units of X must be given up to produce one more unit of Y while keeping all production factors the same. Overall, MRT is a valuable tool for calculating the results of resource distribution and understanding the real costs associated with choices in production.
Formula and Calculation of the Marginal Rate of Transformation (MRT)
Knowing the MRT formula is very important for correctly doing the marginal rate of transformation calculation. Let us see the formula in detail below:
MRT=MCx/MCy
Here,
MCx represents the cost of producing an additional unit of good X
MCy represents the impact of producing good Y when reducing its quantity.
In simpler terms, the MRT shows the ratio of how much good Y needs to be sacrificed to produce one more unit of good X.
To break it down further, the MRT is a measure of opportunity cost – it helps us understand the difference between goods X and Y. By comparing the marginal costs of producing each good, we gain knowledge of the success of the resource distribution process.
The formula basically captures the cost of gaining one unit of a good in terms of what must be given up in the production of another. So, the MRT acts as a numerical guide, presenting the price, in terms of production resources, for making choices between different goods.
Here is the marginal rate of transformation example:
Imagine a scenario where a company currently manufactures 200 widgets and 80 gadgets.
Let's look at the Marginal Rate of Transformation (MRT) as the company decides to increase widget production by 15 units to 215 widgets. They also decrease gadget production by 8 units to 72 gadgets.
In this context, we can assess the MRT by calculating:
the change in the number of gadgets (ΔY = -8)
the change in the number of widgets (ΔX = 15)
Applying the formula (MRT = ΔY/ΔX), we find that (MRT = -8/15 approx -0.53).
The negative MRT value of approximately -0.53 indicates that widget production should be increased by 15 units. The company must sacrifice 8 gadgets to maintain the overall production level.
How Marginal Rate of Transformation Works
The marginal rate of transformation (MRT) is a valuable tool for businesses, offering information on the trade-offs between different goods in production. It helps identify the increase in units of one good produced by reducing the production of another. This complex relationship is important in understanding the opportunity cost linked with production choices.
The MRT becomes a key factor when a business considers options for switching one item for another.
It measures the similar exchange rate between goods, showing how many units of one can be produced by sacrificing the production of another.
It's important to note that the MRT must remain constant while looking for perfect substitute products, providing a solid starting point for decision-making.
The focus on the supply side of the production process makes the MRT a key factor for businesses aiming to reduce costs and improve their use of resources.
The Difference Between the MRT and the Marginal Rate of Substitution (MRS)
The table below shows the difference between MRT(Marginal Rate of Transformation) and the marginal rate of substitution:
| Concept | MRT (Marginal Rate of Transformation) | MRS (Marginal Rate of Substitution) |
---
| Definition | MRT measures the rate of substituting one good to produce an additional unit of another while keeping overall production constant. | MRS measures the rate at which a consumer will give up one good for another without sacrificing the level of satisfaction or utility. |
| Calculation | MRT is calculated by dividing the change in the quantity of one good by the difference in the quantity of another. | MRS is calculated as the ratio of one good's marginal utility to another's marginal utility. |
| Assumptions | MRT assumes constant returns to scale and perfect compatibility between goods. | MRS assumes decreasing marginal utility, which reduces the desire to substitute goods as more of a good is consumed. |
| Limitations | MRT limitations include: |
incorrect assumptions,
lack of change
failure to consider external factors and broader financial factors. | MRS limitations include:
assuming consumers have complete and similar choices,
overlooking income effects
not considering market trends and prices. |
Limitations of Using the Marginal Rate of Transformation (MRT)
Here are some limitations of using MRT:
1. Simple Assumptions
MRT assumes constant returns and perfect replacement of one good with another, which ignores real-world production challenges.
2. Partial Analysis
MRT focuses only on production, ignoring market trends, prices, and other important political and economic factors.
3. Instability
It is a fixed concept that can create instability or uncertainty as it doesn't consider changes over time or shifting economic conditions.
4. Ignoring External Factors
It fails to consider outside factors like the negative outcomes of production or purchasing activities of third parties, leading to improper resource/inventory allocation or distribution.
Learn More About Marginal Rate of Transformation (MRT) With TranZact
Understanding the Marginal Rate of Transformation (MRT) provides valuable information into the complex nature of resource allocation or distribution. While MRT provides a valuable indicator for understanding the balance between goods in production, it is important to understand its limitations, including oversimplified assumptions and non-flexible nature. To improve your understanding of MRT and its real-world applications, consider exploring further with TranZact.
By exploring the details of MRT within the broader business setting, TranZact offers powerful tools and platforms for gaining practical guidance in improving resource management decisions. Boost your knowledge and decision-making skills with TranZact's resources on the Marginal Rate of Transformation.
FAQs on Marginal Rate of Transformation (MRT)
Q1. What does negative MRT mean?
A negative Marginal Rate of Transformation indicates a negative relationship between goods. Therefore, as the production of one good rises, the output of the other decreases, showing the underlying imbalance between them.
Q2. What is the connection of MRT with the Production Possibilities Frontier (PPF)?
The MRT is a key concept used to describe the slope of the PPF curve and the trade-offs involved in the production of different goods. They are interconnected in showing the economic choices and effects of resource distribution in an economy.
Q3. Is there a possibility of MRT being negative or zero?
Yes, the Marginal Rate of Transformation can assume negative or zero values. A negative MRT suggests a trade-off where increasing the production of one good comes at the expense of the other. On the other hand, a zero MRT indicates the ability to boost the production of one good without compromising the output of the other.
Q4. What is the marginal rate of transformation and marginal cost?
The MRT measures the rate at which one good must be sacrificed to produce an additional unit of another good while keeping the total level of satisfaction or output constant. Meanwhile, marginal cost represents the additional cost of producing one more unit of a good or service.
Q5. What is the Marginal Rate of Transformation (MRT)?
The Marginal Rate of Transformation (MRT) measures how much of one good must be sacrificed to produce an additional unit of another good while maintaining a constant level of satisfaction or output. It is the ratio of the change in quantity of one good to the change in quantity of the other.
Q6. Why is the Marginal Rate of Transformation important in business?
The MRT is important for understanding the opportunity cost and compromises involved in resource allocation. It helps measure the effectiveness of production possibilities and assists in making informed decisions about how to use resources to increase overall business profit and customer well-being.
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Table of Contents:
What Is the Marginal Rate of Transformation (MRT)?
Formula and Calculation of the Marginal Rate of Transformation (MRT)
How Marginal Rate of Transformation Works
The Difference Between the MRT and the Marginal Rate of Substitution (MRS)
Limitations of Using the Marginal Rate of Transformation (MRT)
Learn More About Marginal Rate of Transformation (MRT) With TranZact
FAQs on Marginal Rate of Transformation (MRT)
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TranZact is a team of IIT & IIM graduates who have developed a GST compliant, cloud-based, inventory management software for SME manufacturers. It digitizes your entire business operations, right from customer inquiry to dispatch. This also streamlines your Inventory, Purchase, Sales & Quotation management processes in a hassle-free user-friendly manner. The software is free to signup and gets implemented within a week. |
188433 | https://canjhealthtechnol.ca/index.php/cjht/article/download/RC1543/2083?inline=1 | Review of Guidelines on Clonidine for Various Indications
CADTH Health Technology Review
Review of Guidelines on Clonidine for Various Indications
Rapid Review
Key Messages
What Is the Issue?
Clonidine is an antihypertensive medication that has been used for a range of health conditions including hypertension, substance use disorders, menopause, restless leg syndrome, migraines, attention-deficit/hyperactivity disorder (ADHD), and Tourette syndrome. The role of clonidine in the treatment of these health conditions is unclear.
What Did We Do?
To inform decisions around the use of clonidine in various health conditions, we sought to identify and summarize recommendations from evidence-based guidelines.
We searched key resources, including journal citation databases, and conducted a focused internet search for relevant evidence published since 2014. One reviewer screened articles for inclusion based on predefined criteria, critically appraised the included guidelines, and narratively summarized the findings.
What Did We Find?
We identified 12 evidence-based guidelines that included recommendations on the use of clonidine. We identified 1 guideline on hypertension, 4 guidelines on substance use disorders, 4 guidelines on menopause, 2 guidelines on restless leg syndrome, and 1 guideline on Tourette syndrome. We did not identify any evidence-based guidelines that included recommendations on the use of clonidine for the treatment of ADHD or migraine prophylaxis.
The included guidelines recommend clonidine for hypertension in pregnant women, management of opioid withdrawal and alcohol withdrawal, and Tourette syndrome.
The recommendations in the guidelines for menopause were mixed. Two guidelines do not recommend clonidine and 2 guidelines recommend clonidine for the treatment of vasomotor symptoms (i.e., hot flashes) of menopause.
One guideline does not recommend the use of clonidine for restless leg syndrome in people who are pregnant or lactating and 1 guideline states that there is insufficient evidence to support or refute the use of clonidine in restless leg syndrome.
What Does It Mean?
The use of clonidine is recommended for some health conditions and is not recommended for others. Due to the inconsistency in recommendations on the use of clonidine for the control of hot flashes in menopause, decision-makers may wish to consider other factors such as patient preferences and availability of other treatment options.
Future evidence-based guidelines that include recommendations on the use of clonidine for the prevention of migraines, treatment of ADHD and the treatment of hypertension in a broader population would help fill the gaps identified in this report.
Abbreviations
AAN
American Academy of Neurology
ADHD
attention-deficit/hyperactivity disorder
ASAM
American Society of Addiction Medicine
CRISM
Canadian Research Initiative in Substance Misuse
DoD
Department of Defense
IRLSSG
International Restless Legs Syndrome Study Group
NAMS
North American Menopause Society
NICE
National Institute of Health and Care Excellence
SOGC
Society of Obstetricians and Gynaecologists of Canada
VA
Veterans Affairs
Context and Policy Issues
What Are the Health Conditions Included in This Review?
Guidelines on hypertension, substance use disorders, menopause, restless leg syndrome, migraine prevention, ADHD, and Tourette syndrome were eligible for inclusion in this review. Hypertension (high blood pressure) is when the pressure in your arteries is consistently too high.1 Treatment can include lifestyle changes and medications (e.g., diuretics, angiotensin-converting enzyme inhibitors, angiotensin II receptor blockers, and calcium channel blockers).2 Substance use disorder is a condition in which there is a problematic pattern of substance use that causes distress or impairs day-to-day functioning.3 Treatment for substance use disorder can include detoxification, cognitive and behavioural therapies, and medications.3 Menopause is a time that marks the end of menstrual cycles and is diagnosed after a person has gone 12 months without a menstrual period.4 A potential symptom of menopause is hot flashes (also known as vasomotor symptoms) which are a sudden feeling of warmth that spreads over the body.5 Treatment for menopause can include hormonal (i.e., estrogen or estrogen and progesterone) and nonhormonal treatments (e.g., diet, exercise, medications).5
Restless leg syndrome is a condition which causes a strong urge to move the legs and usually occurs in the evening or at night when sitting or lying down.6 Treatment for restless leg syndrome can include at-home therapies (e.g., exercise, warm baths, stress reduction) or medications (e.g., iron supplements, antiseizure medications, dopamine agonists, benzodiazepines, opioids).7 A migraine is a headache that can cause severe throbbing pain or a pulsing sensation and is often accompanied by nausea, vomiting, and extreme sensitivity to light and sound.8 Medications aimed at preventing migraines can include blood-pressure lowering medications, antidepressants, antiseizure drugs, and calcitonin gene-related peptides monoclonal antibodies.9 ADHD is a mental health disorder that include symptoms such as difficulty paying attention, hyperactivity, and impulsive behaviour.10 Treatment for ADHD includes medication (stimulants and other medications), education, skills training, and psychological counselling.11 Tourette syndrome is a neurologic disorder that may cause sudden unwanted and uncontrolled repetitive movements or vocal sounds called tics.12 Treatment for Tourette syndrome can include behavioural treatments, psychotherapy, and medications (e.g., dopamine blocking agents, alpha-adrenergic agonists, stimulants, antidepressants).12
What Is Clonidine?
Clonidine is an alpha-2 adrenergic agonist that has antihypertensive (blood pressure lowering) effects.13 Clonidine lowers blood pressure by relaxing the arteries and increasing the blood supply to the heart.13 Common side effects of clonidine may include abdominal pain, headache, hypotension, fatigue, nausea, constipation, dry mouth, sexual dysfunction, dizziness, and sedation.13 There is also potential for rebound hypertension and withdrawal symptoms if clonidine is discontinued abruptly.13 Clonidine is indicated for the treatment of hypertension and should normally be used in patients in whom treatment with a diuretic or beta-blocker was ineffective or associated with unacceptable adverse effects.14 Clonidine is available as tablets for oral administration and the initial dose is 0.1 mg twice daily.14 After 2 to 4 weeks, further increments of 0.1 mg per day may be added until the desired response is achieved.14 The common therapeutic dose ranges from 0.2 mg to 0.6 mg per day.14 When discontinuing clonidine, the dosage should be reduced gradually.14
Why Is It Important to Do This Review?
In addition to its use in hypertension, clonidine has also been used for a range of other health conditions including ADHD, Tourette syndrome, managing withdrawal from opioids, benzodiazepines, and alcohol, restless leg syndrome, migraine prophylaxis, and control of hot flashes in menopause.13 A review of guidelines of clonidine can help decision-making around which indications clonidine should be used for as well as provide guidance on potential dosing and safety considerations.
Objective
The purpose of this report is to summarize and critically appraise evidence-based guidelines regarding the use of clonidine for hypertension, substance use disorders, control of hot flashes in menopause, restless leg syndrome, migraine prophylaxis, ADHD, and Tourette syndrome.
Research Questions
What are the evidence-based guidelines regarding the use of clonidine in adults with hypertension?
What are the evidence-based guidelines regarding the use of clonidine in adults with substance use disorders (i.e., opioids, benzodiazepines, alcohol)?
What are the evidence-based guidelines regarding the use of clonidine for the control of hot flashes in adults with menopause?
What are the evidence-based guidelines regarding the use of clonidine in adults with restless leg syndrome?
What are the evidence-based guidelines regarding the use of clonidine for prevention of migraines in adults?
What are the evidence-based guidelines regarding the use of clonidine in adults with ADHD?
What are the evidence-based guidelines regarding the use of clonidine in adults with Tourette syndrome?
Methods
Literature Search Methods
An information specialist conducted a literature search on key resources including MEDLINE, Embase, PsycInfo, the Cochrane Database of Systematic Reviews, the International HTA Database, the websites of major international health technology agencies, and those in Canada, as well as a focused internet search. The search approach was customized to retrieve a limited set of results, balancing comprehensiveness with relevancy. The search strategy comprised both controlled vocabulary, such as the National Library of Medicine’s MeSH (Medical Subject Headings), and keywords. Search concepts were developed based on the elements of the research questions and selection criteria. The main search concept was clonidine. Search filters were applied to limit retrieval to guidelines. Conference reviews and conference abstracts were excluded. The search was completed on June 12, 2024 and limited to English-language documents published since January 1, 2014.
Selection Criteria and Methods
One reviewer screened citations and selected studies. In the first level of screening, titles and abstracts were reviewed and potentially relevant articles were retrieved and assessed for inclusion. The final selection of full-text articles was based on the inclusion criteria presented in Table 1.
Table 1: Selection Criteria
| Criteria | Description |
--- |
| Population | Adults with hypertension, opioid use disorder, alcohol withdrawal, benzodiazepine withdrawal, menopause, restless leg syndrome, migraines, ADHD, or Tourette syndrome. |
| Intervention | Clonidine |
| Comparator | Not applicable |
| Outcomes | Guidelines regarding the use of clonidine for the treatment of adults with hypertension, opioid use disorder, alcohol withdrawal, benzodiazepine withdrawal, menopause, restless leg syndrome, migraines, ADHD, or Tourette's syndrome (e.g., recommendations regarding appropriate use, dose, and duration of use). |
| Study designs | Evidence-based guidelines. |
ADHD= attention-deficit/hyperactivity disorder.
Exclusion Criteria
Articles were excluded if they did not meet the selection criteria outlined in Table 1, they were duplicate publications, or were published before 2014. Guidelines with unclear methodology were also excluded.
Critical Appraisal of Individual Studies
The included guidelines were critically appraised by 1 reviewer using the Appraisal of Guidelines for Research and Evaluation (AGREE) II instrument15 as a guide.
Summary of Evidence
Quantity of Research Available
A total of 76 citations were identified in the literature search. Following screening of titles and abstracts, 48 citations were excluded and 28 potentially relevant reports from the electronic search were retrieved for full-text review. Ten potentially relevant publications were retrieved from the grey literature search for full-text review. Of these potentially relevant articles, 25 publications were excluded for various reasons, and 13 publications met the inclusion criteria and were included in this report. These comprised 13 reports pertaining to 12 unique evidence-based guidelines. Appendix 1 presents the Preferred Reporting Items for Systematic reviews and Meta-Analyses (PRISMA)16 flow chart of the study selection.
Additional references of potential interest are provided in Appendix 2.
Summary of Guideline Characteristics
This report included 12 evidence-based guidelines17-28 that included recommendations regarding the use of clonidine for various indications.
The Hypertension Canada (2020)17 guideline included recommendations on hypertension. The Canadian Research Initiative in Substance Misuse (CRISM) (2023),18 Veterans Affairs (VA)/Department of Defense (DoD) (2021),19 American Society of Addiction Medicine (ASAM) (2020),20 and Commonwealth of Australia (2014)21 guidelines included recommendations on substance use disorder. The North American Menopause Society (NAMS) (2023),22 Society of Obstetricians and Gynaecologists of Canada (SOGC) (2021),23 National Institute for Health and Care Excellence (NICE) (2019),24 and Endocrine Society (2015)25 guidelines included recommendations on menopause. The American Academy of Neurology (AAN) (2016)26 and International Restless Legs Syndrome Study Group (IRLSSG) (2015)27 guidelines included recommendations on restless leg syndrome. The AAN (2019)28 guideline included recommendations on Tourette syndrome. Two of the guidelines on substance use disorder included recommendations on the management of opioid withdrawal (VA/DoD 19 and Commonwealth of Australia 21) and 2 of the guidelines included recommendations on alcohol withdrawal (CRISM 18 and ASAM 20). We did not identify any evidence-based guidelines that included recommendations on the use of clonidine for the prevention of migraines or treatment of ADHD that met our inclusion criteria.
The guideline development groups were from Canada,17,18,23 the US,19,20 Australia,21 the UK,24 North America,22 or were international.25-28
Additional details regarding the characteristics of included guidelines are provided in Appendix 3.
Summary of Critical Appraisal
All the included guidelines had clear objectives, guideline questions, and target populations. Only 3 of the guidelines18,19,24 specifically sought the views and preferences of the target populations and 1 of the guidelines20 posted the recommendations for public feedback. Therefore, the recommendations in some of the guidelines may not adequately reflect the values and preferences of patients. Systematic methods were used to search for evidence in 7 of the included guidelines.18-20,24-26,28 In 4 of the guidelines,17,21-23 they state that literature searches were conducted, however, limited details on the search (e.g., databases searched, time frame, screening methods, and so forth.) were provided. The IRLSSG (2015)27 guideline only performed a search in a single database and therefore relevant evidence may have been missed due to lack of a comprehensive search strategy.
There was an explicit link between the recommendations and supporting evidence in 9 of the guidelines.17-19,22,24-28 The SOGC (2021)23 and ASAM (2020)20 guidelines did not include discussion of the evidence that supports the recommendations. The Commonwealth of Australia (2014)21 guideline stated that the approach of using clonidine and other medications to control symptoms of opioid withdrawal is well supported by evidence, however, do not include a description of this evidence. Providing clear descriptions of the evidence used to inform the recommendations increases transparency in the recommendation development process.
The competing interests of the guideline development group were disclosed in all the guidelines. The SOGC (2021)23 and ASAM (2020)20 guidelines did not include statements about any funding received for the development of the guideline.
Additional details regarding the strengths and limitations of included guidelines are provided in Appendix 4.
Summary of Findings
An overview of the recommendations and supporting evidence regarding the use of clonidine for the health conditions covered in the included guidelines is provided in Table 2.
Appendix 5 presents the recommendation in the included guidelines.
Table 2: Overview of Included Guidelines
| Guideline (Year) | Target Population | Recommendation(s) Strength of Recommendation(s) | Supporting Evidence Quality of Evidence | Possible Side Effects | Dosing |
--- --- --- |
| Hypertension |
| Hypertension Canada (2020)17 | Adults and children at risk of or with hypertension | Clonidine can be considered as a second-line drug for antihypertensive therapy for pregnant women with chronic hypertension, gestational hypertension, or preeclampsia (SBP measurements of 140 mm Hg or DBP measurements of 90 mm Hg). Grade D | Based on expert opinion alone. | NR | NR |
| Substance Use Disorders |
| CRISM (2023)18 | High-risk drinking and alcohol use disorder in youth (aged 11 to 25 years) and adults | Clinicians should consider offering clonidine for withdrawal management in an outpatient setting (e.g., primary care, virtual) for patients at low risk of severe complications of alcohol withdrawal (e.g., PAWSS <4). Strong | 2 RCTs that reported clonidine is as effective as chlordiazepoxide in the management of mild to moderate withdrawal symptoms with better control of sympathetic symptoms and reductions in patient anxiety. Low | Hypotension, dry mouth, dizziness, fatigue, headache, nausea, vomiting, constipation, malaise, sleep disorder, sedation, and erectile dysfunction. | Starting dose: 0.1 to 0.2 mg BID Titration: Add 0.2 mg OD PRN Final dose: 0.1 mg to 0.6 mg BID |
| VA/DoD (2021)19 | Adults with a diagnosis of substance use disorder | Clonidine is suggested as a second-line drug for opioid withdrawal management for patients with opioid use disorder when methadone and buprenorphine are contraindicated, unacceptable, or unavailable. Weak | 1 SR that included comparisons of alpha2-adrenergic agonists vs. placebo, methadone, or another alpha2-adrenergic agonist. 1 SR that included comparisons of buprenorphine vs clonidine, lofexidine, or methadone. The work group considered that the benefits of improved withdrawal symptoms outweighed potential harms, and that patient values and preferences varied somewhat. Low | Hypotension | NR |
| ASAM (2020)20 | Adults with a diagnosis of alcohol withdrawal | Clonidine can be used to control autonomic hyperactivity and anxiety when symptoms are not controlled by benzodiazepines alone. Clonidine should not be used alone to prevent or treat withdrawal-related seizures or delirium. Recommendation strength NR | NR | NR | NR |
| Commonwealth of Australia (2014)21 | People who are opioid dependent | 2 approaches are recommended for management of opioid withdrawal: 1. Abrupt cessation of opioid use and relief of symptoms using non-opioid drugs (e.g., benzodiazepines, nonsteroidal anti-inflammatory drugs, antiemetics, clonidine, antispasmodic drugs). 2.Short course of reducing doses of buprenorphine. Recommendation strength NR | The guideline authors state that both approaches are well supported by evidence, however, the use of buprenorphine to manage withdrawal is associated with significantly better relief of withdrawal than clonidine and supplementary medications (1 SR). 4 stars (body of evidence can be trusted to guide practice) | Hypotension | Test dose: 150 mcg to check for hypotensive effects. Treatment dose: 12 to 15 mcg/ kg/day in 4 divided doses. Doses are tapered and then ceased 7 to 10 days after cessation of opioids. |
| Menopause |
| NAMS (2023)22 | People experiencing menopause | Clonidine is not recommended for the treatment of vasomotor symptoms of menopause. Recommendation strength NR | 1 SR that reported that clonidine is modestly more beneficial than placebo at reducing vasomotor symptoms. 2 SRs that reported that clonidine is less beneficial than SSRIs, SNRIs, and gabapentin at reducing vasomotor symptoms. Levels 1 to 3 (good and consistent scientific evidence, limited or inconsistent scientific evidence, and consensus and expert opinion). | Hypotension, lightheadedness, headache, dry mouth, dizziness, sedation, and constipation. Sudden cessation can lead to significant increases in blood pressure. | NR |
| SOGC (2021)23 | Perimenopausal and postmenopausal women | Clonidine is a nonhormonal option for refractory vasomotor symptoms. Conditional | Supporting evidence NR. Moderate | NR | NR |
| NICE (2019)24 | Menopausal women and women with premature ovarian insufficiency | 1. Give information to menopausal women and their family about nonhormonal treatments (e.g., clonidine) for menopausal symptoms Recommendation Strength NR 2. Do not routinely offer clonidine as first-line treatment for vasomotor symptoms alone. Strong | Clonidine was not included in the NMA conducted for the guideline due to the way outcomes were reported in studies of clonidine. Therefore, the relative effectiveness of clonidine in relieving short-term symptoms for women in menopause could not be estimated. Evidence quality NR | NR | NR |
| Endocrine Society (2015)25 | Menopausal and postmenopausal women | A trial of clonidine is suggested for women seeking relief of moderate to severe vasomotor symptoms who do not respond to or are intolerant of SSRIs/SNRIs, gabapentin, or pregabalin. Weak | Several RCTs that demonstrated that clonidine reduces hot flashes but is less effective than SSRI/SNRIs, gabapentin, and pregabalin, and is associated with more side effects. Low | Lightheadedness, hypotension, headache, and constipation. Sudden cessation can be associated with significant increases in blood pressure. | NR |
| Restless Leg Syndrome |
| AAN (2016)26 | Adults with restless leg syndrome | There is insufficient evidence to support or refute the use of clonidine for the treatment of restless leg syndrome. Level U (insufficient evidence) | 1 crossover trial of clonidine vs placebo in which patients reported less paresthesia, motor restlessness, and daytime fatigue during the clonidine treatment arm. There was no difference in periodic limb movement index between groups. Class III (low quality) | Hypotension, decreased cognition, dry mouth, and sleepiness. | NR |
| IRLSSG (2015)27 | People with restless leg syndrome who are pregnant or lactating | Clonidine should probably not be considered for people with restless leg syndrome who are pregnant or lactating. Level 4 (evidence for risk/ ineffectiveness) | A small RCT found limited efficacy of clonidine for restless leg syndrome. Clonidine passes on to breast milk. Evidence quality NR | NR | NR |
| Tourette Syndrome |
| AAN (2019)28 | Children and adults with Tourette syndrome or a chronic tic disorder | 1. Physicians should counsel individuals with tics and ADHD that alpha2 adrenergic agonists (e.g., clonidine) may provide benefit for both conditions Level B (moderate) 2. Alpha2 adrenergic agonists should be prescribed for the treatment of tics when the benefits outweigh the risks Level B (moderate) 3. Physicians must counsel patients on the common side effects of alpha2 adrenergic agonists, including sedation. Level A (strong) 4. Heart rate and blood pressure must be monitored in patients with tics treated with alpha2 adrenergic agonists. Level A (strong) 5. Alpha2 adrenergic agonists must be gradually tapered to avoid rebound hypertension. Level A (strong) | 1. One study reported reduced tic severity in children with tics and a comorbid diagnosis of ADHD who received clonidine plus methylphenidate vs placebo. Class I (highest quality) 2. Three studies reported reduced tic severity in people with tics receiving clonidine vs placebo. Class I and II 3. Two studies reported that sedation was more common in people with tics receiving clonidine vs placebo. 1 SR of alpha2 adrenergic agonists for ADHD in children and adolescents demonstrated hypotension, bradycardia, and sedation with these agents. Class I and II 4. One SR of alpha2 adrenergic agonists for ADHD in children and adolescents demonstrated hypotension, and bradycardia with these agents Evidence quality NR 5. NR | Hypotension, bradycardia, and sedation. Abrupt cessation of alpha2 adrenergic agonists may cause rebound hypertension. | NR |
AAN= American Academy of Neurology; ADHD= attention-deficit/hyperactivity disorder; ASAM= American Society of Addiction Medicine; BID= twice a day; CRISM= Canadian Research Initiative in Substance Misuse; DoD= Department of Defense; DBP= diastolic blood pressure; IRLSSG= International Restless Legs Syndrome Study Group; NAMS= North American Menopause Society; NICE= National Institute for Health and Care Excellence; NMA= network meta-analysis; NR= not reported; OD= once daily; PAWSS= Prediction of Alcohol Withdrawal Severity Scale; PRN= as needed; RCT= randomized controlled trial; SBP= systolic blood pressure; SNRI= serotonin-norepinephrine reuptake inhibitor; SOGC= Society of Obstetricians and Gynaecologists of Canada; SR= systematic review; SSRI= selective serotonin reuptake inhibitor; VA= Veterans Affairs.
Limitations
Some of the included guidelines are limited by the evidence identified to inform the recommendations. The supporting evidence for some of the recommendations was rated as low quality by the guideline authors or the recommendations were based on expert opinion. Additionally, we cannot determine whether the literature searches conducted to inform 4 of the guidelines17,21-23 were comprehensive as very limited details were provided. The IRLSSG (2015)27 guideline only searched a single database and may have missed relevant evidence.
We did not identify any evidence-based guidelines on the use of clonidine for the prevention of migraines or treatment of ADHD that met our inclusion criteria. However, the AAN (2019)28 guideline included a recommendation on the use of clonidine for patients with comorbid ADHD and tics. Most of the included guidelines did not include recommendations or guidance on clonidine dosing. The Hypertension Canada (2020)17 guideline included a recommendation for the use of clonidine in pregnant women however, it did not include recommendations on the use of clonidine for the treatment of hypertension in other populations.
There were inconsistencies in some of the recommendations in the included guidelines on menopause. The NAMS (2023)22 and NICE (2019)24 guidelines do not recommend clonidine for the treatment of vasomotor symptoms of menopause, whereas the SOGC (2021)23 and Endocrine Society (2015)25 guidelines included positive recommendations for clonidine for the treatment of vasomotor symptoms.
Conclusions and Implications for Decision- or Policy-Making
We included 12 evidence-based guidelines on the use of clonidine for various health conditions in this report. These comprised 1 guideline on hypertension,17 4 guidelines on substance use disorders (2 on opioid withdrawal19,21 and 2 on alcohol withdrawal18,20), 4 guidelines on the treatment of vasomotor symptoms of menopause,22-25 2 guidelines on restless leg syndrome,26,27 and 1 guideline on Tourette syndrome.28
The Hypertension Canada (2020)17 guideline recommends considering clonidine as a second-line option for pregnant women with chronic hypertension, gestational hypertension, or preeclampsia.
The 4 guidelines on substance use disorders all included positive recommendations for the use of clonidine. The CRISM (2023)18 guideline recommends offering clonidine for withdrawal management in an outpatient setting for patients at low risk of severe complications of alcohol withdrawal. The ASAM (2020)20 guideline recommends clonidine for the control of autonomic hyperactivity and anxiety when symptoms are not controlled by benzodiazepines alone. The ASAM (2020)20 guideline does not recommend clonidine be used alone to prevent or treat withdrawal-related seizures or delirium. The VA/DoD (2021)19 guideline suggests clonidine as a second-line drug for opioid withdrawal management when methadone and buprenorphine are contraindicated, unacceptable, or unavailable. The Commonwealth of Australia (2014)21 guideline recommends abrupt cessation of opioid use and control of symptoms using non-opioid drugs such as clonidine as an option for the management of opioid withdrawal.
The recommendations on the use of clonidine for the treatment of vasomotor symptoms of menopause were mixed. The NAMS (2023)22 guideline does not recommend clonidine for the treatment of vasomotor symptoms of menopause. The NICE (2019)24 guideline does not recommend routinely offering clonidine as first-line treatment for vasomotor symptoms alone. The SOGC (2021)23 guideline recommends clonidine as a nonhormonal option for refractory vasomotor symptoms. The Endocrine Society (2015)25 guideline suggests a trial of clonidine for women with moderate to severe vasomotor symptoms who do not respond to or are intolerant of SSRIs/SNRIs, gabapentin, or pregabalin.
Neither of the guidelines for restless leg syndrome recommend the use of clonidine. The AAN (2016)26 guideline states that there is insufficient evidence to support or refute the use of clonidine for the treatment of restless leg syndrome. The IRLSSG (2015)27 guideline states that clonidine should probably not be considered for people with restless leg syndrome who are pregnant or lactating.
The AAN (2019)28 guideline includes several recommendations on the use of clonidine for Tourette syndrome. The guideline recommends the use of alpha2 adrenergic agonists (e.g., clonidine) for the treatment of tics when the benefits outweigh the risks.28 The guideline also recommends counselling patients on common side effects and monitoring heart rate and blood pressure of patients treated with alpha2 adrenergic agonists.28
Future evidence-based guidelines that include recommendations on the use of clonidine for the prevention of migraines, treatment of ADHD and the treatment of hypertension in a broader population would help fill the gaps identified in this report. Due to the inconsistency in recommendations on the use of clonidine for the treatment of vasomotor symptoms of menopause, decision-makers may wish to consider other factors such as patient preferences and availability of other treatment options when making decisions around the use of clonidine for this indication.
References
1.Mayo Clinic. High blood pressure (hypertension). 2024; Accessed 2024 Jul 8.
2.Mayo Clinic. High blood pressure (hypertension): Diagnosis & treatment. 2024; Accessed 2024 Jul 8.
3.Cleveland Clinic. Substance Use Disorder (SUD). 2022; Accessed 2024 Jul 8.
4.Mayo Clinic. Menopause. 2023; Accessed 2024 Jul 8.
5.Cleveland Clinic. Menopause. 2024; Accessed 2024 Jul 8.
6.Mayo Clinic. Restless legs syndrome. 2024; Accessed 2024 Jul 8.
7.Cleveland Clinic. Restless legs syndrome. 2023; Accessed 2024 Jul 8.
8.Mayo Clinic. Migraine. 2023; Accessed 2024 Jul 8.
9.Mayo Clinic. Migraine: Diagnosis & treatment 2023; Accessed 2024 Jul 8.
10.Mayo Clinic. Adult attention-deficit/hyperactivity disorder (ADHD). 2023; Accessed 2024 Jul 8.
11.Mayo Clinic. Adult attention-deficit/hyperactivity disorder (ADHD): Diagnosis & treatment. 2023; Accessed 2024 Jul 8.
12.National Institute of Neurological Disorders and Stroke. Tourette syndrome. Accessed 2024 Jul 8.
13.Yasaei R, Saadabadi A. Clonidine. Treasure Island (FL): StatPearls Publishing; 2023: Accessed 2024 Jul 8.
14.Clonidine (clonidine hydrochloride): 0.1 mg and 0.2 mg tablets [product monograph]. Saint-Laurent (QC): Sivem Pharmaceuticals ULC; 2023 Jun 1.
15.Agree Next Steps Consortium. The AGREE II Instrument. Hamilton (ON): AGREE Enterprise; 2017: Accessed 2024 Jul 3.
16.Liberati A, Altman DG, Tetzlaff J, et al. The PRISMA statement for reporting systematic reviews and meta-analyses of studies that evaluate health care interventions: explanation and elaboration. J Clin Epidemiol.2009;62(10):e1-e34. PubMed
17.Rabi DM, McBrien KA, Sapir-Pichhadze R, et al. Hypertension Canada's 2020 comprehensive guidelines for the prevention, diagnosis, risk assessment, and treatment of hypertension in adults and children. Can J Cardiol.2020;36(5):596-624. PubMed
18.Wood E, Bright J, Hsu K, et al. Canadian guideline for the clinical management of high-risk drinking and alcohol use disorder. CMAJ . 2023;195(40):E1364-E1379. PubMed
19.U.S. Department of Veterans Affairs. VA/DoD Clinical Practice Guidelines: Management of substance use disorder (SUD). 2021; Accessed 2024 Jul 3.
20.American Society of Addiction Medicine. The ASAM clinical practice guideline on alcohol withdrawal management. 2020; Accessed 2024 Jul 3.
21.Australian Government Department of Health and Aged Care. National guidelines for medication-assisted treatment of opioid dependence. 2014; Accessed 2024 Jul 3.
22.“The Nonhormone Therapy Position Statement of The North American Menopause Society” Advisory Panel. The 2023 nonhormone therapy position statement of The North American Menopause Society. Menopause.2023;30(6):573-590. PubMed
23.Jacobson M, Mills K, Graves G, Wolfman W, Fortier M.Guideline no. 422f: Menopause and breast cancer. J Obstet Gynaecol Can.2021;43(12):1450-1456 e1451. PubMed
24.National Institute for Health and Care Excellence. Menopause: Diagnosis and management. (NICE guideline NG23). 2019: Accessed 2024 Jul 3.
25.Stuenkel CA, Davis SR, Gompel A, et al. Treatment of symptoms of the menopause: An Endocrine Society Clinical Practice Guideline. J Clin Endocrinol Metab.2015;100(11):3975-4011. PubMed
26.Winkelman JW, Armstrong MJ, Allen RP, et al. Practice guideline summary: Treatment of restless legs syndrome in adults: Report of the Guideline Development, Dissemination, and Implementation Subcommittee of the American Academy of Neurology. Neurology.2016;87(24):2585-2593. PubMed
27.Picchietti DL, Hensley JG, Bainbridge JL, et al. Consensus clinical practice guidelines for the diagnosis and treatment of restless legs syndrome/Willis-Ekbom disease during pregnancy and lactation. Sleep Med Rev.2015;22:64-77. PubMed
28.Pringsheim T, Okun MS, Muller-Vahl K, et al. Practice guideline recommendations summary: Treatment of tics in people with Tourette syndrome and chronic tic disorders. Neurology.2019;92(19):896-906. PubMed
29.Pringsheim T, Holler-Managan Y, Okun MS, et al. Comprehensive systematic review summary: Treatment of tics in people with Tourette syndrome and chronic tic disorders. Neurology.2019;92(19):907-915. PubMed
Appendix 1: Selection of Included Studies
Figure 1: Preferred Reporting Items for Systematic reviews and Meta-Analyses PRISMA16 Flow Chart of Study Selection
Appendix 2: References of Potential Interest
Note this appendix has not been copy-edited.
Guidelines (Unclear Methodology)
BC Guidelines. High-Risk Drinking and Alcohol Use Disorder. 2024;
Canadian ADHD Resource Alliance. Canadian ADHD Practice Guidelines, 4.1 Edition, Toronto (ON); CADDRA, 2020. Available from
BC Guidelines. Opioid Use Disorder - Diagnosis and Management in Primary Care. 2018;
Review Articles
Mattes JA. Treating ADHD in Prison: Focus on Alpha-2 Agonists (Clonidine and Guanfacine). J Am Acad Psychiatry Law. 2016;44(2):151-157. PubMed
Appendix 3: Characteristics of Included Publications
Table 3: Characteristics of Included Guideline – Hypertension
| Intended users, target population | Intervention(s) and major outcomes considered | Evidence collection, synthesis, and quality assessment | Recommendations for development and evaluation | Guideline validation |
--- ---
| Hypertension Canada (2020)17 |
| Intended Users: clinicians Target Population: adults and children at risk of or with hypertension | Interventions: pharmacotherapy, health behaviours Outcomes: NR | A medical librarian conducted comprehensive literature searches for 16 subgroups that represent distinct areas of hypertension. The literature was reviewed independently by subgroup members in a standardized manner. | The guideline panel consisted of content and methodological experts divided into 16 subgroups. The subgroups developed new or revised existing recommendations based on the identified evidence. Recommendations were reviewed and assigned a grade by a methodological expert from a central review committee. Recommendations were assigned a grade from D (lowest) to A (highest) based on the strength and quality of the clinical evidence. Members of the Hypertension Canada Guidelines Committee voted on the draft recommendations with >70% support required for approval. | Feedback is received from end users to improve guideline processes and content and address identified needs. |
NR= not reported.
Note this table has not been copy-edited.
Table 4: Characteristics of Included Guidelines – Substance Use Disorders
| Intended users, target population | Intervention(s) and major outcomes considered | Evidence collection, synthesis, and quality assessment | Recommendations for development and evaluation | Guideline validation |
--- ---
| CRISM (2023)18 |
| Intended Users: health care professionals in primary care and community-based settings, policy-makers developing health system interventions and people with alcohol use problems, their families and other affected populations seeking direction on evidence-based care Target Population: high-risk drinking and alcohol use disorder in youth (aged 11 to 25 years) and adults | Interventions: screening tools, risk assessment tools, pharmacotherapies, psychosocial treatment interventions Outcomes: NR | A systematic literature search was performed to update a previous guideline on alcohol use disorder. A contracted information specialist performed the literature search in multiple databases. Two authors independently screened and identified eligible studies. Study quality was assessed using validated tools by one author. | Evidence summaries from the previous guideline were updated and provided to working groups. The working groups decided through consensus whether recommendations would be accepted without modification, adapted, or removed. The full committee then reviewed the recommendations, incorporated feedback, and approved revisions by consensus. | 13 relevant experts and interested parties from Canada and international jurisdictions reviewed the draft guideline. Feedback from external reviewers was incorporated, and the committee accepted the final version through consensus. |
| VA/DoD (2021)19 |
| Intended users: providers (e.g., physicians, physician assistants, nurse practitioners, nurses, psychologists, social workers, pharmacists, addiction counsellors, chaplains, nutritionists, dieticians, emergency care providers, behavioural health providers) Target population: veterans, active-duty Service Members, or non–active-duty Service Members >18 years old, as well as other adults >18 years old who are eligible for care in the VA and/or DoD health care delivery systems, who have symptoms and/or a diagnosis of substance use disorder, including AUD, OUD, sedative-hypnotic use disorder, stimulant use disorder, or cannabis use disorder | Interventions: pharmacotherapies, psychosocial interventions Outcomes: consumption/ abstinence/ frequency of use, retention/ duration in treatment, withdrawal symptoms, QoL, mortality, overdoses, suicide ideation or attempt. | A systematic review was conducted in multiple databases, including RCTs or other systematic reviews. Evidence was assessed using the GRADE methodology. | A development meeting was held to develop the recommendations. A work group interpreted the systematic review findings and developed the recommendations. When appropriate, recommendations from the previous guideline version were carried forward and modified as necessary. Recommendations were rated based on a modified GRADE and USPSTF methodology. Ratings were based on the quality of the evidence base, associated benefits and harms, patient values and preferences, and other implications. | A draft of the guideline was sent to experts from VA and DoD health care systems and outside organizations for external review. The work group considered all feedback and modified the guideline where appropriate, in line with the evidence. |
| ASAM (2020)20 |
| Intended Users: clinicians, nurse practitioners, physician assistants, and pharmacists who provide alcohol withdrawal management in specialty and nonspeciality addiction treatment settings (including primary care and intensive care and surgery units in hospitals). Target Population: Adults (≥18 years) with a diagnosis of alcohol withdrawal | Interventions: pharmacological (excluding those that are not widely available in the US) and nonpharmacological interventions Outcomes: severity of withdrawal syndrome, treatment completion, transfer to a more intensive level of care, incidence of seizure, delirium, death and adverse events, and linkage to long-term AUD treatment. | A systematic literature review was conducted in multiple databases that included all levels of published literature, including nonrandomized studies and case studies. Additionally, a search for grey literature was conducted. Two independent reviewers screened abstracts and full texts of articles for inclusion. The quality of the included studies was evaluated using validated tools. Systematic reviews of RCTs and guidelines based on systematic reviews were included. Where these study designs were not available, lower-quality evidence sources were included. | The guideline committee and project team developed the draft recommendations using RAM, which involves multiple rounds of rating and a face-to-face meeting. Committee members rated the appropriateness of each guideline statement on a scale from 1 to 9. A statement was deemed appropriate if the median rating was in the 7 to 9 range, and no more than one-third of the experts rated it outside that range. | The guideline was sent to interested parties and posted publicly for external review. Feedback was incorporated as appropriate and in line with the evidence. |
| Commonwealth of Australia (2014)21 |
| Intended users: generalist service providers Target population: people who are opioid dependent | Interventions: pharmacotherapies Outcomes: NR | Literature searches were conducted to identify systematic reviews and clinical trials. | A list of issues to be addressed in the guidelines was developed based on previous guidelines and the authors' experience. Evidence statements were designed for the list of issues. The recommendations were graded using a 4-star rating system. 4 stars (body of evidence can be trusted to guide practice) 3 stars (body of evidence can be trusted to guide practice in most situations) 2 stars (body of evidence provides some support for recommendation[s] but care should be taken in its application) 1 star (body of evidence is weak, and recommendation must be applied with caution). | NR |
ASAM= American Society of Addiction Medicine; AUD= alcohol use disorder; CRISM= Canadian Research Initiative in Substance Misuse; DoD= Department of Defense; GRADE= Grading of Recommendations Assessment, Development and Evaluation; NR= not reported; OUD= opioid use disorder; QoL= quality of life; RAM= RAND/UCLA Appropriateness Method; RCT= randomized controlled trial; USPSTF= US Preventive Services Task Force; VA= Veterans Affairs.
Note this table has not been copy-edited.
Table 5: Characteristics of Included Guideline – Menopause
| Intended users, target population | Intervention(s) and major outcomes considered | Evidence collection, synthesis, and quality assessment | Recommendations for development and evaluation | Guideline validation |
--- ---
| NAMS (2023)22 |
| Intended Users: NR Target Population: people experiencing menopause | Interventions: lifestyle interventions, mind-body techniques, prescription therapies, dietary supplements, and acupuncture, other treatments, and technologies Outcomes: NR | An extensive review of the literature was conducted. | The guideline was written based on the literature review and was submitted to and approved by the NAMS board of trustees. The guideline panel assessed the available literature to develop the recommendations. The level of evidence was assigned a rating from 1 to 3. level 1 (good and consistent scientific evidence) level 2 (limited or inconsistent scientific evidence) level 3 (consensus and expert opinion). | NR |
| SOGC (2021)23 |
| Intended Users: physicians, including gynecologists, obstetricians, family physicians, internists, and emergency medicine specialists; nurses, including registered nurses and nurse practitioners; pharmacists; medical trainees, including medical students, residents, fellows; and other providers of health care for the target population Target Population: perimenopausal and postmenopausal women | Interventions: hormone therapy and nonhormonal options Outcomes: NR | A literature search was conducted in multiple databases. The quality of evidence was rated using the GRADE methodology (ratings range from very low to high). | Details on the development of the recommendations were not provided. The strength of recommendations was rated using the GRADE methodology (strong or conditional). | NR |
| NICE (2019)24 |
| Intended users: health care professionals who care for women in menopause, women in menopause, and their families and carers Target population: menopausal women, women with premature ovarian insufficiency | Interventions: hormonal pharmaceutical treatments, nonhormonal pharmaceutical treatments, nonpharmaceutical treatments, psychological therapy Outcomes: frequency of hot flushes (including night sweats), frequency of sexual activity, psychological symptoms, anxiety, low mood (not clinical depression), musculoskeletal symptoms, safety outcomes, discontinuation, vaginal bleeding | Systematic literature searches were conducted in multiple databases to identify all published clinical evidence (RCTs, nonrandomized trials, and observational studies) relevant to the review questions. Titles and abstracts were screened, and full texts of relevant studies were obtained and screened based on prespecified inclusion and exclusion criteria. Relevant studies were critically appraised using the checklists specified in the NICE guidelines manual. Evidence summaries were generated, and data were combined in meta-analysis where appropriate. Evidence was assessed using the GRADE methodology (ratings range from very low to high). | The guideline development group drafted recommendations based on their interpretation of the relevant evidence. Considerations included balancing the benefits, harms, and costs of different options. | The guideline was subject to a 6-week public consultation and feedback period. |
| Endocrine Society (2015)25 |
| Intended users: health care professionals Target population: menopausal and postmenopausal women | Interventions: hormonal and nonhormonal therapies Outcomes: NR | The guideline task force commissioned 3 systematic reviews on hormone therapies. Other existing meta-analyses and trials were also considered. The quality of evidence was assessed using the GRADE methodology (ratings range from very low to high). | The task force followed the GRADE methodology to develop the recommendations. Consensus on recommendations was determined through email communications, conference calls, and 1 face-to-face meeting. The phrases “we recommend” and “we recommend against” and the number 1 were used for strong recommendations. The phrases “we suggest” and “we suggest against” and the number 2 were used for weak recommendations. | The Australasian Menopause Society, the British Menopause Society, the European Menopause and Andropause Society, the European Society of Endocrinology, and the International Menopause Society (co-sponsors of the guideline) reviewed and commented on the draft. |
GRADE= Grading of Recommendations, Development and Evaluation; NAMS= North American Menopause Society; NICE= National Institute for Health and Care Excellence; NR= not reported; SOGC= Society of Obstetricians and Gynaecologists of Canada.
Note this table has not been copy-edited.
Table 6: Characteristics of Included Guidelines – Restless Leg Syndrome
| Intended users, target population | Intervention(s) and major outcomes considered | Evidence collection, synthesis, and quality assessment | Recommendations for development and evaluation | Guideline validation |
--- ---
| AAN (2016)26 |
| Intended users: clinicians Target population: adults with restless leg syndrome | Interventions: pharmacologic and nonpharmacologic therapies Outcomes: symptoms and clinical consequences of restless leg syndrome (disturbed sleep, periodic limb movements of sleep, depression/anxiety, QoL) | The guideline panel developed the clinical question and search terms. An independent medical librarian performed a systematic literature search in multiple databases. The guideline panel chair reviewed abstracts for inclusion. Two panel members individually reviewed the full articles for inclusion and assessed the quality of evidence and risk of bias. Identified studies were rated on a scale from class IV (lowest quality) to class I (highest quality). Meta-analyses were performed when needed. Class I: High-quality RCT Class II: Lower-quality RCT or prospective matched cohort Class III: All other controlled trials Class IV: Studies not meeting other criteria, including consensus or expert opinion | The guideline panel based recommendations strictly on the identified evidence and did not use expert opinion. The panel considered statistical significance, clinical significance, and precision when drawing conclusions from the evidence. Recommendations were assigned a level based on the quality of evidence. Level A: strong evidence Level B: moderate evidence Level C: weak evidence Level U: insufficient evidence | NR |
| IRLSSG (2015)27 |
| Intended users: health practitioners Target population: people with restless leg syndrome/ Willis-Ekbom disease who are pregnant or lactating. | Interventions: pharmacologic and nonpharmacologic therapies Outcomes: NR | A literature review was conducted in 1 database. | The guideline committee held monthly teleconferences and a single face-to-face meeting. Consensus questions were agreed upon and discussed. The committee used an integrative approach to develop the recommendations considering the existing literature and their clinical experience. Recommendations were rated from 1 to 5: 1. Recommended (high level of evidence for safety/effectiveness) 2. May be considered (evidence for safety effectiveness) 3. Insufficient evidence to reach consensus 4. Probably should not be considered (evidence for risk/ineffectiveness) 5. Not recommended (high level of evidence for risk/ ineffectiveness) | NR |
AAN= American Academy of Neurology; IRLSSG= International Restless Legs Syndrome Study Group; NR= not reported; QoL= quality of life.
Note this table has not been copy-edited.
Table 7: Characteristics of Included Guidelines – Tourette Syndrome
| Intended users, target population | Intervention(s) and major outcomes considered | Evidence collection, synthesis, and quality assessment | Recommendations development and evaluation | Guideline validation |
--- ---
| AAN (2019)28 |
| Intended users: clinicians Target population: children and adults with Tourette syndrome or a chronic tic disorder | Interventions: medical, behavioural, and neurostimulation interventions Outcomes: tic severity and tic-related impairment | A systematic review that included systematic reviews and RCTs was conducted in multiple databases.29 Two reviewers independently screened articles for inclusion. Two panel members rated the class of evidence for each article according to the AAN classification scheme on a scale from class IV (lowest quality) to class I (highest quality). Meta-analysis was performed when appropriate. Class I: High-quality RCT Class II: Lower-quality RCT or prospective matched cohort Class III: All other controlled trials Class IV: Studies that were not in a representative population or had no measures of effectiveness | A modified form of the GRADE process was used to develop conclusions. Four types of premises were used to support recommendations: evidence-based conclusion from the systematic review, generally accepted principals of care, strong evidence from related conditions, and deductive inferences from other premises. The guideline panel assigned each recommendation a level of A, B, or C. Level A: “must,” strongest level, high confidence in the evidence, high magnitude of benefit and low risk Level B: “should” Level C: “may,” lowest level | NR |
AAN= American Academy of Neurology; GRADE= Grading of Recommendations Assessment, Development and Evaluation; NR= not reported; RCT= randomized controlled trial.
Note this table has not been copy-edited.
Appendix 4: Critical Appraisal of Included Publications
Note this appendix has not been copy-edited.
Table 8: Strengths and Limitations of Guidelines Using AGREE II15 – Part 1
| Item | CRISM (2023)18 | NAMS (2023)22 | SOGC (2021)23 | VA/ DoD (2021)19 | ASAM (2020)20 | Hypertension Canada (2020)17 |
--- --- ---
| Domain 1: scope and purpose |
| 1. The overall objective(s) of the guideline is (are) specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| 2. The health question(s) covered by the guideline is (are) specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| 3. The population (patients, public, etc.) to whom the guideline is meant to apply is specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| Domain 2: stakeholder involvement |
| 4. The guideline development group includes individuals from all relevant professional groups | Yes | Yes | Unclear | Yes | Yes | Yes |
| 5. The views and preferences of the target population (patients, public, etc.) have been sought. | Yes | No | No | Yes | Partially | No |
| 6. The target users of the guideline are clearly defined. | Yes | No | Yes | Yes | Yes | Yes |
| Domain 3: rigour of development |
| 7. Systematic methods were used to search for evidence. | Yes | Unclear | Unclear | Yes | Yes | Unclear |
| 8. The criteria for selecting the evidence are clearly described. | Yes | No | No | Yes | Yes | No |
| 9. The strengths and limitations of the body of evidence are clearly described. | Yes | Partially | Yes | Yes | No | Yes |
| 10. The methods for formulating the recommendations are clearly described. | Yes | Partially | No | Yes | Yes | Yes |
| 11. The health benefits, side effects, and risks have been considered in formulating the recommendations. | Yes | Yes | Yes | Yes | Yes | Yes |
| 12. There is an explicit link between the recommendations and the supporting evidence. | Yes | Yes | No | Yes | No | Yes |
| 13. The guideline has been externally reviewed by experts before its publication. | Yes | No | No | Yes | Yes | No |
| 14. A procedure for updating the guideline is provided. | Partially | No | Yes | Yes | No | Yes |
| Domain 4: clarity of presentation |
| 15. The recommendations are specific and unambiguous. | Yes | Yes | Yes | Yes | Yes | Yes |
| 16. The different options for management of the condition or health issue are clearly presented. | Yes | Yes | Yes | Yes | Yes | Yes |
| 17. Key recommendations are easily identifiable. | Yes | Yes | Yes | Yes | Yes | Yes |
| Domain 5: applicability |
| 18. The guideline describes facilitators and barriers to its application. | Yes | No | No | No | No | No |
| 19. The guideline provides advice and/or tools on how the recommendations can be put into practice. | Yes | No | No | Yes | Yes | Yes |
| 20. The potential resource implications of applying the recommendations have been considered. | No | No | No | Yes | No | No |
| 21. The guideline presents monitoring and/or auditing criteria. | No | No | No | No | No | No |
| Domain 6: editorial independence |
| 22. The views of the funding body have not influenced the content of the guideline. | Yes | Yes | Unclear | Yes | Unclear | Yes |
| 23. Competing interests of guideline development group members have been recorded and addressed. | Yes | Yes | Partially | Yes | Yes | Yes |
AGREE II= Appraisal of Guidelines for Research and Evaluation II; ASAM= American Society of Addiction Medicine; CRISM= Canadian Research Initiative in Substance Misuse; DoD= Department of Defense; NAMS= North American Menopause Society; SOGC= Society of Obstetricians and Gynaecologists of Canada; VA= Veterans Affairs.
Table 9: Strengths and Limitations of Guidelines Using AGREE II15 – Part 2
| Item | AAN (2019)28 | NICE (2019)24 | AAN (2016)26 | Endocrine Society (2015)25 | IRLSSG (2015)27 | Commonwealth of Australia (2014)21 |
--- --- ---
| Domain 1: scope and purpose |
| 1. The overall objective(s) of the guideline is (are) specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| 2. The health question(s) covered by the guideline is (are) specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| 3. The population (patients, public, etc.) to whom the guideline is meant to apply is specifically described. | Yes | Yes | Yes | Yes | Yes | Yes |
| Domain 2: stakeholder involvement |
| 4. The guideline development group includes individuals from all relevant professional groups. | Yes | Yes | Yes | Yes | Yes | Yes |
| 5. The views and preferences of the target population (patients, public, etc.) have been sought. | No | Yes | No | No | No | No |
| 6. The target users of the guideline are clearly defined. | Yes | Yes | Yes | Yes | Yes | Yes |
| Domain 3: rigour of development |
| 7. Systematic methods were used to search for evidence. | Yes | Yes | Yes | Yes | Partially | Unclear |
| 8. The criteria for selecting the evidence are clearly described. | Yes | Yes | Yes | Yes | Partially | No |
| 9. The strengths and limitations of the body of evidence are clearly described. | Yes | Yes | Yes | Yes | Yes | Yes |
| 10. The methods for formulating the recommendations are clearly described. | Yes | Yes | Yes | Yes | Yes | Partially |
| 11. The health benefits, side effects, and risks have been considered in formulating the recommendations. | Yes | Yes | Yes | Yes | Yes | Yes |
| 12. There is an explicit link between the recommendations and the supporting evidence. | Yes | Yes | Yes | Yes | Yes | Partially |
| 13. The guideline has been externally reviewed by experts before its publication. | No | No | No | Yes | No | Yes |
| 14. A procedure for updating the guideline is provided. | No | Yes | No | No | No | No |
| Domain 4: clarity of presentation |
| 15. The recommendations are specific and unambiguous. | Yes | Yes | Yes | Yes | Yes | Yes |
| 16. The different options for management of the condition or health issue are clearly presented. | Yes | Yes | Yes | Yes | Yes | Yes |
| 17. Key recommendations are easily identifiable. | Yes | Yes | Yes | Yes | Yes | No |
| Domain 5: applicability |
| 18. The guideline describes facilitators and barriers to its application. | No | Yes | No | No | No | No |
| 19. The guideline provides advice and/or tools on how the recommendations can be put into practice. | No | Yes | No | Yes | Partially | Yes |
| 20. The potential resource implications of applying the recommendations have been considered. | No | Yes | No | No | No | Yes |
| 21. The guideline presents monitoring and/or auditing criteria. | No | No | No | No | No | No |
| Domain 6: editorial independence |
| 22. The views of the funding body have not influenced the content of the guideline. | Yes | Yes | Yes | Yes | Yes | Yes |
| 23. Competing interests of guideline development group members have been recorded and addressed. | Yes | Yes | Yes | Yes | Yes | Partially |
AAN= American Academy of Neurology; AGREE II= Appraisal of Guidelines for Research and Evaluation II; IRLSSG= International Restless Legs Syndrome Study Group; NICE= National Institute for Health and Care Excellence.
Appendix 5: Recommendations in Included Guidelines
Note this appendix has not been copy-edited.
Table 10: Summary of Recommendations – Hypertension
| Recommendation | Quality of evidence and strength of recommendation |
--- |
| Hypertension Canada (2020)17 |
| “Antihypertensive therapy is recommended for average SBP measurements of 140 mm Hg or DBP measurements of 90 mm Hg in pregnant women with chronic hypertension, gestational hypertension, or preeclampsia. Initial antihypertensive therapy should be monotherapy from the following first-line drugs: oral labetalol, oral methyldopa, long-acting oral nifedipine, or other oral b-blockers (acebutolol, metoprolol, pindolol, and propranolol) Other antihypertensive drugs can be considered as second-line drugs including clonidine, hydralazine, and thiazide diuretics. (p.620)”17 Supporting evidence: NA | Grade D (based on expert opinion alone) |
DBP= diastolic blood pressure; NA= not applicable; SBP= systolic blood pressure.
Table 11: Summary of Recommendations – Substance Use Disorders
| Recommendations and supporting evidence | Quality of evidence and strength of recommendations |
--- |
| CRISM (2023)18 |
| “For patients at low risk of severe complications of alcohol withdrawal (e.g., PAWSS < 4), clinicians should consider offering nonbenzodiazepine medications, such as gabapentin, carbamazepine or clonidine for withdrawal management in an outpatient setting (e.g., primary care, virtual). (p. E1368).”18 Supporting evidence: 2 RCTs reported clonidine is as effective as chlordiazepoxide in the management of mild to moderate withdrawal symptoms with better control of sympathetic symptoms and reductions in patient anxiety. | Certainty of evidence: Low Strength of recommendation: Strong |
| VA/DoD (2021)19 |
| “For patients with opioid use disorder for whom withdrawal management is indicated and for whom methadone and buprenorphine are contraindicated, unacceptable, or unavailable, we suggest offering clonidine or lofexidine as a second-line agent for opioid withdrawal management (p. 31)”19 Supporting evidence: 1 SR included comparisons of alpha2-adrenergic agonists vs. placebo, methadone, or another alpha2-adrenergic agonist. 1 SR that included comparisons of buprenorphine vs clonidine, lofexidine, or methadone. The work group considered that the benefits of improved withdrawal symptoms outweighed potential harms (minimal side effects and adverse events [hypotension for lofexidine and clonidine]) and that patient values and preferences varied somewhat. | Quality of evidence: Low Strength of recommendation: Weak |
| ASAM (2020)20 |
| Ambulatory and inpatient management of alcohol withdrawal: “Alpha2-adrenergic agonists such as clonidine can be used as an adjunct to benzodiazepine therapy to control autonomic hyperactivity and anxiety when symptoms are not controlled by benzodiazepines alone. They should not be used alone to prevent or treat withdrawal-related seizures or delirium. (p. 9)”20 Supporting evidence: NR | Quality of evidence: NR Strength of recommendation: NR |
| Commonwealth of Australia (2014)21 |
| “Two distinct medication approaches are recommended for the management of opioid withdrawal: Abrupt cessation of opioid use and symptom amelioration using non-opioid drugs (usually benzodiazepines, nonsteroidal anti-inflammatory drugs, antiemetics, clonidine, antispasmodic drugs [such as hyoscine butylbromide]) for relief of symptoms; Short course (usually less than 1 month) of reducing doses of buprenorphine. (p. 19)”21 Supporting evidence: The guideline authors stated that both approaches are well supported by evidence. They also stated that the use of buprenorphine to manage withdrawal is associated with significantly better relief of withdrawal than clonidine and supplementary medications (1 SR). The SR also reported that completion of withdrawal is significantly more likely when managed with buprenorphine vs clonidine. Buprenorphine is associated with fewer adverse effects and fewer withdrawals from treatment due to adverse effects than clonidine. | Quality of evidence: 4 stars (body of evidence can be trusted to guide practice) Strength of recommendation: NR |
ASAM= American Society of Addiction Medicine; CRISM= Canadian Research Initiative in Substance Misuse; DoD= Department of Defense; NR= not reported; PAWSS= Prediction of Alcohol Withdrawal Severity Scale; RCT= randomized controlled trial; SR= systematic review; VA= Veterans Affairs.
Table 12: Summary of Recommendations – Menopause
| Recommendations and supporting evidence | Quality of evidence and strength of recommendations |
--- |
| NAMS (2023)22 |
| “Because there are other more effective therapies with fewer AEs, clonidine is not recommended. (p. 578)”22 Supporting evidence: 1 SR reported that clonidine is modestly more beneficial than placebo at reducing vasomotor symptoms. 2 SRs reported that clonidine is less beneficial than SSRIs, SNRIs, and gabapentin at reducing vasomotor symptoms. The guideline authors stated that clonidine is used infrequently due to AEs, including hypotension, lightheadedness, headache, dry mouth, dizziness, sedation, and constipation. They also stated that sudden cessation of clonidine can lead to significant elevation of blood pressure. | Quality of evidence: Levels 1 to 3 (good and consistent scientific evidence, limited or inconsistent scientific evidence, and consensus and expert opinion) Strength of recommendation: NR |
| SOGC (2021)23 |
| “Paroxetine, gabapentin, oxybutynin, and clonidine are nonhormonal options for refractory vasomotor symptoms. Paroxetine should be used with caution in patients receiving tamoxifen (p. 1451)”23 Supporting evidence: NR | Quality of evidence: Moderate Strength of recommendation: Conditional |
| NICE (2019)24 |
| “Give information to menopausal women and their family members or carers (as appropriate) about the following types of treatment for menopausal symptoms: hormonal, for example hormone replacement therapy nonhormonal, for example clonidine nonpharmaceutical, for example cognitive behavioural therapy. (p. 7)”24 Supporting evidence: Clonidine was not included in the network meta-analysis conducted for the guideline due to how outcomes were reported in studies of clonidine. Therefore, the relative effectiveness of clonidine in relieving short-term symptoms for women in menopause could not be estimated. In the absence of this data, the guideline development group recognized the importance of clonidine for the treatment of some women in menopause. | Quality of evidence: NR Strength of recommendation: NR |
| “Do not routinely offer SSRIs, SNRIs or clonidine as first-line treatment for vasomotor symptoms alone. (p. 8)”24 Supporting evidence: Clonidine was not included in the network meta-analysis conducted for the guideline due to how outcomes were reported in studies of clonidine. Therefore, the relative effectiveness of clonidine in relieving short-term symptoms for women in menopause could not be estimated. | Quality of evidence: NR Strength of recommendation: Strong |
| Endocrine Society (2015)25 |
| “For those women seeking relief of moderate to severe vasomotor symptoms who are not responding to or tolerating the nonhormonal prescription therapies SSRIs/SNRIs or gabapentin or pregabalin, we suggest a trial of clonidine (if there are no contraindications). (p. 3995)”25 Supporting evidence: Several RCTs demonstrated that clonidine reduces hot flashes. However, it is less effective than SSRI/SNRIs, gabapentin, and pregabalin and is associated with more side effects. | Quality of evidence: Low Strength of recommendation: Weak |
AE= adverse event; NAMS= North American Menopause Society; NICE= National Institute for Health and Care Excellence; NR= not reported; RCT= randomized controlled trial; SNRI= serotonin-norepinephrine reuptake inhibitor; SOGC= Society of Obstetricians and Gynaecologists of Canada; SSRI= selective serotonin reuptake inhibitor.
Table 13: Summary of Recommendations – Restless Leg Syndrome
| Recommendations and supporting evidence | Quality of evidence and strength of recommendations |
--- |
| AAN (2016)26 |
| “There is insufficient evidence to support or refute the use of gabapentin, iron sucrose, oxycodone, clonazepam, bupropion, clonidine, selenium, rifaximin, botulinum neurotoxin, valproic acid, carbamazepine, or valerian in the treatment of restless leg syndrome (p. 2590)”26 Supporting evidence: 1 crossover trial of clonidine vs placebo in which patients reported less paresthesia, motor restlessness, and daytime fatigue during the clonidine treatment arm. There was no difference in periodic limb movement index between groups. The most common adverse events were hypotension, decreased cognition, dry mouth, and sleepiness. | Quality of evidence: Class III Strength of recommendation: Level U (insufficient evidence) |
| IRLSSG (2015)27 |
| Clonidine should probably not be considered for people with restless leg syndrome who are pregnant or lactating. Supporting evidence: A small RCT found limited efficacy of clonidine for restless leg syndrome. Clonidine passes on to breast milk. | Level 4 (evidence for risk/ ineffectiveness) |
AAN= American Academy of Neurology; IRLSSG= International Restless Legs Syndrome Study Group; RCT= randomized controlled trial.
Table 14: Summary of Recommendations – Tourette Syndrome
| Recommendations and supporting evidence | Quality of evidence and strength of recommendations |
--- |
| AAN (2019)28 |
| “Physicians should counsel individuals with tics and comorbid ADHD that alpha2 adrenergic agonists may provide benefit for both conditions (p. 900)”28 Supporting evidence: 1 study that reported reduced tic severity in children with tics and a comorbid diagnosis of ADHD who received clonidine plus methylphenidate vs placebo.29 | Quality of evidence: Class I Strength of recommendation: Level B |
| “Physicians should prescribe alpha2 adrenergic agonists for the treatment of tics when the benefits of treatment outweigh the risks (p. 900)”28 Supporting evidence: 3 studies that reported reduced tic severity in people with tics receiving clonidine vs placebo.29 | Quality of evidence: Class I and II Strength of recommendation: Level B |
| “Physicians must counsel patients regarding common side effects of alpha2 adrenergic agonists, including sedation (p. 900)”28 Supporting evidence: 2 studies reported that sedation was more common in people with tics receiving clonidine vs placebo.29 1 SR of alpha2 adrenergic agonists for ADHD in children and adolescents demonstrated hypotension, bradycardia, and sedation with these agents.28 | Quality of evidence: Class I and II Strength of recommendation: Level A |
| “Physicians must monitor heart rate and blood pressure in patients with tics treated with alpha2 adrenergic agonists (p. 900)”28 Supporting evidence: 1 SR of alpha2 adrenergic agonists for ADHD in children and adolescents demonstrated hypotension and bradycardia with these agents.28 | Quality of evidence: NR Strength of recommendation: Level A |
| “Physicians discontinuing alpha2 adrenergic agonists must gradually taper them to avoid rebound hypertension (p. 900)”28 Supporting evidence: NR. | Quality of evidence: NR Strength of recommendation: Level A |
AAN= American Academy of Neurology; ADHD= attention-deficit/hyperactivity disorder; NR= not reported; SR= systematic review.
ISSN:2563-6596
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188434 | https://www.cuemath.com/one-eighty-degree-rotation-formula/ | Formula For 180 Degree Rotation
Before learning the formula for 180-degree rotation, let us recall what is 180 degrees rotation. A point in the coordinate geometry can be rotated through 180 degrees about the origin, by making an arc of radius equal to the distance between the coordinates of the given point and the origin, subtending an angle of 180 degrees at the origin. We have to rotate the point about the origin with respect to its position in the cartesian plane. It can be well understood in the following section of the formula for 180-degree rotation.
What is the Formula for 180 Degree Rotation?
The formula for 180-degree rotation of a given value can be expressed as if R(x, y) is a point that needs to be rotated about the origin, then coordinates of this point after the rotation will be just of the opposite signs of the original coordinates. i.e., the coordinates of the point after 180-degree rotation are:
R'= (-x, -y)
Let us apply the formula for 180-degree rotation in the following solved examples.
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Examples Using Formula for 180 Degree Rotation
Example 1: Rotate the following points by 180 degrees: (i) A(3,4) (ii) B(2,-7) (iii) C(-5, -1).
Solution:
To find: Rotate the given points by 180 degrees.Given: A(3,4), B(2.-7), C(-5,-1)
Using formula for 180 degree rotation,R(x,y) ⇒ R'(-x,-y)
(i). A(3,4) ⇒ A’(-3,-4)(ii). B(2,-7) ⇒ B’(-2,7)(iii).C(-5,-1) ⇒ C’(5,1)
Answer: A’(-3,-4), B’(-2,7), and C’(5,1) are the 180 degrees rotated points of A(3,4), B(2.-7), and C(-5,-1)
Example 2: Rotate a line AB having ends as A(4,5) and B(-1,2) by 180 degrees.
Solution:
To find: Rotate a line AB by 180 degrees.GivenLine AB, A(4,5), B(-1,2)
Using formula for 180 degree rotation,
R(x,y) ⇒ R'(-x,-y)
For A(4,5) ⇒ A’(-4,-5)For B(-1,2) ⇒ B’(1,-2)
Answer: The coordinates of the new line A’B’ is A’(-4,-5) and B’(1,-2)
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188435 | https://chem.libretexts.org/Under_Construction/Purgatory/Introductory_Chemistry_at_Solano_College_2022/09%3A_The_Mole/9.09%3A_Gas_Density | 9.9: Gas Density - Chemistry LibreTexts
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9: The Mole
Introductory Chemistry at Solano College 2022
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Fri, 26 Aug 2022 23:11:41 GMT
9.9: Gas Density
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[ "article:topic", "showtoc:no", "transcluded:yes", "Gas Density", "source-chem-53772", "program:ck12", "license:ck12", "authorname:ck12", "source@ ]
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Contents
1. Home
2. Under Construction
3. Purgatory
4. Introductory Chemistry at Solano College 2022
5. 9: The Mole
6. 9.9: Gas Density
Expand/collapse global location
Introductory Chemistry at Solano College 2022
Front Matter
1: Introduction to Chemistry
2: Numbers From Measurements
3: Unit Systems and Dimensional Analysis
4: Matter and Change
5: Atoms, Molecules, and Subatomic Particles
6: Electronic Structure and Chemical Periodicity
7: Chemical Bonds
8: Chemical Nomenclature
9: The Mole
10: Chemical Calculations Involving Chemical Equations
11: States of Matter
12: Gas Laws
13: Solutions
14: Acids and Bases
Back Matter
9.9: Gas Density
Last updated Aug 26, 2022
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9.8: Conversions Between Moles and Gas Volume
9.10: Mole Road Map
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Table of contents
1. Why does carbon dioxide sink in air?
2. Gas Density
1. Example 9.9.1: Gas Density
1. Solution
2. Step 1: List the known quantities and plan the problem.
3. Known
4. Unknown
5. Step 2: Calculate.
6. Step 3: Think about your result.
2. Example 9.9.2: Molar Mass from Gas Density:_Molar_Mass_from_Gas_Density)
1. Solution
2. Known
3. Unknown
4. Step 2: Calculate.
5. Step 3: Think about your result.
Summary
Review
Figure 9.9.1 (CK-12 Curriculum Materials license; CK-12 Foundation via CK-12 Foundation)
Why does carbon dioxide sink in air?
When we run a reaction to produce a gas, we expect it to rise into the air. Many students have done experiments where gases such as hydrogen are formed. The gas can be trapped in a test tube held upside-down over the reaction. Carbon dioxide, on the other hand, sinks when it is released. Carbon dioxide has a density greater than air, so it will not rise like the hydrogen gas.
Gas Density
As you know, density is defined as the mass per unit volume of a substance. Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass. A gas with a small molar mass will have a lower density than a gas with a large molar mass. Gas densities are typically reported in g/L. Gas density can be calculated from molar mass and molar volume.
Figure 9.9.2: Balloons filled with helium gas float in air because the density of helium is less than the density of air. (Public Domain; Photographer: Warren Denning, courtesy of the Pioneer Balloon Company via Wikipedia)
Example 9.9.1: Gas Density
What is the density of nitrogen gas at STP?
Solution
Step 1: List the known quantities and plan the problem.
Known
NA 2=28.02 g/mol
1 mol=22.4 L
Unknown
density = ? g/L
Molar mass divided by molar volume yields the gas density at STP.
Step 2: Calculate.
28.02 g 1 mol×1 mol 22.4 L=1.25 g/L
When set up with a conversion factor, the mol unit cancels, leaving g/L as the unit in the result.
Step 3: Think about your result.
The molar mass of nitrogen is slightly larger than molar volume, so the density is slightly greater than 1 g/L.
Alternatively, the molar mass of a gas can be determined if the density of the gas at STP is known.
Example 9.9.2: Molar Mass from Gas Density
What is the molar mass of a gas whose density is 0.761 g/L at STP?
Solution
Step 1: List the known quantities and plan the problem.
Known
NA 2=28.02 g/mol
1 mol=22.4 L
Unknown
molar mass = ? g/mol
Molar mass is equal to density multiplied by molar volume.
Step 2: Calculate.
0.761 g 1 L×22.4 L 1 mol=17.0 g/mol
Step 3: Think about your result.
Because the density of the gas is less than 1 g/L, the molar mass is less than 22.4.
Summary
Calculations are described showing conversions between molar mass and density for gases.
Review
How is density calculated?
How is molar mass calculated?
What would be the volume of 3.5 moles of a gas?
This page titled 9.9: Gas Density is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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9.10: Mole Road Map
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188436 | https://www.wyzant.com/resources/answers/866582/write-an-equation-in-slope-intercept-form-for-the-line-with-the-given-slope | Write an equation in slope intercept form for the line with the given slope that contains the given point. Slope = -3; (4,0) | Wyzant Ask An Expert
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Algebra
Carl M.
asked • 09/24/21
Write an equation in slope intercept form for the line with the given slope that contains the given point. Slope = -3; (4,0)
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Slope Intercept Form of a straight line: y = mx + b
mis the slope
bis the y-intercept
(4,0) is the coordinate the line must go through.
For EXAMPLE,y = 2x + 1
The slope (m) is 2.
A coordinate that this line goes through is (1,3). Coordinates are in this form (x,y).
1 is the x, and 3 is the y.
To prove this....
3 = 2(1) + 1
3 = 3
To answer this question...
If the slope is -3, and it must go through (4,0)...
y = -3x + b
To solve "b", plug in the coordinates.
0 = -3(4) + b
0 = -12 + b
0 + 12 = -12 + b +12(if you add 12 on one side, you must do the same on the other side of the = sign)
12 = b
THEREFORE...
y = -3x +12
Check your answer by plugging in the coordinates.
0 = -3(4) + 12
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188437 | https://math.stackexchange.com/questions/3993570/is-the-following-definition-of-an-elliptic-curve-correct | abstract algebra - Is the following definition of an elliptic curve correct? - Mathematics Stack Exchange
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Is the following definition of an elliptic curve correct?
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\begingroup
Im new to algebraic geometry so I want to make sure im getting my definitions right. I know there are a few ways to state what an elliptic curve is (ex a smooth projective curve of genus one with distinguished K-rational point). But I am wondering if the following is equivalent. For simplicity lets just work over \mathbb{C}.:
\textbf{Definition:} An elliptic curve E is a non-singular projective curve in \mathbb{P}^2 of the form E: Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3
I am wondering if this is sufficient to define an elliptic curve?
abstract-algebra
number-theory
algebraic-geometry
elliptic-curves
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edited Jan 23, 2021 at 20:50
BagggggsBagggggs
asked Jan 21, 2021 at 0:53
BagggggsBagggggs
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3 \begingroup Yes, this is enough. The equivalence is shown in Silverman "The arithmetic of ellipic curves" chapter 3. Your first definition needs the added statement "with a fixed K-rational point"\endgroup Mummy the turkey –Mummy the turkey 2021-01-21 01:02:08 +00:00 Commented Jan 21, 2021 at 1:02
2 \begingroup A truly pedantic quibble, you should say "in" \mathbb{P}^2 and not over \mathbb{P}^2, which means something else in algebraic geometry.\endgroup hunter –hunter 2021-01-21 17:19:22 +00:00 Commented Jan 21, 2021 at 17:19
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\begingroup
This is subtle. It is true that every elliptic curve can be written in this form, and that every smooth projective curve of this form has genus 1.
But an elliptic curve is not a smooth projective curve of genus 1. An elliptic curve (over a field K) is a smooth projective curve of genus 1 together with a distinguished K-rational point, and this is key to the theory because it's the point that forms the identity for the group law. You need to say something about what this distinguished point is: with an equation of this form it's conventional to take it to be the point at infinity, with coordinates (X : Y : Z) = (0 : 1 : 0). But this needs to be said explicitly in the definition; it is in fact possible to pick another point, which changes the group law. This is important for the following reasons among others:
There are smooth projective curves of genus 1 over non-algebraically closed fields K which have no K-rational points, and hence there is no choice of point which turns them into elliptic curves.
Elliptic curves have endomorphisms and automorphisms, and these are required to preserve the distinguished point (which turns out to imply that they preserve the group law). If you don't keep this in mind you will be confused when you read statements about endomorphisms and automorphisms of elliptic curves in the literature (which in fact happened recently on MO). For example, endomorphisms of an elliptic curve form a ring, but only if you require that endomorphisms preserve the distinguished point.
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edited Jan 21, 2021 at 4:54
answered Jan 21, 2021 at 4:49
Qiaochu YuanQiaochu Yuan
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1 \begingroup It's worth pointing out that by the rigidity lemma, the isomorphism class of elliptic curve doesn't actually depend on the chosen base point, assuming one exists. Namely, if C is a smooth connected curve of genus 1 over k and p_1,p_2 in C(k) then there exists unique group multiplications \mu_i on C such that p_i is the identity element. But, the map C\to C:x\mapsto x-p_1+p_2 is an isomorphism between them.\endgroup Alex Youcis –Alex Youcis 2021-01-21 05:34:53 +00:00 Commented Jan 21, 2021 at 5:34
1 \begingroup@Alex: thanks. Can you also comment on smooth projective curves of genus 1 with no K-rational points? I'm guessing the correct statement here is that they are all torsors over elliptic curves, and even more specifically torsors over their Jacobians, but I wasn't able to track down a resource saying this explicitly. (Maybe it just follows straightforwardly from Galois descent?)\endgroup Qiaochu Yuan –Qiaochu Yuan 2021-01-21 05:39:31 +00:00 Commented Jan 21, 2021 at 5:39
1 \begingroup Since quasi-projective varieties are a stack for the etale topology one knows that twists of an elliptic curve (E,0), thought of as an undecorated curve, can be identified with \mathrm{Aut}(E)-torsors. But, \mathrm{Aut}(E)=E\rtimes \mathrm{Aut}(E,0) and so we see that one has an inclusion of E-torsors into \mathrm{Aut}(E)-torsors=twists of E. But, since E is itself nice (e.g. smooth and proper) one knows from abstract nonsense that E-torsors correspond to E-bundles--curves C with a simply transitive action of E as a scheme. Since such C are evidently twists of E\endgroup Alex Youcis –Alex Youcis 2021-01-21 06:05:55 +00:00 Commented Jan 21, 2021 at 6:05
1 \begingroup one obtains the natural inclusion of E-torsors into twists for E. One can easily see that if C is a smooth proper geometrically connected curve with no rational point, then it's a twist of \mathrm{Pic}^0(C) and, in fact, can be easily seen to be a \mathrm{Pic}^0(C)-bundle. This is all covered, in less fancy language, in Section X.2 of Silverman's first book.\endgroup Alex Youcis –Alex Youcis 2021-01-21 06:08:26 +00:00 Commented Jan 21, 2021 at 6:08
\begingroup I'm glad this got accepted now. It's a much better answer than mine :-).\endgroup AAR –AAR 2021-01-21 17:18:24 +00:00 Commented Jan 21, 2021 at 17:18
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\begingroup
Yes, this will do. The abstract definition of an elliptic curve is that it is nonsingular of genus 1 as you mentioned. If your curve is cut out by a degree d equation in \Bbb{P}^2, the degree-genus formula gives g=\frac{(d-1)(d-2)}{2}. Hence, since you have a cubic C, g(C)=\frac{2}{2}=1. Conversely, you can show that any abstract genus 1 curve over an algebraically closed field k has an embedding into projective space \Bbb{P}^2 which realizes it as a degree 3 curve.
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188438 | https://www.london.ac.uk/news-events/blogs/weird-tales-winter-nights | Weird tales for winter nights | University of London
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Senate House Library
Weird tales for winter nights
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Weird tales for winter nights
Date
25 Oct 2022
Written by Emma Fitzpatrick, Serial and Digital Resources Coordinator
Senate House Library celebrates Halloween and looks ahead to the long, winter nights by taking a closer looks at items from the British Library Tales Of The Weird series held in our collections
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The nights are drawing in and soon it will be Halloween. Let’s celebrate the spookiest of seasons by taking a closer look at the British Library’s Tales Of The Weird series and the books from this series which Senate House Library holds in its Modern Collections.
British Library Tales Of The Weird
The British Library first started publishing their Tales Of The Weird series in 2018, following on from the success of the British Library’s crime fiction imprint, the British Library Crime Classics. Both series aim to make some of the rarer material from the British Library’s collections available to new audiences by re-publishing them in an affordable paperback form. Many of the books in the British Library Tales Of The Weird series are anthologies of short stories, although some works from this series do include novellas and other longer forms of fiction. Each book focuses on a particular author or a theme, such as botanical themed weird fiction or weird fiction by women. Each book includes a short but well written introduction to each theme, which provides information to help both seasoned readers of the weird and new readers to the genre to get the most out of each volume. Short author biographies also introduce each work in the collection. These well edited and entertaining story collections from the British Library Tales Of The Weird series make the perfect companions for a long winter evening.
A closer look at some of the books from this series which you can find in Senate House Library’s Collections
Evil roots : killer tales of the botanical gothic / edited by Daisy Butcher
Evil Roots showcases an incredible variety of plant-based gothic fiction from the nineteenth and early twentieth centuries. A new reader of botanical gothic fiction might be tempted to think that the “killer plant” story is cliched. However, this book proves that idea wrong with a well chosen selection of short stories which ably illustrates just how menacing plants can be in the right author’s hands. From a vine holding onto past wrongs in Charlotte Perkins Gillman’s Giant Wisteria to a plant with uncomfortably human characteristics in Emma Vane’s Moaning Lily, this collection brings together some of the best examples of a small but fascinating sub-genre of gothic fiction.
Tales of the tattooed : an anthology of ink / edited by John Miller
Tales of the tattooed brings together thirteen short stories which explore the art and history of the tattoo. Some of the stories in this collection, including W.W. Jacobs’ A Marked Man, reflect well known cultures of tattooing such as the long association between sailors and tattoos. While others reveal some of the more surprising parts of the history of tattooing, including the fashion for tattoos amongst wealthy socialites which swept Europe and America towards the end of the nineteenth and early twentieth centuries. This fashion for tattoos, which included well to do tattooed women as well as men, caused some controversy which is explored in Branded by Albert Payson Terhune. This book presents a varied selection of stories on an unusual and fascinating topic, with an introduction by John Millar which gives the reader a concise and informative summary of some of the diverse cultures of tattooing from around the world and their history.
The ghost slayers : classic tales of occult detection / edited by Mike Ashley
The Ghost Slayers combines the strange allure of weird fiction with the thrill of detection. This short stories in this anthology introduce us to several detectives who fearlessly confront the strange and arcane, including William Hope Hodgson’s Carnacki the Ghost-Finder and Joseph Payne Brennan’s Lucius Leffing, the Psychic Investigator. As with many of the books in this series, The Ghost Slayers contains some stories by well know authors and other stories which are quite rare, including The Case Of The Fortunate Youth by Moray Dalton. This story was originally published in The Premier Magazine, copies of which are now extremely rare. Its republication in The Ghost Slayers makes this compelling tale available for a new set of readers to enjoy. This book is full of entertaining stories, offering something for both lovers of weird fiction and detective fiction fans.
Into the London fog : eerie tales from the weird city / edited by Elizabeth Dearnley
Into the London Fog presents a collection of stories which explore the weird side of the big city. As Elizabeth Dearnley so eloquently puts it in her excellent introduction to this book, these weird tales focus on “the contrast between civilised, brightly lit modernity and darker, atavistic fears about what might lie just beyond the glow of the streetlamp.” Each story is set in a different part of London, taking us on a tour of the city where the familiar becomes strange, mirrors hold the key to unsolved mysteries and portals to magical lands appear amongst the quiet streets of Stoke Newington. This anthology includes The Lodger by Marie Belloc Lowndes, which was influenced by the unsolved murders committed by Jack the Ripper and which reflects on the dangers that come from living so closely with strangers in a crowded urban environment. The works in this anthology, both fictional and non-fictional, offer us a new way of looking at the city that many of us know so well. Do we dare step out into the London fog?
Explore more gothic and supernatural fiction at Senate House Library
Works of gothic and supernatural literature, such as these British Library Tales of the Weird books, are strongly represented in Senate House Library’s collections. You can find more information about our supernatural fiction holdings and other works in our collections related to the paranormal, the occult and the magical by looking at our Research strengths of our collection website - The paranormal, the occult and the magical.
If you are interested in learning about our other areas of collection strength, you can learn more about them on our Research strengths of our collections website.
This page was last updated on 11 January 2024
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188439 | https://pmc.ncbi.nlm.nih.gov/articles/PMC1235277/ | Genomic Divergences between Humans and Other Hominoids and the Effective Population Size of the Common Ancestor of Humans and Chimpanzees - PMC
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Am J Hum Genet
. 2001 Jan 15;68(2):444–456. doi: 10.1086/318206
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Genomic Divergences between Humans and Other Hominoids and the Effective Population Size of the Common Ancestor of Humans and Chimpanzees
Feng-Chi Chen
Feng-Chi Chen
1 Department of Life Science, National Tsing Hua University, Taiwan, and 2 Department of Ecology and Evolution, University of Chicago, Chicago
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1,,, Wen-Hsiung Li
Wen-Hsiung Li
1 Department of Life Science, National Tsing Hua University, Taiwan, and 2 Department of Ecology and Evolution, University of Chicago, Chicago
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1 Department of Life Science, National Tsing Hua University, Taiwan, and 2 Department of Ecology and Evolution, University of Chicago, Chicago
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Address for correspondence and reprints: Dr. Wen-Hsiung Li, Department of Ecology and Evolution, University of Chicago, 1101 East 57th Street, Chicago, IL 60637. E-mail: whli@uchicago.edu
Visiting student at the Department of Ecology and Evolution, University of Chicago.
Received 2000 Nov 13; Accepted 2000 Dec 8; Issue date 2001 Feb.
© 2001 by The American Society of Human Genetics. All rights reserved.
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PMCID: PMC1235277 PMID: 11170892
Abstract
To study the genomic divergences among hominoids and to estimate the effective population size of the common ancestor of humans and chimpanzees, we selected 53 autosomal intergenic nonrepetitive DNA segments from the human genome and sequenced them in a human, a chimpanzee, a gorilla, and an orangutan. The average sequence divergence was only 1.24% ± 0.07% for the human-chimpanzee pair, 1.62% ± 0.08% for the human-gorilla pair, and 1.63% ± 0.08% for the chimpanzee-gorilla pair. These estimates, which were confirmed by additional data from GenBank, are substantially lower than previous ones, which included repetitive sequences and might have been based on less-accurate sequence data. The average sequence divergences between orangutans and humans, chimpanzees, and gorillas were 3.08% ± 0.11%, 3.12% ± 0.11%, and 3.09% ± 0.11%, respectively, which also are substantially lower than previous estimates. The sequence divergences in other regions between hominoids were estimated from extensive data in GenBank and the literature, and Alu s showed the highest divergence, followed in order by Y-linked noncoding regions, pseudogenes, autosomal intergenic regions, X-linked noncoding regions, synonymous sites, introns, and nonsynonymous sites. The neighbor-joining tree derived from the concatenated sequence of the 53 segments—24,234 bp in length—supports the Homo-Pan clade with a 100% bootstrap value. However, when each segment is analyzed separately, 22 of the 53 segments (∼42%) give a tree that is incongruent with the species tree, suggesting a large effective population size (N e) of the common ancestor of Homo and Pan. Indeed, a parsimony analysis of the 53 segments and 37 protein-coding genes leads to an estimate of N e = 52,000 to 96,000. As this estimate is 5 to 9 times larger than the long-term effective population size of humans (∼10,000) estimated from various genetic polymorphism data, the human lineage apparently had experienced a large reduction in effective population size after its separation from the chimpanzee lineage. Our analysis assumes a molecular clock, which is in fact supported by the sequence data used. Taking the orangutan speciation date as 12 to 16 million years ago, we obtain an estimate of 4.6 to 6.2 million years for the Homo-Pan divergence and an estimate of 6.2 to 8.4 million years for the gorilla speciation date, suggesting that the gorilla lineage branched off 1.6 to 2.2 million years earlier than did the human-chimpanzee divergence.
Introduction
The degree of sequence divergence between the human (Homo sapiens) and chimpanzee (Pan troglodytes) genomes has been a subject of numerous studies (e.g., King and Wilson 1975; Sibley and Ahlquist 1987; Goodman et al. 1990), and it has been commonly thought that the two genomes differ by ∼1.6%. However, as this estimate was based mainly on DNA hybridization data (Sibley and Ahlquist 1987) and the DNA sequence data from the η-globin pseudogene region (Bailey et al. 1991), it may not represent an accurate estimate of the average divergence between the two genomes. Indeed, large variation in sequence divergence is often seen among genomic regions. For example, the last intron of the ZFY gene shows only 0.69% divergence between human and chimpanzee (Dorit et al. 1995), whereas the olfactory receptor OR1D3P pseudogene shows a divergence of 3.04% (Glusman et al. 2000). The divergences between the genomes of other hominoids are much less well studied. To have reliable estimates of the average divergences between hominoid genomes, sequence data from many genomic regions are needed.
The divergence dates between human and other hominoids also have received much attention (e.g., Sarich and Wilson 1967; Sibley and Ahlquist 1987; Takahata et al. 1995; Ruvolo 1997). However, there is still much uncertainty about these dates, especially the internodal time span between the human-chimpanzee divergence and the branch node of the gorilla lineage. For example, according to Horai et al.’s study of mitochondrial DNA sequences (1992), the divergence between human and chimpanzee occurred 4.7 ± 0.5 million years ago, whereas the gorilla lineage branched off 7.7 ± 0.7 million years ago, so the time span between the two speciation events was as long as 3 million years, ∼60% of the Homo-Pan divergence time. In contrast, in Bailey et al.’s study of the η-globin pseudogene region (1991), the internodal time span was only ∼10% of the divergence time between human and chimpanzee. Which is closer to the truth? As will be explained later, this internodal time span is useful for estimating the effective size of the ancestral population before the human-chimpanzee divergence.
There has also been considerable interest in the demographic history of hominoids (e.g., Takahata 1990; Rogers and Harpending 1992; Ruvolo 1997). Of particular interest is the effective size (N e) of the ancestral population before the human-chimpanzee divergence (Takahata et al. 1995; Ruvolo 1997), because it may tell us whether there has been a significant reduction in population size in the human lineage since its separation from the chimpanzee lineage. Current data are not sufficient for a reliable estimate of N e; for this purpose, a fairly large number of independent loci from the human, chimpanzee, and gorilla genomes are needed.
To study the above issues, we selected 53 autosomal intergenic nonrepetitive regions throughout the human genome and obtained the homologous sequences of each region from a human, a chimpanzee, a gorilla, and an orangutan. We chose noncoding regions because they are not directly subject to natural selection and therefore can more accurately trace the history of evolution than can coding regions. We also collected and analyzed many data sets from GenBank and the literature. Some of the conclusions derived from these extensive data are substantially different from previous ones.
Material and Methods
Selection of Regions
All of the 53 DNA segments studied were chosen to avoid coding regions. They were selected randomly, with reference to the Genome Channel. We searched for autosomal contigs without any known registered GenBank genes and scanned each contig with the GRAIL and GeneScan programs to detect potential genes. From each contig used, a segment of 2 to 20 kb was chosen, at least 5 kb away from any potential genes in both directions. Then the segment was masked using the RepeatMasker program. All repetitive elements were excluded and the remaining segment, if long enough (⩾800 bp), was used for PCR primer design and amplification.
DNA Samples
DNA samples from one Asian male human (Homo sapiens), one chimpanzee (Pan troglodytes), one gorilla (Gorilla gorilla) and one orangutan (Pongo pygmaeus) were used in this study. The protocol for taking blood samples for extracting DNA has been approved by the institutional review board of the University of Chicago.
PCR Amplification and Sequencing
Touch-down PCR (Don et al. 1991) was applied to each selected segment. Then the PCR products were purified with Wizard PCR Preps DNA Purification Resin Kit (Promega). Sequencing reactions were performed according to the ABI Prism BigDye Terminator Sequencing Kits (Applied Biosystems), modified for quarter reaction. The extension products were purified with 50% Sephadex G-50 resin (DNA grade, Pharmacia) and were run on an ABI 377 XL DNA sequencer using 4.25% gels (Sooner Scientific).
ABI DNA Sequence Analysis 3.0 was used for lane tracking and base calling. All segments were sequenced in both directions and were proofread manually with the SeqMan program of DNAStar.
Data Retrieval from GenBank
Homologous introns, pseudogenes, and coding regions of hominoids were retrieved from GenBank using the Hovergen program (Duret et al. 1994). Most of the accession numbers of coding regions were taken from Satta et al. (2000) and also from the Silver Project Home Page. For sequences with more than one haplotype, the ones with the smallest distances were chosen for analyses. The accession numbers of the chromosome 12 contigs taken from GenBank were: AC007286, AC007458, AC005294, and AC011604 (human), and AC007214 and AC006582 (chimpanzee).
Data Analysis
The DAMBE package (Xia 2000) was used for sequence alignment, calculation of genetic distance, and phylogenetic reconstruction. All sequenced autosomal segments were analyzed separately and then were concatenated and analyzed for the overall divergence or phylogenetic reconstruction. For coding regions, the number of substitutions per synonymous site (K S) and the number of substitutions per nonsynonymous site (K A) were calculated by the method of Li (1993) in DAMBE. Introns and pseudogenes were scanned with Repeat Masker to eliminate all repeats before distance calculation. For the chromosome 12 sequences, all repeats were eliminated and then were submitted to the BLAST server to identify homologous regions between the human and chimpanzee sequences. Sequences similar to any known functional genes were excluded before alignment.
Results
Genomic Divergence
Noncoding regions
The sequence divergences among hominoids in the 53 autosomal intergenic DNA segments studied are listed in table 1. The degree of divergence varies among regions. For example, for the human-chimpanzee pair, the divergence ranges from 0% to 2.66%; figure 1 shows that the majority of the divergence values lie between 0.8% and 1.6%. The average divergences are, respectively, 1.24%, 1.62%, and 1.63% for the human-chimpanzee, human-gorilla, and chimpanzee-gorilla pairs, and they are 3.08%, 3.12%, and 3.09% for the orangutan-human, orangutan-chimpanzee, and orangutan-gorilla pairs. Therefore, the gorilla genome is substantially more different from the human genome than is the chimpanzee genome, and the orangutan genome differs by ∼3% from the human, chimpanzee, and gorilla genomes.
Table 1.
Autosomal DNA Segments Sequenced and Pairwise Divergences among Species
Jukes-Cantor Distancea(%)
Chromosome No.Contig No.Length(bp)H-C H-G C-G H-O C-O G-O
Segments supporting the Homo-Pan clade:
1 T2609 467 1.51 1.30 1.51 3.06 3.28 2.61
2 T1251 447 1.58 3.43 3.67 6.08 6.33 6.08
7 T2012 491 1.44 2.48 3.12 2.91 3.33 2.70
8 T1364 479 1.05 2.55 2.12 3.42 2.98 4.08
10 T1412 431 1.41 2.60 2.36 2.84 3.08 2.36
11 T1419 474 1.28 1.06 1.92 3.01 3.90 3.01
12 T1482 366 1.38 1.66 1.94 3.35 3.35 3.07
14 T2963 350.00 1.73 1.73 2.32 2.32 2.91
15 T2265 458.88 1.54 1.54 2.22 3.12 2.44
15 T2266 477 1.27 2.13 2.78 1.91 2.56 3.43
16 598D4 513.59.59 1.18 1.38 1.98 1.58
17 NT0784 464 1.30 1.30 1.52 4.90 5.13 4.90
17 NT0787 451 1.34 1.34.89 3.17 2.71 2.25
17 NT0801 489.62 1.03.82 3.35 3.35 3.35
17 NT0812 336 1.50 2.42 2.42 3.97 4.29 3.04
17 NT1574 359.84 1.41 1.69 2.26 2.55 1.98
17 NT2294 517 1.17 2.76 1.96 3.36 2.76 2.56
17 NT2984 502 2.22 3.05 2.43 3.26 2.84 2.02
17 NT2986 419 1.20.96 1.20 1.93 2.18 1.45
17 NT2988 468 1.29 2.17 1.73 3.05 3.05 3.05
18 NT0864 373 1.63 3.29 2.18 4.99 3.85 4.42
18 NT0866 443 1.60 1.37 1.14 2.29 1.60 1.37
19 NT0953 479.63.84.84 2.55 2.55 2.12
20 NT2012 507.79 1.39 1.39 3.84 3.84 4.05
20 NT2018 535 1.51 1.70 1.70 3.64 3.83 2.08
20 NT2020 449 1.80 1.35 1.80 3.18 3.64 3.18
20 NT2064 513.39 1.38 1.58 2.58 2.78 3.19
20 NT2085 544 1.49.74 1.11 2.24 2.81 1.86
20 NT2352 511 1.38 1.18.98 2.99 2.79 2.59
20 NT2472 454.89 1.33 1.56 5.01 5.25 5.72
20 NT2560 522 1.15 1.54 1.93 3.52 3.92 3.13
Subtotal 31 segments 14,288 1.20±.09 1.71±.11 1.75±.11 3.17±.15 3.28±.15 2.97±.15
Segments supporting the Homo-Gorilla clade:
3 T2659 557.90.90 1.45 1.82 2.00 2.37
9 T1386 406 2.50 1.49 2.50 4.57 4.05 4.57
11 T2224 371.54.00.54 2.19 1.63 2.19
11U73646463 1.53 1.31 2.41 2.64 3.09 3.54
12 T2927 425 1.67 1.67.95 3.86 2.63 3.12
14 T2960 301.33 1.34 1.00 2.36 2.02 3.05
17 NT2987 442 1.37 1.37 2.30 2.77 3.00 3.00
20 NT1636 535 2.66 2.66 1.51 5.83 4.63 4.63
20 NT2019 411.98 1.97 1.72 2.98 2.73 3.49
20 NT2568 488 1.24 1.24.62 2.71 2.08 2.29
Subtotal 10 segments 4,399 1.42±.18 1.40±.18 1.52±.19 3.16±.27 2.83±.26 3.20±.27
Segments supporting the Pan-Gorilla clade:
10 T2207 478.42 1.27.63 3.64 2.99 3.86
10 T2215 514 1.18 1.57 1.57 2.17 2.98 3.18
10 T2891 542.93.74.93 2.63 2.63 2.63
12 T2906 394.25 1.54 1.54 2.06 2.32 3.64
12 T2924 491 1.64 2.06 1.44 2.47 2.89 2.89
17 T0813 444.91 1.36 1.36 2.75 2.75 3.22
17 276-O15 434 1.87 2.34 2.82 3.06 3.30 4.02
18 NT1506 461.44.65 1.09 2.20 2.65 2.87
18 NT1584 482 2.10 1.26 1.26 4.27 4.71 4.05
18 NT2558 446 1.13 2.51 1.36 4.15 3.68 5.10
19 NT0946 320 1.57 3.18 2.21 1.89.94 2.53
20 NT2563 541 1.67.92 1.11 2.05 2.61 1.86
Subtotal 12 segments 5,547 1.18±.15 1.55±.17 1.40±.16 2.79±.23 2.92±.23 3.30±.25
Overall 53 segments 24,234 1.24±.07 1.62±.08 1.63±.08 3.08±.11 3.12±.11 3.09±.11
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a
C= chimpanzee, G= gorilla, H= human, and O= orangutan.
Figure 1.
Open in a new tab
Distributions of Jukes-Cantor distances among intergenic regions, introns, and Alu s between human and chimpanzee
Table 2 shows additional data from GenBank and the literature. The ∼10-kb region on chromosome 22q11.2 shows a human-chimpanzee divergence of 1.35%, which is somewhat higher than the above estimate of 1.24%. However, the two large contigs from chromosome 12 both show a divergence of 1.2%. Thus, the estimate of 1.2% divergence between human and chimpanzee autosomal intergenic nonrepetitive regions appears to be reliable. This estimate is considerably lower than the estimate of 1.6% from the η-globin pseudogene region (Bailey et al. 1991). The 22q11.2 region shows ∼3% divergence between orangutan and chimpanzee, though a somewhat lower divergence (2.8%) between orangutan and human.
Table 2.
Jukes-Cantor Distances (%) among Noncoding Sequences
Jukes-Cantor Distancea(%)
Sequences Length(bp)H-C H-G C-G H-O C-O G-O
53 autosomal segments 24,234 1.24±.07 1.62±.08 1.63±.08 3.08±.11 3.12±.11 3.09±.11
22q11.2b9,772 1.35±.12…c…2.83±.17 3.06±.18…
Chromosome 12 contig 1d58,593 1.19±.05……………
Chromosome 12 contig 2d46,971 1.20±.05……………
Subtotal for autosomes 139,570 1.21±.03……3.01±.10 3.10±.10…
Xq13.3e10,097.92±.10 1.42±.12 1.41±.12 3.00±.18 2.99±.17 2.96±.17
Xc36f14,425 1.32±.10 1.51±.10 1.57±.11………
Subtotal for X chromosome 24,522 1.16±.07 1.47±.08 1.50±.08………
SMCY g4,758 1.68±.19 2.33±.22 2.78±.25 5.63±.35 6.02±.37 6.17±.37
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a
Repeat elements were excluded. C = chimpanzee, G = gorilla, H = human, and O = orangutan.
b
Zhao et al. (2000).
c
Data unavailable.
d
Chromosome 12 sequences from GenBank. Contig 1: chimpanzee contig AC007214; Contig 2: chimpanzee contig AC006582.
e
Kaessmann et al. (1999).
f
Bohossian et al. (2000); c36 = clone 36.
g
Shen et al. (2000).
With respect to the X chromosome, the Xq13.3 region shows 0.92%, 1.42%, and 1.41% divergences for the human-chimpanzee, human-gorilla, and chimpanzee-gorilla pairs, respectively, whereas the Xc36 region (human X chromosome clone 36, contig NT001194) shows 1.32%, 1.51%, and 1.57% divergences for the three comparisons (table 2). The Xq13.3 region has evolved significantly more slowly than the Xc36 region in the human-chimpanzee comparison, though not in the human-gorilla and chimpanzee-gorilla comparisons. The average distances of the two X-chromosome regions are 1.16% (human-chimpanzee), 1.47% (human-gorilla), and 1.50% (chimpanzee-gorilla), which are slightly lower than the average distances for the autosomal noncoding regions studied. A slightly lower average divergence in X-linked sequences than in autosomal sequences has been noted earlier by Jaruzelska et al. (1999) and Nachman and Crowell (2000 a).
For the Y chromosome, the SMCY region shows 1.68%, 2.33%, and 2.78% divergences for the human-chimpanzee, human-gorilla, and chimpanzee-gorilla comparisons, respectively. These figures are the highest among all the noncoding regions studied. The average distance on the Y chromosome between the orangutan and the human, chimpanzee, and gorilla is 5.94%—about twice the values for autosomal and X-linked noncoding regions.
Introns
The distances among hominoid introns are listed in table 3; because the orangutan data are scanty, only the distances between human, chimpanzee, and gorilla are shown. The average distances are 0.93%, 1.23%, and 1.21% for the human-chimpanzee, human-gorilla, and chimpanzee-gorilla pairs, which are slightly lower than those from the 53 intergenic regions. The human-chimpanzee data are more abundant and the total of 33 loci gives an average distance of 1.03%. Among the loci retrieved, the introns in the β-γ-globin region tend to have a higher divergence than the average. The average intron distances for these globins are 1.89%, 2.16%, and 2.06% for the human-chimpanzee, human-gorilla, and chimpanzee-gorilla pairs. High substitution rates for these genes are also observed for orangutan-human, orangutan-chimpanzee, and orangutan-gorilla pairs (the average for the three pairs is 3.85% [data not shown]).
Table 3.
Distances between Human, Chimpanzee, and Gorilla Introns
Jukes-Cantor Distancea(%)
Locus, Introns Source or Accession Numbers Length(bp)H-C H-G C-G
1q 24-25 region Yu et al., in press9,556.54.99.96
α-fetoprotein precursor, I14M10950, U21916, M38272339 1.19 2.70 2.70
β-1,3-galactosyltransferase, I3AB041413, AB041415, AB041416648 1.25.00 1.25
Complement C4, I1–4M14824, Z31603, Z31599555 1.27 1.09.54
Dopamine D2 receptor, I1AF050737, AF005639, AF005640244 1.24 2.08 1.66
ε- globin, I1–2U23824, AJ002051, M81363975 1.24 1.24.62
Fetal G γ-globin, I 2X06490, X03109, X03111856 2.62 3.22 3.71
Ig C α heavy chain constant region, I1 J002201, X53702, X53703162 2.54 3.85 2.54
Interstitial retinol-binding protein 3, I1AF003990, AF003992, AF003994470.43.86 1.29
Natriuretic protein, IA–BAB037521, AB037522, AB037523770 1.31 1.44 1.18
Phenylalanine hydroxylase gene, I1+7AF003965–66b, AF003968–69, AF003971–72 712.84 1.42 1.13
Protamine 1, I1M60331, L14591, L1458791 2.23 1.11 1.11
Protamine 2, I1M60332, X72968, X71336161 4.48 1.89 3.82
RNA helicase 68KD, I2–4AJ010931, AJ010933, AJ010932574 1.23 1.05.88
T cell receptor gv 10, I1X74798, X86558, X86720112 2.75.91 1.82
Transitional protein 2, I1HSU15422, AF215716, AF215718137 1.47 2.98 1.47
Subtotal 16 loci 16,362.93±.08 1.23±.09 1.21±.09
α-2-HS glycoprotein, I1–6AB038689, AB0386905,301 1.14……
β-globin, I1–2L26475, X02345795 2.17……
Blue opsin, I1–4AF039434, AF039435, U538741,472 1.16……
Cytochrome P450, IBM31664, AF123054407.99……
Decay accelarating factor, I1AB003312, AB003313510 1.19……
Dystrophin gene, I44AF085430, AF0854321,450.83……
Fetal A γ-globin, I3–4M91036, X03110948 3.01……
Glycerol kinase, I1–2AF085433, AF085434, AF085435, AF0854361,582.70……
Haptoglobin 1, I1–4M69197, M844621601.50……
Hypoxanthine phosphoribosyl-transferase, I2+8AF085439, AF085440, AF085441, AF0854421,467.96……
Iduronate sulphate sulphatase, I2+5AF011889, AF085447, AF0854481,498.07……
Inteleukin 2 receptor γ chain, I4+5AF085451, AF085452, AF085453, AF085454731.83……
Lipoprotein lipase, I6+9M76722, Z46493, AF071092-4 2,163.93……
Preproinsulin, I1–2V00565, X61089955 2.02……
Pyruvate dehydrogenase, E1 α subunit, I1–3AF125081, AF1250772,875 1.19……
Pyruvate dehydrogenase E1 α subunit, I9–10AF085457, AF0854591,176 1.20……
Subtotal 16 loci 24,931 1.10±.07……
Overall 32 loci 41,293 1.03±.04……
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a
H= human, C= chimpanzee, and G= gorilla.
b
AF003965-66: from accession number AF003965 to AF003966.
Pseudogenes
In GenBank, pseudogene sequences are less abundant than introns. For the seven loci retrieved (table 4) for human, chimpanzee, and gorilla, the average human-chimpanzee, human-gorilla, and chimpanzee-gorilla distances are 1.64%, 1.87%, and 2.14%, respectively. However, 13 additional loci for human and chimpanzee reduce the human-chimpanzee distance to 1.56% (table 4). On the other hand, for the 6 X-linked pseudogenes available for the human-chimpanzee pair, the average distance is 1.47%, which is somewhat lower than the autosomal pseudogene average, in agreement with the conclusion of Nachman and Crowell (2000 a).
Table 4.
Distances between Pseudogenes
Jukes-Cantor Distancea(%)
Pseudogene Accession Numbers Length (bp)H-C H-G C-G
Autosomal:
α-1,2 fucosyltransferaseU17895, AB006612, AB0066111,045 1.74 2.43 2.23
β-globinX02133, X02135, X021342,146 1.46 1.65 2.27
η-globinU01317, K02542, K0254310,159 1.52 1.49 1.84
Olfactory receptor OR1P1PAF087927, AF101743, AF101763990 1.94 3.62 2.26
Olfactory receptor OR3A5PAF087921, AF101735, AF101756560 1.26 3.01 2.82
Olfactory receptor OR3A4PAF087920, AF101734, AF101756806 1.76 2.92 2.27
Olfactory receptor OR1D3PAF087919, AF101733,940 3.04 2.49 4.27
Subtotal 7 loci 16,646 1.64±.10 1.87±.11 2.14±.11
γ-cytoplasmic actinAF196978, AF196999622 1.63……
α-enolaseAF196980, AF1970011,034 1.17……
CAMP-dep protein kinase regulatory subunit RIαAF196994, AF197015445.9……
CAMP-dep protein kinase regulatory subunit RIαAF196995, AF197016424 1.43……
Connexin 43-kD proteinAF196981, AF197002769 1.44……
Lanosterol 1,4–α-demethylaseAF196989, AF197010359.84……
Cytochrome bAF196982, AF197003718.56……
α-1,3-galactosyltransferaseM60263, AF1970071,048.77……
Interferon-induced 56-kD proteinAF196987, AF197008713 1.27……
Malate dehydrogenaseAF196990, AF197011942 1.07……
NADH dehydrogenaseAF196991, AF1970121,150 1.14……
Proliferation-associated geneAF196992, AF197013881 1.72……
GPI-anchor synthesis geneAF196993, AF197014864 1.88……
Overall 20 loci 26,841 1.56±.08……
X-linked:
Adaptor proteinY09846, AF197017961 1.79……
C4-sterol methyl oxidaseAF196983, AF197004854.59……
Elongation factor 1- αAF196984, AF1970051,010 1.30……
Ferritin-like geneAF196985, AF197006551.73……
HTLV-1 enhancer–binding proteinU03712, AF197018929 2.96……
Malate dehydrogenaseAF196990, AF197011942 1.07……
Overall 6 loci 5,247 1.47±.17……
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a
H= human, C= chimpanzee, and G= gorilla.
Alu s
In a chromosome 12 region for the human and chimpanzee pair, 54 homologous Alu sequences are available (GenBank accession numbers AC007214, AC006582, AC005294, AC007458, AC007286, and AC011604). The individual distances are highly variable (fig. 1 c), but the majority of them are higher than the distances in the intergenic regions (fig. 1 a). The average distance is 2%.
Coding regions
The K S and K A values and the amino acid distances (p) among hominoids are shown in table 5. These values fluctuate greatly among genes, presumably because of stochastic effects, variation in mutation rate, and variation in selection pressure. For example, the K S values between human and chimpanzee range from 0.00% to 4.02%, and the K A values from 0.00% to 3.68%. The average K S values are 1.11% for human-chimpanzee, 1.48% for human-gorilla, and 1.64% for chimpanzee-gorilla. The K S values between orangutan and human, chimpanzee, and gorilla are 2.98%, 3.05%, and 2.95%, respectively. The average K A values are 0.80%, 0.93%, 0.90%, 1.96%, 1.93, and 1.77% for the human-chimpanzee, human-gorilla, chimpanzee-gorilla, human-orangutan, chimpanzee-orangutan, and gorilla-orangutan pairs, respectively. On the other hand, the average amino acid divergences for the six pairs above are 1.34%, 1.58%, 1.65%, 3.60%, 3.63%, and 3.45%, which are slightly higher than the respective intergenic distances.
Table 5.
Coding-Region Distances between Human, Chimpanzee, Gorilla, and Orangutan
K S (%)/K A (%)/AA Distance (%)a
Locus Length(bp)H-C H-G C-G H-O C-O G-O
Atrophin-related protein DRPLA 858.00/.51/.00 1.92/.87/1.41 1.92/1.02/1.41 2.71/.72/2.11 2.71/.88/2.11 2.26/1.24/2.11
Brain natriuretic protein 1,140.73/1.51/1.32 1.35/1.33/1.84.53/1.32/1.57 3.54/3.07/4.74 2.87/2.79/5.25 3.39/2.53/4.99
BRCA 1 3,423.29/1.03/2.19.86/.93/2.02.74/.59/1.23 2.03/1.85/3.77 1.90/1.41/2.89 2.49/1.41/2.89
CC chemokine receptor 5 1,056 1.08/.56/.57 1.75/.74/1.14 2.30/.41/.57 3.44/.59/.85 2.85/.26/.28 3.53/.15/.28
Complement C5 α receptor 1,023.53/.43/.88 3.21/.40/.88 3.23/.83/1.76 4.49/1.84/3.82 4.79/2.40/4.71 7.04/1.80/3.82
Connexin-36 709 2.15/.00/.00 2.15/.00/.00.80/.00/.00 3.63/.00/.00 2.89/.00/.00 2.89/.00/.00
Cytochrome oxidase subunit 4 435.00/.65/.00.00/1.02/.69.00/.37/.69.64/2.51/3.47.64/1.85/3.47.64/2.23/4.17
Eosinophil cationic protein 483.85/.83/1.88 1.09/1.00/1.88 2.22/1.16/2.50 1.42/7.53/13.75 2.01/7.89/13.75 2.76/7.90/14.38
Eosinophil-derived neurotoxin 486 2.65/.24/.62 3.26/.32/.62 3.72/.48/1.24 7.96/3.20/6.83 7.55/3.46/7.45 8.79/3.48/7.45
α-1,2-fucosyl-transferase 1,047 1.85/2.63/4.35 1.43/1.81/2.32 2.27/2.03/4.35 3.13/2.78/5.51 3.62/3.39/7.54 3.73/2.26/5.80
ε-globin 444.65/.00/.00.65/.00/.00.00/.00/.00 1.31/.62/1.36 1.98/.62/1.36 1.98/.62/1.36
γ-globin 888 1.85/.00/.00.00/.35/.68 1.85/.35/.68 2.27/.21/2.72 2.29/1.21/2.72 2.26/1.21/2.72
5-hydroxy-tryptamine receptor 1A 1,260.34/.22/.72.22/.45/.24.57/.25/.48.44/1.20/.72.85/.82/.95.67/1.01/.48
5-hydroxy-tryptamine receptor 1F 1,116.00/.29/.00.24/.72/.81.24/.43/.81.49/.64/.81.49/.35/.81.73/.49/1.62
5-hydroxy-tryptamine receptor 2A 732 3.22/.21/.41 2.72/.00/.00 2.58/.21/.41 2.59/.16/.41 2.99/.37/.82 1.97/.16/.41
Ig α heavy chain constant region 1,059 3.03/2.87/5.14 1.38/2.13/2.57 3.13/3.62/6.57 4.69/5.21/8.29 5.61/6.22/1.57 4.37/5.27/8.86
Igκ constant region 252.00/.00/.00 3.60/2.36/4.76 3.60/2.36/4.76 4.85/7.88/15.48 4.85/7.88/15.48 6.15/8.05/15.48
Interleukin-8 receptor type A 1,089.25/.64/1.14 1.25/1.40/2.57.99/1.03/2.29 1.25/1.40/2.57.99/1.03/2.29.00/.00/.00
Interleukin-8 receptor type B 1,056.56/.82/.57 2.60/1.06/2.55 2.60/1.30/2.55 3.30/1.53/4.26 3.24/1.95/4.26 2.88/1.87/3.98
Leptin 431 1.71/.75/.69 1.31/.37/1.38 1.71/.66/.69.91/.36/2.76 1.45/.84/3.42.92/.52/4.14
Lysozyme 447 2.85/.00/.00 1.26/.35/.68.92/.42/.68 1.91/.97/2.03 1.28/.97/2.03.46/1.42/2.70
β 2-microglobin 369.78/.70/1.68.78/.70/1.68.00/.00/.00 1.59/2.14/5.04.78/.75/3.36.78/.75/3.36
β-nerve growth factor 721 2.48/.60/3.17 2.92/.60/2.38 1.24/.43/1.85 4.13/.82/6.08 2.42/.65/5.29 2.86/.65/5.03
N-formyl peptide receptor 1,044 1.68/.42/.87 2.26/1.04/2.31 2.26/1.22/2.60 3.38/1.31/2.89 2.81/1.19/2.60 3.97/1.58/3.47
N-formyl peptide receptor-like 2 1,047 1.22/.48/1.15 2.76/1.19/2.58 3.34/.90/2.01 5.41/2.15/4.87 6.61/1.87/4.30 2.95/1.51/3.44
Low-affinity N-formyl peptide receptor 1,044 1.36/.51/.86 1.09/.53/.86 2.48/.77/1.72 4.82/1.72/3.16 6.19/2.00/4.02 5.41/1.48/2.87
Olfactory receptor 93 987 1.17/1.56/3.40 3.08/2.14/4.32 3.07/1.12/2.46 5.39/3.25/5.86 5.09/2.22/2.46 5.80/2.14/4.62
Homeobox protein OTX 1 729.00/.00/.00.37/.58/.00.37/.58/.00.74/1.02/.41.74/1.02/.41.37/.37/.41
Homeobox protein OTX 2 342.40/.24/.00.00/.00/.00.40/.24/.00.00/.48/.00.40/.71/.00.00/.48/.00
Protamine 1 156 4.02/3.68/10.81 1.30/3.31/8.11 2.57/2.59/8.11 6.76/9.95/18.92 11.46/11.00/24.32 8.31/1.46/21.62
Protamine 2 309 2.37/3.64/6.86 1.57/3.86/6.86.79/2.08/3.92 6.24/6.14/11.76 6.22/5.51/10.78 5.35/6.98/12.75
RNase k6 453 1.28/.34/.67 1.95/.28/.67.64/.60/1.33 5.38/.85/1.33 4.02/1.20/2.00 4.69/1.11/2.00
Voltage-gated Na+channel α subunit 1,092 1.13/.00/.00.75/.00/.00.76/.00/.00 1.13/.14/.27 1.02/.14/.27.76/.14/.27
T cell receptor gv10 357.83/1.30/2.52.83/.86/1.68 1.71/.43/.84 6.53/4.58/9.24 7.60/4.11/8.40 5.62/3.65/7.56
Urate oxidase 162.00/.95/.00 3.24/.00/1.92 3.24/.95/1.92.93/3.82/5.88.93/4.83/6.00 4.13/3.82/7.84
Zinc finger protein 75 273.00/.00/.00.00/.00/.00.00/.00/.00 4.44/6.26/17.78 4.44/6.26/17.78 4.44/6.26/17.78
Zinc finger protein 80 822 2.24/1.41/3.30 1.44/2.27/4.76 2.61/2.84/6.23 3.94/1.85/4.03 5.19/2.62/5.86 2.81/2.70/5.86
Overall 29,342 1.11/.80/1.34 1.48/.93/1.58 1.64/.90/1.65 2.98/1.96/3.60 3.05/1.93/3.63 2.95/1.77/3.45
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a
K S = number of substitutions per synonymous site; K A = number of substitutions per nonsynonymous site; AA distance = amino acid difference per residue site. C= chimpanzee, G= gorilla, H= human, and O= orangutan.
Comparison of different regions
Figure 1 shows the distributions of distances between human and chimpanzee for the autosomal intergenic regions, introns, and Alu s. The distributions are approximately normal for the intergenic regions and introns. About 60% of the intergenic distances and ∼56% of the intron distances fall in between 0.8% to 1.6%. The Alu distribution is more dispersed and only 50% of the distances fall in between 0.4% and 2.0%, but the majority of the Alu s have diverged more than the intergenic regions and introns.
On average, the Alu sequences give the largest distances, followed in order by pseudogenes, intergenic regions, synonymous sites, introns, and nonsynonymous sites. For example, for the human-chimpanzee pair, the average distances are 2% (Alu s), 1.56% (pseudogenes), 1.24% (autosomal intergenic regions), 1.11% (synonymous), 1.03% (introns), and 0.80% (nonsynonymous). This order is consistent with Li’s (1997) conclusion.
Phylogeny
When the 53 autosomal segments are considered together (concatenated), the neighbor-joining tree (Saitou and Nei 1987) supports the Homo-Pan clade with a 100% bootstrap value (see the topology in fig. 2). When each segment is considered individually, 31 segments support the Homo-Pan clade, 10 support the Homo-Gorilla clade, and 12 support the Pan-Gorilla clade. From these data, we can compute the likelihood ratio of the Homo-Pan clade to the trichotomy (null) hypothesis (Wu 1991):
where a, b, and c are the numbers of loci supporting topology A (Homo-Pan), topology B (Homo-Gorilla), and topology C (Pan-Gorilla), respectively, and n is the total number of loci studied. For n=53, a=31, b=10, and c=12, we have R=1,105.8, which is much larger than the threshold value (17.2 in table 3 of Wu 1991), so the probability for accepting the Homo-Pan clade is practically 1. In addition, the Homo-Pan clade is also supported by the coding region data set (table 5). Thus, in agreement with the studies of Ruvolo (1997) and Satta et al. (2000), there is very strong support for the Homo-Pan clade.
Figure 2.
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Phylogeny of hominoids. The branch lengths (Jukes-Cantor distances) are computed under the assumption of rate constancy and used for estimating divergence dates.
Molecular Clock
The sequence data are also useful for testing the molecular clock (rate constancy) hypothesis among the hominoids. For this purpose we can use Wu and Li’s (1985) relative rate test. This test provides the mean and standard error (SE) of the rate difference between two lineages, using a third (outgroup) lineage as a reference; when the mean/SE ratio is ⩾2, the difference is significant at the 5% level. For the 31 intergenic segments that support the Homo-Pan clade (table 1), the gorilla can be used as a reference to test the rate difference between the human and chimpanzee lineages. Since the average distances for the 31 segments are 1.71% and 1.75% for the human-gorilla pair and the chimpanzee-gorilla pair, respectively, the difference between the two distances is clearly not significant and the molecular clock holds. If the orangutan is used as the reference, the distances for the orangutan-human, orangutan-chimpanzee, and orangutan-gorilla pairs are 3.17%, 3.28%, and 2.97%, respectively, and the only significant difference among the three values is between the second and the third (0.31% ± 0.12%, computed from Wu and Li’s formula), implying a significantly slower rate in the gorilla lineage than the chimpanzee lineage. However, this is the only significant difference among all the comparisons in table 3. Indeed, when the 53 segments are considered together and the orangutan is used as the reference, rate constancy is found to hold for the three lineages, because the three distances are very similar (3.08%, 3.12%, and 3.09%; see table 1).
Table 2 provides more data for testing the molecular-clock hypothesis. For the 22q11.2 region, the distance between chimpanzee and orangutan is significantly larger than that between human and orangutan (3.06%-2.83%=0.23%; SE = 0.11%). However, when all the autosomal sequence data are considered together, the difference becomes nonsignificant. For the Y chromosome region (SMCY), the human lineage has evolved significant more slowly than the chimpanzee and gorilla lineages. For the coding regions shown in table 5, the synonymous distances between orangutan and the other three species are 2.98%, 3.05%, and 2.95%, and the nonsynonymous distances are 1.96%, 1.93%, and 1.77%, none of which (except for the difference between 1.96% and 1.77%) deviates significantly from an equal rate of evolution among the human, chimpanzee, and gorilla lineages. So, overall rate constancy holds well, except that the SMCY region has evolved more slowly in the human lineage than in the chimpanzee and gorilla lineage.
Divergence Times
We are interested in estimating the date for the Homo-Pan divergence (T HC) and the date for the gorilla divergence (T G). The sequence data from the 53 autosomal intergenic segments in table 1 and the synonymous distances in table 5 are suitable for this purpose because, for these two data sets, rate constancy seems to hold among the human, chimpanzee, and gorilla lineages (see above). Assuming rate constancy, we compute the branch lengths in figure 2 for the data set of 53 intergenic regions. Assuming that the speciation time (T) of Pongo is 12 to 16 million years ago (Goodman et al. 1998; D. Pilbeam, personal communication), we obtain the Homo-Pan divergence time as T HC = (0.62/1.55)T = 4.8 to 6.4 million years and the gorilla divergence time as T G = (0.82/1.55)T = 6.3 to 8.5 million years.
The internodal time span (T IN) between the gorilla speciation and the Homo-Pan common ancestor is 0.20/0.62 = 32% of the divergence time between the human and chimpanzee lineage or T IN = (6.3 to 8.5)−(4.8 to 6.4) = 1.5 to 2.1 million years. For the synonymous distances in table 5, the estimates become T HC = 4.5 to 5.9 million years, T G = 6.3 to 8.3 million years, and T IN = 1.8 to 2.4 million years. These estimates are very similar to those from the 53 intergenic regions. Taking average of the two sets of estimates, we obtain T HC = 4.6 to 6.2 million years, T G = 6.2 to 8.4 million years, and T IN = 1.6 to 2.2 million years.
Effective Size of the Ancestral Population
When three species are fairly closely related to each other, the tree obtained from a set of DNA sequence data (known as the gene tree) may not be congruent with the true tree that represents the two speciation events (known as the species tree). Hudson (1983) and Nei (1986) showed that the probability for the gene tree obtained from a set of sequence data to be congruent with the species tree is given by
where t=T IN is the internodal time span between the two speciation events and is expressed in units of 2 N e generations, where N e is the effective size of the population in the time span between the two speciation events. Formula (1) implicitly assumes that an incongruent gene tree can arise because of the sharing of an ancestral polymorphism between species 1 (or 2) and species 3, though species 1 and 2 are more closely related to each other. It depends on N e, because the smaller the N e, the faster the decrease with time in the probability for species 1 (or 2) to share a polymorphism with species 3. Applying Wu’s (1991) maximum-likelihood estimation procedure to a set of data from independent loci, we can equate P with the proportion of loci that support the species tree. For example, if a loci among the n loci studied support the species tree, then the maximum-likelihood estimate of P is a/n. If t and P are both known, one can estimate N e.
Because formula (1) assumes that an incongruent tree arose due to sharing of ancestral polymorphism between “wrong” species, a parsimony analysis is more appropriate than a distance analysis. Of the 53 intergenic segments, 24 segments support the Homo-Pan clade, 7 support the Homo-Gorilla clade, 2 support the Pan-Gorilla clade, and 20 segments give no resolution (i.e., they do not support any of three alternative trees). For the coding loci listed in table 5, the corresponding numbers are 12, 3, 4, and 16. In this analysis, P1 and P2 are pooled together as one locus because they are linked, and so are ε-globin and γ-globin; therefore, there are only 35 “independent loci” instead of 37. Taking the two sets of data together and excluding loci that give no resolution, we have a=24+12=36, n=33+19=52, and P=36/52=69%. From this value and formula (1) we estimate t as t=-ln[(3/2)(1-P)]=0.766(2 N e generations). We estimated above that the internodal time span is t = T IN = 1.6 to 2.2 million years. Assuming a generation time of g=15 to 20 years, we obtain N e=t/(2×0.766 g)≈52,000 to 96,000. If we use the neighbor-joining method, we obtain a considerably larger estimate of N e. For example, for the 53 intergenic segments, we obtain a=31, n=53, P=.59 and N e = 84,000 to 150,000.
Discussion
Genomic Divergences
We have seen that among the types of sequences included in this study, Alu s have, on average, evolved at the highest rate. This is because Alu sequences are not subject to functional constraints, and they contain many CpG dinucleotides, which have a mutation rate about 10 times higher than the genomic average, because of the strong tendency for the C in CpG to mutate to T (Labuda and Striker 1989; Nachman and Crowell 2000 a). In fact, there is a 62% correlation between the rate of substitution in an Alu (fig. 1) and the number of changes at CpG dinucleotide sites in the sequence (data not shown). This is a good example supporting the ideas that the mutation rate in a region may depend on its sequence context and that, when the functional constraints in a sequence are removed, the sequence may evolve at a higher rate than the genomic average.
We noted that pseudogenes show the second-highest rate among the types of autosomal sequences included in this study. Like Alu s, a pseudogene may also contain more CpG dinucleotides than the average for noncoding regions, though not at a frequency as high as that in Alu s. To see if this is, in fact, the case, we computed the CpG frequencies in the 53 intergenic segments included in table 1 and in the 37 genes included in table 5 (table 6). The frequency of CpG in a sequence is computed as the number of CpGs in the sequence, divided by the length of the sequence, minus 1, and the expected frequency of CpG is computed as f C f G, where f C and f G are, respectively, the frequencies of C and G in the sequence. We note that the observed frequency of CpG is much lower in the 53 (noncoding) segments (0.69%) than in the gene sequences (2.77%). The difference is highly significant, even when the expected frequencies are taken into account (table 6). A pseudogene may also contain some other sequence contexts that can confer a higher-than-average mutation rate.
Table 6.
Frequencies of CpG Dinucleotide in Coding and Noncoding Regions[Note]
CpG Frequency
Region No. of CpG Dinucleotides Observed Observed Expected
37 genes 781.0277.0674
53 noncoding segments 169.0069.0426
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Note.— χ 2=41.3, P=.0001.
The above observations suggest that the mutation rate in a functional region often may be higher than the average mutation rate in its nearby introns because of its sequence context, which has been maintained by functional constraints. Thus, the observation of a slightly higher substitution rate at synonymous sites than in introns might be due, in part, to a slightly higher rate of mutation in coding regions than in introns, though it probably also indicates slightly stronger functional constraints in introns than at synonymous sites. The observation that both introns and synonymous sites have on average evolved more slowly than intergenic regions suggests that both introns and synonymous sites are subject to some functional constraints.
For the above reasons, intergenic regions are more suitable than pseudogenes, introns, and synonymous and nonsynonymous sites for estimation of the degree of sequence divergence between hominoid genomes. In fact, the extensive data from intergenic regions suggest that the human and chimpanzee genomes differ by only ∼1.2%, rather than the 1.6% divergence estimated from the η- globin pseudogene region. The η-globin pseudogene region also shows 3.7%, 4.9%, and 4.4% divergences for the orangutan-human, orangutan-chimpanzee, and orangutan-gorilla pairs, which are considerably higher than the ∼3% divergence for these species pairs estimated from the 53 intergenic regions in this study. However, it should be emphasized that our aim is to estimate the genomic divergences among the hominoids in unique noncoding regions. The divergences in repetitive sequences among these genomes might be substantially higher because of a higher mutation rate and frequent deletion and insertion events. In particular, as Alu s have a substantially higher rate of mutation than the average genome (see above), the high content (∼10%) of Alu s in the hominoid genome should have accelerated the divergence between the human and chimpanzee genomes. This might partly explain the higher estimates of genomic divergence between the human and chimpanzee genomes in the literature. Although the DNA hybridization study by Sibley and Alhquist (1987) tried to exclude rapidly reassociating DNA, this procedure is unlikely to delete all repetitive elements, because such elements are very abundant and are highly dispersed in the hominoid genome.
Molecular Clocks and Divergence Dates
There has been strong evidence supporting the hominoid slowdown hypothesis (Goodman 1961), which postulates that the rate of molecular evolution has become slower in the hominoid (apes and humans) lineage since its separation from the Old World monkey lineage (for a review, see Li 1997). For some regions, a further slowdown has occurred in the human and chimpanzee lineages, in comparison to the rate in the gorilla lineage. These include the η-globin pseudogene region (Bailey et al. 1991; Graur and Li 2000), the Xq13.3 region (table 2), the last intron of the ZFX gene (Jaruzelska et al. 1999), and introns 7 and 44 of the Duchene muscular dystrophy gene (Nachman and Crowell 2000 b). However, for some regions, the rate in the gorilla lineage is significantly slower than those in the human and chimpanzee lineages. For example, as noted above, the average rate for the first 31 segments in table 1 is significantly lower in the gorilla lineage than in the chimpanzee lineage, and the average nonsynonymous rate for the genes in table 5 is significantly lower in the gorilla lineage than in the human lineage. So, there is no strong trend toward a slowdown in the human or chimpanzee lineage. In fact, the data in tables 1 and 5 show that the molecular-clock hypothesis, on average, holds well among the human, chimpanzee, and gorilla lineages for the intergenic regions and the synonymous sites. This observation suggests that the generation time effect is weak and often may not be discernable. This suggestion seems reasonable, in view of the fact that the generation time in the human lineage was only slightly longer than those in the chimpanzee and gorilla lineages (see later). For Y-linked sequences, the mutation rate is higher, and the generation time effect is easier to detect. In fact, the SMCY region is seen to have evolved faster in the chimpanzee lineage than in the human lineage (see table 2).
As the molecular-clock hypothesis seems to hold well for the 53 intergenic regions and for the synonymous sites used in this study, these regions are suitable for estimating the divergence dates among the human, chimpanzee, and gorilla lineages. The only uncertain aspect is the date of the orangutan speciation event. If our assumption of 12 to 16 million years is close to the true date, then our estimates should be reliable. Our estimates are similar to those of Goodman et al. (1998) of 7 million and 6 million years ago for the gorilla branching node and the human-chimpanzee divergence, respectively. In any event, our data suggest that the internodal time span between the human-chimpanzee divergence and the gorilla speciation event is about one-third of the divergence time between the human and chimpanzee lineages. This estimate is independent of the calibration of the molecular clock and is in between the estimates of 60% from the mitochondrial data and 10% from the η-globin pseudogene sequence data.
Population-Size Estimation
Generation time
The generation time is an important factor in our estimation of the ancestral population size. According to Nowak and Paradiso (1983), both sexes of chimpanzees reach puberty at age 7 years. But females usually do not give birth until they are 13 years old, and males are not totally integrated into the social hierarchy until they are 15 years old. Reproductive capability in females can last at least until age 40 years (Nowak and Paradiso 1983), and chimpanzees can live to age 50 or even 60 years. This is concordant with Reynolds and Reynolds’s observation (1965) that wild chimpanzees could live to age >40 years and remain healthy. On the other hand, gorillas reach sexual maturity at age 8 years for females and 11 years for males (Nowak and Paradiso 1983). However, females in the wild usually give birth for the first time at age ∼10 years and live to age 30 to 40 years in the wild. In summary, chimpanzees and gorillas start to reproduce a couple of years earlier than humans and have a somewhat shorter lifespan than humans. Although the generation time in a modern human society can be >30 years (Tremblay and Vézina 2000; Sigurdadóttir et al. 2000), the generation time in the long history of human evolution is commonly taken to be 20 years (e.g., Nei and Graur 1984). For this reason and in light of the data cited above, we assume a generation time of 15 to 20 years for the common ancestor of chimpanzees and humans.
Ancestral population size
We have followed Ruvolo (1997) in the use of multiple data sets to estimate the effective size (N e) of the ancestral population before the human-chimpanzee divergence. Because of limited data availability, Ruvolo used only 14 independent coding loci. Among them, 11 loci supported the Homo-Pan clade, and Ruvolo obtained an estimate of N e = 35,000 to 65,000. Her estimate is considerably lower than ours. There are two possible reasons for the difference. First, the number of loci used in Ruvolo’s study was small, so the estimate had a large standard error. Second, as Ruvolo (1997) pointed out, the 14 loci included one mitochondrial locus, one X-linked locus, and one Y-linked locus, all of which tend to give a lower estimate of N e; this is because the effective population sizes for the mtDNA, a Y-linked locus, and an X-linked locus are only N e/4, N e/4, and 3 N e/4, respectively, instead of N e for an autosomal locus. If we exclude these three loci, the proportion of loci supporting the Homo-Pan clade decreases from 11/14=0.79 to 9/11=0.73, which is not significantly different from our value of 0.69.
Note that formula (1) assumes no new mutations. This assumption tends to overestimate P, because new mutations in the internodal time span would produce shared polymorphisms between species 1 and 2. For this reason formula (1) tends to underestimate N e. However, because the number of loci used is still small, our estimate of N e = 52,000 to 96,000 should be taken with caution.
This caution notwithstanding, as our estimate of the effective population size of the common ancestor of human and chimpanzee is about 5 to 9 times higher than the estimate (∼10,000) of the effective population size of humans from various genetic polymorphism data (e.g., Nei and Graur 1984; Takahata 1993; Zhao et al. 2000), the human lineage apparently has undergone a significant reduction in effective population size since its separation from the chimpanzee lineage. It is not clear why this reduction has occurred, but one possibility might be that the human lineage has gone through many local extinction and recolonization events; such events can greatly reduce the effective size of a species (Maruyama and Kimura 1980).
Acknowledgments
This study was supported by NIH grants GM55759 and GM30998. We thank Ning Yu for help, and two anonymous reviewers, J. Hey, M. Jensen-Seaman, D. Pilbeam, M. Ruvolo, and N. Takahata for suggestions.
Electronic-Database Information
Accession numbers and URLs for data in this article are as follows:
BLAST server,
Genome Channel, (for 53 DNA segments studied)
RepeatMasker,
Silver Project Home Page, (for accession numbers of coding regions)
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Articles from American Journal of Human Genetics are provided here courtesy of American Society of Human Genetics
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188440 | https://app.myeducator.com/reader/web/1421a/10/am6yn/ | Skip to book tools
10.3 Odds and Probability
Logistic regression estimates the probability of belonging to each class. In the binary case we get an estimate of the probability of the outcome variable being true. We also get the probability of the outcome variable being not true.
Logistic Regression is designed to produce the odds of a record belonging to a class. These odds can then be converted to probabilities. Before explaining more about how logistic regression works, let’s review how odds are similar but different from probabilities.
Formulas for Odds and Probability
Assume that in a sample of one thousand people, 400 have new phones. The remaining 600 do not have new phones. Calculate the odds of a person having a new phone. Then calculate the probability of a person having a new phone.
The components to calculate odds and probabilities are the same. They are just combined in a different way:
Since odds and probabilities are ratios, the numbers in the ratio can be simplied by dividing by a common denominator. In the phone example, the ratio simplies to 2 for, 3 against, and 5 total.
Odds are a ratio of the number of fors (or successes) to the number of againsts (or failures). The equation for odds is for:against (e.g., 1:6). In the phone example, the odds of having a new phone are 2:3. Another common way odds are expressed is 2 to 3 odds for having a phone.
Data mining tools, however, express odds by converting the ratio to a single number. This is done by dividing the for number on the left and the against number on the right by the number on the right. For example, 2:3 can be converted to single number form by dividing both sides by the number on the right 2/3:3/3 = .667:1. The one on the right side of the colon is implied, so this simplies to 0.6667, which is the single number form of the odds. As another example odds of 6:1 can be converted into single number form of 6. 1 to 1 odds are even odds, which mean the sample is evenly split between fors and againsts.
The equation for probability is #for / (#for + #against), which is also expressed as #successes / (#successes + #failures). For the phone example above, the probability of a person having a new phone is 2/5 = 0.40.
Converting between Odds and Probability
The examples below show how to convert from odds to probability and from probability to odds.
The following video walks through the process of achieving probabilities through logistic regression.
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188441 | https://www.upperelementarysnapshots.com/2017/12/comparison-problems-using-tape-diagrams.html | The Teacher Studio
Comparison Problems: Using Tape Diagrams to Represent Math Thinking
Word problems are a key part of rigorous standards everywhere, but teachers and students alike often dread tackling them. Even textbooks sometimes save them for the bottom few problems on a page or a separate lesson toward the end of a chapter. The thing is--word problems (at least good ones!) are the "real world" part of math. One type of word problems, comparison problems, can be particularly challenging for students. Let's look at some teaching tips that might make them more accessible!
1. Critical Reading of Math Problems
As teachers, we should always be striving to help our students understand that the skills we are teaching are them are FOREVER...not just to complete a math page or worksheet. One skill that we really want to make sure our students understand is the need to critically read math problems to figure out what is being asked, what information is given, and to make a plan for solving. So often we do the thinking for our students. Just look at how many of our math books are organized. A lesson entitled "Solving Addition Stories" doesn't leave much room for student thinking, does it? Providing students with a constantly spiraling variety of problems forces them to think for themselves, learn to look for key information in problems, and make solution decisions accordingly. One idea? Use highlighters to find important information. Underline the question. One thing I do NOT recommend? Looking for key words like "fewer" or "total". These words may seem like a quick fix for students...but they can lead them down the wrong path. How? How about this problem... "Larry has 14 baseball cards. This is 25 fewer than his sister Becca has. How many cards does Becca have?" By teaching "fewer" as a signal to subtract, a student will certainly not think through this problem correctly!
2. Visualization and Modeling
One strategy that can really help students make sense of problems is to be able to visualize and draw models of different problem types. Comparison problems--sometimes represented with "tape" or "strip" diagrams are a GREAT way to help students visualize the math! I thought I'd share a few ways that these can be super helpful for students--whether used as whole class lessons or for intervention groups.
Using a blank "template" of a tape diagram with manipulatives can really help students see the comparison that appears in a problem. This diagram shows a visual representation of the following problem:
"Bill has 13 candies. If Sarah has 5 more than Bill, how many candies will Sarah have?"
Now...this is a VERY simple example and is perfect to use with intermediate students who really need to back up and see the actual manipulation of the numbers.
As students get more adept at these problems, you might see that a sketch with only numbers placed in the diagram is appropriate. Check out this lesson where we "filled" a diagram and then brainstormed a ton of different questions that could work with this problem. Again, part of making sense of problems is realizing that the QUESTION matters...and we need to always be looking for what that question is!
Another strategy to get students really visualizing is to take a blank diagram and create different stories to go along with them...help students get flexible with their thinking.
Once you have worked with the class on this--send them loose to try some on their own! Share with partners or put some great examples under the document camera to look at all the different ways that students found to visualize this math situation.
3. Tips for Additive Comparison Problems
The beauty of comparison problems is that they actually get students thinking about all FOUR operations--and how addition and subtraction are related and how multiplication and division are related--and even how those inverse operations can be used in solving. Although we may call some problems "additive" comparison, the reality is that the information can often be represented by subtraction as well. These problems really form the foundation for later work in algebra, so it's so important that we help students recognize what is happening in these problems--and to find different ways to model them.
4. Tips for Multiplicative Problems
As students develop their math skills, we know they need to be able to deal with "groups of" and division problems as well. This also ties directly to fraction work ("1/3 of ..." as opposed to "3 groups of..."). Helping students SEE the difference between 3 groups of five and 3 MORE than five is such a valuable exercise.
Interested in trying the activities pictured above? Here you go!
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Math, Problem Solving, The Teacher Studio
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All content © Upper Elementary Snapshots • Template by Georgia Lou Studios • Blog Design by Chalk & Apples Design |
188442 | https://edu.rsc.org/download?ac=511946 | 1 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Identifying ions Practical video Supporting resources Registered charity number: 207890 Contents Teacher notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 How to use this video . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Notes on running the practical experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Key terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Common misconceptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Intended outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Additional resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Pause-and-think questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Pause-and-think questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Follow-up worksheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Follow-up worksheet: answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources 1 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Teacher notes These resources support the practical video Identifying ions, available here: rsc.li/3dhnn5B The value of experiencing live practical work cannot be overstated. Numerous studies provide evidence of its value in terms of learner engagement, understanding, results and the likelihood of continuing to study chemistry or work in a related field. This video can be used to complement live practical work, as well as helping learners to understand the methods, equipment and skills when they cannot access the lab. How to use this video The video and additional resources are designed to be used flexibly, but some suggestions follow. Flipped learning Learners view the video ahead of the live practical lesson to help it run more smoothly and keep objectives in focus. This may also help build confidence for some learners and improve their outcomes in the lesson. Use questions from the set provided as part of the preparation task. Consolidation and revision Learners view the video after the practical – this may be directly after the lesson or learners can return to it as part of revision for examinations. Revisiting a practical with a different focus A practical experiment can support many learning outcomes. Focussing on just one or two of those in a lesson will help ensure that the aims are achieved. The video could be used to revisit the experiment with a different focus. Home learning Whether it is remote teaching, homework, or individual learner absence, the video provides an opportunity to engage with a practical experiment and the associated skills when learners are not in the lab. Other tips • Provide your own commentary Mute the voice over and provide your own commentary. This will allow you to better engage with learners and adapt to the needs and objectives of your lesson. • Use questions A set of pause-and-think questions are provided in two formats, one for teacher-led questions and discussion and a student worksheet which can be used independently by learners. Select from these or create your own questions to help engage learners and target specific aims. Notes on running the practical experiment Technician notes including the equipment list and safety notes are available as a separate document here: rsc.li/3dhnn5B. If you are planning to carry out the practical in the classroom, you will need to carry out your own risk assessment.
The flame tests (wooded splint method) take about 10 minutes to carry out and it is safe for learners to work in pairs.
The microscale sodium hydroxide test for positive ions reaction takes around 10 minutes and it is safe for learners to work in pairs. This should be carried out on either the printable sheet (in this booklet) or on the integrated instruction sheet. In both cases you will need to either put the printed sheet into a plastic wallet or laminate it. TIP: Printing the microscale sheet/integrated instructions onto buff coloured paper will make it easier to see when a white precipitate has been formed.
TIP: Once completed learners can take a photo of their results using a mobile phone or tablet; so that they can clear away immediately. This will help avoid learners spilling chemicals on their results table. 2 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Allow 30 minutes for the testing of negative ions, 10 minutes for each test. You may wish to carry out this experiment in a separate lesson. This will give you time to go over the theory in the same lesson.
Once all the tests have been completed, provide the class with an unknown solution X and ask the students to identify either the positive ion present, the negative ion present or both. Further practical activities and context Details of alternative methods to carry out the flame tests can be accessed from rsc.li/2ZpnROO. This activity makes a good plenary or starter; asking students to name the metal ion as they see the different flame colours. The flame tests infographics is a great reference sheet and also explains the chemistry behind the flame colours. The chemistry of fireworks infographic and sparklers infographic provides a nice teaching context. You may also be interested in the Exhibition Chemistry video Rainbow flame demonstration, which provides a wonderful display of colours as well as further details of the chemistry of flame tests. Integrated instructions Printable integrated instructions are provided for learners. These are available as a separate download here: rsc.li/3dhnn5B. Integrated instructions use clear numbering, arrows and simple pictograms, like an eye to show where observations are required. These have been developed using cognitive load theory. Integrated instructions remove unnecessary information, and therefore reduce extraneous load on students, which increases their working memory capacity to think about what they are doing and why. Read more about the use of integrated instructions here: rsc.li/2SdSqkQ.
Results tables Printable results tables have been provided for the three sets of tests. For the microscale tests for metal ions there is a table to perform the experiment on and one to record results.
Key terms Learners will need to have a clear understanding of the following scientific terminology: Ion – a positively or negatively charged particle. Metal halide – general term used to describe the group of ionic compounds that form when a metal reacts with a halogen.
Ionic compound – a compound made up of oppositely charged ions. Ionic compounds are held together by strong electrostatic forces between oppositely charged ions. These forces are called ionic bonds.
Aqueous solution – a solution where the solvent is water. So an aqueous solution of sodium hydroxide contains Na+ ions, OH- ions and H2O molecules.
Precipitate – a solid that forms from ions in an aqueous solution. The precipitate is insoluble in water.
Ionic equation – a symbol equation which focuses on the ions that react together and ignore the ones that do not take part in the reaction (spectator ions).
Prior knowledge Learners should be able to recall: • Particles can be atoms, molecules or ions. • An ion is a positively or negatively charged particle. • An atom or a molecule can lose or gain electron(s) to form an ion. • When an atom/molecule gains negatively charged electron(s), a negative ion is formed. When an atom/ molecule loses negatively charged electron(s), a positive ion is formed. • A solution is formed when a solute (salt) is dissolved in a solvent (water).
Learners should be confident writing word and symbol equations There are some questions included which ask learners to balance symbol equations and write ionic equations. Depending on where the identification of ions comes in your scheme of work your learners may not have come across this yet. Adapt the questions to make them relevant to the stage and level that you are at. Some of the challenge tasks require learners to use and apply their knowledge from other topics.
3 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Common misconceptions 1. When an atom/molecule loses negatively charged electron(s), a positive ion is formed. This is something learners often struggle with later on in their studies. Introducing the electron now, before learners meet the other sub-atomic particles, can help to embed the idea that the loss of electrons results in a positively charged ion, and may help reduce confusion later on.
2. As learners develop their understanding of chemical bonding further, it is common for students to refer to ionic compounds as molecules or to refer to intermolecular forces when explaining properties of ionic compounds. To avoid these misconceptions introduce and emphasise the correct use of the terms ‘ion’ and ‘molecule’ from the outset.
3. Learners often find solution chemistry challenging as they fail to appreciate that as well as the ions taking place in the precipitate reaction, both water molecules and spectator ions are also present. Diagnostic multiple-choice questions are a great way to explore learners’ reasoning behind their answers. You can read more about diagnostic questioning here: rsc.li/3u1kED3.
Intended outcomes It is important that the purpose of each practical is clear from the outset, defining the intended learning outcomes helps to consolidate this. Outcomes can be categorised as hands on, what learners are going to do with objects, and minds on, what learners are going to do with ideas to show their understanding. We have offered some differentiated suggestions for this practical. You may wish to focus on just one or two, or make amendments based your learners’ own needs. (Read more at rsc.li/2JMvKa5.) Consider how you can share outcomes and evaluation with learners, empowering them to direct their own learning. Hands on Minds on Effective at a lower level Students correctly: • Follow instructions • Make careful observations • Carry out a flame test • Carry out tests for negative ions • Carry out the sodium hydroxide test for positive ions Students can: • Correctly record test results in a table • Use the results to identify an unknown sample Effective at a higher level Students correctly: • Plan and carry out a series of tests to identify an unknown sample Students can: • Explain why different metals have different flame colours • Write ionic equations for the sodium hydroxide tests and halide tests How to use the additional resources Using the pause-and-think questions Pause-and-think questions are supplied in two formats: a teacher version for ‘live’ questioning and a student version which can be used during independent study. The time stamps allow you to pause the video when presenting to a class, or learners to use for active revision. The questions have been put into four sections: general questions, flame tests, testing for positive ions and testing for negative ions. Teacher version The questions are presented in a table and you can choose to use as many as appropriate for your class and the learning objectives. Some questions have two timestamps to allow you to adapt the questions for different classes or scenarios. Pause the videos at the earlier timestamp to ask a question before the answer is given, useful for revision or to challenge learners. Pause at the later timestamp to ask a question reflectively and assess whether learners have understood what they have just heard or seen. This would be useful when introducing a topic, in a flipped learning scenario or when additional support and encouragement is needed. 4 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Think about how you will ask for responses. Variation may help to increase engagement – learners could write and hold up short answers; more complex questions could be discussed in groups. Not all answers to questions are included in the video. Some of the questions will draw on prior learning or extend learners’ thinking beyond the video content. Student version The same questions are offered as a printable worksheet for learners. Use in situations where there is not a teacher present to guide discussion during the video, for example homework, revision or remote learning. Using the structure strips Writing about chemistry encourages learners to reflect on their understanding, formulate new ideas and make links between ideas in new ways. Learners also need to practice for longer-answer questions in examinations. Structure strips provide scaffolded prompts and help overcome ‘fear of the blank page’. The learner sticks the strip into the margin of their exercise book or onto an A4 sheet of paper and writes alongside it. Use this long-answer question to consolidate learning after the practical and/or for revision. (Read more at rsc.li/2P0JDlW.) Long-answer question: A sample of an unknown ionic solution has been collected for analysis. The sample is colourless. Describe a series of qualitative tests that could be used to identify the unknown ions in the sample. In your plan you must include instructions for carrying out the tests and the expected results.
Using Johnstone’s triangle Johnstone’s triangle helps learners to understand what going on in a chemical reaction and gain a deeper understanding. It does this by helping learners to make links between three different levels of representation: the macroscopic, symbolic and sub-microscopic levels. The macroscopic is the ‘seen’ level, e.g. practical investigation or observations, which can be described. The symbolic level is the unseen: how we represent the macroscopic through word or symbol equation. The sub-microscopic is unseen, at the atomic level, and includes explanatory models. (Read more at rsc.li/2XhYN9Q.) As part of the additional resources we have included a completed example of Johnstone’s triangle for the sodium hydroxide test for iron(iii); an example for learners to work through for copper(ii), and a template so you can set your own questions (editable file at rsc.li/3dhnn5B). In the example provided we have used a very simple model at the sub-microscopic level. When discussing this model with your learners, you could ask them how to improve it, for example, by showing the water molecules present and the spectator ions.
If your learners are not familiar with using Johnstone’s triangle, then it is recommended that you take the approach ‘I try’ to introduce the triangle; ‘let’s try’ to work through an example together. Finally, ‘you try’ where learners work through an example on their own. When completing the ‘sub-microscopic’ level it is helpful if learners have access to modelling resources such as molymods, modelling clay, ionic jigsaw etc. Using the follow-up worksheet A practical skills worksheet has been included as part of the additional resources. The first section provides structured questions at a support level for learners, the second section provides more challenging applied questions. This worksheet could be used to follow up the practical activity, for example as homework or a revision exercise.
5 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Pause-and-think questions Teacher version Timestamp(s) Question Answer/discussion points 00:22 What is the difference between qualitative analysis and quantitative analysis? Give examples of each.
Qualitative analysis is where we can only identify which ions or molecules are present in an unknown sample. Chemical tests are an example of qualitative analysis.
Quantitative analysis is where we can identify the amount of the substance present, eg the actual concentration of an ion. Titrations are an example of quantitative analysis. 00:33 00:45 What is an ion?
Describe how a positive and negative ion is formed. A positively or negatively charged particle.
A positive ion is formed when a particle loses electrons.
A negative ion is formed when a particle gains electrons.
01:12 What will we be looking for during the chemical tests? How will you know if a test has a positive result?
Signs that a chemical change has taken place, eg colour change, formation of a gas, temperature change, formation of a solid or a precipitate.
Consider the observed result. If there is only one possible ion with this result then the result will be positive, eg a blue precipitate with sodium hydroxide indicates the presence of the copper ion. However if there are several possibilities, eg a white precipitate is formed with sodium hydroxide then you will need to do further tests until there is only one possible answer.
Flame tests 01:22 01:29 What are flame tests used to identify?
Metal ions.
01:57 Why can’t we use a dry splint for a flame test?
A dry splint itself will burn and affect the colour that we see. A soaked splint will not burn immediately so we will only see the flame colour associated with the metal ions initially.
02:32 Why do we test distilled water first. What does this show?
This is a control. It shows that the distilled water that the splints were soaked in does not affect the flame colour. 02:52 02:58 Record the flame colour for lithium in the table provided.
Lithium Li+ crimson 03:09 03:15 Record the flame colour for potassium in the table provided.
Potassium K+ lilac 03:30 03:42 Record the flame colour for calcium in the table provided.
Calcium Ca2+ orange-red 03:52 03:56 Record the flame colour for copper in the table provided.
Copper Cu2+ green 04:11 04:14 Record the flame colour for copper in the table provided.
Sodium Na+ yellow-orange 6 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources 04:25 Check your results or, if you haven’t yet done so, record the flame colours in your results table.
Lithium Li+ crimson Potassium K+ lilac Calcium Ca2+ orange-red Copper Cu2+ green Sodium Na+ yellow-orange 04:36 Challenge: Suggest a reason why different metals have different flame colours.
They have different electronic configurations. For more information, see the links in the teacher notes.
Testing for positive ions 04:55 What are the benefits of microscale chemistry?
Benefits include: • sustainability • safety • more careful observations can be made Read more about the benefits of microscale: rsc.li/2ZtlkTM.
05:17 What colour is sodium hydroxide?
Colourless 05:30 Name the ions present in sodium hydroxide solution.
Sodium ion, Na+ Hydroxide ion, OH- (also some H+ and OH- from the water) 05:47 06:00 Name the green precipitate formed when sodium hydroxide is added to iron (ii) ions.
Iron(ii) hydroxide 06:04 06:58 Record the results shown with each metal ion in your results table.
Iron(ii) Fe2+ green precipitate Iron(iii) Fe3+ rust precipitate Copper(ii) Cu2+ blue precipitate Aluminium Al3+ white precipitate Calcium Ca2+ white precipitate Magnesium Mg2+ white precipitate 06:58 Name the products formed in the reactions and write the ionic equations. Iron(iii) hydroxide Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s) Copper(ii) hydroxide Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s) Aluminium hydroxide Al3+(aq) + 3OH-(aq) → Al(OH)3(s) Calcium hydroxide Ca2+(aq) + 2OH-(aq) → Ca(OH)2(s) Magnesium hydroxide Mg2+(aq) + 2OH-(aq) → Mg(OH)2(s) 07:43 Describe the test for aluminium. What does a positive result look like?
Add excess sodium hydroxide to the white precipitate. If the solid disappears then the aluminium ion is present.
07:50 Both magnesium ions and calcium ions form a white precipitate with sodium hydroxide. Suggest a further test you could do to distinguish between the two metal ions.
Carry out a flame test. If an orange-red flame is observed then the calcium ion is present. (Magnesium has no flame colour in a flame test.) 7 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Testing for negative ions 08:28 Why do you think that the limewater is put into a separate test tube at the start of the carbonate test?
The lime water is to test the gas being produced. If it was mixed with the original solution you may not be able to observe any colour change taking place.
08:34 What does limewater test for?
Carbon dioxide.
09:10 Why does the gas not escape the test tube to the surroundings?
Carbon dioxide is more dense than air. Therefore most of it remains in the test tube because diffusion will be slow.
09:29 09:33 What has happened to the limewater?
Name the gas being produced.
The limewater has gone cloudy.
Carbon dioxide gas is being produced.
09:37 09:43 What is the test for carbonate ions? What does a positive result look like?
Add hydrochloric acid and test any gas being produced by bubbling it through limewater. If the limewater goes cloudy it is a carbonate.
09:37 09:50 Write a word and symbol equation for the reaction of sodium carbonate with hydrochloric acid.
What type of reaction is this?
Sodium carbonate + hydrochloric acid → Sodium chloride + water + carbon dioxide Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) A neutralisation reaction.
10:00 10:10 Why do you think dilute acid is added at the start of the sulfate test?
To remove any ions that might interfere with a positive result.
10:25 10:28 Write a word and symbol equation for the reaction of sodium sulfate with barium chloride.
What type of reaction is this?
Sodium sulfate + barium chloride → sodium chloride + barium sulfate Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s) A double displacement reaction.
10:25 10:34 What is the test for sulfate ions? What does a positive result look like?
Add a few drops of hydrochloric acid and barium chloride to the sample. If a white precipitate forms then a sulfate is present.
10:40 10:46 Write the symbols for the halide ions in your results table: Chloride Bromide Iodide Cl- Br- I-11:03 What was added to the test tube to test for the presence of halides?
Nitric acid and sliver nitrate solution.
11:28 11:47 Record the results in your results table.
Cl- white precipitate is formed Br- cream precipitate is formed I- yellow precipitate is formed 11:40 Write the ionic equations for the formation of silver bromide and silver iodide.
Br-(aq) + Ag+(aq) → AgBr(s) I-(aq) + Ag+(aq) → AgI(s) 12:08 12:15 Unknown solution B, shown in the video, is blue.
What test would you use to confirm the identity of the positive ion?
Since the solution is blue, I would suspect that a copper ion was present so I would do either the sodium hydroxide test or the flame test first. 8 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Pause-and-think questions Student version Pause the video at the time stated to test or revise your knowledge of these practical experiments.
Time Question 00:22 What is the difference between qualitative analysis and quantitative analysis? 00:33 What is an ion? Describe how a positive and negative ion are formed. 01:12 What will we be looking for during the chemical tests? How will you know if a test has a positive result? Flame tests 01:22 What are flame tests used to identify? 02:32 Why do we test distilled water first? What does this show? Record the symbol and flame colours for each metal ion in the table below: Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium 02:58 Lithium.
03:09 Potassium.
03:30 Calcium.
03:52 Copper.
04:11 Sodium.
9 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Testing for positive ions 05:17 What colour is sodium hydroxide? 05:30 Name the ions present in sodium hydroxide solutions. 05:47 Name the green precipitate formed when sodium hydroxide is added to iron (ii) ions 06:04 Record the results shown with each metal ion in the results table below: Positive ion Symbol Observation when added to sodium hydroxide solution Observation with excess sodium hydroxide solution Iron(ii) Iron(iii) Copper(ii) Aluminium Calcium Magnesium 06:58 Name the products formed in the reactions and complete the ionic equations: Iron(iii) hydroxide: Fe3+(aq) + 3OH-(aq) → __ Copper(ii) hydroxide: _________ + 2OH-(aq) → ______________ __________________: Al3+(aq) + 3OH-(aq) → ________________ __________________: Ca2+(aq) + ________ → ______________ Magnesium hydroxide: _________ + __________ → ____________ 07:43 Describe the test for aluminium. What does a positive test look like? 07:50 Both magnesium ions and calcium ions form a white precipitate with sodium hydroxide. Suggest a further test you could do to distinguish between the two metal ions. Testing for negative ions Negative ion Symbol Test Observation Carbonate Sulfate 08:28 Why do you think that the limewater is put into a separate test tube at the start of the carbonate test? 08:34 What does limewater test for? 10 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources 09:29 What has happened to the limewater? 09:33 Name the gas produced. 09:37 Complete the word and symbol equation: Sodium carbonate + hydrochloric acid → __________________ + ________________ + ___________________ Na2CO3(aq) +___________ → ___________ + H2O(l) + ___________ What type of reaction is this? 09:43 Complete the table above with the test and positive result for carbonate ions.
09:58 Why do you think dilute acid is added at the start of the sulfate test? 10:25 Complete the table above with the test and positive result for sulfate ions.
10:25 Complete the word and symbol equation: Sodium sulfate + barium chloride → ________________ + ________________ Na2SO4(aq) + BaCl2(aq) → _____________ + _____________ 10:28 What type of reaction is this? 10:38 Add the symbols for the halide ions to the results table below.
11:28 Record the results for the halide tests in the results table.
Negative ion Symbol Test Observation Halide Chloride Bromide Iodide 11:40 Write the ionic equations for the formation of silver bromide and silver iodide. _________ + _________ → __________ _________ + _________ → __________ 12:08 Unknown solution B is blue. Which metal ion would you expect it to contain? Which test would you do use to confirm you prediction? 11 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Identifying ions Structure strip Identifying ions Structure strip Identifying ions Structure strip Identifying ions Structure strip Identifying ions Structure strip What is a qualitative test and what is it used for?
What is a qualitative test and what is it used for?
What is a qualitative test and what is it used for?
What is a qualitative test and what is it used for?
What is a qualitative test and what is it used for?
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
What is the limitation of the sodium hydroxide test?
What is the limitation of the sodium hydroxide test?
What is the limitation of the sodium hydroxide test?
What is the limitation of the sodium hydroxide test?
What is the limitation of the sodium hydroxide test?
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Describe the sulfate test.
What would a positive test result look like?
Describe the sulfate test.
What would a positive test result look like?
Describe the sulfate test.
What would a positive test result look like?
Describe the sulfate test.
What would a positive test result look like?
Describe the sulfate test.
What would a positive test result look like?
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
12 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Structure strip: suggested answer content Identifying ions Structure strip What is a qualitative test and what is it used for?
A qualitative test is used to identify the chemical composition of an unknown sample. A positive test result, such as a colour change, will confirm that a particular substance is present.
The unknown colourless sample is an ionic solution, so we are going to use a series of qualitative tests to identify the positive and negative ions.
Describe the hydroxide test for positive ions.
What would a positive test result look like? Give an example.
There are two tests that can be used to determine the positive ion. One of these tests is the hydroxide test: • Add 2 drops of sodium hydroxide to 2 drops of the unknown solution. • A coloured precipitate will form. • Observe the colour of the precipitate. This will identify the possible positive metal ion in the unknown solution. For example, a rust coloured precipitate would be a positive result for iron(iii) ions.
What is the limitation of the sodium hydroxide test?
There is more than one metal ion that will produce a white precipitate. If a white precipitate is formed and it does not dissolve in excess sodium hydroxide then further tests will need to be carried out to determine whether the unknown solution contains calcium or magnesium.
Describe the flame test to confirm the positive ion.
What would a positive test result look like?
Give an example.
A further test to confirm the positive ion present is a flame test. The flame test could also be used to distinguish between calcium and magnesium.
• Dip a pre-soaked splint in the unknown solution. • Hold it in a roaring blue Bunsen flame. • Observe the colour of the flame.
The flame colour will identify the positive metal ion present in the unknown solution. For example, a lilac flame would be a positive result for potassium.
Describe the series of tests used for negative ions.
Describe the carbonate test.
What would a positive test result look like?
Now, we will use a series of tests to identify the negative ion. The carbonate test • Put 1 cm3 of the unknown solution into a test tube. • Put 1 cm3 of limewater into a separate test tube. • Add a few drops of hydrochloric acid (HCl). If it fizzes then the carbonate ion is present. • Confirm the presence of carbonate by bubbling the gas through the limewater. If a carbonate ion is present the limewater will turn cloudy.
Describe the sulfate test.
What would a positive test result look like?
The sulfate test • Put 1 cm3 of the unknown solution into a test tube. • Add a few drops of hydrochloric acid HCl and then add barium chloride (BaCl2). • If a white precipitate appears, then the sulfate ion is present.
Describe the halide test.
What would a positive test result look like for chloride, bromide and iodide?
The halide test • Put 1 cm3 of the unknown solution into a test tube. • Add a few drops of nitric acid (HNO3) and silver nitrate (AgNO3). • If a white precipitate forms then the chloride ion is present. If the precipitate is cream, then the bromide ion is present, if it is yellow then the iodide ion is present.
13 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Follow-up worksheet 1. A student carried out some flame tests.
Complete the flame test results table. The first row has been done for you.
Metal ion Symbol Observation: flame colour Potassium K+ Lilac Sodium Li+ Crimson Calcium Cu2+ Green Unknown solution X _______________________ Yellow 2. Describe how you would test for positive ions using the sodium hydroxide chemical test. 3. Complete the sodium hydroxide test results table.
Positive ion Symbol Observation when added to sodium hydroxide solution Iron(ii) Fe2+ Brown precipitate Copper(ii) Blue precipitate 4. Complete the equations, using the example to help. Iron(ii) + hydroxide → iron(ii) hydroxide Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) (a) Iron(iii) + hydroxide → __________________________ Fe3+(aq) + __OH-(aq) → Fe(OH)3() (b) Copper(ii) + hydroxide → __ (ii) hydroxide Cu2+(aq) + __OH-(aq) → Cu(OH)2() 5. A student carried out some tests to identify the ions present in an unknown solution. After adding a few drops of dilute nitric acid and silver nitrate to the sample, a cream coloured precipitate appeared. The student concluded that a _____ ion was present in the solution.
14 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Challenge 6. A student was asked to identify an unknown sample. She carried out some tests and here are the results.
(a) Complete the table.
Test Result Conclusion Add 2 drops of sodium hydroxide to 2 drops of solution A white precipitate is formed Add excess sodium hydroxide to the drop No change to the white precipitate Flame test An orange-red flame is observed Add a few drops of hydrochloric acid to sample No changes observed Add a few drops of hydrochloric acid then a few drops of barium chloride to sample No change observed Add a few drops of nitric acid then a few drops of silver nitrate to sample A white precipitate is formed (b) The unknown sample is (c) Complete the ionic equations for the positive results. __( ) + _ OH-(aq) → _( ) Ag+(aq) + _ ( ) → ______( ) 7. A sample of an unknown ionic solution has been collected for analysis. The sample is colourless. Describe a series of qualitative tests that could be used to identify the unknown ions in the sample. In your plan you must include instructions for carrying out the tests and the expected results. This question has a structure strip to support your written answer. Find more resources to support you here rsc.li/3a7LS37.
15 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Follow-up worksheet: answers 1. A student carried out some flame tests.
Complete the flame test results table. The first row has been done for you.
Metal ion Symbol Observation: flame colour Potassium K+ Lilac Sodium Na+ Yellow Lithium Li+ Crimson Calcium Ca2+ Orange-red Copper Cu2+ Green Unknown solution X Sodium Na+ Yellow 2. Describe how you would test for positive ions using the sodium hydroxide chemical test. Add 2 drops of sodium hydroxide to 2 drops of an unknown solution and observe.
3. Complete the sodium hydroxide test results table.
Positive ion Symbol Observation when added to sodium hydroxide solution Iron(ii) Fe2+ Green precipitate Iron(iii) Fe3+ Brown precipitate Copper(ii) Cu2+ Blue precipitate 4. Complete the equations, using the example to help. Iron(ii) + hydroxide → iron(ii) hydroxide Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) (a) Iron(iii) + hydroxide → iron(iii) hydroxide Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s) (b) Copper(ii) + hydroxide → copper(ii) hydroxide Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s) 5. A student carried out some tests to identify the ions present in an unknown solution. After adding a few drops of dilute nitric acid and silver nitrate to the sample, a cream coloured precipitate appeared. The student concluded that a Chloride (or Cl-) ion was present in the solution.
16 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Challenge 6. A student was asked to identify an unknown sample. She carried out some tests and here are the results.
(a) Complete the table.
Test Result Conclusion Add 2 drops of sodium hydroxide to 2 drops of solution A white precipitate is formed Al3+, Ca2+ or Mg2+ ions could be present Add excess sodium hydroxide to the drop No change to the white precipitate Ca2+ or Mg2+ ions could be present Flame test An orange-red flame is observed Ca2+ ion is present Add a few drops of hydrochloric acid to sample No changes observed The carbonate or CO3 2- ion is not present Add a few drops of hydrochloric acid then a few drops of barium chloride to sample No change observed The sulfate or SO4 2- ion is not present Add a few drops of nitric acid then a few drops of silver nitrate to sample A white precipitate is formed The chloride or Cl- ion is present (b) The unknown sample is Calcium chloride (or CaCl2) (c) Complete the ionic equations for the positive results. Ca2+(aq) + 2OH-(aq) → Ca(OH)2(s) Ag+(aq) + Cl-(aq) → AgCl(s) 7. A sample of an unknown ionic solution has been collected for analysis. The sample is colourless. Describe a series of qualitative tests that could be used to identify the unknown ions in the sample. In your plan you must include instructions for carrying out the tests and the expected results. This question has a structure strip see ‘suggested answer content’ (rsc.li/3a7LS37).
17 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Results tables for flame tests Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium Metal ion Symbol Observation: flame colour Lithium Potassium Calcium Copper Sodium 18 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Microscale reactions of positive ions with sodium hydroxide Positive ion Symbol Positive ion solution and sodium hydroxide solution Iron(ii) Fe2+ Iron(iii) Fe3+ Copper(ii) Cu2+ Aluminium Al3+ Calcium Ca2+ Magnesium Mg2+ 19 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Results table for positve ion tests Positive ion Symbol Observation when added to sodium hydroxide solution Observation with excess sodium hydroxide solution Iron(ii) Iron(iii) Copper(ii) Aluminium Calcium Magnesium Positive ion Symbol Observation when added to sodium hydroxide solution Observation with excess sodium hydroxide solution Iron(ii) Iron(iii) Copper(ii) Aluminium Calcium Magnesium 20 © 2021 Royal Society of Chemistry Identifying ions | Supporting resources Results tables for negative ion tests Negative ion Symbol Test Observation Carbonate Sulfate Halide Chloride Bromide Iodide Negative ion Symbol Test Observation Carbonate Sulfate Halide Chloride Bromide Iodide |
188443 | https://www.ck12.org/flexi/cbse-math/ordered-pairs-in-four-quadrants/how-can-a-rhombus-be-demonstrated-in-coordinate-geometry/ | How can a rhombus be demonstrated in coordinate geometry?
Flexi Says:
To demonstrate a rhombus in coordinate geometry, you can plot its vertices on a coordinate plane and show that it satisfies the properties of a rhombus.
A rhombus is a quadrilateral with all four sides of equal length and opposite sides parallel. The steps to verify if the coordinates form a rhombus are given below:
Step 1: Calculate the side lengths using the distance formula.
Step 2: Verify if all the side lengths calculated are equal to each other.
Step 3: Calculate the slope of opposite sides of the figure.
Step 4: Verify if the opposite sides are parallel by checking if their slopes are equal.
If all these conditions are satisfied, the quadrilateral is a rhombus.
Let's look at an example.
Prove or disprove that the quadrilateral defined by the points , , , is a rhombus.
First, plot the points.
This shape appears to be a rhombus. To prove that it is a rhombus, remember that the definition of a rhombus is a quadrilateral with four congruent sides. Therefore, to prove it is a rhombus you must verify that all sides are the same length. You can use the distance formula or the Pythagorean Theorem to do this.
Even though the shape looked like a rhombus, its four sides are not congruent. Therefore, this is NOT a rhombus.
Related questions:
How can one prove a square in coordinate geometry?Prove that V and A_4 are the proper normal subgroups of S_4.
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188444 | https://math.stackexchange.com/questions/501660/is-there-a-way-to-get-trig-functions-without-a-calculator | trigonometry - Is there a way to get trig functions without a calculator? - Mathematics Stack Exchange
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Is there a way to get trig functions without a calculator?
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In school, we just started learning about trigonometry, and I was wondering: is there a way to find the sine, cosine, tangent, cosecant, secant, and cotangent of a single angle without using a calculator?
Sometimes I don't feel right when I can't do things out myself and let a machine do it when I can't.
Or, if you could redirect me to a place that explains how to do it, please do so.
My dad said there isn't, but I just had to make sure.
Thanks.
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edited Nov 5, 2022 at 13:41
Jam
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asked Sep 22, 2013 at 20:51
Jonathan LamJonathan Lam
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11 Yes. When I was in school, we used lookup tables known as log books.copper.hat –copper.hat 2013-09-22 20:53:33 +00:00 Commented Sep 22, 2013 at 20:53
1 Some angles are easy to get an exact answer, some are not. All can be approximated by their Taylor Polynomials though...Jemmy –Jemmy 2013-09-22 20:53:39 +00:00 Commented Sep 22, 2013 at 20:53
4 You can always work out an answer, or a close approximation, it just depends on how much time and energy you have. The calculator works out approximations (good ones, of course) for many angles.copper.hat –copper.hat 2013-09-22 20:54:56 +00:00 Commented Sep 22, 2013 at 20:54
9 If there weren't a way to evaluate trig functions without a calculator, then how would anyone be able to program the calculators?Dan –Dan 2015-10-18 17:43:26 +00:00 Commented Oct 18, 2015 at 17:43
4 "My dad said there isn't" Superstitions like this amaze me. Obviously there were trigonometric tables before there were calculators, and obviously calculators were created by humans, who could not have created them without knowing some way to compute the functions that they compute.Michael Hardy –Michael Hardy 2020-08-06 23:48:14 +00:00 Commented Aug 6, 2020 at 23:48
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Congratulations! You've stumbled in to a very interesting question!
In higher mathematics, we often notice that some things which are really easy to talk about but difficult to express rigorously have a property which is really easy to express rigorously but something that we probably wouldn't have thought of to begin with.
The trig functions are one of these things. With (a lot of) effort, you can show that
sin x=x−x 3 6+x 5 120−x 7 5040+x 9 362880−⋯sinx=x−x 3 6+x 5 120−x 7 5040+x 9 362880−⋯
where the patterns of increasing the powers of x x by 2 2, and switching between ++ and −− signs continues forever. (The denominators also have a pattern: take the power that x x is raised to in the term and multiply it by all of the smaller numbers down to 1 1; that is the number in the denominator). Note that you have to use radians for this exact formula to work; of course you could come up with one for degrees as well.
When you start realizing that circles are actually quite tricky objects to define, formulas like that one start to look more appealing. I have had multiple mathematics textbooks take this infinitely long expression as the definition of the sine function. (It turns out to be the same thing as the circle definition, but… well, circles get complicated.)
Of course, we can't sit around multiply and add for the rest of our lives just to compute sin 1 1, but we can just cut off the operations after a couple terms. If you go out to the x 7 x 7 term, you can guarantee that your answer is accurate to at least 3 decimal places as long as you use angles between −π 2−π 2 and π 2 π 2. (These are the only angles you really need, if you get rid of multiples of π π properly.)
The cosine formula, in case you are interested, is similar:
cos x=1−x 2 2+x 4 24−x 6 720+x 8 40320−⋯cosx=1−x 2 2+x 4 24−x 6 720+x 8 40320−⋯
The internet has formulas for the other trig functions, but you can always just combine these.
As copper.hat says, there are also these large books where people did the calculations once and wrote them down so that nobody would have to do them again. Of course, these were made long before computers existed; nobody makes them anymore! But somebody from your parents' or grandparents' generation probably still has one sitting in their house.
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answered Sep 22, 2013 at 21:23
Eric StuckyEric Stucky
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1 I have a log book on my shelves. They were supplied at public examinations (intermediate & leaving certificate, and matriculation exams back then) in Ireland. The also included many useful mensuration formulae & trigonometric identities, which were a boon for those like myself that have difficult committing such things to memory!copper.hat –copper.hat 2013-09-23 01:29:05 +00:00 Commented Sep 23, 2013 at 1:29
1 And lots and lots more lovely expansions at the DLMF. This book is the successor to the well-beloved Abramowitz and Stegun, which included expansions such as this as well as log and other tables.Norman Gray –Norman Gray 2015-04-09 23:37:00 +00:00 Commented Apr 9, 2015 at 23:37
1 Well, close enough to zero. But ok, with order 8 and playing symetries, it's working.Fabrice NEYRET –Fabrice NEYRET 2015-10-10 17:40:27 +00:00 Commented Oct 10, 2015 at 17:40
1 @FabriceNEYRET: I think you might have posted this in the wrong thread. (What are you saying is zero?)Eric Stucky –Eric Stucky 2015-10-11 01:19:32 +00:00 Commented Oct 11, 2015 at 1:19
1 A Taylor series around zero is good not too far to zero or you have to use very high level polynomials that get unconvenient for manual computation.Fabrice NEYRET –Fabrice NEYRET 2015-10-11 07:15:48 +00:00 Commented Oct 11, 2015 at 7:15
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Use Taylor Series:
sin x=x−x 3 3!+x 5 5!+⋯=∑n=0∞(−1)n(2 n+1)!x 2 n+1 sinx=x−x 3 3!+x 5 5!+⋯=∑n=0∞(−1)n(2 n+1)!x 2 n+1
cos x=1−x 2 2!+x 4 4!+⋯=∑n=0∞(−1)n(2 n)!x 2 n cosx=1−x 2 2!+x 4 4!+⋯=∑n=0∞(−1)n(2 n)!x 2 n
For others you can look here
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edited Aug 6, 2020 at 23:49
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answered Sep 22, 2013 at 20:56
Stefan4024Stefan4024
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Tailored Taylor
You can use Taylor but first you need to pack your angle into the region x 1=0,2 π x 1=0,2 π. simply by x mod 2 π x mod 2 π
Once you are there if x 1>π x 1>π take the result as sin(x 1)=−sin(x 1−π)sin(x 1)=−sin(x 1−π) reducing it to x 2=0,π.x 2=0,π.
Now if x 2>π 2 x 2>π 2 calculate the result as sin(x 2)=sin(π−x 2)sin(x 2)=sin(π−x 2).
So all this above is easily shifting it all to x 3=0,π 2 x 3=0,π 2
If needed use further
sin(x)=2 sin(x 2)cos(x 2)sin(x)=2 sin(x 2)cos(x 2)
with
cos(x 2)=1−sin(x 2)2−−−−−−−−−−√cos(x 2)=1−sin(x 2)2
in case x 3>1.x 3>1.
All this is making the Taylor expansion much more accurate within a lesser number of steps as x n x n in it for x<1 x<1 will rapidly go to 0 0 besides the help from factorial. Next you represent Taylor series of sin(x)sin(x) in a much more handy way
sin(x)=x(1−x 2 3⋅2(1−x 2 5⋅4(1−x 2 7⋅6(…sin(x)=x(1−x 2 3⋅2(1−x 2 5⋅4(1−x 2 7⋅6(…
Notice that x 2 x 2 is repeating. So choose k k as big as you are willing to calculate and have
f(0)=1−x 2(2 k+1)⋅2 k f(0)=1−x 2(2 k+1)⋅2 k
f(n)=1−x 2(2(k−n)+1)⋅2(k−n)f(n−1)f(n)=1−x 2(2(k−n)+1)⋅2(k−n)f(n−1)
Finally
sin(x)=x f(k−1)sin(x)=x f(k−1)
Binary ladders
For this method as well you need to bring the angle as much down as you can as explained above.
If you do not want to deal with division, otherwise you can use tan(x)tan(x), the task is possible with multiplication only. Take the small m m and
sin(m)≈m=x 0 sin(m)≈m=x 0
cos(m)≈1−m 2 2=y 0 cos(m)≈1−m 2 2=y 0
Then have:
M 2 0=[y 0−x 0 x 0 y 0]M 2 0=[y 0 x 0−x 0 y 0]
M 2 k+1=(M 2 k)2 M 2 k+1=(M 2 k)2
this is just based on duplication formula for sin(x)sin(x) and cos(x)cos(x)
Now it is up to you what small m m you will use as a reference. It can be for example 1 2 10 1 2 10 or 0.00001 0.00001 or any other small number. Smaller it is, a better precision you have.
Now you find the integer n n so that n m≤x<(n+1)m n m≤x<(n+1)m
The game can start.
Write n n in binary expansion.
∑d=1 m 2 k d=n∑d=1 m 2 k d=n
Using
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
cos(x+y)=cos(x)cos(y)−sin(x)sin(y)cos(x+y)=cos(x)cos(y)−sin(x)sin(y)
which is in our case
[y p−x p x p y p][y q−x q x q y q][y p x p−x p y p][y q x q−x q y q]
since we are dealing with evaluation of sin(x),cos(x)sin(x),cos(x) all the time you additionally multiply
M 2 k 1 M 2 k 2…M 2 k m M 2 k 1 M 2 k 2…M 2 k m
where
∑d=1 m 2 k d=n∑d=1 m 2 k d=n
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edited Aug 25, 2018 at 7:22
answered Aug 24, 2018 at 21:31
user1224955 user1224955
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To this very interesting answer I'd like to add that the technique illustrated can also be seen as using complex numbers, because matrices such as M 2 0 M 2 0 are equivalent to y 0−i x 0 y 0−i x 0 under multiplication, addition, and so on. See for example this answer.pglpm –pglpm 2021-02-10 09:21:28 +00:00 Commented Feb 10, 2021 at 9:21
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Bhaskara's approximation (Wikipedia) gives an approximation for sin x∘sinx∘ with less than 0.0016 0.0016 error for 0≤x≤180 0≤x≤180.
sin x∘≈4 x(180−x)40500−x(180−x)sinx∘≈4 x(180−x)40500−x(180−x)
The red curve is the approximation, barely seen:
Here's the difference between the formula and the sin function (maximum at x=11.544):
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edited Nov 27, 2019 at 18:54
answered Nov 19, 2019 at 20:43
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This formula is amazing. To me, "without a calculator" means I need to compute by hand. The error is less than 0.001630. The answer computed is the same for rounding up to 2 decimal spaces. For Taylor's Series up to first 3 terms, the error is less than 0.004525. So I think the approximation of this formula is quite well and it is acceptable. Moreover, I have tried using this formula to compute some values for sine function by hand and also using computer to check the error and answer. And I like this formula more than Taylor's Series since it is easier and fast to compute by hand.Tim –Tim 2023-07-04 05:46:28 +00:00 Commented Jul 4, 2023 at 5:46
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The wikipedia article gives some infinite series, which are probably what your calculator uses. The formulae for sine and cosine are the ones to focus on first. They converge very quickly, but you have to realise that the angles are measured in radians, where 2 π 2 π radians =360∘=360∘. If you do the conversion, you'll be able to calculate quite quickly for yourself.
There are connections to a lot of beautiful and clever maths to be discovered, which explain why all this works. You have asked a great question. Keep going with the answer - there are more dimensions to it than you will see on the surface.
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answered Sep 22, 2013 at 21:01
Mark BennetMark Bennet
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Approximate the Taylor series. In Taylor series we have to use the angle in radians and by converting it into degrees and by making some approximations we can get a simple formulas like sin X=0.017∗X sinX=0.017∗X for X<33 X<33 degrees and sin X=0.016∗X sinX=0.016∗X for 33<X<45 33<X<45
cos X=1−0.000145 X 2 cosX=1−0.000145 X 2 for X<45 X<45 degrees
By using these two formulas we can calculate any sin and cos functions for any degrees by using methods sin(90+X)sin(90+X) ,sin(90−X)sin(90−X), cos(270+X)cos(270+X) like...
which will give minimum 98% accuracy.
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edited Jul 17, 2018 at 13:52
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answered May 29, 2015 at 5:56
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Long before there were power series, in the second century A.D., Ptolemy, a man who wrote in Greek and probably lived in Alexandria, created a table of values of what amounts to the sine function.
See this page.
Chapter 10 of Book I of the ''Almagest'' presents geometric theorems used for computing chords. Ptolemy used geometric reasoning based on Proposition 10 of Book XIII of Euclid's Elements to find the chords of 72∘72∘ and 36∘.36∘. That Proposition states that if an equilateral pentagon is inscribed in a circle, then the area of the square on the side of the pentagon equals the sum of the areas of the squares on the sides of the hexagon and the decagon inscribed in the same circle.
He used Ptolemy's theorem on quadrilaterals inscribed in a circle to derive formulas for the chord of a half-arc, the chord of the sum of two arcs, and the chord of a difference of two arcs. The theorem states that for a quadrilateral inscribed in a circle, the product of the lengths of the diagonals equals the sum of the products of the two pairs of lengths of opposite sides. The derivations of trigonometric identities rely on a cyclic quadrilateral in which one side is a diameter of the circle.
To find the chords of arcs of 1∘1∘ and (1 2)∘(1 2)∘ he used approximations based on Aristarchus's inequality. The inequality states that for arcs α α and β,β, if 0<β<α<90∘,0<β<α<90∘, then
sin α sin β<α β<tan α tan β.sinα sinβ<α β<tanα tanβ.
Ptolemy showed that for arcs of 1∘1∘ and (1 2)∘,(1 2)∘, the approximations correctly give the first two sexigesimal places after the integer part.
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answered Sep 28, 2021 at 17:01
Michael HardyMichael Hardy
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I like continued fractions. For example, if x is in (-pi/2, pi/2), as mentioned above for power series,
sin(x) = x/1+ x^2/6+ x^2, which matches the first 8 terms of the Taylor series given above.
tan(x) = x/1- x^2/3- x^2/5- x^2/7. This also matches the first 8 terms of the Taylor series for tan(x).
These can be easily converted into rational functions (a polynomial divided by another polynomial) so that only one division is required.
Continued fractions are always worth a try because often, when they match the first few terms of the power series, the remaining terms of their power series are very close to those of the desired function. That is way better than setting those terms to zero, as when you truncate the power series.
Use of a calculator with CAS is extremely helpful in converting power series into continued fractions. Mine is an HP-Prime.
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edited Apr 15, 2020 at 5:48
answered Apr 15, 2020 at 5:39
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188445 | https://www.sciencedirect.com/science/article/abs/pii/S0950705125006501 | Efficient black-box adversarial attacks via alternate query and boundary augmentation - ScienceDirect
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Knowledge-Based Systems
Volume 319, 15 June 2025, 113604
Efficient black-box adversarial attacks via alternate query and boundary augmentation
Author links open overlay panel Jiatian Pi a 1, Fusen Wen b 1, Fen Xia c, Ning Jiang c, Haiying Wu c, Qiao Liu a
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Abstract
Most existing query-based black-box attacks use surrogate models as transferable priors to improve query efficiency. However, these methods still suffer from high query times and complexity due to the following three reasons. First, they usually use a transfer-based strategy to find a starting point, which is not conducive to fast optimization. Second, most of them exploit transferable priors in a complex way that severely constrains query efficiency. Third, their performance usually depends on the number of surrogate models and the more surrogate models, the better the performance. To this end, we propose an optimization framework based on fusion attack and boundary augmentation, which make full use of transfer prior and query feedback to achieve a more effective and efficient attack. Specifically, we first use the surrogate model to conduct a warm-up attack guided by query feedback, which provides a better starting point for fast optimization. Then, we introduce a data-augmentation-based transferable attack into query-based method for alternative query. Since the alternate attack framework can quickly find out the adversarial area of the target model, it improves the query efficiency. Finally, we design a decision boundary enhancement strategy to make the decision boundary of the model more diverse. This strategy can reduce the number of surrogate models used yet still achieve competitive performance. To validate the effectiveness of the proposed method, we conduct experiments with three victim models on the ImageNet dataset. Extensive experiment results show that our method achieves favorable performance against the state-of-the-art methods. While the proposed method gets a 100% attack success rate, the query times can be reduced by several orders of magnitude.
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Keywords
Black-box attack
Query feedback
Transferable prior
Random search
1. Introduction
In recent years, deep neural networks have achieved great success in various vision tasks, such as image classification, , object detection , autonomous driving, and face recognition . As a result, attacking these deep networks has become an interesting topic.
Since Szegedy et al. find that a small search to an input image can cause a deep neural network to misrecognize and propose the concept of adversarial attacks, a lot of progress has been made in adversarial attacks. There are two kinds of adversarial attacks: white-box attacks and black-box attacks. In the context of white-box attacks , , the structure, parameters, and even gradient information of the target model are available, making it easy for an adversary to use those information for attacking the model. Unlike white-box attacks, in black-box attacks, only a small amount of information about the target model is accessible or not at all. This setting makes black-box attacks more practical. Black-box attacks can be divided into query-based attacks and transfer-based attacks. Transfer-based attacks , first generate adversarial examples on the white-box model and then transfer them into the target model. Such methods fail to achieve a high attack success rate, mainly because the surrogate model is quite different from the target model. Query-based attacks can be further divided into score-based , , , and decision-based attacks. Score-based attacks usually perform a gradient estimation and then use the approximate gradient to perform white-box attacks. These methods achieve a high attack success rate, but approximating gradient in high-dimensional spaces is not trivial, and results in expensive query costs.
Inspired by the high consistency among the visual models, some other score-based methods , , improve the efficiency of the query by introducing surrogate models as transfer priors without gradient estimation. Most of these methods use query feedback to find or train an alternative model and require that the decision boundary of the alternative model is closer to the target model. Although these methods have made some progress, they have not fully exploited the relationship between transfer prior and query feedback, resulting in an inefficient query due to the following reasons. (1) Most of them use the transfer-based strategy to find a starting point. Although the starting point found by this way is closer to the decision boundary of the target model than the original example, it suffers from overfitting and inaccurate gradient problems. These problems cause subsequent attacks to pay more attention to the non-adversarial regions of the model, resulting in more query times. (2) They usually first fine-tune the surrogate model with query feedback and then combine the new transfer priors with white-box attack methods. However, these methods consume a lot of computing resources in the process of fine-tuning, resulting in a low query efficiency. (3) Although the introduction of transfer priors can solve the problem of low query efficiency to some extent, these methods usually require to use of a large number of surrogate models. The more surrogate models, the better the performance. However, using too many surrogate models consumes significant computational resources, thus reducing query efficiency.
To address the above-mentioned problems, we propose an optimization framework using fusion attack and boundary augmentation, which combines query feedback and transferable priors more reasonably and efficiently. The proposed method can reduce the number of query times without losing or even improving the attack success rate. Specifically, our method contains three components. First, we use the surrogate model guided by query feedback to find a good starting point for the warm-up attack. Different from the transfer-based strategy, our method can find a starting point that is closer to the target. This means that starting from this point, we need fewer queries to find the adversarial region of the target model. Second, we introduce a data-augmentation-based transferable attack method DST , , into the query-based method SimBA-ODS for alternate queries. Using this alternate query way can effectively avoid falling into local minima and improve query efficiency. Third, we propose a decision boundary augmentation strategy that uses a set of random factors to increase the diversity of the model’s decision boundary. The proposed strategy can generate a large number of virtual decision boundaries to reduce the dependence on the number of surrogate models. To validate the effectiveness of the proposed method, we conduct an ablation study and compare with state-of-the-art methods on the ImageNet dataset. Extensive experiment results demonstrate that the proposed method achieves favorable attack performance.
We summarize the contributions of this paper as follows.
•We propose a query-based attack method that combines query feedback and transferable priors more efficiently. The proposed method contains three novel components: a more accurate starting point searching method, a fusion alternate attack model, and a model boundary random augmentation strategy.
•We propose a data-augmentation-based fusion alternate attack method and a boundary augmentation strategy, which can significantly improve the query efficiency and ensure the effectiveness of the attack.
•Extensive experimental results demonstrate that our method achieves a favorable success rate against the state-of-the-art methods while requiring lower query times.
The rest of the paper is organized as follows. We first introduce some very related attack methods in Section 2. Then, we present the details of the proposed method in Section 3. Next, we conduct a series of validated experiments and give some analysis in Section 4. Finally, we draw a briefly conclusion in Section 5.
2. Related work
Black-box attacks can be roughly divided into transfer-based attacks and query-based attacks. In this section, we mainly introduce these two kinds of black-box attack methods.
2.1. Preliminaries
Adversarial examples can be generated by adding imperceptible noise to clean examples. For a model f, given a clean image x and true label y, the procedure described above for attacking can be expressed as follows: (1)F i n d x a d v,a r g m a x(f(x a d v))≠y,s.t.‖x a d v−x‖p≤ζ,where ‖⋅‖p denotes the l p norm, ζ is the pre-set max searching budget, x a d v is the generated adversarial example. In order to generate adversarial examples that meet the above conditions, Goodfellow et al. first proposed the Fast Gradient Sign Method (FGSM). This method assumes that the loss function is linear or at least locally linear, adding searchings along the direction of gradient ascent to enhance the loss function. The formula for FGSM is as follows: (2)x a d v=x+α⋅s i g n(∇x L(x,y)),where α denotes the searching step size, and ∇x L(x,y) denotes the searching direction. Its search boundary is governed by the l∞ norm. This classic attack method requires obtaining the actual gradient of the target model. In the context of a more realistic black-box attack, the search direction∇x L(x,y) is usually replaced by an approximate gradient.
2.2. Transferability-based black-box attacks
The transferability of adversarial examples means that adversarial examples generated by attacking one model can effectively attack other models under the same task. The early transferability-based black-box attack techniques , , , first attacked the surrogate model with the same task as the target model employing the white-box approach to generate adversarial examples. Then they attacked the black-box target model using the adversarial examples that were produced. However, the surrogate model and the target model differ greatly in terms of structure, training data, etc., which results in a poor attack success rate. To address these issues, existing methods improve the transferability of adversarial examples from both data and model perspectives.
Model augmentation. In order to improve the transferability of adversarial examples, the most direct idea is to integrate multiple models. This method was first proposed by Yanpei Liu et al. , who fused the output confidence scores of different surrogate models by weighted average. Specifically, it can be expressed as, given a set of models f n(x) and a set of weights w n, the ensemble model F(x) is the weighted average of this set of models, that is, F(x)=∑i=0 n w i f i(x). Because of its simplicity and effectiveness, it quickly became a basic operation. After that, a method of combining logits and losses appeared. Although integrating numerous surrogate models to enhance transferability is more effective than using only one surrogate model in terms of attack efficiency, the number of models is constrained by computational resources. In addition to integrating multiple surrogate models, some methods generate adversarial examples by augmenting a single model. The Ghost Network proposed by Yingwei Li et al. generates multiple variant models by adding operations such as dropout and perturbed residual connection (Skip Connection) to a single model, and then integrates these models to generate adversarial examples. The Skip Gradient Method(SGM) proposed by Dongxian Wu et al. obtains more transferable adversarial examples by changing the weight of the residual module. Inspired by this, we pay attention to the influence of the residual module on the decision boundary of the model and introduce a set of random factors to increase the diversity of the decision boundary of the model.
Data augmentation. Another branch of research involves data augmentation to increase the transferability of adversarial examples. The Translation-invariant (TI) technique uses a sequence of transformed images from a spatial domain perspective to enhance the transferability of adversarial examples. Combined with Iterative FGM, the update formula is: (3)x adv t+1=x a d v t+α⋅s i g n(W∗∇x L(x a d v t,y)),where W is a smoothing kernel, and ∗ denotes convolution. Diverse Input Iterative 19 technology improves the generalization ability of adversarial examples by randomly resizing and filling with zero elements. Combined with Iterative FGM, the update formula is: (4)x a d v t+1=C l i p x ɛ{x a d v t+α⋅s i g n(∇x L(T(x a d v t;p),y t r u e;θ))},T(x a d v t;p)=T(x a d v t),p x a d v t,1−p,where α denotes the attack step length, θ represents the model parameters. Scale-Invariant 20 introduces the scale invariance of the deep learning model to optimize the adversarial searching of the input image. Combined with Iterative FGM, the update formula is: (5)x a d v t+1=x a d v t+α⋅s i g n(∇x L(S(x a d v t),y t r u e)).We combine these three techniques into a transfer-based attack method DST, which can help share a portion of queries.
2.3. Query-based black-box attacks
Query-based attacks can be divided into score-based and decision-based attacks according to the type of query feedback. A decision-based attack , exploits only the output labels of the target model. Simply put, the attacker first obtains an initial adversarial example with a large search value and then searches for a smaller search value near the model decision boundary (the boundary between the adversarial area and the non-adversarial area) to obtain the final adversarial example. Although decision-based attack methods are closer to real attack scenarios, the adversarial examples generated by this method are not necessarily effective due to over-reliance on the accuracy of decision boundary estimation, so we focus on score-based attacks. Score-based attacks are able to obtain the output confidence score of the target model. Score-based attacks can be further separated into prior-free attacks and prior-based attacks based on whether or not prior information is used.
Prior-free attack. In the branch of prior-free attacks, the zero-order optimization (ZOO) method was the first to propose the use of finite difference methods for gradient estimation. Despite its effectiveness, its query complexity is proportional to the dimension of the input space, in particular, a high-dimensional input space will result in an exceptionally large number of queries. Afterward, the improved method reduces the number of queries by reducing dimensionality. Bhagoji et al. reduces the cost of the query through two strategies of random feature extraction and principal component analysis. Tu et al. search in the latent space of the autoencoder–decoder to speed up the estimation of gradients. To put it simply, the above ZOO method mainly queries the target model to obtain the output confidence score, then performs gradient estimation, and finally uses the estimated gradient combined with the white-box attack method to generate adversarial examples. Some attack methods re-examine the generation process of adversarial examples from other perspectives. For example, the Natural Evolutionary Strategies(NES) method estimates gradients via a natural evolution strategy. NAttack represents the generation of adversarial examples as a deterministic probability distribution from which potential adversarial examples can be identified. Unlike previous methods, Simple Black-Box Adversarial Attack (SimBA) does not require gradient estimation, but uses a greedy search strategy to update adversarial examples.
Prior-based attack. Bandits notes the presence of data-dependent and time-dependent priors on gradient distributions and integrates them into the spherical gradient estimation framework. The time-dependent prior means that the historical gradient can be used as the prior information of the current gradient. Data-dependent prior means that the gradient around a certain coordinate point in the input image can be used as its prior information. This prior information is mostly used in the gradient estimation framework and needs to be obtained by querying the target model. Although efficient, the high cost of query times is still a challenging problem. As mentioned in Bandits, algorithms such as NES are almost optimal without any prior information, but adding prior information can greatly improve query efficiency. Therefore, we focus on query black-box attacks based on prior information.
In contrast, some methods introduce surrogate model gradients as transferable priors and exploit them through white-box methods. However, unlike the prior mentioned in the Bandits method, the gradient of the surrogate model is used as the prior information of the query algorithm, which is directly integrated into the query-based attack algorithm. Many studies have demonstrated the effectiveness of this prior, which does not require querying the target model. Prior-guided random gradient-free (P-RGF) integrates transferable priors into the gradient estimation framework, achieving a higher attack success rate with fewer queries. TRansferable EMbedding based Black-box Attack (TREMBA) attempts to perform efficient searches in low-dimensional embedding space. Specifically, the method first trains an encoder–decoder based on a large number of pre-trained models, which can generate effective adversarial searchings to the target network in the low-dimensional embedding space. NES is then applied to search for adversarial examples of the target network in this low-dimensional embedding space. Another research is the Hybrid Batch Attack proposed by Suya et al. which combines the transfer-based attack method with the gradient estimation-based attack method, uses the seed-first strategy to implement batch attacks, and improves query efficiency. Simulator Attack utilizes meta-learning to reverse engineer an alternative to the black-box model. During the attack process, the decision boundary of any target model can be simulated by fine-tuning the surrogate model with query feedback. However, the process of training the surrogate model with this method is very complicated and consumes a lot of computing resources. At the same time, this experiment only performs well on small datasets and is not suitable for practical scenarios. SimBA-ODS performs random sampling in the output space through the Output Diversified Sampling (ODS) , which greatly reduces the sampling dimension compared with the traditional random sampling of the input space and can significantly improve the attack efficiency. In our study, we choose SimBA-ODS as the optimization method. Most of the above methods are optimized by methods such as query feedback or meta-learning to find alternative models to black-box models. They all ignore the importance of the transferable prior itself. In contrast to existing approaches, our suggested framework, which combines query feedback with transferable prior in an organic way rather than through over-optimization, significantly increases query efficiency, even if our technique also includes a surrogate model as a transferable prior.
3. Method
We propose a novel attack framework based on fusion attack and boundary augmentation that can efficiently utilize query feedback and transferable priors. The algorithm framework is shown in Fig. 1. Our algorithm consists of three parts: a more accurate starting point search method, a fusion-alternating attack framework, and a model-boundary random augmentation strategy. In this section, we describe each part in detail. Section 3.1 gives an overview of the algorithm framework. Section 3.2 briefly introduces the baseline method SimBA-ODS, which is just a variant of the SimBA method. In Section 3.3, we first describe how to find more accurate starting points. Then, Section 3.4 introduces our proposed fusion-alternating attack framework. Finally, in Section 3.5, the model boundary stochastic augmentation strategy is elaborated. The pseudo-code of this algorithm is shown in Algorithm 1.
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Fig. 1. Framework of the proposed attack method which consists of an accurate starting point searching method, a fusion alternate attack model, and a model boundary random augmentation strategy. γ∗ is a set of random factors that adjust the decision boundary of the model. The original image arrives at a better query starting point x c u r r after the warm-up attack (pink part). From this point, execute the n q query attack (blue part). If the current image is not successfully attacked and the iteration termination condition is not reached, continue to execute n t DST attacks (green part). When the target model misclassifies the current image or the number of queries exceeds the maximum limit, it will stop continuing to iterate, otherwise continue to alternate attacks.
3.1. Overview
Our algorithm framework can be divided into two phases: warm-up attack and alternate attack. The warm-up attack provides a great starting point for subsequent alternating attacks. The alternate attack process alternately queries the surrogate model and the target model to obtain the final adversarial examples. In the warm-up phase, the clean example x first queries the target model to get the confidence score of each category, and then uses the transferable prior to getting x c u r r under the guidance of the confidence score. x c u r r is the better starting point we found. In the alternating attack phase, we introduce a set of random factors r∗=[r 1,r 2,…,r k], and randomly select a set from them at each iteration to adjust the decision boundary of the surrogate model. where k represents the number of skip connections in the surrogate model. During the alternate attack process, the current example x c u r r first executes n q times of SimBA-ODS attacks, and if x t e m p is not yet an adversarial example, it turns to execute n t times of DST attacks. The example x t e m p after performing the alternation attack can successfully fool the target model with a very small number of queries.
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3.2. Baseline method SimBA-ODS
Our strategy is based on random search, a well-known iterative optimization technique first proposed by Rastrigin in 1963. The main idea behind this method is to randomly sample an updated direction in each iteration, and if it improves the objective function, add it to the current sample x′. Despite its simplicity, random search works well in a variety of situations and is independent of the gradient data from the objective function. The method SimBA proposed by Guo et al. was the first to use random search techniques in the field of adversarial attacks. We employ a SimBA variation, SimBA-ODS, which replaces orthonormal basis vectors with ODS sampling, as the optimization method in this section. Given a clean image x, a classifier f and the diversity directions w d∈R C for random sampling from a uniform distribution over [−1,1]C, the definition of ODS direction d O D S as follows: (6)d O D S(x,f,w d)=∇x(w d T f(x))‖∇x(w d T f(x))‖2.
The key technology in the SimBA-ODS is the ODS algorithm which finds a better search space. This search space is low-dimensional and more likely to exist adversarial examples. According to Eq. (6), we cannot directly calculate the ODS searching d O D S(x,f,w d), because the gradient of the target model f is unknown, so a surrogate model is introduced. The original paper stated that multiple surrogate models can make the attack stronger, so the algorithm has a certain dependence on the number of surrogate models. Simultaneously, the algorithm randomly selects a surrogate model and directly uses its gradient as the initial search direction. Due to the large discrepancy between the surrogate model and the target model, this initial searching direction makes subsequent iterations more likely to point to non-adversarial regions of the target model, resulting in substantial query costs. In order to solve the above problems, we carefully design the initial searching direction combined with query feedback, and at the same time introduce a fusion alternating attack framework and model boundary random enhancement strategy in the subsequent attack stage to improve query efficiency.
3.3. Accurate starting point searching
Good initialization is very important when applying neural networks because the objective function is non-convex. On the other hand, since we know nothing about the structure of the search space, many classic random search algorithms such as genetic algorithm and stochastic gradient descent choose to start from a random position to eliminate the preference of the search process. The difference is that we carefully set this random point. We reconsidered the problem of minimizing the number of queries without losing the attack success rate, and believed that a good starting point must be able to accurately guide the subsequent query optimization path. First, construct a vector of the size of the output dimension filled with 0 and 1 through the confidence score obtained from the query, set the largest class (untargeted scenario) or target class (targeted scenario) except the true class to 1, and set the rest of the class to 0. Second, the gradient direction guided by this vector is obtained through an integrated surrogate model. Finally, the original example conducts a one-step SimBA attack in this direction. This means that the attack direction in the warm-up attack phase is determined, not random. This not only effectively combines query feedback and transferable priors, but also obviously provides a better starting point for subsequent query attacks. In query-based black-box attacks, better starting points can be interpreted as points at which candidate examples can cross the decision boundary of the target model at a faster rate (fewer number of queries).
Although these techniques , , also consider the starting point’s significance, they consistently opt for the transfer-based black-box attack technique. Despite being closer to the decision boundary of the target model than the original examples, the candidate adversarial examples generated entirely by the white-box model are far from the target adversarial examples. In order to uncover candidate adversarial examples nearer the decision boundary, we combine query feedback and transferable priors in the warm-up attack phase in an effort to make the most of their use. Fig. 2 shows the difference in attack performance between the general approach and ours.
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Fig. 2. Schematic diagram of accurate starting point searching. The gray part represents the correctly recognized region of the target model, that is, the non-adversarial region of the image. The black border of the gray part represents the decision boundary of the target model, and the part outside the decision boundary represents the adversarial region of the image. Examples of successful attacks (green dots) in the left figure represent the attack process of the general method. They tend to require more iterative steps for the attack to be successful. The successful attack example in the right figure represents the attack process of our proposed method. Starting from our well-designed starting point q t r u e can make the current example quickly point to the adversarial region of the target model. Among them, q 1 and q 2 indicate the direction of white-box attack using two surrogate models under the guidance of query feedback, and q t r u e is calculated by q 1 and q 2.
3.4. Fusion alternate attack
The fusion alternation attack part can be further divided into transfer-based and query-based parts. The DST approach does not query the target model at each iteration, whereas SimBA-ODS does, despite the fact that SimBA-ODS is not quite a query-based approach. Therefore, we roughly divide the alternation attack process into the above two parts. Alternate attack performs n q times of SimBA-ODS attacks first and then performs n t times of DST attacks. The SimBA-ODS approach is detailed in depth in Section 3.1 and will not be repeated here.
Our optimization algorithm is based on a greedy search strategy, and the nature of each decision point taking a local optimum leads to the lack of a technical mechanism for the algorithm to jump out of the local minimum. This means that once the algorithm falls into a local minimum, it is difficult to escape, which is one of the reasons for the low attack success rate. Consequently, it may be more difficult for the optimization process to get stuck in local minima if a transfer-based approach is added to the query-based attack. Meanwhile, the introduction of transfer-based attacks can also help to share part of the queries. A general solution to transfer-based adversarial attacks is to use a white-box model to generate adversarial examples. The classic algorithm is the single-step iterative attack method FGSM proposed by Goodfellow in 2015. Afterward, Kurakin et al. proposed a multi-step iterative attack method I-FGSM , and the experimental results show that it has a stronger white-box attack capability than FGSM. Their iterative formulation is shown as follows: (7)x a d v t+1=C l i p x,ɛ(x a d v t+1+α⋅s i g n(∇x L(f(x a d v t+1)),y)),s.t.‖x a d v t+1−x a d v t+1‖p≤δ,where C l i p x,ɛ(⋅) denotes that x a d v is guaranteed to be in the range (0,255). its searching boundaries can be controlled using either l 2 or l∞ norm. We use the following equation to optimize the adversarial searching. (8)x a d v t+1=C li p x,ɛ(x a d v t+α⋅s i g n(∇x L(f(D(x a d v t))),y)),where D denotes the data transformation method DST we introduced.
In transfer adversarial attacks, a series of I-FGSM-based methods aim to address a well-recognized problem, namely, how to improve the transferability of adversarial examples. Compared with FGSM, I-FGSM has a better attack success rate, but I-FGSM is easy to overfit the specific surrogate model during the iterative process, making it difficult to transfer to other models. Data augmentation is considered to be an effective way to avoid network overfitting and an effective strategy to improve the transferability of adversarial attack tasks. Given a clean image x of width and height, we first scale and transform the image by a certain value. The current image is then randomly resized with probabilityp and zero-pixel padding is applied at the edges. It is worth noting that we optimize the adversarial searching on four different scales of 0.25, 0.5, 0.75, and 1 when performing scaling. Finally, we apply a translation transformation to the current image. Therefore, the introduced data augmentation method can be expressed as: (9)D(x)=W∗∑i=1 4 P a d d i n g(R e s i z e(s i(x))),p x,1−p,where W is a smoothing kernel, and ∗ denotes convolution.
Accurate starting point search and fusion alternating attack strategies both reduce the number of iterations of the optimization process by integrating query feedback with transferable priors. This allows our proposed method to successfully generate adversarial examples with fewer queries, making full use of transferable information and indirectly enhancing the transferability of adversarial examples.
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Fig. 3. Schematic diagram of model boundary enhancement. As shown in the figure above, different squares in the middle of the model represent different data distributions. We use a random factor to move the different distributions toward the decision boundary of different optimal surrogate models in the iterative process, which is more likely to deceive the target model.
3.5. Model boundary random augmentation
The dependence of existing methods on the number of surrogate models is a common problem, and we believe that how to achieve the attack effect of a large number of model ensembles with a small number of models is a problem worth exploring. SGM points out that the gradient of skip connections can improve the attack success rate of black-box adversarial examples, and introduces an attenuation factor to reduce the gradient of the residual module to improve the transferability of adversarial examples. Inspired by this, we introduce a set of random factors to solve the above problems. Different from previous model integration methods, we consider this problem from the essence of the problem, that is, to achieve a large number of model integration effects by increasing the diversity of model decision boundaries. SGM uses fixed factors for all residual modules. SGM points out that skip connections are easier to attack and expose more transferable information, so a fixed attenuation factor is introduced to all residual modules to reduce the gradient of the residual modules. However, the residual modules in the model have different network depths and different model parameters. Attenuating all residual modules to the same extent during the search process for generating adversarial examples may prevent the utilization of the transfer information exposed by the corresponding skip connections at various depths, such as shallow features and deep features.
In order to ensure the diversity of model decision boundaries, we introduce a set of random factors γ = [γ 1, γ 2, …, γ k ] ∈[0.8,1]k, where k represents the number of residual module layers. First, according to the heuristic method, the initialization of this set of random factors is controlled within a specific range [0.8, 1], and then Gaussian noise is added to this set of random factors to generate multiple sets of variable random factors. When computing the gradient optimization searching after backpropagation, a set of factors is randomly selected to adjust the decision boundary of the model. Our model augmentation strategy reduces the number of required surrogate models, in other words, we exploit transferable priors more fully. Given a clean example x and a DNN model f, we write the update direction chosen by the ODS during the alternate attack process as follows: (10)d O D S(x,f,w d,γ∗)=w d T(∇x f(x,γ∗))‖∇x(w d T f(x,γ∗))‖2,∇x f(x,γ∗)=∂L∂z k∏i=1 k−1(γ i+1∗⋅∂f i+1∂z i+1)∂z 0∂x,where Z i(i=1,2,3,…,k) denotes the output of the i th residual layer and f i+1 indicates the i+1-th residual layer. The gradient of the DST method is calculated as in Eq. (10).
It is worth noting that our strategy does not actually change the decision boundary of the surrogate model but virtually increases the model boundary. In the iterative process, our ensemble strategy allows different inputs to move towards the boundary of their corresponding optimal surrogate model, which can be understood as a virtual enhancement of the model boundary. Fig. 3 is a visual description of the process. By using more of the gradient of the skip connection and less of the gradient of the residual module, our suggested model boundary random augmentation produces the effect of virtual enhancement of the model boundary. The focus on the skip connection that exposes more transferable information directly improves the transferable of adversarial examples
4. Experiment
We conduct a series of experiments to validate the high efficiency of our method. The experimental setup is discussed in Section 4.1, including the test dataset, model, and parameter settings. Ablation experiments are then carried out in Section 4.2 to show the effect of different parameter settings and parts of the algorithm on the experimental results. Finally, Section 4.3 shows the untargeted attack performance of all methods under the normal target model and defensive target model respectively. It also demonstrates the targeted attack performance of our method under the normal target model.
4.1. Experiment setting
Dataset and models. We first conducted experiments on the ImageNet dataset. Measuring the attack efficiency under this dataset is an essential way to assess the performance of different attack algorithms. Following previous work , we randomly select 1000 test images in the validation set of the ImageNet dataset to conduct untargeted attack experiments under the normal models and the defense models. At the same time, 100 images are randomly selected from the verification set to conduct targeted attack experiments under the normal models, and the image size is uniformly 299 × 299 × 3. When conducting experiments with targeted attacks, we randomly select classes other than ground truth labels among the 1000 classes as target classes. We choose VGG-16, ResNet-50, Inception-V3 as the victim model, and ResNet-101, DensNet-121 as the surrogate model. All the above models are pre-trained on ImageNet. We implemented the algorithm using PyTorch . We also conduct attacks on the defense models, including JPEG compression , randomization, and guided denoiser . It is worth noting that in this section we use Inception-V3 as the backbone model for JPEG compression and randomization.
To further verify the effectiveness of the proposed method, we also conduct attacks on TinyImageNet dataset . In contrast to the 224 × 224 or 299 × 299 ImageNet datasets, the 64 × 64 TinyImageNet dataset has a lower image resolution and less detail, even though it is based on a subset of the ImageNet dataset. Following previous studies , 1,000 tested images are randomly selected from validation sets for evaluation. We choose ResNeXt-101(32 × 4d), ResNeXt-101 (64 × 4d) as the victim model, and VGG-16, DensNet-121 as the surrogate model. All the above models are pre-trained on TinyImageNet.
Parameters setting. We evaluated our method under the l 2 norm bound when the dataset is ImageNet, and typically limit the maximum searching budgets to δ=0.001×3×299×299≈16.37. And evaluated under the l∞ norm limitation when the dataset is TinyImageNet, the maximum search budget is typically set at δ=0.001×3×64×64≈3.51. We set the maximum query budget to 10,000 for untargeted attacks and 30,000 for targeted attacks in this study. To evaluate the efficacy of our strategy, we provide the attack success rate (ASR), average query numbers (Avg.Q), and median query numbers (Med.Q). When it comes to hyperparameters, we set the step size ϵ to 0.1, query iterations n q to 5, and DST iterations n t to 10. n q and n t are adjustable hyperparameters.
4.2. Ablation study
Ablation studies are performed to verify the effectiveness of various parts of the algorithm and determine the impact of key parameters.
Accurate starting point searching. Our method incorporates query feedback at the initial stage to find better query starting points. In order to verify the effectiveness of our method, based on the experimental settings in Section 4.1, we designed two comparative experiments: random starting point and finding the starting point by transferable prior. The experimental results are shown in Fig. 4. The “No-warm” method starts updating searching from random Gaussian noise, and the “Transfer” method uses the gradient direction of the surrogate model as the update direction for the current example. From Fig. 4, we can draw the following conclusions: (1) The setting of the starting point mainly has a certain impact on the number of queries, but has little impact on the success rate of the attack; (2) By comparing several methods horizontally, it can be concluded that the current example starts to search from our well-designed starting point, and the attack can be successful only by consuming a small number of queries.
Fusion alternate attack. During the attack, we introduce the transfer-based attack method DST to perform alternate queries. To verify the effectiveness of our method, we conduct ablation experiments without introducing the DST method under the experimental settings in Section 4.1. The experimental results are shown in Table 1, where the “No-DST” method means that no DST method is introduced. It can be seen from Table 1 that the introduction of transferable priors in the attack process can not only greatly reduce the number of queries, but also improve the attack success rate. When the random search process of the query falls into a local optimum, the transfer method can help it escape from the local minimum point, which is one of the main reasons for the significant improvement in attack performance. At the same time, DST helps share a portion of queries, so the median and mean numbers of queries are greatly reduced.
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Fig. 4. Comparison of different starting points. An untargeted attack under the l 2 norm is performed on the ImageNet dataset. The left figure shows the average query counts of the three methods under different target models, the middle figure shows the median query counts and the right figure shows the attack success rate.
Model boundary random augmentation. In order to reduce the dependence of the algorithm on the number of surrogate models, we introduce a set of random factors to adjust the decision boundary of the model. In order to verify the effectiveness of the method, we set up 2 surrogate models (ResNet-101, DenseNet-121), 4 surrogate models (VGG19-BN, ResNet-34, DenseNet-121, MobileNet-V2) and 8 surrogate models Model comparison experiments. The latter adds AlexNet, GoogleNet, VGG13-BN and SqueezeNet-1.1 to the four models. All models are from Pytorch Torchvision. The experimental results are shown in Fig. 5. Comparing the experimental results of 2, 4, and 8 surrogate models without adding the model boundary random enhancement strategy, it can be concluded that the more models, the better the attack effect, especially the average number of queries. Comparing the experimental results before and after the model boundary random enhancement strategy, it can be seen that our enhancement strategy can achieve or approach the effect of a large number of surrogate model integration attacks. Even query efficiency has been improved to a certain extent. For example, when attacking the ResNet model, the “Ours ensemble” method outperforms the “No ours ensemble” method on the mean and median of queries under 8 surrogate models.
Table 1. Alternating ablation study. The effect of introducing transfer-based attack methods on the experimental results is compared by performing untargeted attacks under the l 2 norm on the ImageNet dataset.
| Method | VGG-16 | ResNet-50 | Inception-V3 |
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| Empty Cell | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q |
| No-DST | 100% | 53.9 | 9 | 100% | 25.1 | 8 | 98.40% | 143.5 | 23 |
| Ours | 100% | 6.5 | 6 | 100% | 6.7 | 6 | 100% | 10.5 | 7 |
Tuning attack intervaln qandn t. In our attack framework, the candidate adversarial examples after the warm-up attack first perform n q times of SimBA-ODS attack, and then perform n t times of DST attack. To find the optimal alternation interval, we conduct ablation experiments with different intervals under the experimental setup in Section 4.1. The target model we choose is VGG-16, and the surrogate models are ResNet-101 and DenseNet-121. The experimental results are shown in Table 2. From Table 2, we can draw the following conclusions: (1) Different intervals mainly have a certain degree of impact on query efficiency, that is, the best result (n q = 5, n t = 10) and the worst result (n q = 20, n t = 5) have a difference of 12 in the average number of queries. (2) The best experimental results show that transfer-based attacks are twice as many as query-based attacks, indicating that the DST attack we introduced can well help the query optimization process jump out of the local minimum area.
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Fig. 5. We compare the performance of our algorithm with different numbers of surrogate models by conducting untargeted attacks on the ImageNet dataset under the l 2 norm.
4.3. Comparison with state-of-the-art methods
Compared methods. We selected several classic methods from several different types of query attack methods for comparison. Including prior-free attack methods, such as NES , ZOO , AutoZoom , NAttack , RGF , Bandits T and Bandits T D use time- and data-dependent priors in gradient distributions, P-RGF , SimBA-ODS and GFCS use surrogate models as priors, and the transfer-based approach DST. In the DST method, we set the iteration to stop when the searching amount reaches the attack budget δ. Among them, SimBA-ODS is the baseline method. For the above methods, we use the officially released code and fix the original paper parameters unchanged.
Untargeted attack results under the normal models. First, we conduct experiments as usual on the target model without defense strategies. While this setting has been considered to be a simple problem, when we restrict other variables slightly (such as the number of surrogate models), gaps between the algorithms can be seen. We conduct experiments under the model settings described in Section 4.1. Table 3 shows the experimental results of all methods on the ImageNet dataset. The results show that: (1) Our method is able to reduce the average number of queries and the median number of queries to single digits, achieving a qualitative leap compared to the baseline method SimBA-ODS. (2) Compared with other methods, our method greatly reduces the number of queries while increasing the attack success rate to 100%, even under the more challenging Inception-V3. The improvement in the attack success rate is mainly due to the fact that the transfer-based attack method DST performs alternate queries during the query process, which largely prevents the query process from falling into a local minimum. At the same time, our method carefully designs the starting point of the query and cooperates with the optimization framework of the alternate query, which greatly reduces the number of queries. Fig. 6 shows the average number of queries to achieve different expected success rates. From Fig. 6, we can see that our method is more effective. Fig. 7 shows the adversarial examples generated by our method. The first row is the original image, the second row is the noise generated by the algorithm, and the third row is the generated adversarial example.
Table 2. Tuning attack intervaln qandn t. Attack performance(ASR, Avg.Q and Med.Q) of different attack intervals on VGG16.
| Interval | n q = 5 | n q = 10 | n q = 20 |
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| Empty Cell | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q |
| n t = 5 | 100% | 12.6 | 7.0 | 100% | 15.6 | 9.0 | 100% | 19.0 | 9.0 |
| n t= 10 | 100% | 6.5 | 6.0 | 100% | 13.2 | 9.0 | 100% | 17.0 | 9.0 |
We note that GFCS also adopts the ODS strategy. The method first simply performs gradient sampling of the surrogate model and then performs diversified sampling of the ODS output. Since different vision models under the same task have similar decision boundaries, the method performs well on simple tasks, i.e., the method achieves extremely low median queries under all three target models. As can be seen from the large difference between the average number of queries and the median under Inception-V3 in Table 3, the method still requires a large number of queries in more challenging target models and difficult examples. Although the authors are aware of the importance of the surrogate model gradient, it is still underutilized and overly dependent on the number of surrogate models. Our approach differs in that it uses transfer prior and query feedback reasonably in the initial stage as well as during the attack process, and it also introduces a set of stochastic factors to weaken the algorithm’s reliance on the number of surrogate models.
Table 3. Untargeted attack performance on ImageNet. Average number of queries(Avg.Q), median number of queries(Med.Q) and attack success rate(ASR) on ImageNet, under l 2 norm and a maximum query number of 10,000.
| Methods | VGG-16 | ResNet-50 | Inception-V3 |
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| Empty Cell | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q |
| NES | 98.7% | 1103.0 | 816.0 | 98.4% | 988.0 | 714.0 | 95.5% | 1752.0 | 1071.0 |
| AutoZoom | 96.3% | 1589.0 | 949.0 | 94.8% | 2065.0 | 1223.0 | 85.4% | 2443.0 | 1847.0 |
| NAttack | 99.6% | 593.0 | 357.0 | 99.5% | 535.0 | 357.0 | 98.2% | 1020.0 | 510.0 |
| RGF | 99.8% | 749.0 | 561.0 | 99.6% | 673.0 | 510.0 | 97.7% | 1309.0 | 816.0 |
| P-RGF | 100.0% | 468.7 | 204.0 | 100.0% | 234.6 | 153.0 | 100.0% | 1377.9 | 714.0 |
| Bandits T | 94.0% | 584.0 | 225.0 | 96.2% | 1076.0 | 446.0 | 92.4% | 1560.0 | 810.0 |
| Bandits T D | 94.9% | 278.0 | 82.0 | 96.8% | 512.0 | 195.0 | 97.2% | 874.0 | 352.0 |
| DST | 76.2% | – | – | 73.3% | – | – | 7.0% | – | – |
| SimBA-ODS | 99.8% | 436.4 | 166.5 | 99.7% | 432.6 | 163.0 | 96.5% | 1028.2 | 380.0 |
| GFCS | 100.0% | 17.2 | 5.0 | 100.0% | 31.4 | 6.0 | 98.5% | 404.5 | 18.0 |
| Ours | 100.0% | 6.5 | 6.0 | 100.0% | 6.7 | 6.0 | 100.0% | 10.5 | 7.0 |
Untargeted attack results under the defensive models. As can be seen from the results in Table 4, our method still exhibits great query efficiency compared with other methods. Obviously, in the face of a powerful defense model, the general black-box attack method greatly reduces the query efficiency. However, our method can still achieve a 100% attack success rate on JPEG Compression and Guided Denoiser, and it is rare that it can be done with very few queries.
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Fig. 6. Comparisons of the average query at different success rates under the untargeted l 2 norm attack.
Targeted attack results under the normal models. The experimental results are shown in Table 5. For several other comparative attack methods, it takes at least 1249 attacks in targeted attacks to achieve a 100% attack success rate. Even the GFCS method requires 754 queries to achieve a 99% attack success rate. In contrast, our proposed method only needs about 11 queries to achieve a 100% attack success rate, outperforming all compared methods.
Table 4. Untargeted attack performance over defensive methods. Attack success rate(ASR) and average number of queries(Avg.Q) on ImageNet, under l 2 norm and a maximum query number of 10,000.
| Methods | JPEG compression | Randomization | Guided denoiser |
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| Empty Cell | ASR | Avg.Q | ASR | Avg.Q | ASR | Avg.Q |
| NES | 47.3% | 3114.0 | 23.2% | 3632.0 | 48.0% | 3633.0 |
| SPSA | 40.0% | 2744.0 | 9.6% | 3254.0 | 46.0% | 3526.0 |
| RGF | 41.5% | 3126.0 | 19.5% | 3259.0 | 50.3% | 3569.0 |
| P-RGF | 61.4% | 2419.0 | 60.4% | 2153.0 | 51.4% | 2858.0 |
| Bandits T D | 95.8% | 1086.7 | – | – | 20.3% | 759.6 |
| Square | 98.8% | 342.3 | – | – | 98.2% | 392.6 |
| SimBA | 96.0% | 762.8 | – | – | 98.0% | 971.6 |
| DST | 24.3% | – | 22.5% | – | 32.9% | – |
| GFCS | 96.0% | 1498.2 | 92.1% | 2064.7 | 94.5% | 632.2 |
| Ours | 100.0% | 9.4 | 99.8% | 10.7 | 100.0% | 7.8 |
In addition, from Table 3, Table 4, and Table 5, we are surprised to find that the existing methods make the query times of the test images present a long-tail distribution. That is, it is easy to be successfully attacked for easy samples or simple models but performs poorly for hard samples or complex models. Our proposed method does not suffer from this problem and thus is more practical in real-world scenarios.
Untargeted attack results under TinyImageNet dataset. The experimental results are shown in Table 6. It can be seen from the results in Table 6 that our method has a very large advantage in attack efficiency compared with other methods. That is, we can achieve a 100% attack success rate on the black box target model with single-digit access.
Table 5. Targeted attack performance on ImageNet. Average number of queries(Avg.Q), median number of queries(Med.Q) and attack success rate(ASR) on ImageNet, under l 2 norm and a maximum query number of 30,000.
| Methods | Inception-V3 | VGG-16 | ResNet-50 |
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| Empty Cell | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q | ASR | Avg.Q | Med.Q |
| ZOO | 78.0% | 2.11 × 10 6 | – | – | – | – | – | – | – |
| AutoZoom | 100.0% | 14 228.0 | – | – | – | – | – | – | – |
| NAttack | 100.0% | 14 229.0 | – | | – | – | – | – | – |
| Bandits | 100.0% | 25 341.0 | – | – | – | – | – | – | – |
| P-RGF | 84.0% | 5834.6 | 4729.0 | 100.0% | 2402.6 | 1858.0 | 100.0% | 1249.2 | 736.0 |
| SimBA-ODS | 73.0% | 7607.6 | 5218.0 | 98.0% | 3316.0 | 2061.0 | 97.0% | 3077.9 | 1882.0 |
| GFCS | 76.0% | 9692.8 | 2777.0 | 99.0% | 1547.4 | 402.0 | 99.0% | 754.1 | 116.0 |
| Ours | 100.0% | 12.0 | 8.0 | 100.0% | 11.0 | 8.0 | 100.0% | 10.5 | 8.0 |
We also discuss the additional time complexity besides black-box query. Under the experimental setting of Section 4.1, the running time of NES, RGF, P-RGF, Bandits, and Ours methods attacking the target ResNet101-32 × 4d model is 1.23 h, 1.96 h, 3.73 h, 2.08 h, and 0.22 h respectively. This demonstrates the effectiveness of our suggested approach once again. This is mostly because our approach utilizes the transferable information as efficiently as possible during both the alternating attack stage and the warm-up attack stage. The current example can quickly approach the adversarial region of the model thanks to the combination of transferable prior and query feedback.
Table 6. Untargeted attack performance on TinyImageNet. Average number of queries(Avg.Q), median number of queries(Med.Q) and attack success rate(ASR) on TinyImageNet, under l∞ norm and a maximum query number of 10,000.R32:ResNeXt-101 (32 × 4d), R64: ResNeXt-101 (64 × 4d).
| Methods | ASR | Avg.Q | Med.Q |
--- --- |
| Empty Cell | R 32 | R 64 | R 32 | R 64 | R 32 | R 64 |
| NES | 45.30% | 45.50% | 2104.0 | 2078.0 | 765.0 | 816.0 |
| RGF | 85.30% | 87.40% | 2088.0 | 2087.0 | 1280.0 | 1305.0 |
| P-RGF | 83.90% | 85.90% | 1583.0 | 1581.0 | 657.0 | 690.0 |
| Meta Attack | 33.80% | 36% | 4101.0 | 4012.0 | 3712.0 | 3649.0 |
| Bandits | 94.10% | 95.30% | 1737.0 | 1662.0 | 954.0 | 1014.0 |
| Simulator Attack | 96.80% | 97.90% | 1380.0 | 1445.0 | 850.0 | 878.0 |
| Ours | 100% | 100% | 5.7 | 5.7 | 6.0 | 5.0 |
1. Download: Download high-res image (1018KB)
2. Download: Download full-size image
Fig. 7. Adversarial examples generated by our method.
5. Conclusions
In this paper, we propose a novel black-box query optimization framework, which can make full use of transfer prior and query feedback more effectively and efficiently. The proposed framework contains three novel components: a more accurate starting point searching method, a fusion alternate attack model, and a model boundary random augmentation strategy. The proposed starting point searching method conducts a warm-up attack guided by query feedback, and therefore provides a more accurate starting point for fast optimization. What is more, the proposed fusion attack mechanism reduces the possibility of subsequent queries getting stuck in local minima. Furthermore, the proposed boundary random augmentation strategy reduces the number of surrogate models used yet still achieves good performance. These three modules solve the query and calculation efficiency problems faced by different stages in the adversarial attack framework, thus greatly improving the overall efficiency of the attack. Extensive experimental results on the ImageNet dataset demonstrate that the proposed method achieves favorable performance.
CRediT authorship contribution statement
Jiatian Pi: Writing – original draft, Funding acquisition, Data curation, Conceptualization. Fusen Wen: Writing – original draft, Validation, Conceptualization. Fen Xia: Software, Resources, Investigation. Ning Jiang: Software, Resources, Project administration. Haiying Wu: Software, Resources, Project administration. Qiao Liu: Writing – review & editing, Writing – original draft, Supervision, Methodology.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgments
This research is supported by the Natural Science Foundation of Chongqing, China (Grant No. CSTB2024NSCQ-LZX0039), by the National Natural Science Foundation of China (Grant No. 62302073), by the Science and Technology Research Program of Chongqing Municipal Education Commission (Grant No. KJZD-K202200501), and by the Chongqing Education Commission Key ProjectKJZD-K202114802.
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Data availability
Data will be made available on request.
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188446 | https://calculator.academy/price-per-pound-calculator/ | Skip to content
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Price Per Pound Formula
The following formula is used to calculate the price per pound.
PPP = TP / TW
Formula source: University of Minnesota Extension — “Unit Pricing” (section “Fresh produce,” step 2)
Where PPP is the price per pound
TP is the total price per unit
TW is the total weight per unit
Price Per Pound Definition
A price per pound is the total monetary price per weight in pounds of an object or product.
Price Per Pound Example
How to calculate a price per pound?
The first step in calculating a price per pound is to determine the total price per unit. To do this, take the total price and divide it by the total number of units.
For this example, the price per unit is found to be $5.00
Next, determine the total weight per unit. Similar to the first step, the weight per unit can be found by dividing the total weight by the total number of units.
In this case, this yields a weight of 2lbs per unit.
Finally, calculate the price per pound using the formula above:
PPP = TP / TW
PPP = $5.00 / 2
PPP = $2.50 / lb
FAQ
Why is a price per pound used?
Price per pound is used to easily compare the cost of different items, regardless of their size or weight. This allows consumers to make informed decisions about which items to purchase and how much of them to buy.
How is a price per pound calculated?
Price per pound is calculated by dividing the cost of an item by its weight in pounds. For example, if a bag of apples costs $4 and weighs 2 pounds, the price per pound of apples would be $2.
How does the price per pound affect our purchases?
Price per pound affects our purchases by allowing us to compare the cost of different items and make informed decisions about which items to purchase and how much of them to buy. It also helps us control portion sizes and make informed decisions about buying in bulk.
Can the price per pound vary for the same item at different stores?
Yes, the price per pound can vary for the same item at different stores. It’s important to compare prices at different stores to find the best deal.
How does buying in bulk affect the price per pound?
Buying in bulk can often result in a lower overall cost, but it’s essential to consider the price per pound when making this decision. The price per pound may be lower for a larger item, but the overall cost may still be higher.
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When comparing prices per pound, it’s essential to consider the quality and freshness of the item, as well as any additional costs, such as packaging or shipping.
How does the weight of an item affect the price per pound?
The weight of an item directly affects the price per pound. A heavier item will result in a lower price per pound, while a lighter item will result in a higher price per pound.
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188447 | https://numeracyliteracy.com/happy-and-unhappy-numbers/ | Happy and Unhappy Numbers
A number (positive integer) is called a happy number when it is replaced by the sum of the squares of its digits on a repeated basis until the sum of the squares of its digits equals to 1. The numbers for which the process of summing up of the squares of its digits ends in 1 are happy numbers. On the other hand, numbers that don’t end in 1 are called unhappy or sad numbers.
Let’s look at the examples below to understand the concept behind it better.
Example 1:- 23 is a happy number. Let’s see how.
23 = 22 + 32 = 4 + 9 = 13 (sum of the squares of its digits)
13 = 12 + 32 = 1 + 9 = 10 (sum of the squares of the digits)
10 = 12 + 02 = 1 (sum of the squares of the digits)
In the above example, we noticed that, when the number 23 is replaced by the sum of the squares of its digits repeatedly, it ends in 1. Therefore, 23 is a happy number.
Example 2:- 36 is an unhappy number. Let’s see how. 36 = 32 + 62 = 9 + 36 = 45
45 = 42 + 52 = 16 + 25 = 41
41 = 42 + 12 = 16 + 1 = 17
17 = 12 + 72 = 1 + 49 =50
50 = 52 + 02 = 25 + 0 = 25
25 = 22 + 52 = 4 + 25 = 29
29 = 22 + 92 = 4 + 81 = 85
85 = 82 + 52 = 64 + 25 = 89
89 = 8 + 9 = 64 + 81 = 145
145 = 1 + 4 + 5 = 1 + 16 + 25 = 42
42 = 4 + 2 = 16 + 4 = 20
20 = 2 + 0 = 4 + 0 = 4
Here, the number 36 loops endlessly in a cycle which doesn’t end in 1. Therefore 36 is an unhappy or sad number.
Properties of Happy and unhappy numbers:-
If a number is happy, then all the numbers in its sequence (Obtained by the sum of the squares of the digits), are also happy numbers. Look at the example 1. Here, since 23 is a happy number, all other numbers in its sequence, i.e. 13, 10 and 1 are also happy numbers.
If a number is unhappy, all the numbers in its sequence (obtained by the sum of the squares of the digits) are also unhappy numbers. In example 2, since 36 is an unhappy number, all the numbers in its sequence like, 45, 41, 17, 50, 25, 29, 85, 89, 145, 42, 20 and 4 are also unhappy numbers.
The happiness of a number is unaffected, if the digits are rearranged in any manner.
The happiness of a number is also unaffected, if one or more zeros are inserted anywhere in the number or removed from anywhere in the number.
There are infinitely many happy numbers.
There are infinitely many unhappy numbers.
Any permutation or arrangement of the digits of a happy or unhappy number is also a happy or unhappy number.
Happy numbers between 1 to 100
1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100
Happy numbers between 101 to 500
103, 109, 129, 130, 133, 139, 167, 176, 188, 190, 192, 193, 203, 208, 219, 226, 230, 236, 239, 262, 263, 280, 291, 293, 301, 302, 310, 313, 319, 320, 326, 329, 331, 338, 356, 362, 365, 367, 368, 376, 379, 383, 386, 391, 392, 397, 404, 409, 440, 446, 464, 469, 478, 487, 490, 496
Happy numbers between 501 to 1000
536, 556, 563, 565, 566, 608, 617, 622, 623, 632, 635, 637, 638, 644, 649, 653, 655, 656, 665, 671, 673, 680, 683, 694, 700, 709, 716, 736, 739, 748, 761, 763, 784, 790, 793, 802, 806, 818, 820, 833, 836, 847, 860, 863, 874, 881, 888, 899, 901, 904, 907, 910, 912, 913, 921, 923, 931, 932, 937, 940, 946, 964, 970, 973, 989, 998, 1000
HAPPY LEARNING……………………………..
← Using 4s to make numbers 1 to 10
Weak Prime Numbers →
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188448 | https://www.khanacademy.org/science/ap-physics-2/ap-geometric-optics/ap-mirrors/v/refraction-and-snell-s-law | Refraction and Snell's law (video) | Khan Academy
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AP®︎/College Physics 2
Course: AP®︎/College Physics 2>Unit 5
Lesson 3: Refraction
Refraction and Snell's law
Refraction through glass slab
Snell's law example
Snell's law of refraction
Reflection and refraction questions
Total internal reflection
Science>
AP®︎/College Physics 2>
Geometric optics>
Refraction
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Refraction and Snell's law
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Refraction and Snell's Law.Created by Sal Khan.
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Darmon 7 years ago Posted 7 years ago. Direct link to Darmon's post “A merely theoretical post...” more A merely theoretical postulation: if we were to create a beam of light that is comprised of a single stream of photons, would this light beam undergo refraction? I ask this question because the theory explaining the occurrence of refraction always relies on the fact that one portion of a light beam will hit the second medium before another portion, implying that the light ray must have "width" to undergo refraction.
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Andrew M 7 years ago Posted 7 years ago. Direct link to Andrew M's post “It's not merely theoretic...” more It's not merely theoretical, it's a question testable by experiment, and the answer is yes, it will refract. You can see a demonstration of behavior similar to (but not the same as) this, in this video, where a beam of single photons diffracts. Diffraction is not refraction but it is another wavelike behavior that seems like it should be impossible for particles to do, but they do.
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AMR™ 12 years ago Posted 12 years ago. Direct link to AMR™'s post “What if the medium is vic...” more What if the medium is vice versa ,first the slower medium and then passes through faster one? What would happen actually..??/// I am little bit of curious to know it...
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AKSHAI 12 years ago Posted 12 years ago. Direct link to AKSHAI's post “if light travels from a d...” more if light travels from a denser medium to a rarer medium then it actually bends away from the normal as the speed of light increases as it enters the rarer(hence faster) medium.
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udaygsneaker 5 years ago Posted 5 years ago. Direct link to udaygsneaker's post “When light travels from a...” more When light travels from a denser medium to a rarer medium, it gains velocity, but to gain velocity wouldn't the photon need to gain energy to increases its velocity? If yes then from where would it get the energy?
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Charles LaCour 5 years ago Posted 5 years ago. Direct link to Charles LaCour's post “This gets a bit complex. ...” more This gets a bit complex. Light waves or photons are expressions of vibrations in the electromagnetic field. Light interacts with electrically charged objects. When light interacts with a material like glass it interacts with the electromagnetic field in the glass which causes interference between the electromagnetic oscillations of light and the charged particles in the material. The resulting fluctuation in the electromagnetic field has a slower phase propagation than the light wave had. When the fluctuation gets to the edge of the glass the charged particles in the glass no longer interfere with the lights fluctuation so the speed of its phase propagation is back to its normal speed. There is no need for a increase or decrease in energy for this to happen.
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The Sky Rider 3 years ago Posted 3 years ago. Direct link to The Sky Rider's post “Doesn't this mean that it...” more Doesn't this mean that it is technically possibly to create a infinite loop of light?, by altering the mediums to constantly turn the light in the direction of the circle used to create the loop?
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arjunan 2 years ago Posted 2 years ago. Direct link to arjunan's post “then does a light ray vio...” more then does a light ray violate the law of rectilinear movement of a light ray during refraction?
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Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “Light still moves in a st...” more Light still moves in a strait line in a constant medium. When a light ray encounters a change in what it is traveling through this can affect its speed and direction.
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Mahadev 10 years ago Posted 10 years ago. Direct link to Mahadev's post “Why is the speed of light...” more Why is the speed of light different in different mediums?Does this have anything to do with the mediums density?
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Mr. Burton 10 years ago Posted 10 years ago. Direct link to Mr. Burton's post “Somewhat. Even light will...” more Somewhat. Even light will interact with matter in some way. Because of this, some mediums are easier to traverse than others.
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Akshit Pathania 11 years ago Posted 11 years ago. Direct link to Akshit Pathania's post “Is it okay if I write Sne...” more Is it okay if I write Snell's law as
sin i/ sin r = k
with i being angle of incidence, r being angle of refraction and k being a constant?
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roht2326 11 years ago Posted 11 years ago. Direct link to roht2326's post “Yes, if k=nr/ni, the rati...” more Yes, if k=nr/ni, the ratio of the refractive indices. However, usually you want to find r, so a more convenient equation would be r = arcsin( ni sin(I) / nr ).
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Drashti 8 years ago Posted 8 years ago. Direct link to Drashti's post “Why the ray of light inci...” more Why the ray of light incident normally on the surface of mediums does not suffer refraction but passes undeviated? I request you to answer this question by some logical means and not by theoretical way (Snell's law).
A satisfactory answer to this question would be a great help. Thank you in advance!
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Mark Geary 8 years ago Posted 8 years ago. Direct link to Mark Geary's post “Refraction is a change in...” more Refraction is a change in a wave's direction due to a change in speed. If the wave impinges the surface normally, the change in speed is applied equally across the wavefront, yielding no change in direction. The wave slows down (or speeds up, depending on the relative indices of refraction), but does not change direction.
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ayesharashid011 9 years ago Posted 9 years ago. Direct link to ayesharashid011's post “what does it mean when we...” more what does it mean when we say: a perfectly smooth reflecting surface cannot be seen.
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Andrew M 9 years ago Posted 9 years ago. Direct link to Andrew M's post “what would you see if it ...” more what would you see if it was perfectly smooth and reflected every single bit of light?
Hint: can you see a mirror? Are you sure?
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santhosh prabahar 8 years ago Posted 8 years ago. Direct link to santhosh prabahar's post “As light don't pass throu...” more As light don't pass through opaque objects , can we say that those objects have a refractive index of infinity?
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Video transcript
In the last couple of videos we talked about reflection. And that's just the idea of the light rays bouncing off of a surface. And if the surface is smooth, the incident angle is going to be the same thing as the reflected angle. We saw that before, and those angles are measured relative to a perpendicular. So that angle right there is going to be the same as that angle right there. That's essentially what we learned the last couple of videos. What we want to cover in this video is when the light actually doesn't just bounce off of a surface but starts going through a different medium. So in this situation, we will be dealing with refraction. Refraction. Refraction, you still have the light coming in to the interface between the two surfaces. So let's say--so that's the perpendicular right there, actually let me continue the perpendicular all the way down like that. And let's say we have the incident light ray coming in at some, at some angle theta 1, just like that...what will happen--and so let's say that this up here, this is a vacuum. Light travels the fastest in a vacuum. In a vacuum. There's nothing there, no air, no water, no nothing, that's where the light travels the fastest. And let's say that this medium down here, I don't know, let's say it's water. Let's say that this is water. All of this. This was all water over here. This was all vacuum right up here. So what will happen, and actually, that's kind of an unrealistic-- well, just for the sake of argument, let's say we have water going right up against a vacuum. This isn't something you would normally just see in nature but let's just think about it a little bit. Normally, the water, since there's no pressure, it would evaporate and all the rest. But for the sake of argument, let's just say that this is a medium where light will travel slower. What you're going to have is is this ray is actually going to switch direction, it's actually going to bend. Instead of continuing to go in that same direction, it's going to bend a little bit. It's going to go down, in that direction just like that. And this angle right here, theta 2, is the refraction. That's the refraction angle. Refraction angle. Or angle of refraction. This is the incident angle, or angle of incidence, and this is the refraction angle. Once again, against that perpendicular. And before I give you the actual equation of how these two things relate and how they're related to the speed of light in these two media-- and just remember, once again, you're never going to have vacuum against water, the water would evaporate because there's no pressure on it and all of that type of thing. But just to--before I go into the math of actually how to figure out these angles relative to the velocities of light in the different media I want to give you an intuitive understanding of not why it bends, 'cause I'm not telling you actually how light works this is really more of an observed property and light, as we'll learn, as we do more and more videos about it, can get pretty confusing. Sometimes you want to treat it as a ray, sometimes you want to treat it as a wave, sometimes you want to treat it as a photon. But when you think about refraction I actually like to think of it as kind of a, as a bit of a vehicle, and to imagine that, let's imagine that I had a car. So let me draw a car. So we're looking at the top of a car. So this is the passenger compartment, and it has four wheels on the car. We're looking at it from above. And let's say it's traveling on a road. It's traveling on a road. On a road, the tires can get good traction. The car can move pretty efficiently, and it's about to reach an interface it's about to reach an interface where the road ends and it will have to travel on mud. It will have to travel on mud. Now on mud, obviously, the tires' traction will not be as good. The car will not be able to travel as fast. So what's going to happen? Assuming that the car, the steering wheel isn't telling it to turn or anything, the car would just go straight in this direction. But what happens right when--which wheels are going to reach the mud first? Well, this wheel. This wheel is going to reach the mud first. So what's going to happen? There's going to be some point in time where the car is right over here. Where it's right over here. Where these wheels are still on the road, this wheel is in the mud, and that wheel is about to reach the mud. Now in this situation, what would the car do? What would the car do? And assuming the engine is revving and the wheels are turning, at the exact same speed the entire time of the simulation. Well all of a sudden, as soon as this wheel hits the medium, it's going to slow down. This is going to slow down. But these guys are still on the road. So they're still going to be faster. So the right side of the car is going to move faster than the left side of the car. So what's going to happen? You see this all the time. If the right side of you is moving faster than the left side of you, you're going to turn, and that's exactly what's going to happen to the car. The car is going to turn. It's going to turn in that direction. And so once it gets to the medium, it will now travel, it will now turn-- from the point of the view from the car it's turning to the right. But it will now travel in this direction. It will be turned when it gets to that interface. Now obviously light doesn't have wheels, and it doesn't deal with mud. But it's the same general idea. When I'm traveling from a faster medium to a slower medium, you can kind of imagine the wheels on that light on this side of it, closer to the vertical, hit the medium first, slow down, so light turns to the right. If you were going the other way, if I had light coming out of the slow medium, so let's imagine it this way. Let's have light coming out of the slow medium. And if we use the car analogy, in this situation, the left side of the car is going to-- so if the car is right over here, the left side of the car is going to come out first so it's going to move faster now. So the car is going to turn to the right, just like that. So hopefully, hopefully this gives you a gut sense of just how to figure out which direction the light's going to bend if you just wanted an intuitive sense. And to get to the next level, there's actually something called Snell's Law. Snell's Law. Snell's Law. And all this is saying is that this angle-- so let me write it down here--so let's say that this velocity right here is velocity 2 this velocity up here was velocity 1, going back to the original. Actually, let me draw another diagram, just to clean it up. And also that vacuum-water interface example, I'm not enjoying it, just because it's a very unnatural interface to actually have in nature. So maybe it's vacuum and glass. That's something that actually would exist. So let's say we're doing that. So this isn't water, this is glass. Let me redraw it. And I'll draw the angles bigger. So let me draw a perpendicular. And so I have our incident ray, so in the vacuum it's traveling at v1--and in the case of a vacuum, it's actually going at the speed of light, or the speed of light in a vacuum, which is c, or 300,000 kilometers per second, or 300 million meters per second--let me write that-- so c is the speed of light in a vacuum, and that is equal to 300-- it's not exactly 300, I'm not going into significant digits-- this is true to three significant digits--300 million meters per second. This is light in a vacuum. Light in vacuum. And I don't mean the thing that you use to clean your carpet with, I mean an area of space that has nothing in it. No air, no gas, no molecules, nothing in it. That is a pure vacuum and that's how fast light will travel. Now it's travelling really fast there, and let's say that--and this applies to any two mediums-- but let's say it gets to glass here, and in glass it travels slower, and we know for our example, this side of the car is going to get to the slower medium first so it's going to turn in this direction. So it's going to go like this. We call this v2. Maybe I'll draw it--if you wanted to view these as vectors, maybe I should draw it as a smaller vector v2, just like that. And the angle of incidence is theta 1. And the angle of refraction is theta 2. And Snell's Law just tells us the ratio between v2 and the sin-- remember Soh Cah Toa, basic trig function-- and the sin of the angle of refraction is going to be equal to the ratio of v1 and the angle--the sin of the angle of incidence. Sin of theta 1. Now if this looks confusing at all, we're going to apply it a bunch in the next couple of videos. But I want to show you also that there's many many ways to view Snell's Law. You may or may not be familiar with the idea of an index of refraction. So let me write that down. Index of refraction. Index, or refraction index. And it's defined for any medium, for any material. There's an index of refraction for vacuum, for air, for water. For any material that people have measured it for. And they usually specify it as n. And it is defined as the speed of light in a vacuum That's c. Divided by the velocity of light in that medium. So in our example right here, we could rewrite this. We could rewrite this in terms of index of refraction. Let me do that actually. Just cause that's sometimes the more typical way of viewing Snell's Law. So I could solve for v here if I--one thing I could do is just--if n is equal to c divided by v then v is going to be equal to c divided by n. And I can multiply both sides by v if you don't see how I got there. The intermediary step is, multiply both sides times v, you get v times n is equal to c, and then you divide both sides by n, you get v is equal to c over n. So I can rewrite Snell's Law over here as instead of having v2 there, I could write instead of writing v2 there I could write the speed of light divided by the refraction index for this material right here. So I'll call that n2. Right, this is material 2, material 2 right over there. Right, that's the same thing as v2 over the sin of theta 2 is equal to v1 is the same thing as c divided by n1 over sin of theta 1. And then we could do a little bit of simplification here, we can multiple both sides of this equation--well, let's do a couple of things. Let's-- Actually, the simplest thing to do is actually take the reciprocal of both sides. So let me just do that. So let me take the reciprocal of both sides, and you get sin of theta 2 over cn2 is equal to sin of theta 1 over c over n1. And now let's multiply the numerator and denominator of this left side by n2. So if we multiply n2 over n2. We're not changing it, this is really just going to be 1, but this guy and this guy are going to cancel out. And let's do the same thing over here, multiply the numerator and the denominator by n1, so n1 over n1. That guy, that guy, and that guy are going to cancel out. And so we get n2 sin of theta 2 over c is equal to n1 sin of theta 1 over c. And now we can just multiply both sides of this equation by c and we get the form of Snell's Law that some books will show you, which is the refraction index for the slower medium, or for the second medium, the one that we're entering, times the index of the sin of the index of refraction is equal to the refraction index for the first medium times the sin of the angle of incidence. The incident angle. So this is another version right here This is another version right there of Snell's Law. Let me copy and paste that. And if this is confusing to you, and I'm guessing that it might be, especially if this is the first time you're seeing it, we're going to apply this in a bunch of videos, in the next few videos, but I really just want to make sure, I really just want to make sure you're comfortable with it. So these are both equivalent forms of Snell's Law. One deals with the velocities, directly deals with the velocities, right over here, the ratio of the velocity to the sin of the incident or refraction angle and here it uses the index of refraction. And the index of refraction really just tells you it's just the ratio of the speed of light to the actual velocity. So something where light travels really slowly where light travels really slowly, this will be a smaller number. And if this is a smaller number, this is a larger number. And we actually see it here. And you're going to see a little tidbit of the next video right over here. But here's a bunch of refraction indices for different materials. It's obviously 1 for a vacuum, because for a vacuum you have the refraction index is going to be c divided by the speed of light in that material. Well, in a vacuum it's traveling at c. So it's going to be 1. So that's where that came from. And you can see in air, the speed is only slightly smaller, this number's only going to be slightly smaller than the speed of light in a vacuum. So in air, it's still pretty close to a vacuum. But then for a diamond, it's traveling a lot slower. Light is travelling a lot slower in a diamond than it is in a vacuum. Anyway, I'll leave you there, we're going to do a couple more videos, we're going to do more examples using Snell's Law. Hopefully you got the basic idea of refraction. And in the next video, I'll actually use this graphic right here to help us visualize why it looks like the straw got bent.
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Top Voted |
188449 | https://mathsinteresting.quora.com/Find-all-positive-integer-solutions-to-math-x-2-xy-y-2-13-math | Something went wrong. Wait a moment and try again.
Maths Interesting
A space is about all about interesting problems in mathematics. All can learn, ask, answer and contribute.
Find all positive integer solutions to
x
2
−
x
y
+
y
2
=
13
?
2 Answers
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Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
2y
Completing the square and then multiplying both sides by 4 yields
(2x−y)2+3y2=52.
The second term implies that y∈{1,2,3,4}.
Substituting these into the equation above and noting that we need x∈N, we conclude that
(x,y)=(4,1),(4,3),(3,4),(1,4).
Ramakrishnan Parthasarathy
Forgot more Math than I once knew in a non-tech career!
·
2y
x=y does not offer an integer solution for x2=13 or y2=13
Arbitrarily choosing x>y,
x(x−y)=13−y2
13−y2>0∈Z+ when y=1,2,3
y=1⟹x(x−1)=12, which has a solution x=4∈Z+
y=2⟹x(x−2)=9 , which has no solution ∈Z+
y=3⟹x(x−3)=4 , which has a solution x=4∈Z+
So, the solutions are [x,y]=[4,1] and [4,3] and, accounting for y>x , the flipped versions [1,4] and [3,4]
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PART IIICapacity and Powers
Marginal note:Capacity of a corporation
15 (1) A corporation has the capacity and, subject to this Act, the rights, powers and privileges of a natural person.
Marginal note:Idem
(2) A corporation may carry on business throughout Canada.
Marginal note:Extra-territorial capacity
(3) A corporation has the capacity to carry on its business, conduct its affairs and exercise its powers in any jurisdiction outside Canada to the extent that the laws of such jurisdiction permit.
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Marginal note:Powers of a corporation
16 (1) It is not necessary for a by-law to be passed in order to confer any particular power on the corporation or its directors.
Marginal note:Restricted business or powers
(2) A corporation shall not carry on any business or exercise any power that it is restricted by its articles from carrying on or exercising, nor shall the corporation exercise any of its powers in a manner contrary to its articles.
Marginal note:Rights preserved
(3) No act of a corporation, including any transfer of property to or by a corporation, is invalid by reason only that the act or transfer is contrary to its articles or this Act.
1974-75-76, c. 33, s. 16
1978-79, c. 9, s. 1(F)
Marginal note:No constructive notice
17 No person is affected by or is deemed to have notice or knowledge of the contents of a document concerning a corporation by reason only that the document has been filed by the Director or is available for inspection at an office of the corporation.
1974-75-76, c. 33, s. 17
1978-79, c. 9, s. 1(F)
Marginal note:Authority of directors, officers and agents
18 (1) No corporation and no guarantor of an obligation of a corporation may assert against a person dealing with the corporation or against a person who acquired rights from the corporation that
(a) the articles, by-laws and any unanimous shareholder agreement have not been complied with;
(b) the persons named in the most recent notice sent to the Director under section 106 or 113 are not the directors of the corporation;
(c) the place named in the most recent notice sent to the Director under section 19 is not the registered office of the corporation;
(d) a person held out by a corporation as a director, officer, agent or mandatary of the corporation has not been duly appointed or has no authority to exercise the powers and perform the duties that are customary in the business of the corporation or usual for a director, officer, agent or mandatary;
(e) a document issued by any director, officer, agent or mandatary of a corporation with actual or usual authority to issue the document is not valid or genuine; or
(f) a sale, lease or exchange of property referred to in subsection 189(3) was not authorized.
Marginal note:Exception
(2) Subsection (1) does not apply in respect of a person who has, or ought to have, knowledge of a situation described in that subsection by virtue of their relationship to the corporation.
R.S., 1985, c. C-44, s. 18
2001, c. 14, s. 8
2011, c. 21, s. 15(E)
Previous Version
PART IVRegistered Office and Records
Marginal note:Registered office
19 (1) A corporation shall at all times have a registered office in the province in Canada specified in its articles.
Marginal note:Notice of registered office
(2) A notice of registered office in the form that the Director fixes shall be sent to the Director together with any articles that designate or change the province where the registered office of the corporation is located.
Marginal note:Change of address
(3) The directors of a corporation may change the place and address of the registered office within the province specified in the articles.
Marginal note:Notice of change of address
(4) A corporation shall send to the Director, within fifteen days of any change of address of its registered office, a notice in the form that the Director fixes and the Director shall file it.
R.S., 1985, c. C-44, s. 19
2001, c. 14, s. 9
2018, c. 8, s. 6(F)
Previous Version
Marginal note:Corporate records
20 (1) A corporation shall prepare and maintain, at its registered office or at any other place in Canada designated by the directors, records containing
(a) the articles and the by-laws, and all amendments thereto, and a copy of any unanimous shareholder agreement;
(b) minutes of meetings and resolutions of shareholders;
(c) copies of all notices required by section 106 or 113; and
(d) a securities register that complies with section 50.
Marginal note:Directors records
(2) In addition to the records described in subsection (1), a corporation shall prepare and maintain adequate accounting records and records containing minutes of meetings and resolutions of the directors and any committee thereof.
Marginal note:Retention of accounting records
(2.1) Subject to any other Act of Parliament and to any Act of the legislature of a province that provides for a longer retention period, a corporation shall retain the accounting records referred to in subsection (2) for a period of six years after the end of the financial year to which the records relate.
Marginal note:Records of continued corporations
(3) For the purposes of paragraph (1)(b) and subsection (2), where a body corporate is continued under this Act, “records” includes similar records required by law to be maintained by the body corporate before it was so continued.
Marginal note:Place of directors records
(4) The records described in subsection (2) shall be kept at the registered office of the corporation or at such other place as the directors think fit and shall at all reasonable times be open to inspection by the directors.
Marginal note:Records in Canada
(5) If accounting records of a corporation are kept outside Canada, accounting records adequate to enable the directors to ascertain the financial position of the corporation with reasonable accuracy on a quarterly basis shall be kept at the registered office or any other place in Canada designated by the directors.
Marginal note:When records or registers kept outside Canada
(5.1) Despite subsections (1) and (5), but subject to the Income Tax Act, the Excise Tax Act, the Customs Act and any other Act administered by the Minister of National Revenue, a corporation may keep all or any of its corporate records and accounting records referred to in subsection (1) or (2) at a place outside Canada, if
(a) the records are available for inspection, by means of a computer terminal or other technology, during regular office hours at the registered office or any other place in Canada designated by the directors; and
(b) the corporation provides the technical assistance to facilitate an inspection referred to in paragraph (a).
Marginal note:Offence
(6) A corporation that, without reasonable cause, fails to comply with this section is guilty of an offence and liable on summary conviction to a fine not exceeding five thousand dollars.
R.S., 1985, c. C-44, s. 20
1994, c. 24, s. 8
2001, c. 14, s. 10
Marginal note:Access to corporate records
21 (1) Subject to subsection (1.1), shareholders and creditors of a corporation, their personal representatives and the Director may examine the records described in subsection 20(1) during the usual business hours of the corporation, and may take extracts from the records, free of charge, and, if the corporation is a distributing corporation, any other person may do so on payment of a reasonable fee.
Marginal note:Requirement for affidavit — securities register
(1.1) Any person described in subsection (1) who wishes to examine the securities register of a distributing corporation must first make a request to the corporation or its agent or mandatary, accompanied by an affidavit referred to in subsection (7). On receipt of the affidavit, the corporation or its agent or mandatary shall allow the applicant access to the securities register during the corporation’s usual business hours, and, on payment of a reasonable fee, provide the applicant with an extract from the securities register.
Marginal note:Copies of corporate records
(2) A shareholder of a corporation is entitled on request and without charge to one copy of the articles and by-laws and of any unanimous shareholder agreement.
Marginal note:Shareholder lists
(3) Shareholders and creditors of a corporation, their personal representatives, the Director and, if the corporation is a distributing corporation, any other person, on payment of a reasonable fee and on sending to a corporation or its agent or mandatary the affidavit referred to in subsection (7), may on application require the corporation or its agent or mandatary to provide within 10 days after the receipt of the affidavit a list (in this section referred to as the “basic list”) made up to a date not more than 10 days before the date of receipt of the affidavit setting out the names of the shareholders of the corporation, the number of shares owned by each shareholder and the address of each shareholder as shown on the records of the corporation.
Marginal note:Supplemental lists
(4) A person requiring a corporation to provide a basic list may, by stating in the affidavit referred to in subsection (3) that they require supplemental lists, require the corporation or its agent or mandatary on payment of a reasonable fee to provide supplemental lists setting out any changes from the basic list in the names or addresses of the shareholders and the number of shares owned by each shareholder for each business day following the date the basic list is made up to.
Marginal note:When supplemental lists to be provided
(5) The corporation or its agent or mandatary shall provide a supplemental list required under subsection (4)
(a) on the date the basic list is furnished, where the information relates to changes that took place prior to that date; and
(b) on the business day following the day to which the supplemental list relates, where the information relates to changes that take place on or after the date the basic list is furnished.
Marginal note:Holders of options
(6) A person requiring a corporation to furnish a basic list or a supplemental list may also require the corporation to include in that list the name and address of any known holder of an option or right to acquire shares of the corporation.
Marginal note:Contents of affidavit
(7) The affidavit required under subsection (1.1) or (3) shall state
(a) the name and address of the applicant;
(b) the name and address for service of the body corporate, if the applicant is a body corporate; and
(c) that the basic list and any supplemental lists obtained pursuant to subsection (4) or the information contained in the securities register obtained pursuant to subsection (1.1), as the case may be, will not be used except as permitted under subsection (9).
Marginal note:Idem
(8) If the applicant is a body corporate, the affidavit shall be made by a director or officer of the body corporate.
Marginal note:Use of information or shareholder list
(9) A list of shareholders or information from a securities register obtained under this section shall not be used by any person except in connection with
(a) an effort to influence the voting of shareholders of the corporation;
(b) an offer to acquire securities of the corporation; or
(c) any other matter relating to the affairs of the corporation.
Marginal note:Offence
(10) A person who, without reasonable cause, contravenes this section is guilty of an offence and liable on summary conviction to a fine not exceeding five thousand dollars or to imprisonment for a term not exceeding six months or to both.
R.S., 1985, c. C-44, s. 21
2001, c. 14, ss. 11, 135(E)
2011, c. 21, s. 16(E)
Previous Version
Marginal note:Register
21.1 (1) The corporation shall prepare and maintain, at its registered office or at any other place in Canada designated by the directors, a register of individuals with significant control over the corporation that contains
(a) for each individual with significant control,
(i) their name and date of birth,
(ii) their residential address, and
(iii) their address for service, if it has been provided to the corporation;
(a.1) the citizenship of each individual with significant control;
(b) the jurisdiction of residence for tax purposes of each individual with significant control;
(c) the day on which each individual became or ceased to be an individual with significant control, as the case may be;
(d) a description of how each individual is an individual with significant control over the corporation, including, as applicable, a description of their interests and rights in respect of shares of the corporation;
(e) any other prescribed information; and
(f) a description of each step taken in accordance with subsection (2).
Marginal note:Updating of information
(2) The corporation shall, at the following times, take reasonable steps to ensure that it has identified all individuals with significant control over the corporation and that the information in the register is accurate, complete and up-to-date:
(a) at least once during each financial year of the corporation;
(b) on the request of the Director; and
(c) at the times provided for in the regulations.
Marginal note:Recording of information
(3) If the corporation becomes aware of any information referred to in paragraphs (1)(a) to (e) as a result of steps taken in accordance with subsection (2) or through any other means, the corporation shall record that information in the register within 15 days of becoming aware of it.
Marginal note:Information from shareholders
(4) If the corporation requests information referred to in any of paragraphs (1)(a) to (e) from one of its shareholders, the shareholder shall, to the best of their knowledge, reply accurately and completely as soon as feasible.
Marginal note:Disposal of personal information
(5) Within one year after the sixth anniversary of the day on which an individual ceases to be an individual with significant control over the corporation, the corporation shall — subject to any other Act of Parliament and to any Act of the legislature of a province that provides for a longer retention period — dispose of any of that individual’s personal information, as defined in subsection 2(1) of the Personal Information Protection and Electronic Documents Act, that is recorded in the register.
Marginal note:Offence
(6) A corporation that, without reasonable cause, contravenes this section is guilty of an offence and liable on summary conviction to a fine not exceeding $100,000.
Marginal note:Non-application
(7) This section does not apply to a corporation
(a) that is a reporting issuer or an émetteur assujetti under an Act of the legislature of a province relating to the regulation of securities;
(b) any of the securities of which are listed and posted for trading on a designated stock exchange, as defined in subsection 248(1) of the Income Tax Act; or
(c) that is a member of a prescribed class.
2018, c. 27, s. 183
## 2018, c. 27, s. 183
The following provision is not in force.
183 The Act is amended by adding the following after section 21:
Marginal note:Register
21.1 (1) The corporation shall prepare and maintain, at its registered office or at any other place in Canada designated by the directors, a register of individuals with significant control over the corporation that contains
(a) the names, the dates of birth and the latest known address of each individual with significant control;
(b) the jurisdiction of residence for tax purposes of each individual with significant control;
(c) the day on which each individual became or ceased to be an individual with significant control, as the case may be;
(d) a description of how each individual is an individual with significant control over the corporation, including, as applicable, a description of their interests and rights in respect of shares of the corporation;
(e) any other prescribed information; and
(f) a description of each step taken in accordance with subsection (2).
Marginal note:Updating of information
(2) At least once during each financial year of the corporation, the corporation shall take reasonable steps to ensure that it has identified all individuals with significant control over the corporation and that the information in the register is accurate, complete and up-to-date.
+ Marginal note:Recording of information
(3) If the corporation becomes aware of any information referred to in paragraphs (1)(a) to (e) as a result of steps taken in accordance with subsection (2) or through any other means, the corporation shall record that information in the register within 15 days of becoming aware of it.
+ Marginal note:Information from shareholders
(4) If the corporation requests information referred to in any of paragraphs (1)(a) to (e) from one of its shareholders, the shareholder shall, to the best of their knowledge, reply accurately and completely as soon as feasible.
+ Marginal note:Disposal of personal information
(5) Within one year after the sixth anniversary of the day on which an individual ceases to be an individual with significant control over the corporation, the corporation shall — subject to any other Act of Parliament and to any Act of the legislature of a province that provides for a longer retention period — dispose of any of that individual’s personal information, as defined in subsection 2(1) of the Personal Information Protection and Electronic Documents Act, that is recorded in the register.
+ Marginal note:Offence
(6) A corporation that, without reasonable cause, contravenes this section is guilty of an offence and liable on summary conviction to a fine not exceeding five thousand dollars.
+ Marginal note:Non-application
(7) This section does not apply to a corporation that
(a) is a reporting issuer or an émetteur assujetti under an Act of the legislature of a province relating to the regulation of securities;
(b) is listed on a designated stock exchange, as defined in subsection 248(1) of the Income Tax Act; or
(c) is a member of a prescribed class.
Marginal note:Inability to identify individuals
21.2 A corporation to which section 21.1 applies shall take prescribed steps if it is unable to identify any individuals with significant control over the corporation.
Marginal note:Disclosure to Director
21.3 (1) A corporation to which section 21.1 applies shall disclose to the Director, on request, any information in its register of individuals with significant control.
Marginal note:Access — affidavit
(2) Shareholders and creditors of the corporation or their personal representatives, on sending to the corporation or its agent or mandatary the affidavit referred to in subsection (3), may on application require the corporation or its agent or mandatary to allow the applicant access to the register of the corporation referred to in subsection 21.1(1) during the usual business hours of the corporation and, on payment of a reasonable fee, provide the applicant with an extract from that register.
+ Marginal note:Affidavit
(3) The affidavit required under subsection (2) shall contain
(a) the name and address of the applicant;
(b) the name and address for service of the body corporate, if the applicant is a body corporate; and
(c) a statement that any information obtained under subsection (2) will not be used except as permitted under subsection (5).
Marginal note:Application by body corporate
(4) If the applicant is a body corporate, the affidavit shall be made by a director or officer of the body corporate.
+ Marginal note:Use of information
(5) Information obtained under subsection (2) shall not be used by any person except in connection with
(a) an effort to influence the voting of shareholders of the corporation;
(b) an offer to acquire securities of the corporation; or
(c) any other matter relating to the affairs of the corporation.
Marginal note:Offence
(6) A person who, without reasonable cause, contravenes subsection (5) is guilty of an offence and liable on summary conviction to a fine not exceeding five thousand dollars or to imprisonment for a term not exceeding six months, or to both.
Marginal note:Offence — preparation and maintenance of register
21.4 (1) Every director or officer of a corporation who knowingly authorizes, permits or acquiesces in the contravention of subsection 21.1(1) by that corporation commits an offence, whether or not the corporation has been prosecuted or convicted.
Marginal note:Offence — recording of false or misleading information
(2) Every director or officer of a corporation who knowingly records or knowingly authorizes, permits or acquiesces in the recording of false or misleading information in the register of the corporation referred to in subsection 21.1(1) commits an offence.
+ Marginal note:Offence — provision of false or misleading information
(3) Every director or officer of a corporation who knowingly provides or knowingly authorizes, permits or acquiesces in the provision to any person or entity of false or misleading information in relation to the register of the corporation referred to in subsection 21.1(1) commits an offence.
+ Marginal note:Offence — subsection 21.1(4)
(4) Every shareholder who knowingly contravenes subsection 21.1(4) commits an offence.
+ Marginal note:Penalty
(5) A person who commits an offence under any of subsections (1) to (4) is liable on summary conviction to a fine not exceeding $200,000 or to imprisonment for a term not exceeding six months, or to both.
2022, c. 10, s. 430
2023, c. 29, s. 1
2024, c. 17, s. 232
Previous Version
Date modified:
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Linear Dependency
Real Vector Spaces
Field in Algebra
Linear Algebra Course
Vectors (mathematics)
Linear Independence
Fields (mathematics)
5
What is the condition that vectors (a1, a2) and (b1, b2) in V2(F) are linearly dependent?
Peter Shea
B. Sc in Mathematics & Computer Science, Monash University (Graduated 1972)
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Author has 5.2K answers and 1.2M answer views
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4y
This requires some calculation 3d, but in 2d, it just means that one is a multiple of the other. Just scale one of them so that one of its components is the same as the equivalent component (e.g. x-value) as the other, and see if its y-value also matches. Equivalently, construct the determinant with row 1 containing (a1 a2) and row 2 containing (b1 b2). If the value of the determinant is 0, the vectors are linearly dependent.
Ali Reza
PHD from K.N.Toosi University of Technology (Graduated 2019)
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Author has 958 answers and 452.2K answer views
·
4y
Let A=(a1, a2) and B=(b1, b2) be two vectors. Then A and B are linearly dependent if and only if one of them is a scaler multiple of the other. That is A=cB ,for example. In the plane R^2 ,this is equivalent that A and B are parallel. Also we can use the determinant defined by two vectors A , B (you can define the determinant by consider both of A, B in rows or columns by any ordering). If this determinant is zero then A, B are linearly dependent and if this determinant is not zero then A, B are linearly independent .
Alexander Farrugia
Ph.D. in Mathematics, University of Malta (Graduated 2016)
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Author has 3.2K answers and 27.5M answer views
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9y
Related
How can I prove that the vectors
2
i
+
j
−
3
k
,
i
−
4
k
and
4
i
+
3
j
−
k
are linearly dependent?
One way to do this is to show that the determinant of
⎛⎜⎝214103−3−4−1⎞⎟⎠
is zero. The matrix above was created by writing the coefficients of i in the first row, the coefficients of j in the second row, and the coefficients of k in the third row.
Since
∣∣
∣∣214103−3−4−1∣∣
∣∣=2×∣∣∣03−4−1∣∣∣−1×∣∣∣13−3−1∣∣∣+4×∣∣∣10−3−4∣∣∣
=2((0×−1)−(3×−4))−((1×−1)−(3×−3))+4((1×−4)−(−3\ti
One way to do this is to show that the determinant of
⎛⎜⎝214103−3−4−1⎞⎟⎠
is zero. The matrix above was created by writing the coefficients of i in the first row, the coefficients of j in the second row, and the coefficients of k in the third row.
Since
∣∣
∣∣214103−3−4−1∣∣
∣∣=2×∣∣∣03−4−1∣∣∣−1×∣∣∣13−3−1∣∣∣+4×∣∣∣10−3−4∣∣∣
=2((0×−1)−(3×−4))−((1×−1)−(3×−3))+4((1×−4)−(−3×0))
=2×12−8+4×−4=0
the vectors are linearly dependent.
Another way to show this is to find a suitable linear combination of one of the vectors in terms of the other two. For example, the third vector can be written in terms of the other two as
4i+3j−k=3(2i+j−3k)−2(i−4k)
so the three vectors are linearly dependent.
Related questions
If {v1, v2, …, vn} is a linearly dependent set of vectors, how do you prove one of these vectors is a linear combination of the other?
If F is the field of real numbers, how do you prove that the vectors (a1,a2) and (b1,b2) in V_2(F) are linearly dependent if a2b2-a2b1 = 0?
I have two paired vectors, A= [a1,a2,..., a100], B= [b1,b2,..., b100]. The two vectors are paired. However, a1, a2, ..., a100 are not independent on each other. It is the same for the elements in vector B. What test should I use to compare the mean of A with the mean of B?
) Let b1 = (1, 1, 0), b2 = (1, -1, 0) and b3 = (2, 2k, k). For what value of k the vectors b1, b2 and b3 are linearly dependent?
How many vectors are in {a1, a2, a3}?
Richard Goldstone
PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995)
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Author has 1.8K answers and 3.9M answer views
·
Sep 2
Related
If vector A is equal to (a1, a2) and vector B is equal to (b1, b2), then vector A .vector B is equal to a1b1 + a2b2, why?
You are asking why the dot product is defined the way it is, i.e. why
⎡⎢
⎢
⎢
⎢⎣a1a2⋮an⎤⎥
⎥
⎥
⎥⎦⋅⎡⎢
⎢
⎢
⎢⎣b1b2⋮bn⎤⎥
⎥
⎥
⎥⎦=a1b1+a2b2+⋯+anbn.
Historically, the dot product formula appeared for R3 as part of the result of multiplying two pure quaternions. It was part of a bigger quaternion product expression until Oliver Heaviside separated the scalars in the formula into the dot product and the cross product, initiating the way vector analysis is mostly done nowadays, although more conceptual methods,
You are asking why the dot product is defined the way it is, i.e. why
⎡⎢
⎢
⎢
⎢⎣a1a2⋮an⎤⎥
⎥
⎥
⎥⎦⋅⎡⎢
⎢
⎢
⎢⎣b1b2⋮bn⎤⎥
⎥
⎥
⎥⎦=a1b1+a2b2+⋯+anbn.
Historically, the dot product formula appeared for R3 as part of the result of multiplying two pure quaternions. It was part of a bigger quaternion product expression until Oliver Heaviside separated the scalars in the formula into the dot product and the cross product, initiating the way vector analysis is mostly done nowadays, although more conceptual methods, some returning to the quaternionic perspective, are gaining in currency.
The historical development is not satisfying for learners who know nothing of quaternions and may never know about them, so we’ve invented various other more appealing explanations which are not, however, historical. Here is one such account.
One way to appreciate the definition of the dot product is to begin with the question of when two vectors in R2 are perpendicular. The answer for R2 goes back to high-school Algebra I. Suppose we have
v=[v1v2],w=[w1w2].
The vector v lies on the line y=(v2/v1)x and the vector w lies on the line y=(w2/w1)x. The vectors are perpendicular if and only if the lines are pendicular, and from Algebra I we know the lines are perpendicular if and only if their slopes are negative reciprocals, i.e. if and only if
w2w1=−v1v2⟺v1w1+v2w2=0,
so the characteristic dot product expression appears as an algebraic condition for two vectors in R2 to be perpendicular.
When the boxed expression is not zero, does it carry any information about the vectors in question? In a special case it does; when v=w, the boxed expression becomes v21+v22, which by the Pythagorean theorem is the square of the magnitude of v.
Emboldened by these observations, suppose we define v⊙w=v1w1+v2w2. Then we have and
Is there more information in than this? First, it is easy to check from the boxed definition above that
for all
for all and all
for all (and a corresponding distributive law for )
Using Item 3, the distributive law, just as in Algebra I, we find that
The tail-to head convention for vector addition indicates that the vectors and can be viewed as the sides of a triangle. For this triangle, the Law of Cosines says that
Comparing this with the previous expression for shows that we must have
and so our provisional in is in fact the usual dot product which provides not only information about perpendicularity but more generally information about the angle between any two vectors in via the formula
We did all this in We can’t make exactly the same arguments in because we don’t have the negative reciprocal slope characterization of perpendicular lines once the dimension is three or greater. However, if we simply define then all the statements and arguments above work in the same way, and we end up with which then indicates that thereby promoting the original condition to all
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Assistant
Bot
·
1y
In the vector space [math]V_2(F)[/math], which consists of 2-dimensional vectors over a field [math]F[/math], the vectors [math]\mathbf{a} = (a_1, a_2)[/math] and [math]\mathbf{b} = (b_1, b_2)[/math] are linearly dependent if there exist scalars [math]c_1[/math] and [math]c_2[/math], not both zero, such that:
[math]c_1 \mathbf{a} + c_2 \mathbf{b} = \mathbf{0}[/math]
This can be expressed in terms of the components of the vectors:
[math]c_1 (a_1, a_2) + c_2 (b_1, b_2) = (0, 0)[/math]
This leads to the following system of equations:
[math]c_1 a_1 + c_2 b_1 = 0[/math]
[math]c_1 a_2 + c_2 b_2 = 0[/math]
The vectors [math]\mathbf{a}[/math] and [math]\mathbf{b}[/math] are linearly dependent if one of the vectors can be expressed as a scalar multiple of the ot
In the vector space [math]V_2(F)[/math], which consists of 2-dimensional vectors over a field [math]F[/math], the vectors [math]\mathbf{a} = (a_1, a_2)[/math] and [math]\mathbf{b} = (b_1, b_2)[/math] are linearly dependent if there exist scalars [math]c_1[/math] and [math]c_2[/math], not both zero, such that:
[math]c_1 \mathbf{a} + c_2 \mathbf{b} = \mathbf{0}[/math]
This can be expressed in terms of the components of the vectors:
[math]c_1 (a_1, a_2) + c_2 (b_1, b_2) = (0, 0)[/math]
This leads to the following system of equations:
[math]c_1 a_1 + c_2 b_1 = 0[/math]
[math]c_1 a_2 + c_2 b_2 = 0[/math]
The vectors [math]\mathbf{a}[/math] and [math]\mathbf{b}[/math] are linearly dependent if one of the vectors can be expressed as a scalar multiple of the other. This occurs if and only if the determinant of the matrix formed by placing these vectors as columns is zero:
[math]\text{det}\begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix} = a_1 b_2 - a_2 b_1 = 0[/math]
Thus, the condition for linear dependence is:
[math]a_1 b_2 - a_2 b_1 = 0[/math]
If this condition holds, it indicates that the vectors are linearly dependent. If it does not hold, they are linearly independent.
Fernando Pepes
Lives in Vila Velha, ES, Brazil
·
4y
Related
How would one prove that for vectors a, b, c ∈ R3 if the vectors a × b, b ×c, c × a are linearly dependent, then they are all propotional?
I'm sorry to tell you but I do not have enough lnowledge to solve this with elegance. Though, think about what is a vectorial product. What does it happen to it?
The vectorial product creates a different vector that is orthogonal to the planet created by the two original vectors; So, returning to the original problem:
[math]a,b,c[/math] will form (in a general case) three different planes; The vectors [math]a \times c, b \times c, a \times b [/math]will represent the orientation of these planes, since they will be orthogonal to them. Then, if these three vectors are linear dependent (which is the same to say that they ar
I'm sorry to tell you but I do not have enough lnowledge to solve this with elegance. Though, think about what is a vectorial product. What does it happen to it?
The vectorial product creates a different vector that is orthogonal to the planet created by the two original vectors; So, returning to the original problem:
[math]a,b,c[/math] will form (in a general case) three different planes; The vectors [math]a \times c, b \times c, a \times b [/math]will represent the orientation of these planes, since they will be orthogonal to them. Then, if these three vectors are linear dependent (which is the same to say that they are a linear combination of one another, you just need to consider some variables [math]\alpha, \beta, \gamma [/math]that will be the values of the referred linear combination. If you know how to calculate all vectors and relate them, it will lead to the conclusion that probably the angle between the original vectors has to be 0 or even п. I think the difficult part is the understanding of the problem, and I'm not so used to the LATEX commands, then I'll just shed some light on it.
Feel free to tell me if anything that is here is inaccurate or wrong. I will be grateful. ☆
Related questions
What is the real life application of linear dependent vector?
Is it possible that vectors v1, v2, and v3 are linearly dependent, but the vectors w1 v1 v2 w2 v2 v3 and w3 v3 v1 are Linearly independent?
Suppose S is a set of n-linearly independent vectors that span the n-dimensional vector space V. Can you prove that S is a basis for V?
Let, W = {(a1, a2, a3, a4, a5) ∈ R 5 : a2 = a3 = a4 & a1 + a5 = 0}. Is W a subspace of R 5 over the field R? If yes then what is dim(W)?
If
[math]\vec{0}[/math]
is included, are all vectors in R^n linearly dependent?
Richard Goldstone
PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995)
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Author has 1.8K answers and 3.9M answer views
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Updated 4y
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How would one prove that for vectors a, b, c ∈ R3 if the vectors a × b, b ×c, c × a are linearly dependent, then they are all propotional?
Oops, I misread the statement as claiming that [math]a,[/math] [math]b[/math], and [math]c[/math] are proportional. Most of what I wrote works but some editing is required.
The main point is the
Lemma.
The cross products [math]a \times b,[/math] [math]b \times c,[/math] and [math]c \times a[/math] are linearly dependent if and only if [math]a, b,[/math] and [math]c[/math] are linearly dependent.
Proof.
Suppose
[math]\alpha(a \times b) + \beta(b \times c) + \gamma(c \times a) = \vec{0}.[/math]
We can’t have all three of [math]\alpha, \beta,[/math] and [math]\gamma[/math] equal to zero. If, say, [math]\beta \ne 0,[/math] then dot both sides with [math]a[/math] to get
[math]\beta a \cdot(b \times c)=0.[/math]
But [math]a \cdot (b \times c)=\det(a,b,c),[/math] where [math]a, b, c[/math] are the rows of the mat
Oops, I misread the statement as claiming that [math]a,[/math] [math]b[/math], and [math]c[/math] are proportional. Most of what I wrote works but some editing is required.
The main point is the
Lemma.
The cross products [math]a \times b,[/math] [math]b \times c,[/math] and [math]c \times a[/math] are linearly dependent if and only if [math]a, b,[/math] and [math]c[/math] are linearly dependent.
Proof.
Suppose
[math]\alpha(a \times b) + \beta(b \times c) + \gamma(c \times a) = \vec{0}.[/math]
We can’t have all three of [math]\alpha, \beta,[/math] and [math]\gamma[/math] equal to zero. If, say, [math]\beta \ne 0,[/math] then dot both sides with [math]a[/math] to get
[math]\beta a \cdot(b \times c)=0.[/math]
But [math]a \cdot (b \times c)=\det(a,b,c),[/math] where [math]a, b, c[/math] are the rows of the matrix whose determinant is to be found, so we have [math]\det(a,b,c)=0[/math] and this is equivalent to [math]a, b,[/math] and [math]c[/math] being linearly dependent.
Analogous arguments work for [math]\alpha \ne 0[/math] and [math]\gamma \ne 0.[/math]
This shows that if the three cross products are linearly dependent, then [math]a, b,[/math] and [math]c[/math] are linearly dependent. But the converse is immediately obvious from the geometry of the cross product and the fact that linear dependence of [math]a[/math], [math]b[/math], and [math]c[/math] means they are coplanar. This proves the Lemma. [math]\qquad \blacksquare[/math]
With the lemma in hand, the hypotheses imply that [math]a[/math], [math]b[/math], and [math]c[/math] are coplanar; call the plane [math]\Pi.[/math] Then [math]a \times b,[/math] [math]b \times c,[/math] and [math]c \times a[/math] are all perpendicular to [math]\Pi[/math] and so must be proportional. (This argument just repeats the second part of the proof above.)
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Srks Jarrel
Math Ph.D., Arithmetic Geometry Researcher
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Upvoted by
Kostyantyn Mazur
, PhD Mathematics, New York University (2018)
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7y
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Why is a zero vector linearly dependent?
Most of the answers address a slightly different question, but your question is legit as it is. Indeed the zero vector itself is linearly dependent.
Recall that a set of vectors [math]{ \vec{v_1}, \vec{v_2} \dots, \vec{v_n} }[/math] is linearly dependent if you can find constants [math]c_1,\dots, c_n[/math] not all zero such that
[math]c_1 \vec{v_1} + c_2 \vec{v_2} + \dots + c_n \vec{v_n} = \vec{0}[/math]
In other words there is a way to express the zero vector as a linear combination of the vectors [math]\vec{v_1},\vec{v_2},\dots,\vec{v_n}[/math] where at least one coefficient of the vectors in non-zero.
Example 1. The vectors [math]\vec{v_1} = (1,2)[/math]
Most of the answers address a slightly different question, but your question is legit as it is. Indeed the zero vector itself is linearly dependent.
Recall that a set of vectors [math]{ \vec{v_1}, \vec{v_2} \dots, \vec{v_n} }[/math] is linearly dependent if you can find constants [math]c_1,\dots, c_n[/math] not all zero such that
[math]c_1 \vec{v_1} + c_2 \vec{v_2} + \dots + c_n \vec{v_n} = \vec{0}[/math]
In other words there is a way to express the zero vector as a linear combination of the vectors [math]\vec{v_1},\vec{v_2},\dots,\vec{v_n}[/math] where at least one coefficient of the vectors in non-zero.
Example 1. The vectors [math]\vec{v_1} = (1,2)[/math] and [math]\vec{v_2} = (2,4)[/math] are linearly dependent because, if you take [math]c_1 = 2[/math] and [math]c_2 = -1[/math] a quick computation shows that [math]c_1 \vec{v_1} + c_2 \vec{v_2} = (2,4) + (-2,-4) = \vec{0}[/math].
Example 2. The vectors [math]\vec{v_1} = (1,0,0)[/math] and [math]\vec{v_2} = (0,1,0)[/math] are NOT lineaely dependent because for any possible choice of [math]c_1,c_2[/math] the combination [math]c_1\vec{v_1} + c_2\vec{v_2} = (c_1,c_2,0)[/math] is equal to the zero vector only if both [math]c_1=0[/math] and [math]c_2=0[/math]. In other words the only way to express the zero vector as a linear combination of [math]\vec{v_1}[/math] and [math]\vec{v_2}[/math] is the combination in which every constant is equal to zero.
For your question, we need to consider the case in which the set of vectors contains only one vector. So you are asking when a single vector [math]\vec{v}[/math] can be linearly dependent. In this case the previous definitions says that [math]\vec{v}[/math] is linearly dependent only if you can find a non-zero constant [math]c[/math] such that
[math]c \vec{v} = \vec{0}[/math]
But this says that a multiple of [math]\vec{v}[/math] is the zero vector and [math]c\neq 0[/math]. Multiplying on both sides by [math]1/c[/math] this implies that
[math]\vec{v} = \frac{1}{c} \vec{0} = \vec{0}[/math]
So the only vector that is linearly dependent is the zero vector.
Alberto Cid
M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008)
·
Author has 2K answers and 3.8M answer views
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2y
Related
How do you find out if this vector system is linearly dependent:
[math] v_1=(5,4,3), v_2=(3,3,2), v_3=(8,1,3) [/math]
?
Take the determinant of matrix
5 4 3
3 3 2
8 1 3
5•(9–2)—4•(9–16)+3•(3–24)=
=35+28–63 = 0
So they are NOT linearly independent.
Also, solve the system:
[math]x•V_1+y•V_2+z•V_3=0[/math]
Eq1: 5x + 3y + 8z = 0;
Eq2: 4x + 3y + z = 0;
Eq3: 3x + 2y + 3z=0
Substracting Eq1-Eq2:
x+7z = 0 → x = -7z
The equations become:
-35z + 3y + 8z = 0
-28z + 3y + z = 0
-21z + 2y + 3z = 0
All three are the same: y =9z
So, every (-7z, 9z, z) solves it, not only (0, 0, 0). For z = 1 you have: x=—7; y=9; z=1
That is: [math]-7V_1+9V_2+1V_3 = 0[/math]
So: [math]V_3 = 7•V_1-9•V_2[/math]
7•(5, 4, 3)—9•(3, 3, 2) = (8, 1, 3)
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Charles S.
former mathematician, current patent lawyer
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Upvoted by
Tom McFarlane
, M.S. Mathematics, University of Washington (1994) and
Aditya Garg
, M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 7.6K answers and 60.4M answer views
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4y
Related
What are the conditions for vectors to be linearly dependent (linear algebra, matrices, vectors, math)?
I mean, there’s the definition… it kind of speaks for itself. A set of vectors [math]{v_1, v_2, \ldots, v_n}[/math] is linearly dependent if there are constants [math]a_1, \ldots, a_n[/math] that are not all equal to 0, such that the sum [math]a_1 v_1 + a_2 v_2 + \ldots + a_n v_n = 0[/math].
That last equation is sometimes called a “relation.” So a little more succinctly, a set of vectors is linearly dependent if there is a relation between them.
You can filter that definition through other linear algebra vocabulary, but it kind of boils down to the same thing. For example, if you know about the concept of span and dimension and al
I mean, there’s the definition… it kind of speaks for itself. A set of vectors [math]{v_1, v_2, \ldots, v_n}[/math] is linearly dependent if there are constants [math]a_1, \ldots, a_n[/math] that are not all equal to 0, such that the sum [math]a_1 v_1 + a_2 v_2 + \ldots + a_n v_n = 0[/math].
That last equation is sometimes called a “relation.” So a little more succinctly, a set of vectors is linearly dependent if there is a relation between them.
You can filter that definition through other linear algebra vocabulary, but it kind of boils down to the same thing. For example, if you know about the concept of span and dimension and all that, then you can say a set of [math]n[/math] vectors is linearly dependent if their span has dimension less than [math]n[/math].
But in some sense, that’s not a good answer to your question. In order for that answer to “hit,” I suspect you already need to have a pretty good understanding of what linear dependence does and doesn’t mean.
To hopefully give you a better feel, I’ll describe an imprecise way in which vectors can be linearly dependent. But hopefully that imprecise way will ring some bells that the precise definition doesn’t.
What does it mean for a set of vectors to be linearly dependent? It means one or more of those vectors is redundant. Extraneous. Flab. It’s like saying “My male son likes cheese.” You don’t need to say “male,” because “male” is already implied by “son.” If male and son were vectors, then their collective span would be one-dimensional. All sons are male. And, barring some kind of advance in bioengineering I’m unaware of, all males are sons.
Switch back to linear algebra. Often, you’re interested in some kind of phenomenon with respect to a few vectors. Since it’s linear algebra, if the vectors all exhibit the phenomenon you’re interested in, so will any linear combination of them. If this is true, the particular vectors themselves kind of recede in importance, and what you’re really doing is studying the phenomenon on the span of those vectors.
Maybe you want to see if you can extend that phenomenon — to make the scope of vectors that satisfy it “bigger.” How do you do that? You can’t just pick any ole’ vector to test. Even if you haven’t seen that specific vector before, you don’t learn anything new if that vector happens to be in the span of other vectors you’ve previously tested. It’s redundant. That is to say, it’s linearly dependent on the other vectors you’ve seen. In order to learn something new, you have to pick a linearly independent vector and study the phenomenon.
Hopefully that bit of intuition helps. If it strikes you as useful, go back and stare at the precise definitions some more, and try to match up what I’ve said to the actual mathematics.
Brad Moffat
Author has 4K answers and 11.6M answer views
·
5y
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How can we say 3 vectors are linearly dependent?
When a vector is said to be “dependent” on another, it doesn’t mean that it’s value depends on the other in any way.
It means that it’s value might be expressed as a multiple of that other vector, if you wanted to.
Here’s an example :
In the diagram above, I don’t “need” a vector called “3a” in my world (the red one.)
All I need is a scalar called 3 and the vector, “a” that I already had.
3a can be expressed as a combination of a.
Here, the center, dotted vector is a combination of A and B. I don’t need to define a new vector called “dotted line arrow thing,” because it can be represented by the oth
When a vector is said to be “dependent” on another, it doesn’t mean that it’s value depends on the other in any way.
It means that it’s value might be expressed as a multiple of that other vector, if you wanted to.
Here’s an example :
In the diagram above, I don’t “need” a vector called “3a” in my world (the red one.)
All I need is a scalar called 3 and the vector, “a” that I already had.
3a can be expressed as a combination of a.
Here, the center, dotted vector is a combination of A and B. I don’t need to define a new vector called “dotted line arrow thing,” because it can be represented by the other two that I already have.
In this case we can (if we want) say that dotted line vector “depends on” A and B. Or that the dotted line thing is “linearly dependent” on A and B.
What if I had an arrow that was parallel to A. Well, then I’d be back to being just linearly dependent on A and not dependent on B.
(We can still say that it’s dependent on A and B if we really wanted to, because the scalar of B can be zero.)
Now to your question.
Any three vectors in 2D will be linearly dependent. Back to the diagram above. We can just as easily say that B is dependent on A and dotted guy.
Because if I wanted to I could add (-1)A to dotted guy and get B. Remember that -A is just A pointing in the opposite direction. If you walk along dotted guy and then backwards along A guess where you end up?
Here’s the fun part. Let’s say that dotted guy above was in fact our old friend 3a from before. (Pretend that it’s pointing in dotted guy’s direction now.)
Now I can say that B is dependent on A and a.
Can you see why?
If you can make a representation of adding vectors or multiples of vectors, then that answer vector is said to be dependent on the vectors you used.
B = -A + 3a , therefore B is linearly dependent on A and a.
After a little thought you’ll see that any two vectors that aren’t parallel can be used to make to any vector in 2D. Play around with this until you get it :
See the instructions below.
w defined by the basis vectors u and v.
You can choose your two starting vectors, u and v, by moving the big yellow and green dots.
Once you have your two starting vectors, use the sliders to make any other vector.
Nipun Ramakrishnan
Studied Electrical Engineering and Computer Science at University of California, Berkeley (Graduated 2019)
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Author has 276 answers and 5.3M answer views
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7y
Related
How do I determine the projection of vector a = (a1, a2, a3) onto the unit vector along b with b = (b1, b2, b3)?
When you have [math]b[/math] as a unit vector, the scalar projection is simply the dot product of the two vectors and you can scale the unit vector by this value to obtain the projection vector.
But it’s important to understand why this is true.
Think about what the projection actually is for any general two vectors. Personally, I think having the geometric intuition is the best way to think about projections. So let’s take a look:
In this figure we have vectors [math]A[/math] and [math]B[/math] and using trigonometry, our scalar projection actually ends up just being the quantity [math]| \vec A|cos(\theta)[/math]. If we knew the quantity [math]cos(\the[/math]
When you have [math]b[/math] as a unit vector, the scalar projection is simply the dot product of the two vectors and you can scale the unit vector by this value to obtain the projection vector.
But it’s important to understand why this is true.
Think about what the projection actually is for any general two vectors. Personally, I think having the geometric intuition is the best way to think about projections. So let’s take a look:
In this figure we have vectors [math]A[/math] and [math]B[/math] and using trigonometry, our scalar projection actually ends up just being the quantity [math]| \vec A|cos(\theta)[/math]. If we knew the quantity [math]cos(\theta)[/math], this would be an easy problem, right? But we’re out of luck — or are we?
If I give you two vectors [math]a[/math] and [math]b[/math], there’s actually a natural way to calculate the angle between the vectors using the definition of the dot product.
[math]cos(\theta) = \frac{\vec a \cdot \vec b}{|\vec a||\vec b|}[/math]
Now let’s derive the scalar projection for vectors [math]a[/math] and [math]b[/math].
[math]|\vec a|cos(\theta) = |\vec a|\frac{\vec a \cdot \vec b}{|\vec a||\vec b|} = \frac{\vec a \cdot \vec b}{|\vec b|} = \vec a \cdot \vec b[/math]
The last step comes from the fact that we know [math]b[/math] is a unit vector. And as mentioned earlier, the projected vector is [math]c \vec b[/math] where [math]c = \ [/math][math]\vec a \cdot \vec b[/math].
Vector projections are useful in all sorts of cool applications so it’s actually important that you understand them.
Vance Faber
Studied Mathematics
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2.9K answers and 1.8M answer views
·
5y
Related
How do I prove that vectors b1,b2,b3 are also linearly independent when b1=a1,b2,=a1+a2,b3=a1+a2+a3, and vectors a1,a2,a3 are linearly independent?
Solve for the a's in terms of the b's. Then every vector which is a linear combination of the a's will also be a linear combination of the b's. This shows that the spaces spanned by the two sets are identical and must have dimension 3. If 3 vectors span a space of dimension 3, they must be independent.
I should probably add that your question is a particular instance of a much more general property. Suppose we have a set of vectors of rank k (that means they span a space of dimension k). And suppose we multiply each vector by the same invertible matrix A. We will get another set of vectors with
Solve for the a's in terms of the b's. Then every vector which is a linear combination of the a's will also be a linear combination of the b's. This shows that the spaces spanned by the two sets are identical and must have dimension 3. If 3 vectors span a space of dimension 3, they must be independent.
I should probably add that your question is a particular instance of a much more general property. Suppose we have a set of vectors of rank k (that means they span a space of dimension k). And suppose we multiply each vector by the same invertible matrix A. We will get another set of vectors with rank k.
Related questions
If {v1, v2, …, vn} is a linearly dependent set of vectors, how do you prove one of these vectors is a linear combination of the other?
If F is the field of real numbers, how do you prove that the vectors (a1,a2) and (b1,b2) in V_2(F) are linearly dependent if a2b2-a2b1 = 0?
I have two paired vectors, A= [a1,a2,..., a100], B= [b1,b2,..., b100]. The two vectors are paired. However, a1, a2, ..., a100 are not independent on each other. It is the same for the elements in vector B. What test should I use to compare the mean of A with the mean of B?
) Let b1 = (1, 1, 0), b2 = (1, -1, 0) and b3 = (2, 2k, k). For what value of k the vectors b1, b2 and b3 are linearly dependent?
How many vectors are in {a1, a2, a3}?
What is the real life application of linear dependent vector?
Is it possible that vectors v1, v2, and v3 are linearly dependent, but the vectors w1 v1 v2 w2 v2 v3 and w3 v3 v1 are Linearly independent?
Suppose S is a set of n-linearly independent vectors that span the n-dimensional vector space V. Can you prove that S is a basis for V?
Let, W = {(a1, a2, a3, a4, a5) ∈ R 5 : a2 = a3 = a4 & a1 + a5 = 0}. Is W a subspace of R 5 over the field R? If yes then what is dim(W)?
If
is included, are all vectors in R^n linearly dependent?
How do I find the adjoint operator of T defined on space l2 as T (a1, a2.. ) = (0, a1, a2…)? How do I determine whether it is self-adjoint and normal?
What is the basis and dimension of the subspace of
spanned by the set of vectors
?
How would one prove that for vectors a, b, c ∈ R3 if the vectors a × b, b ×c, c × a are linearly dependent, then they are all propotional?
How do you solve this problem? Suppose b1, b2, b3, b4 is a basis of vector space V. Prove that b1 + b2, b2 + b3, b3 + b4, b4 is also a basis of V.
Let
be a list of nonzero linearly dependent vectors. How do I show there is a vector
in the list such that it is in the span of linearly independent vectors
?
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© Quora, Inc. 2025 |
188452 | https://arxiv.org/pdf/2508.00776 | From Dynamic Programs to Greedy Algorithms ∗
Dieter van Melkebeek August 4, 2025
Abstract
We show for several computational problems how classical greedy algorithms for special cases can be derived in a simple way from dynamic programs for the general case: interval scheduling (restricted to unit weights), knapsack (restricted to unit values), and shortest paths (restricted to nonnegative edge lengths). Conceptually, we repeatedly expand the Bellman equations underlying the dynamic program and use straightforward monotonicity properties to figure out which terms yield the optimal value under the respective restrictions. The approach offers an alternative for developing these greedy algorithms in undergraduate algorithms courses and/or for arguing their correctness. In the setting of interval scheduling, it elucidates the change in order from earliest start time first for the memoized dynamic program to earliest finish time first for the greedy algorithm.
1 Introduction
Many an undergraduate algorithms course features Interval Scheduling to illustrate the design paradigms of greed and dynamic programming. The unweighted version admits an appealing greedy strategy that can be shown correct via several types of arguments (charging, exchange, greedy stays ahead). The weighted version exhibits a straightforward principle of optimality and demonstrates one of the measures that can be taken to keep the number of subproblems in a dynamic program small – ordering the input components judiciously. However, the appropriate order for the memoized dynamic program differs from the order in the greedy algorithm. This tends to confuse students and gives some the impression that, even though both algorithms solve the unweighted version, they are not that closely related. Existing expositions convey intuition behind the choice of one or both of the orders, but do not explain where the difference arises and do not highlight the close relationship between the algorithms. This note addresses those shortcomings. We derive the greedy algorithm from the dynamic program by simplifying the latter under the preconditions of the former. The change in order follows naturally. We apply a similar strategy to two other common examples, namely Knapsack , where we derive the greedy algorithm for the unit-value restriction from the dynamic program for the general case, and Shortest Paths , where we derive Dijkstra’s algorithm from the Bellman-Ford dynamic program.
The paradigms. Let us start by reviewing dynamic programming and greed in the context of a discrete optimization problem Π. We can think of an instance x of size n as consisting of n components, x1, . . . , x n,and need to make a decision yi about each component xi so as to maximize or minimize an objective function
fx(y1, . . . , y n) under certain constraints. In a dynamic program , we pick a particular component, say x1. We consider all possible settings v1 for
y1, optimize fx with the additional constraint y1 = v1, and assign y1 to a value v∗
1
that produces the best result. The approach hinges on an efficient reduction from instances with the above additional constraint to proper Π-instances of size smaller than n. Repeated application yields a recursive algorithm for Π that
∗This material is based upon work supported by the U.S. National Science Foundation under Grant No. CCF-2312540.
1
arXiv:2508.00776v1 [cs.DS] 1 Aug 2025
can be captured in the Bellman equations, which express the optimal value for input x as the maximum or minimum of terms involving the optimal values for smaller instances. Correctness typically follows in a streamlined fashion. The name of the game is to ensure that the number of subproblems on any given input
x remains small, i.e., the number of distinct calls to Π that arise throughout the entire recursive process on input x. This often requires reformulating the decisions to choose among, and sometimes also reordering the components. If and when the approach succeeds, memoization yields an efficient top-down algorithm for Π. Alternately, when the subproblems can be described in terms of x and k additional parameters, a
k-dimensional table with the optimal values of the subproblems can be computed iteratively in a bottom-up fashion, after which a second phase retrieves decisions that achieve the optimal values in a top-down fashion. A greedy algorithm also starts by picking a specific component, say x1, but is able to efficiently figure out an optimal setting v∗
1
for y1 without making any recursive calls. It never considers settings for y1 other than v∗
1
and proceeds with the remaining components, resulting in a single iterative phase. The order of the components plays a critical role in getting the approach to work. If and when we can find a suitable order, an optimal setting v∗
i
for yi often follows straightforwardly by taking the settings of the prior components into account and myopically assuming that there are no subsequent components. However, coming up with a suitable order and arguing correctness usually require considerable insight and ingenuity.
Approach and instantiations. Instead of coming up with a greedy order for Π from scratch, we derive it from a dynamic program for a generalization Π ′ of Π. We repeatedly expand the Bellman equations underlying the dynamic program for Π ′ and exploit the preconditions of Π and basic monotonicity properties to simplify the resulting terms and figure out which ones yield the optimal value. We succeed in the following classical settings (see subsequent sections for problem definitions):
◦ Interval Scheduling . We start from a straightforward dynamic program for the weighted version and derive the classical greedy algorithm for the unweighed version. The weighted version represents a rare example of a computational problem where the order of the components in the dynamic program matters; earliest start time first yields the best bound on the number of subproblems. Our derivation of the greedy algorithm for the unweighted version transforms the order in a natural way into earliest finish time first. The monotonicity of the 1-dimensional table of subproblem solution values justifies the transition.
◦ Knapsack . We start from the standard dynamic program for the general problem with integral item weights and arbitrary item values. Compared to Interval Scheduling , the subproblems have an additional parameter, namely, the weight limit of the knapsack. The order of the items plays no role in the analysis of the dynamic program. Our derivation for the case with unit values leads to lightest first as the greedy order. The justification follows from the monotonicity of the 2-dimensional table of subproblem solution values with respect to each parameter separately.
◦ Shortest Paths . We consider the single-source version. In contrast to the previous two settings, the decision for each component is non-binary, namely, the last edge on a shortest path from the source to the component vertex. We start from the Bellman-Ford dynamic program for digraphs with arbitrary edge lengths. Our approach turns the program into Dijkstra’s greedy algorithm under the precondition that the edge lengths are nonnegative. The justification is the same monotonicity property that underpins the traditional correctness proofs of Dijkstra’s algorithm, namely, that extending a path cannot make it shorter.
Correctness considerations. The alternate derivations of the greedy algorithms implicitly yield correct-ness arguments that differ from traditional ones and are more modular. The first two examples are ordinarily proven using a charging argument (one-to-one mappings from the greedy selection to any other valid selec-tion) or an exchange argument (massaging any optimal solution into the greedy solution by a finite sequence of local transformations, each of which maintains validity and optimality). 2No such arguments are known for Dijkstra’s algorithm but, interestingly, all three examples can be proven correct using a classical strategy that is dubbed “greedy stays ahead” in [KT05]. The critical ingredient in such an argument is a quality measure for partial solutions such that, for full solutions, optimal quality implies optimal objective value. The correctness argument then consists of showing that at any point in time, the quality measure of the partial greedy solution is at least as high as of any other partial solution on the same components. If and when applicable, the quality measure formally captures intuition for why the greedy order leads to an optimal solution. Coming up with such a quality measure is often nontrivial. In each of the examples, an appropriate quality measure follows from the function argument to which the monotonicity property is applied.
Didactic considerations. When designing an undergraduate algorithms course, one needs to decide whether to teach dynamic programming before greed (as done in textbooks like [CLRS22, Eri19]) or vice versa (as done in textbooks like [KT05, DPV06, Rou19]). Arguments in favor of the latter include prior familiarity of students with greed as opposed to dynamic programming, and developing simpler algorithms for more restricted settings before algorithms for the general setting. Arguments in favor of dynamic pro-gramming before greed include the structured development and wide applicability of dynamic programming versus the ad-hoc character of greedy algorithms and their limited scope, and that correctness proofs for dynamic programs can be streamlined better and tend to be more straightforward. Our alternate approach can be viewed as another argument in favor of covering dynamic programming first, along with the advice to first develop a correct dynamic program for a problem and subsequently see whether the algorithm can be made faster based on greedy considerations that exploit additional structure in the instances of interest. In my own offerings, I have switched to covering dynamic programming first. I employ the alternate approach, not to derive greedy algorithms from dynamic programs but rather as concrete instantiations of the above advice and to provide alternate correctness arguments in addition to the traditional ones. In the setting of
Interval Scheduling , the explanation of the change in order from earliest start time first to earliest finish time first seems to enlighten students. The alternate view can be helpful even when covering greed first, namely, to reconnect to the greedy algorithms for the special cases after presenting the dynamic programs for the general case.
Related work and open question. The paradigms of dynamic programming and greed are studied in several areas, including algorithms and complexity, operations research and management science, and programming languages. The connections reported in this note ought to be known, but I have been unable to find them in the literature (including textbooks and lecture notes). The coverage in [CLRS22] gets closest in that it also relates dynamic programming and greed for Interval Scheduling and for Knapsack ,but the perspective and connections are quite different from ours. For Interval Scheduling , [CLRS22] studies the unweighted version only, and only considers earliest finish time first, so there is no explanation for the switch in order. Their dynamic program tries every possible interval to reduce an instance into at most two smaller instances; they then argue that picking the first interval is always best. For Knapsack ,[CLRS22] simplifies the problem by allowing fractional items rather than restricting to unit values, and explains why the fractional version allows a greedy approach whereas the original problem does not seem to. The presentation derives the greedy algorithm for the fractional variant from the dynamic program for the proper problem. For Shortest Paths , some connections between Bellman-Ford and Dijkstra are well-known, but our deriva-tive approach seems new. The closest presentations [Den03, Ber17, Sni06] view Dijkstra as a modification of Bellman-Ford where the Bellman equations are applied best first (and each only once) instead of breadth first (and each possibly multiple times). They argue correctness of Dijkstra from scratch rather than deriving it from the correctness of Bellman-Ford. Much of the existing literature on the paradigms aims at precisely formalizing them. The programming languages community has developed very general formalisms within the framework of category theory to synthesize programs from problem specifications according to various algorithmic paradigms. A paper with 3a very similar title to ours [BdM93] builds dynamic programs and greedy algorithms under correctness conditions that take the form of monotonicity requirements in the categorical calculus of relations. The conditions for a greedy algorithm are stricter than for a dynamic program, but merely express the very existence of a local criterion that guarantees optimality, and the construction presupposes knowledge of the local criterion. As such, whereas the generality of the formalism is very suitable for program synthesis, the results provide little clue to an algorithm designer what greedy approaches might work or how to formulate a given computational problem to enable a greedy approach. Proofs of successful instantiations hinge on classical correctness arguments for known greedy algorithms. See Section 5 for more details. In a simpler framework for optimization problems with a decomposable structure, [Lew06] proposes a “canon-ical” way of deriving greedy algorithms from dynamic programs, but without correctness guarantees. The canonical local criterion is the straightforward one induced by the decomposition of the objective function. The framework is easy to apply but is much more limited than the categorical one and suffers from similar drawbacks. It provides no indication what order of components could work, and correctness proofs again boil down to classical ones. Attempts with a more complexity-theoretic angle consider restricted frameworks that capture interesting classes of instantiations, establish lower bounds for algorithms within the framework that solve certain com-putational problems, and show connections between approximability of optimization problems and their ex-pressibility within the framework [KH67, Edm71, KL81, HR85, Hel89, Woe00, BNR03, AB04, DI09, ABB +11, BDI11, JS20]. Although not the focus of this note, a potential research direction is to formally capture our approach as well as greedy-stays-ahead arguments, and to investigate their relationship in general. See also Section 5.
Organization. The next three sections develop the three instantiations mentioned above. Each time we review the standard dynamic program for the general problem and derive from it the well-known greedy algorithm for the restricted problem. Readers who want to quickly get to the core, can skip the reviews and jump to Section 2.2, and then to Sections 3.2 and 4.2. The last section takes a closer look at the various monotonicity requirements involved, which relates to the open question. Sections 2 to 4 each include the formal specifications of the optimization problems involved – the general problem as well as the restricted one. All three problems have value versions (which output the value of an optimal solution) and argument versions (which output a valid setting of y1, . . . , y n that yields the optimal value under fx for a given instance x). As the derivations of the greedy algorithms for the argument versions only rely on the dynamic programs for the value versions, we specify the value versions for the general problems and the argument versions for the restricted problems.
2 Interval Scheduling
Also known as activity selection or scheduling classes, Interval Scheduling aims to select from a given list of weighted intervals a non-overlapping subset of maximum total weight. For reasons of similarity with
Knapsack , we will use the term “value” rather than “weight”. Without loss of generality, the values are nonnegative. In regard to the alternate names, we refer to the start time and finish time of an interval. For ease of exposition, we consider non-empty half-open intervals.
Problem specification 1: Interval Scheduling (value version)
Input: nonempty interval Ii = [ si, f i) ⊆ R with start time si and finish time fi for i ∈ [n]value vi ∈ R≥0 for i ∈ [n]
Output: max P
i∈S
vi : S ⊆ [n] ∧ (∀i̸ = j ∈ S) Ii ∩ Ij = ∅
The input can be viewed as consisting of the components xi = ( si, t i, v i) for i ∈ [n]. We need to make a binary decision yi about each component xi, where yi indicates whether interval Ii is included. Alternately, 4the problem can be modeled as successive non-binary decisions as to which interval to include next (if any). We adopt the former view.
2.1 Dynamic program
We consider component x1, determine the optimum under the two choices for y1, and take the best of the two. For either choice of y1, the optimum value can be expressed in terms of the solution of an instance given by a subsequence of x1, . . . , x n: the subsequence x2, . . . , x n for y1 = 0, and the subsequence of the indices j with I1 ∩ Ij = ∅ for y1 = 1. Repeated application of this reduction yields a recursive algorithm, which we can memoize. Since the order of the components within a sequence does not affect the outcome, the subproblems can all be described by subsets J ⊆ [n]. The crucial question is how many distinct subsets arise throughout the recursion. For an arbitrary order of the components, the number can be 2 Ω( n), as illustrated by the example in Figure 1, in which n = 2 m and the values remain unspecified.
R
I1
Im+1
I2
Im+2
I3
Im+3
. . . Im
. . . I2m
Figure 1: Ordering for Interval Scheduling instance yielding 2 Ω( n) distinct subproblems. At level m of the recursion tree for the instance in Figure 1, every possible subset J of {m + 1 , . . . , m } arises as the remaining subproblem for a valid choice of decisions y1, . . . , y m, namely the choice yi = 0 if m + i̸ ∈ J,and yi = 1 otherwise. Thus, the number of distinct subproblems is at least 2 m = 2 n/ 2.However, when we order the components earliest start time first, every subset J in the recursion tree has the structure of a suffix of [ n], of which there are only n + 1. This is because when we apply the reduction to a subproblem corresponding to the suffix J .
= {i, . . . , n } (initially i = 1), either choice of yi merely removes a prefix from the suffix J: the prefix of length 1 for yi = 0, and the prefix consisting of all intervals with start time before fi for yi = 1. This leads to a 1-dimensional table OPT of subproblem solutions: For i ∈ [n + 1] OPT( i) .
= optimal value for instance xi, . . . , x n. (1) The reduction yields the following Bellman equations: OPT( i) =
max(OPT( i + 1) , v i + OPT(next( i)) for i ∈ [n]0 for i = n + 1 , (2) where next( i) denotes the index of the first component j whose start time sj comes at or after the finish time fi of component i, or n + 1 if no such component exists. In symbols, next( i) .
= min {k ∈ { i, . . . , n + 1 } : sk ≥ fi} , (3) where sn+1
.
= ∞ is used as a sentinel. Note that next( i) can be computed in time Θ(log n) using binary search when the components are ordered by start time. Sorting the components and then evaluating (2) in reverse order yields the correct answer for the value version, OPT(1), in time Θ( n log n). For future reference, here is pseudocode (assuming access to next). 5Algorithm 1 Pseudocode for Interval Scheduling (value version) Sort components such that si ≤ si+1 for each i ∈ [n − 1] OPT( n + 1) ← 0
for i = n down to 1 do
OPT( i) ← max(OPT( i + 1) , v i + OPT(next( i)))
return OPT(1) Once we have computed the OPT table, a solution to the argument version can be retrieved in a second pass over the table, using the Bellman equations to determine which decisions are optimal. The second pass proceeds in the order of the components and takes Θ( n) time, resulting in a total running time of Θ( n log n)for the argument version. For future reference, we again include pseudocode (assuming access to next and OPT).
Algorithm 2 Pseudocode for Interval Scheduling solution retrieval
S ← ∅ ; i ← 1
while i ≤ n do if OPT( i) = OPT( i + 1) then
i ← i + 1
else
S ← S ∪ { i}; i ← next( i)
return S
2.2 Greedy algorithm
We now restrict to the special cases where all values are 1.
Problem specification 2: Unit-Value Interval Scheduling (argument version)
Input: nonempty interval Ii = [ si, f i) ⊆ R with start time si and finish time fi for i ∈ [n]
Output: S ⊆ [n] such that |S| is maximized and ( ∀i̸ = j ∈ S) Ii ∩ Ij = ∅
We derive a greedy algorithm for the unit-value setting from the above dynamic program for the general setting. Consider the components ordered earliest start time first, as in the dynamic program. The recursive cases of the Bellman equations (2) can be simplified using the precondition v ≡ 1 and be rewritten as OPT( i) = max(1 + OPT(next( i)) , OPT( i + 1)) . (4) Repeatedly applying (4) to expand the last OPT term on the right-hand side, and ultimately using the base case OPT( n + 1) = 0 yields OPT( i) = max
1 + OPT(next( i)) ,
1 + OPT(next( i + 1)) ,
1 + OPT(next( i + 2)) , (5) ...1 + OPT(next( n))
.
Each term 1 + OPT(next( j)) in (5) corresponds to the situation where interval Ij is the first interval in the 6given order that is selected from among Ii, . . . , I n. We can rewrite (5) as OPT( i) = 1 + max
i≤j≤n
(OPT(next( j))) (6) = 1 + OPT( min
i≤j≤n
(next( j))) (7) = 1 + OPT(next( i∗)) , (8) where
i∗ .
= arg min
i≤j≤n
next( j). (9) Step (6) follows by extracting the constant 1 and is the one and only step that hinges on the restriction of the special case. Step (7) holds because OPT is non-increasing, a property that follows immediately from its defining equation (1): Increasing the argument of OPT means shrinking the set of components that can be used for the underlying maximization problem. Step (8) merely makes the argument that achieves the maximum explicit, which is what definition (9) of i∗ does. (It is irrelevant how ties for i∗ are broken; for concreteness, we pick the smallest among the tied candidates.) By definition (3), the value of next( j) is determined by the finish time fj of Ij , and is non-decreasing as a function of fj since increasing fj shrinks the set of components k of which we take the minimum. Thus,
i∗ is the (first) index of an interval Ij with the earliest finish time among Ii, . . . , I n. The interpretation of the terms in (5) then tells us that an optimal solution is to select interval Ii∗ and then continue with the remaining intervals that do not overlap with Ii∗ . That is to say, we greedily consider the intervals earliest finish time first, which is exactly what the classical greedy algorithm does. For completeness, we include pseudocode for an efficient implementation.
Algorithm 3 Pseudocode for Unit-Value Interval Scheduling (argument version) Sort components such that fi ≤ fi+1 for each i ∈ [n − 1]
S ← ∅ ; f ← −∞
for i = 1 to n do if si ≥ f then
S ← S ∪ { i}; f ← fi
return S
After the initial sorting, which takes time Θ( n log n), the remainder of the algorithm runs in time Θ( n), resulting in an overall running time of Θ( n log n).
2.3 Order considerations
We end our discussion of Interval Scheduling with a few additional remarks about the different orders involved. The order of components that we mention is always the one for applying the reduction that underlies both paradigms. In the dynamic programming, this order is used in the memoized process and in the second phase (solution retrieval) of the iterative implementation (see Algorithm 2). The first phase of the iterative implementation considers the components in the reverse order (see Algorithm 1). The reverse order of the dynamic program for Interval Scheduling is earliest start time last. Upon reversal of the time line, this becomes earliest finish time first. The orientation of the time line does affect correctness and an increasing loop variable is more natural than a decreasing one. As such, rather than sorting components earliest deadline first as in Algorithm 1, typical implementations of the first phase of the dynamic program for Interval Scheduling sort the components earliest finish time first, which is the greedy order for Unit-Value Interval Scheduling . This agreement is a coincidence. When teaching greed before dynamic programming, picking the greedy order for Unit-Value Interval Scheduling as the order in the first iterative phase of the dynamic program for Interval Scheduling requires justification. 7The justification for the greedy order is correctness, whereas the justification for the order in the dynamic program is efficiency. To complete the picture, we analyze ordering the components earliest finish time first in the dynamic program for Interval Scheduling . The subsets J are no longer merely suffixes of [ n], but still have enough structure to keep the number of subproblems small, albeit larger than for earliest start time first. Each J now has the form {j ∈ { i, . . . , n } : sj ≥ fℓ} for 0 ≤ ℓ < i ≤ n, where f0
.
= −∞ acts as a sentinel. The parameter space is therefore 2-dimensional and of size O(n2). Figure 2 describes instances for which Ω( n2) of the combinations (ℓ, i ) actually occur. This means that the number of subproblems for earliest finish time first is Θ( n2) as opposed to Θ( n) for earliest start time first, resulting in a running time of Θ( n2) in lieu of Θ( n log n).
R
I1
Im+1
I2
Im+2
I3
Im+3
. . . Im
...
I2m
Figure 2: Interval Scheduling instance where earliest start time first yields Θ( n2) distinct subproblems. The instances in Figure 2 have even n = 2 m, intervals ordered earliest finish time first, and unspecified values. For each i ∈ [m], all subproblems with parameters ( ℓ, i ) for 0 ≤ ℓ ≤ i − 1 are present at level i − 1 of the recursion tree. They correspond to subsets of the form J = {i, . . . , m } ∪ { m + ℓ + 1 , . . . , 2m} and occur when we set yℓ = 1 (for ℓ ≥ 1) and yℓ+1 = · · · = yi−1 = 0. There are m+1 2
= Θ( n2) such subproblems in total.
3 Knapsack
Given a list of items characterized by their weights and values, we want to select a subset of maximum total value that does not exceed the weight limit of our knapsack. Weights are intended to be nonnegative; values are nonnegative without loss of generality. In order to make the problem amenable to dynamic programming, we need to constrain either the weights or the values to be discrete. We opt for the weights.
Problem specification 3: Knapsack (value version)
Input: item i with weight wi ∈ Z≥0 and value vi ∈ R≥0 for i ∈ [n]weight limit W ∈ Z≥0
Output: max P
i∈S
vi : S ⊆ [n] ∧ P
i∈S
wi ≤ W
Similar to Interval Scheduling , the input can be viewed as consisting of the components xi = ( wi, v i) for
i ∈ [n] and the additional entity W . We need to make a binary decision yi about each component xi, where
yi indicates whether item i is included in the knapsack. Alternately, the sequence of binary decisions can be replaced by successive non-binary decisions as to which item to include next (if any). We again adopt the former view. 83.1 Dynamic program
Given a decision for the first item, it remains to solve the instance consisting of items 2 through n with weight limit W in case we do not include item 1, and the same instance but with a reduced weight limit of W − w1 in case we include item 1 (which is only an option for w1 ≤ W ). The subproblems that arise through repeated application of the reduction consist of a suffix of the given list of items, and a nonnegative integer at most W as the weight limit. The solutions can be represented in a matrix OPT. For i ∈ [n + 1] and w ∈ { 0, . . . , W }
OPT( i, w ) .
= optimal value for components xi, . . . , x n and weight limit w. (10) The Bellman equations are OPT( i, w ) =
max
OPT( i + 1 , w ), v i + OPT( i + 1 , w − wi)
| {z }
considered only if wi≤w
for i ∈ [n]0 for i = n + 1 .
(11) We evaluate (11) row-by-row from row i = n to row i = 1. This yields the optimal value, OPT(1 , W ), in time Θ( n · W ). A subset S ⊆ [n] realizing the optimal value can be retrieved from the OPT table in a similar fashion as in Algorithm 2, based on (11) in lieu of (2). The retrieval takes an additional Θ( n) steps, resulting in an overall running time of Θ( n · W ) for the argument version of Knapsack .
3.2 Greedy algorithm
We now restrict to the special cases where all values are 1.
Problem specification 4: Unit-Value Knapsack (argument version)
Input: item i with weight wi ∈ Z≥0 for i ∈ [n]weight limit W ∈ Z≥0
Output: S ⊆ [n] such that |S| is maximized and P
i∈S
wi ≤ W
We derive a greedy algorithm for the unit-value setting from the above dynamic program for the general setting. The recursive cases of the Bellman equations (11) can be simplified using the precondition v ≡ 1and be rewritten as OPT( i, w ) = max
1 + OPT( i, w − wi)
| {z }
considered only if wi≤w
, OPT( i + 1)
. (12) Repeatedly applying (12) to expand the last OPT term on the right-hand side and ultimately the base case OPT( n + 1 , ·) = 0 yields OPT( i, w ) = max {1 + OPT( j + 1 , w − wj ) : i ≤ j ≤ n ∧ wj ≤ w} , (13) where the term for j corresponds to the choice of j as the first item from among i, . . . , n to be included. The maximum of the empty set is considered zero, corresponding to the case where none of the items i through
n has weight at most w.By its definition (10), OPT( i, w ) is non-increasing in i and non-decreasing in w. In general, these mono-tonicity properties in the individual arguments are insufficient to determine the optimal choice for j in (13). However, if wi happens to be the minimum of wi, . . . , w n, then OPT( i, w − wi) ≥ OPT( i, w − wj ) ≥ OPT( j, w − wj )9holds for any j ∈ { i, . . . , n }, where the first inequality follows from the monotonicity of OPT with respect to its second argument, and the second inequality from the monotonicity of OPT with respect to its first argument. Thus, if we order the items lightest weight first (with ties broken arbitrarily), j = i always is an optimal choice in (13) (provided any valid choice exists). Here is pseudocode for a straightforward Θ( n log n)implementation of the resulting greedy strategy, using wn+1
.
= ∞ as a sentinel.
Algorithm 4 Pseudocode for Unit-Value Knapsack (argument version) Sort items such that wi ≤ wi+1 for each i ∈ [n − 1]
S ← ∅ ; w ← W ; i ← 1
while wi ≤ w do
S ← S ∪ { i}; w ← w − wi
return S
Due to the sorting, Algorithm 4 takes time Θ( n log n). For completeness, we point out that the running time can be improved to Θ( n). The output S∗ is the longest prefix of the sorted order that has total weight at most W . Thanks to the non-negativity of the weights, the length of the prefix S∗ can be found in linear time using binary search and linear-time selection.
4 Shortest Paths
Given a digraph G, edge lengths, and two vertices, s and t, we want to find a shortest path from s to t in
G, or report that no such path exists. Nonexistence happens if G has no path from s to t at all. When negative edge lengths are allowed, it also happens because there is a path from s to t but every such path can be made shorter. We distinguish between the two cases of nonexistence in the notion of distance.
Definition 5 (distance). Let G = ( V, E ) be a digraph with edge lengths ℓ : E → R, and s, t ∈ V . The distance from s to t is
d(s, t ) .
=
∞ if there is no path from s to t
−∞ if there is a path from s to t but no shortest one min {ℓ(P ) : P path s ⇝ t} otherwise ,
where ℓ(P ) .
= P
e∈P
ℓ(e) denotes the length of the path P (viewed as a sequence of edges) and s ⇝ t denotes that the path starts at s and ends at t.The value version of the problem amounts to computing the distance d(s, t ). Typical algorithms implicitly compute the distance for more pairs of vertices than just ( s, t ). We focus on the single-source variant of the problem, which explicitly outputs the distance for a given source s and all possible targets t.
Problem specification 6: Shortest Paths (value version)
Input: digraph G = ( V, E ), ℓ : E → R, s ∈ V
Output: d(s, t ) for every t ∈ V
In the argument version, we need to additionally output a shortest path from s to t whenever one exists. We impose two natural constraints. First, we want each output path to be simple , i.e., to have no repeated vertices. This is possible because any path P from s to t that repeats a vertex contains a cycle as a subpath. Since d(s, t )̸ = −∞ , the cycle has to have nonnegative length and can be cut out from P without increasing the path length. Starting with an arbitrary shortest path from s to t, a finite number of such cuts yields a simple shortest path, as desired. Second, the principle of optimality implies that any prefix of a shortest path from s to t is a shortest path from s to some (other) vertex. We impose such relationships on the single-source output. More precisely, whenever P is the path output for t then, for any vertex v on the path, the subpath of P from s to v is the path output for v.10 Combined, the two constraints are equivalent to the shortest paths forming a tree with root s and edges oriented away from the root. Such a collection of paths can be described by a predecessor function p :
F \ { s} → F , where F denotes the set of vertices t ∈ V with finite d(s, t ), and p(t) symbols the predecessor of t on the chosen shortest path from s to t. A simple shortest path from s to t can be retrieved in reverse order by starting from t and repeatedly applying p until we reach s.
4.1 Dynamic program
Consider a nonempty path P from s to t. We apply the principle of optimality to the subpath of P without its last edge, and try all possibilities for the last edge. The approach yields the following Bellman equations for the distances d(s, t ) for every vertex t̸ = s that is reachable from s:
d(s, t ) = min {d(v, t ) + ℓ(v, t ) : ( v, t ) ∈ E}. (14) These equations are all we need to know about the dynamic program to derive a greedy algorithm under the precondition of nonnegative edge lengths. The dynamic program, known as Bellman-Ford, computes the distances and solves the argument version in time Θ( n(n + m)), where n .
= |V | and m .
= |E|.
4.2 Greedy algorithm
We now restrict attention to digraphs with nonnegative edge lengths. In this setting, distances cannot be negative. We again consider the single-source variant.
Problem specification 7: Nonnegative Shortest Paths (argument version)
Input: digraph G = ( V, E ), ℓ : E → R≥0, s ∈ V
Output: d(s, t ) for every t ∈ V
predecessor p(v) of v on a simple shortest path s ⇝ t for every s̸ = t ∈ V with finite d(s, t )We derive Dijkstra’s greedy algorithm from the Bellman equations (14) for the distances. We need to determine the order to compute the distances d(s, t ) for all reachable t ∈ V . We start with t = s. Since edge lengths are nonnegative, the empty path is optimal, witnessing that d(s, s ) = 0 and leaving p(s) undefined. In every subsequent step, let T denote the set of vertices t for which we have already determined d(s, t ). If there are no reachable vertices outside of T , we know that all remaining distances are ∞ and can stop. Otherwise, consider any reachable vertex t ∈ T .
= V \ T . For analysis purposes, start from the single term
d(s, t ) at level 0. Obtain each next level from the previous one by rewriting every term of the form d(s, v )for v ∈ T as
d(s, v ) → min {d(s, u ) + ℓ(u, v ) : ( u, v ) ∈ E}
based on (14). At level n, the resulting expression is the minimum of the following terms: (a) d(s, u ) + ℓ(u, t ) for all u ∈ T with ( u, t ) ∈ E. These terms correspond to expansions that reach a vertex
u inside T at the first level. (b) d(s, u ) + ℓ(u, v ) + ℓ(P ) for some ( u, v ) ∈ E with u ∈ T , v ∈ T , and some nontrivial path P : v ⇝ t.These terms correspond to expansions that reach a vertex u inside T but not at the first level. (c) d(s, u )+ ℓ(P ) for some u ∈ T and some path P inside T containing n−1 edges. These terms correspond to expansions where no level reaches a vertex u inside T .The minimum over all reachable t ∈ T is achieved by a term of type (a). This follows from the non-negativity of the edge lengths:
◦ The term d(s, u ) + ℓ(u, v ) + ℓ(P ) of type (b) is at least d(s, u ) + ℓ(u, v ), which is a term of type (a) as
v ∈ T is a possible choice for t in (a). 11 ◦ The term d(s, u )+ ℓ(P ) of type (c) can be ignored because it corresponds to paths with P as a subpath, and P contains a repeated vertex since it has n − 1 edges and lies entirely in the set T ⊆ V \ { s} of size at most n − 1. The above analysis shows that for (u∗, v ∗) .
= arg min
(u,v )∈E∩T×T
(d(s, u ) + ℓ(u, v )) , (15) we can compute d(s, v ∗) from the distances we already know, namely, as d(s, v ∗) = d(s, u ∗) + ℓ(u∗, v ∗). Thus, we pick t = v∗ as the next vertex in the order, extend T with t and set p(v∗) = u∗.To efficiently calculate (15), we maintain the vertices v ∈ T in a priority queue, keyed on λ(v) .
= min( {d(s, u )+
ℓ(u, v ) : (u, v ) ∈ E} ∪ {∞} ). The iteration for t requires one minimum extraction and deg +(t) key updates, where deg +(t) denotes the out-degree of t. In total, we need n .
= |V | minimum extractions and P
t∈V
deg +(t) = |E| .
= m key updates. An implementation based on binary heaps runs in time Θ(( n + m) log n) and an implementation based on Fibonacci heaps in time Θ( m + n log n).
5 Monotonicity Properties
We end with some reflections on the monotonicity conditions we used and compare them with monotonicity requirements in other works, in particular the formulations based on the categorical calculus of relations. The principle of optimality [Bel54] that underlies dynamic programming can be viewed as a monotonicity property: Better solutions for subproblems cannot worsen the solution to the given instance. The framework of [BdM93] contains an explicit requirement that the combining function of the subproblems be monotone with respect to the objective function; under that condition, the framework produces a correct dynamic program. To obtain a correct greedy algorithm, the framework needs an additional requirement, which can also be viewed as a monotonicity condition in the categorical calculus of relations:
S · β ⊆ β · R, (16) where · denotes composition of relations, S captures the local criterion, R the global objective, and β is the prefix relation between partial and full solutions. The inclusion (16) states that if p1 and p2 are partial solutions that agree on all but the last step, and the local criterion prefers p1 over p2, then for any full solution
y2 that extends p2, there exists a full solution y1 extending p1 such that y1 is at least as good as y2 with respect to the global objective. [Cur03] refers to requirement (16) as “better global”, and develops similar results under the weaker “best global” requirement, which only imposes the above condition for prefixes p1 in which the last step is optimal with respect to the local criterion. In fact, an even weaker requirement suffices, namely, imposing the condition on prefixes p1 that are obtained by successive locally-optimal choices. This last requirement merely states that the local criterion guarantees optimality. [Cur03] also defines local variants of “best” and “better”, where the global objective is replaced by a quality measure for partial solutions. They can be viewed as formalizations of greedy-stays-ahead in the general categorical framework of relational calculus. All four variants enable synthesizing a greedy algorithm within the framework, but need a local criterion S that satisfies the requirement (16) or its alternates. The con-struction of S and the proof that it works are left to the algorithm designer, as is how to specify a given computational problem so as to allow such S.Our monotonicity properties differ from the ones in the category theoretic setting. In the context of Interval Scheduling and Knapsack , we need to relate the optimal objective values for different subproblems, e.g., for an instance and the instance with one component less. In contrast, the requirements in the category theoretic setting relate the objective values of different solutions to the same subproblem. 12 For completeness, we mention that monotonicity properties have been used in other contexts to improve dynamic programs, without turning them into greedy algorithms. An example is the monotonicity of the break points in the classical dynamic program for optimal binary search trees, which allows shaving off a linear factor in the running time and relates to totally monotone matrices [Knu71, Yao82, AKM +87, BGLZ09]. Such monotonicity properties differ from ours as they refer to arguments whereas ours refer to values of the objective function.
Acknowledgements
I would like to thank Allan Borodin, Sanjoy Dasgupta, Jon Kleinberg, Tim Roughgarden, and Eva Tardos for feedback and encouragement. I am also grateful to (former) students Tom Watson, Drew Morgan, Nicollas Sdroievski, and Ahmed Shaaban for relevant discussions.
References
[AB04] Spyros Angelopoulos and Allan Borodin. The power of priority algorithms for facility location and set cover. Algorithmica , 40(4):271–291, 2004. [ABB +11] Michael Alekhnovich, Allan Borodin, Joshua Buresh-Oppenheim, Russell Impagliazzo, Avner Magen, and Toniann Pitassi. Toward a model for backtracking and dynamic programming.
Computational Complexity , 20(4):679–740, 2011. [AKM +87] Alok Aggarwal, Maria M. Klawe, Shlomo Moran, Peter W. Shor, and Robert E. Wilber. Geo-metric applications of a matrix-searching algorithm. Algorithmica , 2:195–208, 1987. [BDI11] Joshua Buresh-Oppenheim, Sashka Davis, and Russell Impagliazzo. A stronger model of dynamic programming algorithms. Algorithmica , 60(4):938–968, 2011. [BdM93] Richard S. Bird and Oege de Moor. From dynamic programming to greedy algorithms. In Formal Program Development , pages 43–61, 1993. [Bel54] Richard Bellman. The theory of dynamic programming. Bulletin of the American Mathematical Society , 60(6):503–516, 1954. [Ber17] Dimitri P. Bertsekas. Dynamic programming and optimal control . Athena Scientific, 4th edition, 2017. [BGLZ09] Wolfgang W. Bein, Mordecai J. Golin, Lawrence L. Larmore, and Yan Zhang. The Knuth-Yao quadrangle-inequality speedup is a consequence of total monotonicity. ACM Transaction on Algorithms , 6(1):17:1–17:22, 2009. [BNR03] Allan Borodin, Morten N. Nielsen, and Charles Rackoff. (Incremental) priority algorithms.
Algorithmica , 37(4):295–326, 2003. [CLRS22] Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms . MIT Press, 4th edition, 2022. [Cur03] Sharon A. Curtis. The classification of greedy algorithms. Science of Computer Programming ,49(1):125–157, 2003. [Den03] Eric V. Denardo. Dynamic Programming: Models and Applications . Dover Publications, 2003. [DI09] Sashka Davis and Russell Impagliazzo. Models of greedy algorithms for graph problems. Algo-rithmica , 54(3):269–317, 2009. [DPV06] Sanjoy Dasgupta, Christos H. Papadimitriou, and Umesh V. Vazirani. Algorithms . McGraw-Hill, 2006. 13 [Edm71] Jack Edmonds. Matroids and the greedy algorithm. Mathematical Programming , 1(1):127–136, 1971. [Eri19] Jeff Erickson. Algorithms . Independently published, 2019. [Hel89] Paul Helman. A common schema for dynamic programming and branch and bound algorithms.
Journal of the ACM , 36(1):97–128, 1989. [HR85] Paul Helman and Arnon Rosenthal. A comprehensive model of dynamic programming. SIAM Journal on Algebraic Discrete Methods , 6(2):319–334, 1985. [JS20] Stasys Jukna and Hannes Seiwert. Approximation limitations of pure dynamic programming.
SIAM Journal on Computing , 49(1):170–205, 2020. [KH67] Richard M. Karp and Michael Held. Finite-state processes and dynamic programming. SIAM Journal on Applied Mathematics , 15(3):693–718, 1967. [KL81] Bernhard Korte and L´ aszl´ o Lov´ asz. Mathematical structures underlying greedy algorithms. In Proceedings of the International Fundamentals of Computation Theory Conference (FCT) ,volume 117, pages 205–209, 1981. [Knu71] Donald E. Knuth. Optimum binary search trees. Acta Informatica , 1:14–25, 1971. [KT05] Jon Kleinberg and Eva Tardos. Algorithm Design . Pearson, 2005. [Lew06] Art Lew. Canonical greedy algorithms and dynamic programming. Journal of Control and Cybernetics , 35(3):621–643, 2006. [Rou19] Tim Roughgarden. Algorithms Illuminated . Soundlikeyourself Publishing, 2019. [Sni06] Moshe Sniedovich. Dijkstra’s algorithm revisited: the dynamic programming connexion. Journal of Control and Cybernetics , 35(3):599–620, 2006. [Woe00] Gerhard J. Woeginger. When does a dynamic programming formulation guarantee the existence of a fully polynomial time approximation scheme (FPTAS)? INFORMS Journal on Computing ,12(1):57–74, 2000. [Yao82] F. Frances Yao. Speed-up in dynamic programming. SIAM Journal on Algebraic Discrete Methods , 3(4):532–540, 1982. 14 |
188453 | https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:complex/x9e81a4f98389efdf:fta/a/complex-numbers-faqs | Complex numbers: FAQ (article) | Khan Academy
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Precalculus
Course: Precalculus>Unit 3
Lesson 9: The fundamental theorem of algebra
The Fundamental theorem of Algebra
Quadratics & the Fundamental Theorem of Algebra
Number of possible real roots of a polynomial
Complex numbers: FAQ
Math>
Precalculus>
Complex numbers>
The fundamental theorem of algebra
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Complex numbers: FAQ
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What is the complex plane?
The complex plane is a way of visually representing complex numbers. We plot the real component of the complex number on the x-axis, and the imaginary component on the y-axis.
How do we find the distance and midpoint of complex numbers?
The distance and midpoint of complex numbers are similar to the distance and midpoint of points on a coordinate plane. The distance is how far apart two complex numbers are, and the midpoint is the point that is exactly in between them.
To find the distance between two complex numbers, we use the Pythagorean theorem.
To find the midpoint between two complex numbers, we use the average of their real parts and the average of their imaginary parts.
What is a complex conjugate?
The complex conjugate of a complex number is what we get when we change the sign of the imaginary component. For example, the complex conjugate of 2+3 i is 2−3 i.
Why do we divide complex numbers by using their complex conjugates?
When we multiply a complex number by its complex conjugate, we get a real number. This makes it easier to divide one complex number by another, since we can multiply both the numerator and the denominator by the complex conjugate of the denominator. This way, we make the denominator a real number that we can more easily divide.
What are some identities with complex numbers?
Identities are equations that are always true, no matter what values we plug in for the variables. They are useful for simplifying expressions and solving problems. Some common identities with complex numbers are:
i 2=−1
i 3=−i
i 4=1
(a+b i)2=a 2+2 a b i−b 2
(a+b i)3=a 3+3 a 2 b i−3 a b 2−b 3 i
(a+b i)4=a 4+4 a 3 b i−6 a 2 b 2−4 a b 3 i+b 4
a+b i―=a−b i (this is the complex conjugate)
|a+b i|=a 2+b 2 (this is the modulus or absolute value)
(a+b i)(a−b i)=a 2+b 2
(a+b i)(c+d i)―=(a+b i)―⋅(c+d i)―
|a+b i|⋅|c+d i|=|(a+b i)(c+d i)|
We can use these identities to simplify complex numbers and find their properties, like modulus, argument, and polar form.
What are the modulus and argument of a complex number?
The modulus of a complex number is the distance from the origin to the complex number on the complex plane. The argument is the angle the complex number makes with the positive x-axis.
Why do we multiply and divide complex numbers in polar form?
It's often easier to multiply and divide complex numbers when they are in polar form, rather than in rectangular form. In polar form, we can multiply two complex numbers by multiplying their moduli and adding their arguments, and we can divide two complex numbers by dividing their moduli and subtracting their arguments.
What is the fundamental theorem of algebra?
The fundamental theorem of algebra is a very important and beautiful result that tells us something amazing about complex numbers and polynomials. The fundamental theorem of algebra says that every polynomial of degree n, where n is a positive whole number, has exactly n complex roots, or solutions. A root of a polynomial is a value that makes the polynomial equal to zero, like x=2 for x 2−4=0. A complex root is a root that is a complex number, like x=1+i for x 2−2 x+2=0.
The fundamental theorem of algebra is amazing because it tells us that complex numbers are essential for solving any polynomial equation, no matter how complicated or simple it is. It also tells us that complex numbers are complete, meaning that there is no other type of number that we need to invent to find the roots of any polynomial. It also tells us that if a complex number is a root of a polynomial, then so is its complex conjugate. For example, if 1+i is a root of x 2−2 x+2=0, then so is 1−i.
The fundamental theorem of algebra has many applications in mathematics, science, engineering, and art, where polynomials are used to model and study various phenomena and patterns. For example, polynomials can be used to describe the shape of curves, the motion of objects, the behavior of waves, the distribution of data, the encryption of information, and the creation of fractals.
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Abhinav Karthik 3 years ago Posted 3 years ago. Direct link to Abhinav Karthik's post “The complex conjugate of ...” more The complex conjugate of (a+bi)(c+di)≠ The complex conjugate of (a+bic+di).
It is equal to the [Complex conjugate of ac-bd+i(da+bc)]=ac-bd-adi-bci.
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Jeff Dodds 3 years ago Posted 3 years ago. Direct link to Jeff Dodds's post “Good eye, Abhinav! I fixe...” more Good eye, Abhinav! I fixed a formatting issue -- see the updated identity. It's a cool one...it says that the conjugate of the product of the two complex numbers is equal to the product of their conjugates. Thank you for your comment!
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Ross 2 years ago Posted 2 years ago. Direct link to Ross's post “I think technically below...” more I think technically below is not correct:
a + bi = a − bi
(a - bi) is a conjugate of (a + bi) but it does not equal it.
Maybe rephrase that bullet point to:
(a - bi) is a complex conjugate of (a + bi)
To avoid confusion.
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smhousehold 2 years ago Posted 2 years ago. Direct link to smhousehold's post “I thought so too at first...” more I thought so too at first, but then I noticed a bar symbol above a+b, which means "conjugate." But yeah, a little confusing at first because my eye skipped over the bar symbol like it was a graphical separator.
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danlanx434 10 months ago Posted 10 months ago. Direct link to danlanx434's post “Does x^2 - 2x + 1 disprov...” more Does x^2 - 2x + 1 disprove the fundamental theorem of algebra? The fundamental theorem of algebra says that the quadratic should have 2 roots, but the only root of x^2 - 2x + 1, including complex roots, is x = 1.
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Tanner P 10 months ago Posted 10 months ago. Direct link to Tanner P's post “Good question. It may see...” more Good question. It may seem weird, but the roots don't necessarily have to be unique. The fundamental theorem of algebra says that every nth-degree polynomial will have n roots, including multiplicity. This means that you count repeated roots multiple times. Since x^2 - 2x + 1 = (x-1)(x-1) = (x-1)^2, count x = 1 twice to get 2 roots.
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thelastcylon 7 months ago Posted 7 months ago. Direct link to thelastcylon's post “The biggest question I ha...” more The biggest question I have with respect to complex numbers:
How did I not notice, that my calculator can convert between polar and rectangular form, until after I mastered the unit?
facepalm
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dylan.t.le a year ago Posted a year ago. Direct link to dylan.t.le's post “Just wanted to say someth...” more Just wanted to say something:
When I first looked into multiplying complex numbers, it first appeared that the easier way to multiply or divide complex numbers is in rectangular form.
It appears that (a+bi)/(c+di) or (a+bi)(c+di) would be much easier than in polar form r(cos(A)+isin(A)) k(cos(B)+isin(B)).
I mean, wouldn't it be much simpler to used simple algebra in rectangular form than the littler bit harder trigonometric properties in polar form?
However, when we look into more repeated multiplication other than multiplying just two numbers, we find that multiplying the complex binomials much more time consuming with more and more binomials. Instead, polar form is more consistent and recognizes an easy pattern with more and more repeated multiplication. z^7? (a+bi)^7 is much more time consuming to computate than using De Moivre's formula that z^7 = r^7(cos(nA)+isin(A))
Even with multiple unique complex numbers, in polar form it is much simpler. Just multiply all of the moduluses and add all of the arguments in polar form than using a vary long Distributive Property method for each of the n amounts of complex numbers in rectangular form. Same for dividing.
This is just to the people who are wondering why polar form is easier to multiply and divide than rectangular numbers. I hope it helps!
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EHAN 2 years ago Posted 2 years ago. Direct link to EHAN's post “when they mentioned that ...” more when they mentioned that n should be a positive whole number, I wondered what the graph of x^1.5 or sqrt(x^3) would look like? does anybody know how to plot it? I tried desmos but it wouldn't show anything below the x-axis
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Kaleb Morgan 2 years ago Posted 2 years ago. Direct link to Kaleb Morgan's post “What is the argument of a...” more What is the argument of a complex number?
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kubleeka 2 years ago Posted 2 years ago. Direct link to kubleeka's post “If you plot the number in...” more If you plot the number in the complex plane, then draw a segment from the number to the origin, the argument is the angle between the positive real axis and the segment.
So positive real numbers have an argument of 0, negative reals have an argument of π, i has an argument of π/2, and 1+i has an argument of π/4.
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Sadie Cafagna 4 months ago Posted 4 months ago. Direct link to Sadie Cafagna's post “How do you find complex r...” more How do you find complex roots when the degree is larger than 2?
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Serena 2 months ago Posted 2 months ago. Direct link to Serena's post “under the title: "What ar...” more under the title: "What are some identities with complex numbers?" identity number (bullet point number;) 10 and 11 seem to both equate to the exact same thing, how do you use each of those identities? It's like saying (5)(5)=(5)x(5)? I am wondering if I am missing something?
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Graphing a Linear Equation Using Slope Intercept Form
Now that you've completed a lesson on graphing slope you
are finally ready to graph linear equations.
There are several different ways to graph linear equations. You've already learned how to graph using a table of values. That's okay for the beginner, but it can be a little time consuming.
Using slope intercept form is one of the quickest and easiest ways to graph a linear equation.
Before we begin, I need to introduce a little vocabulary. We are going to talk about x and y intercepts.
An x intercept is the point where your line crosses the x-axis. The y intercept is the point where your line crosses the y-axis.
We are only going to focus on the y intercept in this lesson, but you'll need to know x intercept for later.
Let's take a look at intercepts.
Slope intercept form is used when your linear equation is written in the form:
y = mx + b
x and y are your variables. m will be a numeral, which is your slope. b will also be a numeral and
this is the y-intercept.
In this form only (when your equation is written as y = ....) the coefficient of x is the slope and
the constant is the y intercept.
When an equation is written in this form, you can look at the equation and have enough information to graph the equation.
Take a look...
Slope Intercept Form
Let's look at a few examples and I promise that you'll LOVE this new way of graphing!
Example 1: Graphing Using Slope Intercept Form
Graph the equation: y = 2x + 4
Before we begin, let's identify the slope and y-intercept.
Slope = 2 or 2/1
Y-intercept = 4 or (0,4)
Step 1: Plot the y-intercept on your graph. The y-intercept is 4, so I will plot the point (0,4)
Step 2: From the y-intercept (0,4) use the slope to plot your next point.
The slope is 2, so you will rise 2 (up) and run 1 (to the right).
Step 3: Draw a line through your two points.
This line represents the equation y = 2x + 4.
Every point on this line is a solution to this equation.
Need a little more clarification? No problem, just check out the
following video. Example 1 will be explained again step by step.
Let's take a look at one more example.
In this example, we will graph an equation that has a negative slope.
Example 2: A Negative Slope
Graph the equation, y = -1/3x
Let's first identify the slope and y-intercept.
The slope is -1/3.
The y-intercept is 0. Since there is no number value for b, the y-intercept is 0. This means that the y-intercept is at the origin or (0,0).
Step 1: Plot your y-intercept. (0,0).
Step 2: The slope is -1/3. Therefore, from the y-intercept, we will count down 1 and right 3. Then plot your next point at (-1,3).
Step 3: Draw a line through your two points.
This line represents the equation y = -1/3x.
Every point on this line is a solution to this equation.
Notice that the slope in this equation is negative. This means that our line must be "falling" from left to right.
Always double check your line and your slope. If your slope is positive, then your line should "rise" from left to right. If your slope is negative, then your line should "fall" from left to right.
Here's a quick summary of this lesson:
Rules for Graphing Using Slope Intercept Form
Your y intercept is always the first point that you plot on the line. Your point will always be
(0, b).
Then use your slope to plot your next point.
If you have two points, you can draw a straight line and this is the line that represents your equation.
Any point on that line is a solution to the equation.
Tip: You have to be very accurate in plotting your points and drawing your lines in order to be able
to read your graph to find other solutions!
Home
Graphing Equations
Slope Intercept Form
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188455 | https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/11%3A_Time-Independent_Perturbation_Theory/11.08%3A_Fine_Structure_of_Hydrogen | Skip to main content
11.8: Fine Structure of Hydrogen
Last updated
: Mar 31, 2025
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11.7: Linear Stark Effect
11.9: Zeeman Effect
Page ID
: 15941
Richard Fitzpatrick
University of Texas at Austin
( \newcommand{\kernel}{\mathrm{null}\,})
According to special relativity, the kinetic energy (i.e., the difference between the total energy and the rest mass energy) of a particle of rest mass mm and momentum pp is T=√p2c2+m2c4−mc2.
T=p2c2+m2c4−−−−−−−−−−−√−mc2.(11.8.1)
In the non-relativistic limit p≪mcp≪mc, we can expand the square-root in the previous expression to give T=p22m[1−14(pmc)2+O(pmc)4].
T=p22m[1−14(pmc)2+O(pmc)4].(11.8.2)
Hence, T≃p22m−p48m3c2.
T≃p22m−p48m3c2.(11.8.3)
Of course, we recognize the first term on the right-hand side of this equation as the standard non-relativistic expression for the kinetic energy. The second term is the lowest-order relativistic correction to this energy. Let us consider the effect of this type of correction on the energy levels of a hydrogen atom. So, the unperturbed Hamiltonian is given by Equation ([e12.58]), and the perturbing Hamiltonian takes the form H1=−p48m3ec2.
H1=−p48m3ec2.(11.8.4)
Now, according to standard first-order perturbation theory (see Section 1.4), the lowest-order relativistic correction to the energy of a hydrogen atom state characterized by the standard quantum numbers nn, ll, and mm is given by ΔEnlm=⟨n,l,m|H1|n,l,m⟩=−18m3ec2⟨n,l,m|p4|n,l,m⟩=−18m3ec2⟨n,l,m|p2p2|n,l,m⟩.
ΔEnlm=⟨n,l,m|H1|n,l,m⟩=−18m3ec2⟨n,l,m|p4|n,l,m⟩=−18m3ec2⟨n,l,m|p2p2|n,l,m⟩.
However, Schrödinger’s equation for a unperturbed hydrogen atom can be written p2ψn,l,m=2me(En−V)ψn,l,m,
p2ψn,l,m=2me(En−V)ψn,l,m,(11.8.5)
where V=−e2/(4πϵ0r)V=−e2/(4πϵ0r). Because p2p2 is an Hermitian operator, it follows that ΔEnlm=−12mec2⟨n,l,m|(En−V)2|n,l,m⟩=−12mec2(E2n−2En⟨n,l,m|V|n,l,m⟩+⟨n,l,m|V2|n,l,m⟩)=−12mec2[E2n+2En(e24πϵ0)⟨1r⟩+(e24πϵ0)2⟨1r2⟩].
ΔEnlm=−12mec2⟨n,l,m|(En−V)2|n,l,m⟩=−12mec2(E2n−2En⟨n,l,m|V|n,l,m⟩+⟨n,l,m|V2|n,l,m⟩)=−12mec2[E2n+2En(e24πϵ0)⟨1r⟩+(e24πϵ0)2⟨1r2⟩].
It follows from Equations ([e9.74]) and ([e9.75]) that ΔEnlm=−12mec2[E2n+2En(e24πϵ0)1n2a0+(e24πϵ0)21(l+1/2)n3a20].
ΔEnlm=−12mec2[E2n+2En(e24πϵ0)1n2a0+(e24πϵ0)21(l+1/2)n3a20].
Finally, making use of Equations ([e9.55]), ([e9.56]), and ([e9.57]), the previous expression reduces to
ΔEnlm=Enα2n2(nl+1/2−34),
ΔEnlm=Enα2n2(nl+1/2−34),(11.8.6)
where α=e24πϵ0ℏc≃1137
α=e24πϵ0ℏc≃1137(11.8.7)
is the dimensionless fine structure constant.
Note that the previous derivation implicitly assumes that p4p4 is an Hermitian operator. It turns out that this is not the case for l=0l=0 states. However, somewhat fortuitously, our calculation still gives the correct answer when l=0l=0. Note, also, that we are able to employ non-degenerate perturbation theory in the previous calculation, using the ψnlmψnlm eigenstates, because the perturbing Hamiltonian commutes with both L2L2 and LzLz. It follows that there is no coupling between states with different ll and mm quantum numbers. Hence, all coupled states have different nn quantum numbers, and therefore have different energies.
Now, an electron in a hydrogen atom experiences an electric field E=er4πϵ0r3
due to the charge on the nucleus. However, according to electromagnetic theory, a non-relativistic particle moving in a electric field E with velocity v also experiences an effective magnetic field
B=−v×Ec2
Recall, that an electron possesses a magnetic moment [see Equations ([e10.58]) and ([e10.59])] μ=−emeSdue to its spin angular momentum, S. We, therefore, expect an additional contribution to the Hamiltonian of a hydrogen atom of the form [see Equation ([e10.60a])] H1=−μ⋅B=−e24πϵ0mec2r3v×r⋅S=e24πϵ0m2ec2r3L⋅S,where L=mer×v is the electron’s orbital angular momentum. This effect is known as spin-orbit coupling. It turns out that the previous expression is too large, by a factor 2, due to an obscure relativistic effect known as Thomas precession . Hence, the true spin-orbit correction to the Hamiltonian is
H1=e28πϵ0m2ec2r3L⋅S.
Let us now apply perturbation theory to the hydrogen atom, using the previous expression as the perturbing Hamiltonian.
Now, J=L+S
is the total angular momentum of the system. Hence, J2=L2+S2+2L⋅S,giving L⋅S=12(J2−L2−S2).Recall, from Section [s11.2], that while J2 commutes with both L2 and S2, it does not commute with either Lz or Sz. It follows that the perturbing Hamiltonian ([e12.127]) also commutes with both L2 and S2, but does not commute with either Lz or Sz. Hence, the simultaneous eigenstates of the unperturbed Hamiltonian ([e12.58]) and the perturbing Hamiltonian ([e12.127]) are the same as the simultaneous eigenstates of L2, S2, and J2 discussed in Section [s11.3]. It is important to know this because, according to Section 1.6, we can only safely apply perturbation theory to the simultaneous eigenstates of the unperturbed and perturbing Hamiltonians.
Adopting the notation introduced in Section [s11.3], let ψ(2)l,s;j,mj be a simultaneous eigenstate of L2, S2, J2, and Jz corresponding to the eigenvalues L2ψ(2)l,s;j,mj=l(l+1)ℏ2ψ(2)l,s;j,mj,S2ψ(2)l,s;j,mj=s(s+1)ℏ2ψ(2)l,s;j,mj,J2ψ(2)l,s;j,mj=j(j+1)ℏ2ψ(2)l,s;j,mj,Jzψ(2)l,s;j,mj=mjℏψ(2)l,s;j,mj.
According to standard first-order perturbation theory, the energy-shift induced in such a state by spin-orbit coupling is given by ΔEl,1/2;j,mj=⟨l,1/2;j,mj|H1|l,1/2;j,mj⟩=e216πϵ0m2ec2⟨1,1/2;j,mj|J2−L2−S2r3|l,1/2;j,mj⟩=e2ℏ216πϵ0m2ec2[j(j+1)−l(l+1)−3/4]⟨1r3⟩.Here, we have made use of the fact that s=1/2 for an electron. It follows from Equation ([e9.75a]) that ΔEl,1/2;j,mj=e2ℏ216πϵ0m2ec2a30[j(j+1)−l(l+1)−3/4l(l+1/2)(l+1)n3],where n is the radial quantum number. Finally, making use of Equations ([e9.55]), ([e9.56]), and ([e9.57]), the previous expression reduces to
ΔEl,1/2;j,mj=Enα2n2[n{3/4+l(l+1)−j(j+1)}2l(l+1/2)(l+1)],
where α is the fine structure constant. A comparison of this expression with Equation ([e12.121]) reveals that the energy-shift due to spin-orbit coupling is of the same order of magnitude as that due to the lowest-order relativistic correction to the Hamiltonian. We can add these two corrections together (making use of the fact that j=l±1/2 for a hydrogen atom—see Section [s11.3]) to obtain a net energy-shift of
ΔEl,1/2;j,mj=Enα2n2(nj+1/2−34).
This modification of the energy levels of a hydrogen atom due to a combination of relativity and spin-orbit coupling is known as fine structure.
Now, it is conventional to refer to the energy eigenstates of a hydrogen atom that are also simultaneous eigenstates of J2 as nLj states, where n is the radial quantum number, L=(S,P,D,F,⋯) as l=(0,1,2,3,⋯), and j is the total angular momentum quantum number. Let us examine the effect of the fine structure energy-shift ([e12.138]) on these eigenstates for n=1,2 and 3.
For n=1, in the absence of fine structure, there are two degenerate 1S1/2 states. According to Equation ([e12.138]), the fine structure induced energy-shifts of these two states are the same. Hence, fine structure does not break the degeneracy of the two 1S1/2 states of hydrogen.
For n=2, in the absence of fine structure, there are two 2S1/2 states, two 2P1/2 states, and four 2P3/2 states, all of which are degenerate. According to Equation ([e12.138]), the fine structure induced energy-shifts of the 2S1/2 and 2P1/2 states are the same as one another, but are different from the induced energy-shift of the 2P3/2 states. Hence, fine structure does not break the degeneracy of the 2S1/2 and 2P1/2 states of hydrogen, but does break the degeneracy of these states relative to the 2P3/2 states.
For n=3, in the absence of fine structure, there are two 3S1/2 states, two 3P1/2 states, four 3P3/2 states, four 3D3/2 states, and six 3D5/2 states, all of which are degenerate. According to Equation ([e12.138]), fine structure breaks these states into three groups: the 3S1/2 and 3P1/2 states, the 3P3/2 and 3D3/2 states, and the 3D5/2 states.
The effect of the fine structure energy-shift on the n=1, 2, and 3 energy states of a hydrogen atom is illustrated in Figure below:
Figure 23: Effect of the fine structure energy-shift on the and 3 states of a hydrogen atom. Not to scale.
Note, finally, that although expression ([e12.137]) does not have a well defined value for l=0, when added to expression ([e12.121]) it, somewhat fortuitously, gives rise to an expression ([e12.138]) that is both well-defined and correct when l=0.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
11.7: Linear Stark Effect
11.9: Zeeman Effect |
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2015 AIME II Problems/Problem 5
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (cheese)
6 Solution 5
7 Video Solution
8 See also
Problem
Two unit squares are selected at random without replacement from an grid of unit squares. Find the least positive integer such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than .
Solution 1
Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions . There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is , and the number of ways to pick two squares out of Grid A is . So, the probability that the two chosen squares are adjacent is . We wish to find the smallest positive integer such that , and by inspection the first such is .
Solution 2
Consider a grid, where there are corner squares, edge squares, and center square. A grid has corner squares, edge squares, and center squares. By examining simple cases, we can conclude that for a grid that is , there are always corners squares, edge squares, and center squares.
Each corner square is adjacent to other squares, edge squares to other squares, and center squares to other squares. In the problem, the second square can be any square that is not the first, which means there are to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is .
Simplifying, we get which we can set to be less than . By inspection, we find that the first such is .
Solution 3
There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is multiplied by . Add that to the probability of an edge and an adjacent square( multiplied by ) and the probability of a corner and an adjacent square( multiplied by ) to get . Simplify to get . With some experimentation, we realize that the smallest value of n is .
Solution 4 (cheese)
Notice how a chosen unit square on the grid has 4 vertically & horizontally adjacent squares around it (not counting corners or sides.) That's . Using this, we rewrite as . Notice that the denominator is really close to , and the problem is asking for the least positive integer less than . Therefore, the closest possible estimation is . We can check this by adding in our corners and sides. Easy multiplication and simplification finds us with as the correct answer.
~orenbad
Solution 5
Lets start with some simpler cases. For example, lets use a 2x2. We have horizontally and vertically, for a total of . Then, our possible outcomes is just , so our probability is just . Now, lets do a 3x3. We have that there are horizontally and vertically, for a total of . Our possible outcomes is just . Now, our probability is just . Now, lets try to generalize. For each colum, there is ways, then we multiply that by for our columns. Then we multiply all that by for the rows to get . For our denominator, we have squares, and we need to choose of them, so our probability is: . Lets simplify this. We have: . So, we have this to be less than . So we have: , so the smallest that satisfies this condition is
-jb2015007
Video Solution
~MathProblemSolvingSkills.com
See also
2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4Followed by
Problem 6
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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Plants & Animals for Kids
Differences Between Insects and Spiders: Lesson for Kids
Contributors: Mark Boster
Instructor
Instructor:
Mark Boster
Cite this lesson
Have you ever looked at a cookie and thought you saw chocolate chips, but they were really raisins? Well, sometimes animals look alike, but they are really different. This includes insects and spiders. Do you know the difference?
Table of Contents
MOM!
Arthropods
Insects
Arachnids
Lesson Summary
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QuizCourse15K views
MOM! ----
''Mom, Mom,'' bellowed Ted, ''Eugenia threw a nasty insect on me!''
''I did not,'' said Eugenia. ''It was a spider.''
Is a spider classified as an insect or not? How can you tell?
Arthropods ----------
All bugs are not the same. There is a group of animals called arthropods. Arthropods are invertebrate animals that have a segmented body, jointed limbs, and an outer body that the animals occasionally shed.
Here are a few key points:
All insects are arthropods, but not all arthropods are insects.
Also, all arachnids are arthropods, but not all arthropods are arachnids.
If this confuses you, let's try to un-confuse you with the first diagram. See, insects and arachnids are both arthropods, but not all of the arthropods are arachnids or insects.
Arthropods include insects and arachnids.
Insects -------
Insects are the largest group of arthropods there is, and in fact, the largest group of any animal on Earth. There are over one million different types of insects on Earth. But how can you tell if a bug is an insect?
All insects must have three pairs of legs. That means they have a total of six legs.
Insects can live almost anywhere on Earth.
At some time in their life, an insect must have wings. Maybe not when they are young, but they may have wings when they are adults.
All insects have three parts, or segments, to their bodies.
Parts of an Insect
The head for eating, seeing, smelling, hearing, and any other senses they may have. All the heads of insects have two antennas and two compound eyes. That means they have two eyes that each see many of the same images, which makes it easier to see motion.
The thorax helps insects get around. Their legs are attached to the thorax.
The abdomen is what insects use to reproduce.
Some examples of insects are ants, butterflies, flies, grasshoppers, and lightning bugs.
Arachnids ---------
So, then what is a spider? A spider is in a group of arthropods called arachnids. Arachnids are a group of animals with the following traits:
Arachnids have four pairs of legs, for a total of eight legs. (Remember that insects have three pairs of legs or six legs in total)
Arachnids never have wings. (Insects do at some point in their lives.)
Arachnids do not have antennas. (Insects do!)
Arachnids have two body sections. These are the cephalothorax (head and thorax together), and the abdomen. (Insects have three body sections.)
Parts of an Arachnid
The cephalothorax has the mouth, eyes, and is also where the legs attach.
Like insects, the abdomen is used in mating.
Other than spiders, arachnids also include scorpions, mites, and ticks
Lesson Summary --------------
Arthropods are invertebrate animals that have a segmented body, jointed limbs, and an outer body that occasionally sheds. Arthropods include both insects and spiders. The differences in the two types of animals are found in the number of body sections, the number of legs, and if they have antennas and wings.
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Plants & Animals for Kids
34 chapters
317 lessons
Chapter 1
Animal Habitats & Endangerment for Elementary School
Dolphin Habitats: Lesson for Kids 3:19 min
Why Are Dolphins Endangered? - Lesson for Kids
Horse Habitat Facts: Lesson for Kids 3:21 min
Why Are Cheetahs Endangered? - Lesson for Kids 3:43 min
African Savanna Lesson for Kids: Facts & Habitat 4:00 min
Endangered Species: Lesson for Kids
Endangered Animals in Africa: Lesson for Kids
Endangered Animals Research: Lesson for Kids
Why are Penguins Endangered? - Lesson for Kids
Why are Sharks Endangered? - Lesson for Kids
Why are Lions Endangered? - Lesson for Kids 2:41 min
Why are Asian Elephants Endangered? - Lesson for Kids
Why are African Elephants Endangered? - Lesson for Kids
Endangered Monkeys: Lesson for Kids
Endangered Birds: Lesson for Kids
Endangered Frogs: Lesson for Kids
Endangered Animals In Asia: Lesson for Kids 3:50 min
Why are Sea Turtles Endangered? - Lesson for Kids
Endangered Species in Canada: Lesson for Kids
Why Are Blue Whales Endangered? - Lesson for Kids 3:59 min
Why are Gorillas Endangered? - Lesson for Kids
Why is the Great White Shark Endangered? - Lesson for Kids 3:37 min
Why is the Grey Wolf Endangered? - Lesson for Kids 3:16 min
Why Are Jaguars Endangered? - Lesson for Kids 3:35 min
Why is the Snow Leopard Endangered? - Lesson for Kids
Why are Rhinos Endangered? - Lesson for Kids
Why are Numbats Endangered? - Lesson for Kids
Great White Shark Habitat: Lesson for Kids
Whaling & Overfishing in Antarctica: Lesson for Kids
Extinct Tiger Facts: Lesson for Kids
Bobcat Habitat Facts: Lesson for Kids
Grizzly Bear Habitat: Lesson for Kids
Kangaroo Habitat Facts: Lesson for Kids
Koala Habitat: Lesson for Kids
Snake Habitat: Lesson for Kids 2:48 min
Jellyfish Habitat Facts: Lesson for Kids 3:07 min
Polar Habitats Lesson for Kids: Description & Facts
Chapter 2
Animal Life Cycles for Elementary School
Life Cycle of Dolphins: Lesson for Kids
Life Cycle of a Horse: Lesson for Kids 3:06 min
Life Cycle of a Lion: Lesson for Kids 2:51 min
Tiger Life Cycle: Lesson for Kids 3:18 min
Life Cycle of an Emperor Penguin: Lesson for Kids 3:14 min
African Goliath Beetle Life Cycle: Lesson for Kids
Kakapo Parrot Lesson for Kids: Facts & Life Cycle
Locust Life Cycle: Lesson for Kids
Life Cycle of a Housefly: Lesson for Kids 3:53 min
Life Cycle of a Platypus: Lesson for Kids
Scorpion Life Cycle: Lesson for Kids 3:16 min
Life Cycle of a Crocodile: Lesson for Kids 3:56 min
Hedgehog Life Cycle: Lesson for Kids 3:26 min
Koala Life Cycle: Lesson for Kids 2:49 min
Monkey Life Cycle: Lesson for Kids 2:51 min
Starfish Life Cycle: Lesson for Kids
Wolf Life Cycle: Lesson for Kids 3:12 min
Lifespan of an Alligator: Lesson for Kids
What are the Names of Animal Groups? - Lesson for Kids
Sheep Life Cycle: Lesson for Kids 3:14 min
Bald Eagle Life Cycle: Lesson for Kids
Life Cycle of a Hen: Lesson for Kids 2:34 min
Life Cycle of a Turkey: Lesson for Kids 3:56 min
Chapter 3
Cheetahs for Elementary School
Cheetah Cub Facts: Lesson for Kids 3:11 min
How Do Cheetahs Hunt? - Lesson for Kids
How Do Cheetahs Run So Fast? - Lesson for Kids 2:24 min
Chapter 4
Horses for Elementary School
Pinto Horse Facts: Lesson for Kids
Quarter Horse Facts: Lesson for Kids
Palomino Horse Facts: Lesson for Kids
Appaloosa Horse Facts: Lesson for Kids
Miniature Horse Facts: Lesson for Kids
Clydesdale Horse Facts: Lesson for Kids
Arabian Horse Facts: Lesson for Kids
Chapter 5
Tigers for Elementary School
Snow Tiger Facts: Lesson for Kids 3:40 min
Tiger Teeth Facts: Lesson for Kids
Sumatran Tiger Facts: Lesson for Kids
South Chinese Tiger Facts: Lesson for Kids
Bali Tiger Facts: Lesson for Kids
Indochinese Tiger Facts: Lesson for Kids
Caspian Tiger: Lesson for Kids
Malayan Tiger: Lesson for Kids
Chapter 6
What Animals Eat for Elementary School
Dolphin Diet: Lesson for Kids
What do Horses Eat? - Lesson for Kids 2:32 min
Polar Bear Diet: Lesson for Kids
What Do Crabs Eat? - Lesson for Kids 3:41 min
What Did Dinosaurs Eat? - Lesson for Kids
What Do Hedgehogs Eat? - Lesson for Kids
What Do Hippos Eat? - Lesson for Kids 3:04 min
What Do Kangaroos Eat? - Lesson for Kids 3:55 min
What Do Lizards Eat? - Lesson for Kids 3:05 min
What Do Monkeys Eat? - Lesson for Kids 4:12 min
What Does an Octopus Eat? - Lesson for Kids 3:51 min
What Do Snakes Eat? Lesson for Kids 2:33 min
What Do Starfish Eat? - Lesson for Kids 3:38 min
What Do Jellyfish Eat? - Lesson for Kids 3:32 min
What Do Zebras Eat? - Lesson for Kids 3:58 min
Wood Duck Diet: Lesson for Kids
Electric Eel Diet: Lesson for Kids
African Bush Elephant Diet: Lesson for Kids
Fangtooth Fish Diet: Lesson for Kids
What do Pandas Eat? - Lesson for Kids
What Do Koalas Eat? - Lesson for Kids 3:15 min
What do Wasps Eat? - Lesson for Kids
Chapter 7
Animals of Geographic Locations for Kids
Irish Animals Lesson for Kids
Animals of the Gobi Desert: Lesson for Kids
The American White Pelican: Lesson for Kids
Chapter 8
Large Mammals for Kids
Cape Buffalo Facts: Lesson for Kids
Water Buffalo Facts: Lesson for Kids
Short-Faced Bear Lesson for Kids: Size & Habitat
Short-Faced Bear Habitat: Lesson for Kids
Short-Faced Bear Facts: Lesson for Kids
Animals Related to Dinosaurs: Lesson for Kids 3:54 min
Facts about Dinosaur Eggs: Lesson for Kids
Types of Elephants: Lesson for Kids
Elephant Anatomy: Lesson for Kids
African Bush Elephant Facts: Lesson for Kids
Ibex Facts: Lesson for Kids
Chapter 9
Bugs, Butterflies & Insects for Kids
Stick Bug Facts: Lesson for Kids
Thorn Bug Facts: Lesson for Kids 2:54 min
Leaf Bug Facts: Lesson for Kids 2:57 min
Red Admiral Butterfly Facts: Lesson for Kids
Butterfly Migration: Lesson for Kids
All About Butterflies: Lesson for Kids
Butterfly Metamorphosis: Lesson for Kids
Butterfly Fish Facts: Lesson for Kids
Parts of a Butterfly: Lesson for Kids
Rainforest Butterflies Facts: Lesson for Kids
Owl Butterfly Facts: Lesson for Kids
Redback Spider Facts: Lesson for Kids
Viewing now Differences Between Insects and Spiders: Lesson for Kids
Up next Hawaiian Happy Face Spider Facts: Lesson for Kids 3:44 min Watch next lesson
Trapdoor Spider Facts: Lesson for Kids
Brown Recluse Spider Facts: Lesson for Kids
Garden Spider Facts: Lesson for Kids
Daddy Long Legs Facts: Lesson for Kids
Water Spider Facts: Lesson for Kids
Social Insects Lesson for Kids: Types, Definition & Examples
Queen Ant Lesson for Kids: Facts, Lifespan & Size
Types of Ants: Lesson for Kids
Drone Bee Facts: Lesson for Kids
Drone Bees Lesson for Kids: Lifespan & Job
Drone Bees Job: Lesson for Kids
Jerusalem Cricket Lesson for Kids: Life Cycle & Facts
What is a Minibeast? - Lesson for Kids
Chapter 10
Reptiles for Kids
Where Do Crocodiles Live? - Lesson for Kids 3:14 min
Chinese Alligator Facts: Lesson for Kids
Chinese Alligator Diet: Lesson for Kids
Chinese Alligator Habitat: Lesson for Kids
Differences Between Chinese & American Alligators: Lesson for Kids
Different Types of Crocodiles: Lesson for Kids
Arakan Forest Turtle: Lesson for Kids
Chapter 11
Aquatic Animals & Marine Life for Kids
Leafy & Weedy Sea Dragon Facts: Lesson for Kids
Pilot Whale Facts: Lesson for Kids
Southern Right Whale Facts: Lesson for Kids
Orca Whale Facts: Lesson for Kids
Gray Whale Facts: Lesson for Kids
Whale Sounds: Lesson for Kids
Guppy Facts: Lesson for Kids
Guppy Life Cycle: Lesson for Kids 2:27 min
Chapter 12
Foxes & Wolves for Kids
Bat-Eared Fox Facts: Lesson for Kids
Fox Hunting Facts: Lesson for Kids
Desert Fox Facts: Lesson for Kids
Terry Fox Facts: Lesson for Kids
Ethiopian Wolf Facts: Lesson for Kids
Maned Wolf Facts: Lesson for Kids
Chapter 13
Small Mammals for Kids
Hedgehog Hibernation Facts: Lesson for Kids
Pygmy Hippo Facts: Lesson for Kids
Elephant Shrew Facts: Lesson for Kids
Pygmy Shrew Facts: Lesson for Kids
Chapter 14
Australian Plants and Animals for Kids
Tree-Kangaroo Facts: Lesson for Kids
Facts About Emus: Lesson for Kids
Cassowary Information: Lesson for Kids
Laughing Kookaburra Facts: Lesson for Kids
Moa Facts: Lesson for Kids 3:31 min
Australian Pelican Facts: Lesson for Kids
Rainbow Lorikeet Facts: Lesson for Kids
Chapter 15
Monkeys, Apes, and Chimps for Kids
Snow Monkey Facts: Lesson for Kids
Types of Monkeys: Lesson for Kids
Rainforest Monkey Facts: Lesson for Kids 3:51 min
Vervet Monkey Facts: Lesson for Kids
Proboscis Monkey Facts: Lesson for Kids
Chapter 16
Snakes for Elementary School
Types of Snakes: Lesson for Kids
Garter Snake Facts: Lesson for Kids
Australian Snakes Facts: Lesson for Kids
Green Tree Snake Facts: Lesson for Kids
Sidewinder Snake Facts: Lesson for Kids
Milk Snake Facts: Lesson for Kids
Snake Characteristics: Lesson for Kids
Eastern Brown Snake: Lesson for Kids
Rainforest Snakes Facts: Lesson for Kids 3:52 min
Grass Snake Facts: Lesson for Kids
Viper Snake Facts: Lesson for Kids
Tiger Snake Facts: Lesson for Kids
Chapter 17
Crustaceans for Elementary School
Sea Crab Facts: Lesson for Kids
Crab Predators: Lesson for Kids
Types of Crabs: Lesson for Kids
Japanese Spider Crab Facts: Lesson for Kids
Japanese Spider Crab Habitat: Lesson for Kids
Chapter 18
Birds for Elementary School
What Are Waterfowl? - Lesson for Kids
Wood Duck Facts: Lesson for Kids
Wood Duck Habitat: Lesson for Kids
Duck Anatomy: Lesson for Kids
Types of Ducks in Texas: Lesson for Kids
Types of Ducks in Ohio: Lesson for Kids
Types of Ducks in Florida: Lesson for Kids
Types of Ducks in Michigan: Lesson for Kids
Types of Ducks in California: Lesson for Kids
Webbed Feet Animals: Lesson for Kids 3:43 min
Indian Runner Duck: Lesson for Kids
Yellow Warbler Facts: Lesson for Kids
Cockatoo Facts: Lesson for Kids
Hummingbird Habitat: Lesson for Kids
Life Cycle of a Hummingbird: Lesson for Kids
Bee Hummingbird Facts: Lesson for Kids
Anna's Hummingbird Facts: Lesson for Kids
Ruby-throated Hummingbird Facts: Lesson for Kids
Black-chinned Hummingbird Facts: Lesson for Kids
Violet Sabrewing Hummingbird Facts: Lesson for Kids
Songbird Facts: Lesson for Kids
Stork Facts: Lesson for Kids
What is Ornithology? | Definition & Types 4:15 min
Arctic Tern Facts: Lesson for Kids
Bluebird Facts: Lesson for Kids
Canary Bird Facts: Lesson for Kids
Crow Facts: Lesson for Kids 3:15 min
Cuckoo Bird Information: Lesson for Kids
Duckling Facts: Lesson for Kids 4:08 min
Hen Bird Information: Lesson for Kids 3:02 min
Mallard Duck Facts: Lesson for Kids
Meadowlark Facts: Lesson for Kids
Western Meadowlark Facts: Lesson for Kids
Nightingale Facts: Lesson for Kids
Nuthatch Facts: Lesson for Kids
Chicken Facts: Lesson for Kids
Black Robin Facts: Lesson for Kids
Sparrow Facts: Lesson for Kids 3:32 min
House Sparrow Facts: Lesson for Kids
Chapter 19
Fish for Elementary School
Electric Eel Anatomy: Lesson for Kids
Gulper Eel Lesson for Kids: Size & Facts
Conger Eel Facts: Lesson for Kids
Conger Eel Habitat: Lesson for Kids
Eel Predators: Lesson for Kids
Snake Eel Facts: Lesson for Kids
Slender Giant Moray: Lesson for Kids
Green Moray Eel Facts: Lesson for Kids
Green Moray Eel Habitat: Lesson for Kids
Freshwater Fish: Lesson for Kids
Saltwater Fish: Lesson for Kids
Ocean Fish Lesson for Kids: Types & Names
Fangtooth Fish Facts: Lesson for Kids
Fangtooth Fish Habitat: Lesson for Kids
How Do Fish Breathe Underwater? - Lesson for Kids 3:10 min
What Are the Three Classes of Fish? - Lesson for Kids
Different Kinds of Sharks: Lesson for Kids
What Is Ichthyology? - Lesson for Kids
Cartilaginous Fish Lesson for Kids: Definition & Facts
Chapter 20
Trees for Elementary School
Life Cycle of Trees: Lesson for Kids
Parts of a Tree Lesson for Kids: Functions & Diagram 2:51 min
Deciduous Trees Lesson for Kids: Definition & Facts
Coniferous Trees Lesson for Kids: Types & Facts
Deciduous vs. Coniferous Trees: Lesson for Kids
Beech Trees: Lesson for Kids
Horse Chestnut Tree Facts: Lesson for Kids
Chapter 21
Plants for Elementary School
Types of Plants: Lesson for Kids 3:18 min
Plant Parts: Lesson for Kids 2:59 min
Flowering Plants Lesson for Kids: Definition & Facts
Life Cycle of Flowering Plants: Lesson for Kids
Nonflowering Plants Lesson for Kids: Names & Examples 3:47 min
Chapter 22
Pigs for Elementary School
Pig Life Cycle: Lesson for Kids 3:54 min
Pig Research: Lesson for Kids
Teacup Pig Facts: Lesson for Kids
Wild Pig Facts: Lesson for Kids
Chapter 23
Sea Creatures for Elementary School
Reef Shark Lesson for Kids: Definition & Facts
Sea Monkey Lesson for Kids: Definition & Facts
Sea Cucumber Lesson for Kids: Definition & Facts
Chapter 24
Large Birds for Elementary School
Albatross Facts: Lesson for Kids
Brown Pelican Facts: Lesson for Kids
Rhea Bird Facts: Lesson for Kids
Black Swan Facts: Lesson for Kids
Trumpeter Swan Facts: Lesson for Kids
Chapter 25
Cranes for Elementary School
Crane Fly Facts: Lesson for Kids
Red-Crowned Crane Facts: Lesson for Kids
Crane Bird | Types, Appearance & Examples
Sarus Crane Facts: Lesson for Kids
Sandhill Crane Facts: Lesson for Kids
Whooping Crane: Lesson for Kids
Blue Crane: Lesson for Kids
Chapter 26
Doves for Elementary School
Dove Facts: Lesson for Kids 3:50 min
Mourning Dove Facts: Lesson for Kids
White-Winged Dove Facts: Lesson for Kids
Pigeon Facts: Lesson for Kids
Chapter 27
Egrets for Elementary School
Egret Facts: Lesson for Kids
Great Egret Facts: Lesson for Kids
Snowy Egret Facts: Lesson for Kids
Chapter 28
Geese for Elementary School
Goose Facts: Lesson for Kids
Canada Goose Facts: Lesson for Kids
Greylag Goose: Lesson for Kids
Chapter 29
Gulls for Elementary School
Facts About Seagulls: Lesson for Kids
Herring Gull Facts: Lesson for Kids
California Gull Facts: Lesson for Kids
Chapter 30
Hawks for Elementary School
What do Hawks Eat? | Hunting & Diet 2:23 min
Red-shouldered Hawk Facts: Lesson for Kids
Ferruginous Hawk Facts: Lesson for Kids
Chapter 31
Herons for Elementary School
Heron Facts: Lesson for Kids
Great Blue Heron Facts: Lesson for Kids
Grey Heron Facts: Lesson for Kids
Chapter 32
Rain Forest Birds for Elementary School
Blue Macaw Facts: Lesson for Kids
Blue & Yellow Macaw Facts: Lesson for Kids
Quetzal Facts: Lesson for Kids
Keel-billed Toucan Facts: Lesson for Kids
Toco Toucan Facts: Lesson for Kids
Chapter 33
Falcons for Elementary School
Peregrine Falcon Life Cycle: Lesson for Kids
Falcon Habitat: Lesson for Kids
Falcon Facts: Lesson for Kids
Chapter 34
Woodpeckers for Elementary School
Woodpecker Facts: Lesson for Kids
Downy Woodpecker Facts: Lesson for Kids
Pileated Woodpecker Facts: Lesson for Kids
Ivory-Billed Woodpecker Facts: Lesson for Kids
Red-Headed Woodpecker Facts: Lesson for Kids
Red-Cockaded Woodpecker Facts: Lesson for Kids
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188458 | https://www.geeksforgeeks.org/python/python-program-to-remove-a-specific-digit-from-every-element-of-the-list/ | Python Program to remove a specific digit from every element of the list
Last Updated :
23 Jul, 2025
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Given a list of elements, the task here is to write a Python program that can remove the presence of all a specific digit from every element and then return the resultant list.
Examples:
Input : test_list = [333, 893, 1948, 34, 2346], K = 3
Output : ['', 89, 1948, 4, 246]
Explanation : All occurrences of 3 are removed.
Input : test_list = [345, 893, 1948, 34, 2346], K = 5
Output : [34, 893, 1948, 34, 2346]
Explanation : All occurrences of 5 are removed.
Method 1 : Using loop, str() and join()
In this, we perform the task of reforming elements by converting them to strings and checking for each digit, and ignoring while joining to get new element. Lastly, each element is converted to integer using int().
Example:
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
res = []
for ele in test_list:
# joining using join(),
if list(set(str(ele))) == str(K) and len(set(str(ele))) == 1:
res.append('')
else:
res.append(int(''.join([el for el in str(ele) if int(el) != K])))
printing result
print("Modified List : " + str(res))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time complexity: O(nm), where n is the length of the list "test_list", and m is the average length of the string representation of the elements in "test_list".
Auxiliary Space: O(n), where n is the length of the list "res".
Method 2 : Using list comprehension, int(), str() and join()
Similar to the above method, joining is done using join() and interconversion is performed using int() and str().
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
list comprehension performing task as one liner
res = ['' if list(set(str(ele))) == str(K) and len(set(str(ele))) == 1 else int(
''.join([el for el in str(ele) if int(el) != K])) for ele in test_list]
printing result
print("Modified List : " + str(res))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time Complexity: O(NN), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(N), where n is the number of elements in the list “test_list”.
Method 3: Using replace() method
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
removing specific digit
res = []
for ele in test_list:
x = str(ele).replace(str(K), '')
res.append(int(x))
printing result
print("Modified List : " + str(res))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time complexity: O(nm), where n is the length of the list and m is the length of the largest integer in the list.
Auxiliary space: O(n), since we are creating a new list to store the modified integers.
Method #4: Using map(),lambda functions.
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
removing specific digit
res = list(map(lambda x: str(x).replace(str(K), ''), test_list))
res = list(map(int, res))
printing result
print("Modified List : " + str(res))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Using split(),join() methods
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
list comprehension performing task as one liner
res = []
for i in test_list:
x = str(i)
y = x.split(str(K))
y = "".join(y)
res.append(int(y))
printing result
print("Modified List : " + str(res))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#6: Using Recursion
This code takes a number and a target digit as arguments and removes all occurrences of the target digit from the number using recursion. The function first extracts the right-most digit of the number, removes it, and then checks the remaining number recursively. If the right-most digit matches the target digit, it is skipped; otherwise, it is added back to the remaining number. Finally, the function returns the modified number.
Python3
````
def remove_digit(number, digit):
if number == 0:
return 0
# extract the right-most digit
last_digit = number % 10
# remove the right-most digit and check recursively for the remaining number
remaining_number = remove_digit(number // 10, digit)
# check if the last digit matches the target digit, and return the appropriate value
if last_digit == digit:
return remaining_number
else:
return remaining_number 10 + last_digit
example usage
test_list = [345, 893, 1948, 34, 2346]
K = 3
res = [remove_digit(ele, K) for ele in test_list]
print("Modified List : " + str(res))
This code is contributed by Vinay Pinjala.
````
Output
Modified List : [45, 89, 1948, 4, 246]
Time complexity: O(N)
Where n is the number of elements in the list. This is because we need to iterate through each element in the list and call the function recursively for each sublist.
Auxiliary Space: O(n)
Where n is the number of elements in the list. This is because we need to create a new list to store the modified elements, and each recursive call creates a new stack frame to store the state of the function.
Method #7: Using bitwise operations
Step-by-step approach:
Define a function remove_digit_k that takes two arguments: a list lst and an integer k.
Inside the function, define an empty list res to store the modified elements.
Define a constant digit_mask as the bitwise complement of a number with a single 1-bit in the k-th position (i.e., digit_mask = ~(1 << k)).
Loop through each element ele in the input list lst.
If the element ele is an integer and the k-th digit is not present in its decimal representation (i.e., (ele >> k) & 1 == 0), append it to the res list as-is.
Otherwise, use the bitwise AND operator to clear the k-th bit from the integer ele (i.e., cleared = ele & digit_mask), then convert it to a string and append it to the res list as an integer (i.e., res.append(int(str(cleared)))).
Return the res list.
Method 7: Using string manipulation and list comprehension
In this approach, we convert the list of integers to a list of strings and then use string manipulation to remove the specific digit (K) from each string. Finally, we convert the list of strings back to a list of integers.
Step-by-step approach:
Initialize the list of integers.
Print the original list.
Initialize K.
Convert the list of integers to a list of strings using list comprehension.
Use string manipulation to remove the specific digit (K) from each string using another list comprehension.
Convert the list of modified strings back to a list of integers using list comprehension.
Print the modified list.
Python3
````
initializing list
test_list = [345, 893, 1948, 34, 2346]
printing original list
print("The original list is : " + str(test_list))
initializing K
K = 3
converting the list of integers to a list of strings
str_list = [str(i) for i in test_list]
using string manipulation to remove the specific digit (K) from each string
modified_str_list = [s.replace(str(K), '') for s in str_list]
converting the list of modified strings back to a list of integers
modified_list = [int(s) for s in modified_str_list]
printing the modified list
print("Modified List : " + str(modified_list))
````
Output
The original list is : [345, 893, 1948, 34, 2346]
Modified List : [45, 89, 1948, 4, 246]
Time complexity: O(nk), where n is the length of the list and k is the length of the maximum number in the list.
Auxiliary space: O(nk), as we are creating new lists of strings and integers.
M
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Nash v Inman 2 KB 1
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Nash v Inman 2 KB 1 is a UK Contract Law case concerning the enforceability of contracts with minors for necessaries.
The case summary contains 247 words.
Keywords:
Contract Law – Sale of goods – Necessaries – Minor – Burden of proof – Legal obligations
Facts:
In Nash v Inman , Mr. Nash, a tailor, supplied 13 waistcoats among other items to Mr. Inman, a Cambridge University undergraduate and minor at the time. Mr. Inman subsequently refused to pay for them, leading to Nash's legal action for the recovery of the price of the goods.
Issue:
The central question was whether the clothing supplied to Mr. Inman (minor) fell within the category of necessaries for which a minor is legally obliged to pay.
Held:
The Court defined “necessaries” as goods suitable to the condition in life of a minor and to their actual requirements at the time of sale and delivery. It held that the obligation to prove the necessity of the goods falls upon the plaintiff, Mr. Nash. Since he failed to demonstrate that the clothes were actual necessaries for Mr. Inman, the Court concluded that the contract was not for necessaries and hence was void from the outset.
So, the Court found in favour of Inman, the defendant, on the grounds that the plaintiff could not establish that the clothes delivered were necessaries for the minor.
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Nash v Inman 2 KB 1
Nash v Inman 2 KB 1 is a UK Contract Law case concerning the enforceability of contracts with minors for necessaries.
The case summary contains 247 words.
Keywords:
Contract Law – Sale of goods – Necessaries – Minor – Burden of proof – Legal obligations
Facts:
In Nash v Inman , Mr. Nash, a tailor, supplied 13 waistcoats among other items to Mr. Inman, a Cambridge University undergraduate and minor at the time. Mr. Inman subsequently refused to pay for them, leading to Nash's legal action for the recovery of the price of the goods.
Issue:
The central question was whether the clothing supplied to Mr. Inman (minor) fell within the category of necessaries for which a minor is legally obliged to pay.
Held:
The Court defined “necessaries” as goods suitable to the condition in life of a minor and to their actual requirements at the time of sale and delivery. It held that the obligation to prove the necessity of the goods falls upon the plaintiff, Mr. Nash. Since he failed to demonstrate that the clothes were actual necessaries for Mr. Inman, the Court concluded that the contract was not for necessaries and hence was void from the outset.
So, the Court found in favour of Inman, the defendant, on the grounds that the plaintiff could not establish that the clothes delivered were necessaries for the minor.
Read our notes and case summaries on Contract Law made by First Class law graduates to prepare like a pro for your assessments.
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188460 | https://www.doubtnut.com/qna/121601745 | The equation of a line passing through the point (3,−2,1) with the direction angle is 135∘,60∘,120∘ is
x−3√2=y+2−1=z−11
x−3−1=y+2√2=z−1√2
x+3−√2=y+21=z−11
x+31=y+21=z−1√2
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188461 | https://link.springer.com/chapter/10.1007/978-0-387-68276-1_8 | Advertisement
Geometric Inequalities
Part of the book series:
Springer Series in Statistics ((SSS))
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Abstract
Although the triangle has a limited number of parameters (sides, angles, altitudes, etc.), the range of inequalities among these entities is surprisingly large. Bottema, Djordjević, Janić, Mitrinović, and Vasić (1969), in their book Geometric Inequalities, have collected approx- imately 400 inequalities for the triangle. It is shown in this chapter 5 that majorization provides a unified approach to obtaining many known geometric inequalities. This unification also has the advantage of suggesting new inequalities.
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Authors and Affiliations
Department of Statistics, University of British Columbia, Vancouver, BC V6T 1Z2, Canada
Albert W. Marshall
2781 W. Shore Drive, Lummi Island, WA, 98262, USA
Albert W. Marshall
Department of Statistics, Stanford University, Stanford, CA, 94305, USA
Ingram Olkin
Department of Statistics, University of California, Riverside, CA, 92521, USA
Barry C. Arnold
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Marshall, A.W., Olkin, I., Arnold, B.C. (2010). Geometric Inequalities.
In: Inequalities: Theory of Majorization and Its Applications. Springer Series in Statistics. Springer, New York, NY.
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188462 | https://ajp.amjpathol.org/article/S0002-9440(20)30250-9/fulltext | The Surprising Story of IL-2 - The American Journal of Pathology
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ASIP Gold-Headed Cane Award LectureVolume 190, Issue 9p1776-1781 September 2020 Open Archive
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The Surprising Story of IL-2
From Experimental Models to Clinical Application
Abul K.Abbas
Abul K.Abbas
Correspondence
Address correspondence to Abul K. Abbas, M.B.B.S., Department of Pathology, University of California San Francisco, UCSF, M590, 505 Parnassus Avenue, San Francisco, CA 94143.
abul.abbas@ucsf.edu
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Affiliations & Notes Article Info
Department of Pathology, University of California San Francisco, San Francisco, California
Publication History:
Accepted May 20, 2020
Footnotes:
A.K.A. is the 2021 recipient of the ASIP Gold-Headed Cane Award. This award recognizes significant long-term contributions to the field of pathology, including meritorious research, outstanding teaching, general excellence in the discipline, and demonstrated leadership in the field of pathology.
Disclosures: The author has served as a consultant for Delinia/Celgene and ILTOO-Pharma, which are developing therapeutic agents based on IL-2.
DOI: 10.1016/j.ajpath.2020.05.007 External LinkAlso available on ScienceDirect External Link
Copyright: © 2020 American Society for Investigative Pathology. Published by Elsevier Inc.
User License: Elsevier user license | Elsevier's open access license policy
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Abstract
Discovery and Early History of IL-2
Rediscovering the Function of IL-2
Dual Functions of IL-2
Treg Maintenance by IL-2
Regulating the Dual Action of IL-2
Therapeutic Applications of IL-2
Conclusions
Acknowledgments
References
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Abstract
Discovery and Early History of IL-2
Rediscovering the Function of IL-2
Dual Functions of IL-2
Treg Maintenance by IL-2
Regulating the Dual Action of IL-2
Therapeutic Applications of IL-2
Conclusions
Acknowledgments
References
Article metrics
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Abstract
Equilibrium in the immune system is maintained by a balance between activation, which generates effector and memory cells, and suppression, which is mediated mainly by regulatory T cells. Understanding this balance and how to exploit it therapeutically is one of the dominant themes of modern immunology. The cytokine IL-2 was discovered as a growth factor for T cells and thus a key component of immune activation. It was initially used to boost immune responses in patients with cancer. Studies in experimental models and humans showed that the major function of IL-2 is to maintain functional regulatory T cells, and thus its essential function is in immune suppression. How the same cytokine can serve two opposing roles is a subject of current investigation. Because of these advances, IL-2 is now being tested as a cytokine for suppressing pathologic immune responses in autoimmune diseases and graft rejection. Fully understanding the biology of IL-2 may enable us to custom-design this cytokine for different applications in humans.
Discovery and Early History of IL-2
IL-2 was discovered in the 1970s as a soluble factor that stimulated the proliferation of T cells in vitro and was able to maintain T cells in culture for prolonged times. It was hence called T-cell growth factor (“factor” was the term used for secreted molecules in blood or culture supernatants that were defined according to biological activities but not molecular properties.)1
1.
Morgan, D.A. ∙ Ruscetti, F.W. ∙ Gallo, R.C.
Selective in vitro growth of T lymphocytes from normal human bone marrows
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It was the first cytokine whose encoding gene was cloned and sequenced, and it was designated IL-2.3
3.
Taniguchi, T. ∙ Matsui, H. ∙ Fujita, T. ...
Structure and expression of a cloned cDNA for human interleukin-2
Nature. 1983; 302:305-310
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An interesting fact is that when cytokine genes were molecularly cloned, the cytokines were given the IL designation and numbered in sequence. So why is the first IL #2? It seems some investigators believed that another cytokine stimulated the production of IL-2 and because it was above IL-2 in a functional hierarchy, the designation #1 was held for this supposed factor. However, a lymphocyte-activating factor that induced IL-2 was not identified, and thus the first IL remains #2. The name IL-1 was later given to a cytokine with different functions, and thus the second cytokine to be cloned ended up with the #1.
IL-2 is produced mainly by T cells, particularly CD4+ helper cells. The cellular receptor for IL-2 is a three-chain molecule. The receptor’s signaling function is mediated by the β and γ chains (CD122 and CD132, respectively), which are expressed on T cells (and natural killer cells) constitutively, can be up-regulated by activation, and dimerize to bind IL-2 with low affinity (dissociation constant approximately 1 nmol/L). The α chain (CD25) has no signaling capacity but confers on the receptor the ability to bind IL-2 with high-affinity (dissociation constant approximately 10 pmol/L).4
4.
Spolski, R. ∙ Li, P. ∙ Leonard, W.J.
Biology and regulation of IL-2: from molecular mechanisms to human therapy
Nat Rev Immunol. 2018; 18:648-659
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Signals from this receptor activate the transcription factors STAT5 and NF-κB, and these transcription factors induce the molecules that promote cell survival and proliferation, the two principal actions of IL-2. CD25 is transiently increased upon activation of T cells by antigen and other stimuli, enabling antigen-activated T cells to be the preferential responders to IL-2. Thus, upon antigen stimulation, T cells both produce and respond to IL-2, leading to the preferential expansion of antigen-specific clones. CD25 is also expressed on regulatory T cells (Tregs).
Many studies relying on in vitro experiments and some in which forms of IL-2 were administered or expressed in vivo collectively led to the wide acceptance that IL-2 functioned to stimulate the proliferation of antigen-stimulated T cells. As these clones of T cells expand, they also differentiate into effector and memory cells, providing immediate and long-lived defense against pathogens. The idea that IL-2 was a potent stimulator of immune responses led to clinical trials of IL-2 in patients with metastatic cancer.5
5.
Rosenberg, S.A.
IL-2: the first effective immunotherapy for human cancer
J Immunol. 2014; 192:5451-5458
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Because the half-life of administered IL-2 is very short, large doses had to be given to achieve biologically effective levels, which stimulated the production of many other cytokines, leading to severe toxicities resulting mainly from vascular leakage. As a result, although occasional responses were observed, IL-2 was not adopted as a first-line or adjunct cancer therapy.
Rediscovering the Function of IL-2
The function of IL-2 started to be re-evaluated when IL-2 gene knockout mice were analyzed. It was predicted that in the absence of the cytokine, mice would be immunodeficient, with a reduction in the number of effector and memory T cells. However, mice lacking IL-2 developed massive lymphadenopathy and splenomegaly and manifestations of autoimmune disease, such as colitis and autoimmune hemolytic anemia.6
6.
Sadlack, B. ∙ Merz, H. ∙ Schorle, H. ...
Ulcerative colitis-like disease in mice with a disrupted interleukin-2 gene
Cell. 1993; 75:253-261
Abstract
Full Text (PDF)
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The same phenotype was observed in knockout mice lacking the IL-2 receptor α or β chain.7
7.
Suzuki, H. ∙ Kündig, T.M. ∙ Furlonger, C. ...
Deregulated T cell activation and autoimmunity in mice lacking interleukin-2 receptor beta
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,8
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Willerford, D.M. ∙ Chen, J. ∙ Ferry, J.A. ...
Interleukin-2 receptor α chain regulates the size and content of the peripheral lymphoid compartment
Immunity. 1995; 3:521-530
Abstract
Full Text (PDF)
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PubMed
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In fact, even brief treatment of mice with antibodies that neutralize IL-2 or block the IL-2 receptor resulted in acute autoimmune manifestations.9
9.
McHugh, R.S. ∙ Shevach, E.M.
Cutting edge: depletion of CD4+CD25+ regulatory T cells is necessary, but not sufficient, for induction of organ-specific autoimmune disease
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These analyses were complemented by human studies showing that T cells in patients with lupus and other autoimmune diseases produce reduced amounts of IL-2, not the expected increase of this immune stimulator.11
11.
Alcocer-Varela, J. ∙ Alarcón-Segovia, D.
Decreased production of and response to interleukin-2 by cultured lymphocytes from patients with systemic lupus erythematosus
J Clin Invest. 1982; 69:1388-1392
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Thus, despite initial skepticism, it became clear that a major and essential function of IL-2 is to maintain self-tolerance and prevent autoimmunity, and depletion or decreased production of this cytokine is associated with systemic autoimmunity.
How a major stimulator of immune responses serves to also inhibit or prevent some reactions, particularly those against self-antigens, was puzzling. Initial studies suggested that IL-2 potentiated activation-induced cell death (apoptosis) of lymphocytes,12
12.
Lenardo, M.J.
Interleukin-2 programs mouse alpha beta T lymphocytes for apoptosis
Nature. 1991; 363:858-861
Crossref
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and this action was postulated to be a mechanism for maintaining unresponsiveness (tolerance) to self-antigens. However, there were few in vivo data to confirm this idea, and IL-2 has also been shown to protect T cells from death; thus, the role of IL-2 in T-cell apoptosis remains unclear. A different mechanistic hypothesis came from the observation that normal T cells could suppress the autoimmune potential of T cells from IL-2−/− mice, implying that IL-2 worked in trans: IL-2 from one cell could suppress the reaction of another cell.13
13.
Krämer, S. ∙ Schimpl, A. ∙ Hünig, T.
Immunopathology of interleukin (IL) 2-deficient mice: thymus dependence and suppression by thymus-dependent cells with an intact IL-2 gene
J Exp Med. 1995; 182:1769-1776
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The breakthrough came when, shortly after the findings of the gene knockout mice were reported, it was shown that T cells (which express high levels of CD25) were essential for maintaining self-tolerance, and their depletion resulted in autoimmunity.14
14.
Sakaguchi, S. ∙ Sakaguchi, N. ∙ Asano, M. ...
Immunologic self-tolerance maintained by activated T cells expressing IL-2 receptor alpha-chains (CD25). Breakdown of a single mechanism of self-tolerance causes various autoimmune diseases
J Immunol. 1995; 155:1151-1164
Crossref
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Google Scholar
These T cells are now known as regulatory T cells (Tregs) and are defined by the co-expression of CD25 and the transcription factor forkhead box P3 (FOXP3). In humans, Tregs are more heterogeneous, and some effector cells also express FOXP3 and CD25, at least transiently. Nevertheless, the stable expression of both these molecules remains the clearest marker of Tregs. Collectively, these studies indicated that IL-2 not only stimulated immune responses and generated effector cells but also was required for Treg-mediated suppression of immune responses. Many subsequent studies showed that introducing Tregs into mice lacking IL-2 receptors prevented the development of autoimmunity,15
15.
Malek, T.R. ∙ Yu, A. ∙ Vincek, V. ...
CD4 regulatory T cells prevent lethal autoimmunity in IL-2Rβ-deficient mice. Implications for the nonredundant function of IL-2
Immunity. 2002; 17:167-178
Full Text
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firmly establishing the role of IL-2 in maintaining Tregs. The following sections summarize studies from our laboratory as well as from others to decipher the mechanism by which IL-2 acts on Tregs to maintain self-tolerance and how these actions can be exploited to develop therapeutic strategies.
Dual Functions of IL-2
IL-2 is now known to promote the generation of the complete set of Tregs and to maintain these cells in peripheral tissues.16
16.
Toomer, K.H. ∙ Malek, T.R.
Cytokine signaling in the development and homeostasis of regulatory T cells
Cold Spring Harb Perspect Biol. 2018; 10:a028597
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Our own studies have focused on the role of this cytokine in peripheral Treg maintenance. The diversity of the immune system poses a major challenge to studying antigen-specific immune responses; because the system has to recognize and respond to an almost unlimited number and variety of antigens, very few cells respond to any one antigen, and tracking these cells is technically difficult. We and others have overcome this problem experimentally by generating transgenic mice that express a single T-cell receptor (TCR) specific for a known antigen and either introducing this antigen or transferring the specific T cells into recipients that express the antigen to follow the responses of the T cells. A great advantage of TCR transgenics is that mutations that would cause autoimmunity can be bred into these mice, and the animals do not develop disease unless they are exposed to the antigen that the specific TCR recognizes. For instance, a TCR transgenic mouse strain can be crossed with an IL-2–deficient mouse so that all of the T cells are incapable of producing IL-2. These mice will not by themselves develop autoimmunity because there is no antigen to activate the T cells, but reactions characteristic of autoimmunity will develop if the T cells are exposed to the specific antigen. We have used this strategy to analyze the consequences of exposing mice with T cells specific for the model antigen ovalbumin (Ova) to Ova, either systemically or selectively in the skin, in the context of normal or defective IL-2 genes.
Transfer of Ova-specific wild-type (IL-2 sufficient) CD4+ T cells into mice expressing Ova systemically results in an acute inflammatory disease that peaks in severity within 2 to 3 weeks.17
17.
Knoechel, B. ∙ Lohr, J. ∙ Kahn, E. ...
Sequential development of interleukin 2–dependent effector and regulatory T cells in response to endogenous systemic antigen
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Similarly, transfer of the wild-type T cells into mice expressing Ova in the skin induces acute dermatitis.18
18.
Gratz, I.K. ∙ Rosenblum, M.D. ∙ Maurano, M.M. ...
Cutting edge: self-antigen controls the balance between effector and regulatory T cells in peripheral tissues
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In both cases, the development of the disease is associated with the generation of large numbers of FOXP3– effector T cells that produce inflammatory cytokines such as interferon-γ and IL-17. Remarkably, in both situations, the disease resolves spontaneously, and resolution is associated with a gradual accumulation of FOXP3+ Tregs. This sequence of pathogenic T cells followed by protective Tregs has been observed in other models of inflammatory disease. It is unclear if the Tregs develop from effector cells or from naive T cells that have not fully differentiated into effector cells; distinguishing these possibilities will require single-cell analyses using fate-mapping reporter mice.
This experimental system gave us the opportunity to study the consequences of absence of IL-2. When IL-2–deficient TCR-transgenic T cells were transferred into antigen-expressing recipients, the disease developed more slowly. The most striking difference from wild-type cell transfers, however, was that the disease was progressive, with no resolution, and frequently fatal. The delayed disease was due to reduced generation of effector cells, and the failure to resolve the disease was due to an inability to generate Tregs.
These experimental studies illustrate a key feature of IL-2: the cytokine has dual functions, one being to induce effector T cells (part of the immune response) and the other to generate Tregs (which mediate immunologic control) (Figure 1). The unexpected result is that stimulation of T cells to generate effector cells, which has been the accepted role of IL-2, is a redundant function, because effector cells are generated even in the absence of IL-2, albeit more slowly and in lower numbers (accounting for the delayed inflammatory disease in the mouse models). The obligatory (nonredundant) function of IL-2 is to induce and maintain Tregs. This conclusion could have been predicted from the phenotypes of the knockout mice lacking IL-2 or its receptor. Thus, IL-2 can be considered more of an immunologic control factor than a growth factor.
Figure viewer
Figure 1 Dual activities of IL-2. IL-2 is produced by CD4+ T cells following their activation by antigen-presenting cells (APCs), and it stimulates the proliferation and differentiation of these cells. IL-2 also acts on forkhead box P3-positive (FOXP3+) regulatory T cells (Tregs) to maintain them in a functional state, capable of suppressing the development of effector and memory cells. The dashed arrow indicates that the IL-2 that acts on Tregs is likely derived from conventional CD4+ cells responding to antigens (because Tregs do not produce their own IL-2).
Treg Maintenance by IL-2
The realization that IL-2 is essential for the maintenance of Tregs has raised many questions about the role of this cytokine in T-cell biology. First, similar to the effect of IL-2 on other T cells, it is predictable that it promotes Treg survival. Tregs are not highly proliferative, and therefore inducing cycling of these cells is likely not a critical action of IL-2. Normally, the survival of cells is dependent on signals from growth factors and extracellular proteins, all of which block apoptosis and prevent the death of the cells. To analyze the prosurvival function of IL-2 and to determine if it played other roles, we developed a model in which apoptotic pathways could be manipulated. Mice lacking IL-2 have greatly reduced numbers of FOXP3+ Tregs and develop autoimmune hemolytic anemia. We postulated that in the absence of IL-2, there would be a deficiency of survival signals, leading to apoptosis of IL-2–dependent T cells. The absence of survival signals is typically detected by the cytosolic sensor BCL2 like 11 (Bim, also known as Bcl2l11), which then activates proapoptotic members of the Bcl-2 apoptosis regulator family. By eliminating Bim, we could prevent this pathway of cell death. We showed that if IL-2−/− mice were crossed with Bim−/− (Bcl2l11–/–) mice, Tregs were restored in the double-knockouts, confirming that IL-2 promotes cell survival by preventing Bim-dependent apoptosis, and that in the absence of Bim, IL-2 is dispensable for Treg survival.19
19.
Barron, L. ∙ Dooms, H. ∙ Hoyer, K.K. ...
Cutting edge: mechanisms of IL-2–dependent maintenance of functional regulatory T cells
J Immunol. 2010; 185:6426-6430
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Thus, one key function of IL-2 is to prevent apoptotic death of Tregs. However, even though Tregs were present when apoptosis was eliminated, the autoimmune disease was not prevented. This observation led to the hypothesis that IL-2 not only promoted survival of Tregs but was also required for the functional competence of these cells. Treatment of the double-knockout mice with IL-2 restored the function of the Tregs and prevented the autoimmune disease. Studies by other groups have shown that IL-2 promotes the expression of FOXP3 and the activation of STAT5, the transcription factors that drive much of the suppressive function of Tregs.20
20.
Chinen, T. ∙ Kannan, A.K. ∙ Levine, A.G. ...
An essential role for the IL-2 receptor in Treg cell function
Nat Immunol. 2016; 17:1322-1333
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A second important question arose from the realization that Tregs do not synthesize IL-2. This is inconsistent with the accepted model for IL-2 action, according to which CD4+ T cells both produce and respond to IL-2 via an autocrine pathway. If Tregs cannot make their own IL-2, what is the source of the cytokine for Treg maintenance? We initially developed an experimental model for defining the targets of IL-2 action by transferring FOXP3– TCR-transgenic T cells into a normal mouse, exposing the T cells to their cognate antigen, and evaluating which cells expressed phospho-STAT5, the transcription factor downstream of the IL-2 receptor. The surprising result was that when the conventional CD4+ T cells produced IL-2, the initial responders were not these T cells themselves but FOXP3+ Tregs in the lymphoid organ.21
21.
O'Gorman, W.E. ∙ Dooms, H. ∙ Thorne, S.H. ...
The initial phase of an immune response functions to activate regulatory T cells
J Immunol. 2009; 183:332-339
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Thus, IL-2 made by T cells in an immune response functions to activate adjacent Tregs. These results led to the idea that every immune response is accompanied by concomitant control mechanisms, which prevent both autoimmunity and collateral damage during the response. Detailed imaging analyses subsequently showed that conventional IL-2–producing FOXP3– T cells and IL-2–responsive FOXP3+ Tregs formed a cluster around dendritic cells presenting antigen. This co-localization enables the conventional T cells to produce IL-2 in response to antigen displayed by dendritic cells. These T cells secrete IL-2, which then acts on the nearby Tregs and activates the Tregs to suppress activation of the conventional T cells, in a classical feedback loop.22
22.
Liu, Z. ∙ Gerner, M.Y. ∙ Van Panhuys, N. ...
Immune homeostasis enforced by co-localized effector and regulatory T cells
Nature. 2015; 528:225-230
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More detailed analysis of the source of IL-2 for Treg maintenance has come from mice in which IL-2 is selectively depleted from various cell populations. Although by far the major physiologic source of IL-2 is antigen-activated CD4+ T cells, small amounts of the cytokine may be produced by other cell populations. By removing IL-2 from individual cells, it has been shown that the major source of the cytokine for maintaining Tregs in most secondary lymphoid organs is, predictably, conventional CD4+ T cells.23
23.
Owen, D.L. ∙ Mahmud, S.A. ∙ Vang, K.B. ...
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It is possible, however, that Tregs in different tissue compartments vary in their dependence on IL-2.24
24.
Smigiel, K.S. ∙ Richards, E. ∙ Srivastava, S. ...
CCR7 provides localized access to IL-2 and defines homeostatically distinct regulatory T cell subsets
[Erratum appeared in J Exp Med 2019, 216:1965] J Exp Med. 2014; 211:121-136
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Regulating the Dual Action of IL-2
Because it is now established that IL-2 has two opposing functions (to stimulate effector cell responses and to maintain Tregs), an important question is, how are these activities regulated so IL-2 produces optimal outcomes under different conditions? Although there is no definitive answer to this question, important insights have emerged. Tregs respond to IL-2 at concentrations 10 to 100 times lower than the amount needed to activate conventional FOXP3– T cells.25
25.
Yu, A. ∙ Snowhite, I. ∙ Vendrame, F. ...
Selective IL-2 responsiveness of regulatory T cells through multiple intrinsic mechanisms supports the use of low-dose IL-2 therapy in type 1 diabetes
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This high sensitivity of Tregs to IL-2 is partly because Tregs constitutively express high levels of the high-affinity trimeric IL-2 receptor, whereas conventional T cells express CD25 (which is required for the high-affinity receptor complex) transiently and only after activation. In addition to IL-2 receptor expression, Tregs may have developed signaling pathways that make them highly responsive to the cytokine. T-cell activation is induced by signals generated from at least three sets of stimuli: antigen, co-stimulators, and cytokines. Many of these signals drive the generation of effector and memory T cells, but Tregs have to tune down the pathways that lead to effector responses.26
26.
Yan, D. ∙ Farache, J. ∙ Mingueneau, M. ...
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[Erratum appeared in Proc Natl Acad Sci U S A 2016, 113:E256] Proc Natl Acad Sci U S A. 2015; 112:14942-14947
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It is possible, therefore, that Tregs have evolved to reduce their responsiveness to antigen and co-stimulation and are thus much more dependent on IL-2 for their activation (Figure 2). Another possibility is that Tregs express high levels of tyrosine phosphatases that promote signaling from the IL-2 receptor.25
25.
Yu, A. ∙ Snowhite, I. ∙ Vendrame, F. ...
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Furthermore, because Tregs contain a larger intracellular pool of IL-2 receptor chains, they can rapidly replenish the receptor on the cell surface after it is endocytosed following cytokine binding.27
27.
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Based on these findings, we postulate that conventional T cells respond best to the short-lived, high-level bursts of IL-2 that are produced upon exposure to foreign (eg, microbial) antigens, whereas Tregs are maintained by continuous low-level production of IL-2 in response to self-antigens (or perhaps commensals and other environmental stimuli). This hypothesis is difficult to prove because it is not currently possible to quantify the amount of any cytokine produced in lymphoid organs under different conditions.
Figure viewer
Figure 2 Postulated signaling pathways in conventional and regulatory T cells (Tregs). Naive conventional forkhead box P3-negative (FOXP3–) T lymphocytes (Tconv) integrate signals from multiple receptors to initiate the activation programs that result in cell proliferation and differentiation into effector and memory cells. Tregs may tune down several of these pathways to prevent the generation of effector cells, resulting in a greater dependence on signals from the IL-2 receptor (IL-2R). APC, antigen-presenting cell; Ca++, calcium ions; MAPK, mitogen-activated protein kinases; mTOR, mechanistic target of rapamycin; PI3K, phosphatidylinositol 3-kinase; TCR, T-cell receptor.
Therapeutic Applications of IL-2
Because the first defined function of IL-2 was to stimulate immune responses, it was initially used in patients to boost responses to cancer5
5.
Rosenberg, S.A.
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and in HIV-seropositive patients to enhance T-cell immunity.28
28.
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Immunologic and virologic effects of subcutaneous interleukin 2 in combination with antiretroviral therapy: a randomized controlled trial
JAMA. 2000; 284:183-189
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In neither case was the treatment sufficiently effective and safe to be widely adopted.
With increasing emphasis on the role of IL-2 in controlling immune activation, the emphasis has shifted to using the cytokine to suppress harmful responses. The problem remains, however, that IL-2 has dual functions and can be a stimulator or an inhibitor. The first attempts to selectively exploit its inhibitory function relied on simply reducing the dose of IL-2, because Tregs are much more responsive to the cytokine than are effector cells, and thus at low doses the Tregs should be preferentially activated. The initial proof-of-concept trials showed that low-dose IL-2 was safe and effective in treating vasculitis29
29.
Saadoun, D. ∙ Rosenzwajg, M. ∙ Joly, F. ...
Regulatory T-cell responses to low-dose interleukin-2 in HCV-induced vasculitis
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and chronic graft-versus-host disease.30
30.
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Interleukin-2 and regulatory T cells in graft-versus-host disease
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Subsequent clinical trials have tested the efficacy of low-dose IL-2 in systemic lupus erythematosus, type 1 diabetes, other autoimmune diseases, and allograft rejection (reviewed elsewhere31
31.
Klatzmann, D. ∙ Abbas, A.K.
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). There is, however, concern that simply altering the dose may not be a consistent or reliable way to selectively manipulate the cellular target of IL-2 in patients. If IL-2 administration activates T-cell responses in patients with ongoing autoimmune disease, it could exacerbate the disease. Conversely, if it induces broad immunosuppression, it may make patients susceptible to opportunistic infections. Thus far, these adverse events have not been observed in the ongoing trials. Alternative approaches that are currently being explored include engineering mutated forms of IL-2 that bind preferentially to Tregs, using anti–IL-2 antibodies that selectively target the cytokine to one or the other cell population, and conjugating IL-2 with polyethylene glycol in ways that make it active on conventional T cells or Tregs.
Although we have emphasized the therapeutic potential of the cytokine for treating inflammatory diseases, it is known that blocking the high-affinity IL-2 receptor with an anti-CD25 antibody is an effective treatment for acute graft rejection.32
32.
van den Hoogen, M.W. ∙ Hilbrands, L.B.
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On face value, it seems paradoxical that both administering the cytokine and blocking its action may be useful for controlling different inflammatory reactions. One possibility to reconcile these apparently conflicting observations is that blocking the IL-2 receptor is effective in situations of temporary and strong T-cell activation when CD25 is rapidly but transiently induced on responding T cells, and IL-2 drives the effector response. This would be the situation in acute graft rejection. It would, however, be risky to maintain patients on IL-2 receptor blockade for prolonged periods because this treatment would be expected to deplete Tregs and trigger chronic inflammation. Thus, IL-2 blockade could be a therapy for acute inflammatory reactions and Treg-selective IL-2 for chronic inflammatory diseases.
The next phase of cytokine therapy will likely rely on chemical modifications and combinations. One idea is to use mutants or pegylated forms of IL-2 that preferentially activate effector cells in combination with immune checkpoint blockade to treat cancer patients. Conversely, forms of IL-2 that activate Tregs may be combined with anti-inflammatory agents, such as tumor necrosis factor antagonists, or with tolerogenic peptide administration, to induce maximal immunologic control in autoimmune diseases and for controlling graft rejection.
Conclusions
The evolution of our understanding of IL-2 is a remarkable example of how basic research, mostly in experimental models, informs therapeutic strategies in humans. Starting with the phenotypes of gene knockout mice, and using information from related studies, immunologists have challenged the dogmas about what IL-2 does and have established new paradigms. The dual actions of this cytokine have increased its potential for use in multiple ways. The detailed understanding of the cytokine, its receptor, and signaling pathways holds promise for using structure-based drug design to construct versions of IL-2 that have distinct actions and are useful for treating different types of diseases. The exciting possibility is that similar approaches may be useful for custom-designing the actions and therapeutic effects of other, pleiotropic cytokines. If this potential is realized, the first molecularly defined cytokine will pave the way for fully exploiting this class of molecules as therapeutic agents.
Acknowledgments
I thank Drs. Andrew Lichtman, Jon C. Aster, and Vinay Kumar for their thoughtful reading of this article and their valuable comments. I am deeply indebted to the many students, research fellows, and colleagues who have contributed to the work summarized in this article. Numerous immunologists have made seminal contributions to our understanding of IL-2, and I regret not citing all of their work because of limitations of space.
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188463 | https://www.quora.com/Why-is-it-that-in-determining-the-GCD-the-same-factor-with-the-lowest-power-is-chosen | Something went wrong. Wait a moment and try again.
Greatest Common Factor ( ...
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Greatest Common Divisor
5
Why is it that in determining the GCD, the same factor with the lowest power is chosen?
Roger Pickering
Spent 6 years at 2 universities doing maths
·
Author has 14.8K answers and 5.9M answer views
·
3y
Let's choose a couple of numbers, say m=36 and n=24
Then when we factorise them we get m=4∗9=22∗32 and n=8∗3=23∗3.
Now how many times can we divide each number by 2? We have m=4something and n=8something so the highest power of 2 they have in common is 22 and similarly the highest power of 3 they have in common is 31.
As 2 and 3 are the only common prime factors it follows that the Greatest Common Divisor is 22∗3=12 and with a bit of luck you will see why we chose the smaller powers of 2 and 3.
Related questions
Why do we only choose all the same factors in determining the GCD while in LCM we choose all of the factors?
Can you give an example of finding the lowest common multiple (LCM), highest common factor (HCF), and greatest common divisor (GCD)?
What is the process for determining the numbers included in a lowest common multiple or greatest common factor equation?
What is the difference between the highest common factor (HCF), lowest common multiple (LCM), and greatest common divisor (GCD)?
How do you prove that if
gcd
(
a
,
b
)
=
1
then
gcd
(
a
c
,
b
)
=
gcd
(
c
,
b
)
(elementary number theory, gcd and lcm, math)?
Mark Gritter
recreational mathematician
·
Author has 5.7K answers and 11.7M answer views
·
3y
This only works if we take “factor” to mean “prime factor”. We can show pretty easily that the exponent can’t be larger than the minimum power, and it can’t be smaller than the minimum power, so it must be exactly the minimum power. :)
One way to do that, which is a little nonelementary, is to use Bezout’s identity, which says that the GCD of a and b is the smallest positive integer g which can be written in the form ax+by, with x and y both integers.
So, suppose pk divides both a and b. Say a=pkm and b=pkn for another pair of integers m and n.
Then ax+by=pkmx+pkny=pk(mx+ny)
This only works if we take “factor” to mean “prime factor”. We can show pretty easily that the exponent can’t be larger than the minimum power, and it can’t be smaller than the minimum power, so it must be exactly the minimum power. :)
One way to do that, which is a little nonelementary, is to use Bezout’s identity, which says that the GCD of a and b is the smallest positive integer g which can be written in the form ax+by, with x and y both integers.
So, suppose pk divides both a and b. Say a=pkm and b=pkn for another pair of integers m and n.
Then ax+by=pkmx+pkny=pk(mx+ny). So any candidate for the GCD must have pk as a divisor too. There’s no way to get rid of one of those power of p by multiplying by an integer! That’s the “can’t be any smaller” half.
On the other hand, suppose pk divides a but not b. Then take Bezout’s identity again:
g=pkm+n
and divide both sides by pk:
gpk=m+npk
We agreed that pk does not divide n, so the last term is not an integer, so the right hand side is not an integer. (We can’t add a fraction and an integer and get an integer!) But that means that the left hand side is not an integer, so pk does not divide g. That’s the “can’t be any larger” half.
So the exponent on p in the GCD has to be the minimum of the exponent on p in both a and b.
Roger Pickering
Spent 6 years at 2 universities doing maths
·
Author has 14.8K answers and 5.9M answer views
·
3y
Related
Why do we choose the highest prime factor in determining LCM?
Let's take an example using the two numbers 12 and 18.
We can easily see that the least common multiple is 36=3x12 and 2x18, but why is that?
Well 12=4x3=22x3 and 18=2x32.
The least common multiple has to have factors of 2 and 3, and because 2 squared occurs in one of the numbers that also has to occur in the LCM. Similarly because 3 squared occurs in one of the numbers that also has to occur in the LCM.
All of that depends on the fundamental theorem which says that any Natural Number (other than 1) can be expressed as a product of prime factors and powers of primes. Furthermore that list of
Let's take an example using the two numbers 12 and 18.
We can easily see that the least common multiple is 36=3x12 and 2x18, but why is that?
Well 12=4x3=22x3 and 18=2x32.
The least common multiple has to have factors of 2 and 3, and because 2 squared occurs in one of the numbers that also has to occur in the LCM. Similarly because 3 squared occurs in one of the numbers that also has to occur in the LCM.
All of that depends on the fundamental theorem which says that any Natural Number (other than 1) can be expressed as a product of prime factors and powers of primes. Furthermore that list of prime powers is unique - there is only one such list for each natural number.
Michiel Bogaert
Unable to carry all the maths books he studied
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3.3K answers and 2.8M answer views
·
5y
Related
Is this true: gcd (gcd(a,b),gcd(b,c)) = gcd (a,b, c) where gcd is greatest common divisor?
I don’t know … lets see together, shall we?
The gcd can be easily calculated when you split a number into prime factors
12 = 223
18 = 233
both have 23 → gcd = 6
So, it’s kind of like set theory … but we need to differentiate between numbers
Set( 12 ) = { first 2, second 2, first 3 }
Set( 18 ) = { first 2, first 3, second 3 }
Set( gcd( 12,18 ) ) = { first 2, first 3 }
the gcd is the interection of the sets
As such
gcd(a,b) = A ∩ B
gcd(b,c) = B ∩ C
gcd(gcd(a,b),gcd(b,c)) = (A ∩ B) ∩ (B ∩ C)
while
gcd (a,b,c) = A ∩ B ∩ C
and condisering
(A ∩ B) ∩ (B ∩ C) = A ∩ B ∩ C
they are in fact equal.
A ∩ B = gre
I don’t know … lets see together, shall we?
The gcd can be easily calculated when you split a number into prime factors
12 = 223
18 = 233
both have 23 → gcd = 6
So, it’s kind of like set theory … but we need to differentiate between numbers
Set( 12 ) = { first 2, second 2, first 3 }
Set( 18 ) = { first 2, first 3, second 3 }
Set( gcd( 12,18 ) ) = { first 2, first 3 }
the gcd is the interection of the sets
As such
gcd(a,b) = A ∩ B
gcd(b,c) = B ∩ C
gcd(gcd(a,b),gcd(b,c)) = (A ∩ B) ∩ (B ∩ C)
while
gcd (a,b,c) = A ∩ B ∩ C
and condisering
(A ∩ B) ∩ (B ∩ C) = A ∩ B ∩ C
they are in fact equal.
A ∩ B = green/blue
B ∩ C = green/ pink
green/blue ∩ green/ pink = green = A ∩ B ∩ C
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Related questions
What is the difference between the lowest common factor (LCF) and the highest common factor (HCF)?
What is the full form of GCD?
What is the GCD of (120, 3)?
Is there a number that does not have a greatest common divisor (GCD) or highest common factor (HCF), but still has a lowest common multiple (LCM)?
How do you find the lowest common multiple (LCM), highest common factor (HCF), and greatest common divisor (GCD) of three numbers?
Jayanta Mukherjee
B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990)
·
Author has 42.5K answers and 11M answer views
·
4y
Related
Why do we only choose all the same factors in determining the GCD while in LCM we choose all of the factors?
The operations LCM (Least Common Multiple) and HCF (Highest Common Factor) are applicable on positive integers.
GCD (Greatest Common Divisor) same as HCF.
Suppose, the integers under consideration are a, b, c and d.
First prime factorisation of each of them will be done separately.
As LCM (a, b, c, d) can in no case be smaller than the largest number among a, b, c and d; so, we need to choose all prime factors of them.
As HCF (a, b, c, d) can in no case be larger than the smallest number among a, b, c and d; so, we need to choose only common prime factors of them.
Robert Nichols
Author has 5K answers and 15.5M answer views
·
7y
Related
Why do we use the smallest power of each common prime factor when finding the GCF of 2 or more integers?
For the same reason you only use the primes in common, instead of using all the primes present in either prime factorization.
Example: What is the GCF of 60 and 36?
The prime factorization of 60? 2² x 3 x 5
The prime factorization of 36? 2² x 3²
Notice when we find the GCF we will not include the prime factor 5, because 5 was not present in the prime factorization of 36. By this same logic, we will not select the exponent ² for the prime factor 3, because only 3, not 3² was present in the prime factorization of 60. However, when selecting an exponent for 2, we will select ², because 2² was present
For the same reason you only use the primes in common, instead of using all the primes present in either prime factorization.
Example: What is the GCF of 60 and 36?
The prime factorization of 60? 2² x 3 x 5
The prime factorization of 36? 2² x 3²
Notice when we find the GCF we will not include the prime factor 5, because 5 was not present in the prime factorization of 36. By this same logic, we will not select the exponent ² for the prime factor 3, because only 3, not 3² was present in the prime factorization of 60. However, when selecting an exponent for 2, we will select ², because 2² was present in both 60 and 36.
The GCF of 36 and 60 is 2² x 3, or 12.
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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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Jay S Prajapati
Mathematics
·
4y
Related
What is the algorithm of finding the GCD (greatest Common Divisor)?
This is the algorithm :
Step 1 : Read two positive integers and store them in X and Y.
Step 2 : Divide X by Y. Let the reminder be R and the quotient be Q.
Step 3 : If r is zero then go to step 7.
Step 4: Assign Y to X.
Step 5 : Assign R to Y.
Step 6 : Go to step 2.
Step 7 : Print Y ( the required GCD)
Step 8 : Stop.
Explanation :
I hope it helps. 🙂
This is the algorithm :
Step 1 : Read two positive integers and store them in X and Y.
Step 2 : Divide X by Y. Let the reminder be R and the quotient be Q.
Step 3 : If r is zero then go to step 7.
Step 4: Assign Y to X.
Step 5 : Assign R to Y.
Step 6 : Go to step 2.
Step 7 : Print Y ( the required GCD)
Step 8 : Stop.
Explanation :
I hope it helps. 🙂
Smitaj
Bsc in Physics & Mathematics, Monash University (Graduated 1973)
·
Author has 25.1K answers and 18M answer views
·
Feb 21
Related
What's a power factor and how is it determined?
At ANY INSTANT the voltage applied multiplied by the current at the same instant gives the power.
If the volts are continually altering, as in an AC circuit then it is possible for the average volts to occur at a different time from average current.
If you multiply those volts by that current they are not both happening at the same time. This means you are getting an incorrect reading.
Of course if you take the instantaneous volts and the instantaneous current at each time in an entire cycle you DO get the correct power for how much energy is supplied during the cycle.
If you carefully measure the
At ANY INSTANT the voltage applied multiplied by the current at the same instant gives the power.
If the volts are continually altering, as in an AC circuit then it is possible for the average volts to occur at a different time from average current.
If you multiply those volts by that current they are not both happening at the same time. This means you are getting an incorrect reading.
Of course if you take the instantaneous volts and the instantaneous current at each time in an entire cycle you DO get the correct power for how much energy is supplied during the cycle.
If you carefully measure the ACTUAL power consumed and then calculate the RMS volts RMS amps ( called the APPARENT power) you get two different answers based on the exact moment that these happen.
The ratio of actual power / apparent power is the power factor.
If that factor is 1 then both readings indicate the same.
If the factor is low then the highest current in each cycle is happening at a very different time from the highest volts in each cycle.
In turn this means that a lot more current is flowing than you might think.
So let me take 1 kW at 250 V. According to ohms law you expect 4 A of current.
A power factor of 0.5 means that the equipment would be using 8A of current. But not at the same moment as the volts peak in each cycle.
The power APPEARS to be 250 8 = 2kW but the actual power is only 1kW
Hence the ratio is 1kW/ 2kW = 0.5 and there is your power factor.
It is difficult to determine without experiment. Only possible if you know the resistance, inductive load and capacitive load and you know that the supply is a pure sine wave. But as the inductive load normally changes as the amount of energy used changes then so too does the power factor. Hence experiment is about the only way to KNOW.
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Kaliprasad Tripathy
Software Professional
·
Author has 391 answers and 161.5K answer views
·
3y
Related
Why do you use the highest power of each factor when finding the LCM?
this will become evident from the example as below
say we need to find LCM of 9 and 24
9 = 3 ^2
24 = 3 2^3
now for the LCM it has to be divisible by 9 and 24 . for it to be divisible by 9 we need to take the highest power of 3 that is 3^2 (9 contains the highest power of 3) and no other, if we take a lesser power say 3^0 or 3^1 it shall not be divisible by 9. we do not require a power greater than 3^2 because a smaller power shall do the job and if we use 3^3 we shall get a multiple of LCM.
For it to be divisible by 24 we need to take the highest power of 2that is 2^3,(as 24 contains highest powe
this will become evident from the example as below
say we need to find LCM of 9 and 24
9 = 3 ^2
24 = 3 2^3
now for the LCM it has to be divisible by 9 and 24 . for it to be divisible by 9 we need to take the highest power of 3 that is 3^2 (9 contains the highest power of 3) and no other, if we take a lesser power say 3^0 or 3^1 it shall not be divisible by 9. we do not require a power greater than 3^2 because a smaller power shall do the job and if we use 3^3 we shall get a multiple of LCM.
For it to be divisible by 24 we need to take the highest power of 2that is 2^3,(as 24 contains highest power of 2) if we take a lesser power say 2^0 or 2^1 it shall not be divisible by 24. we do not require a power greater usign the reasoning above
so we get the LCM = 3^2 3^3 = 72
we can formalize above steps(variable in place of constants) and get a proof.
Job Bouwman
PhD candidate in medical imaging (MRI). Former high school math teacher.
·
Upvoted by
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. · Author has 1.1K answers and 7.1M answer views
·
8y
Related
Why does the monomial with the lowest power, not including the constant, dominate near the origin?
I’ll flip coin A once, coin B twice, and coin C thrice. You may keep whichever coins display heads every time they are flipped. Which are you most likely to get?
You expected gain equals:
E=(12)1+(12)2+(12)3
If we do this a lot, the contribution of the first coin will dominate your profit.
Aleš Mihev
M.Sc. in Computer Science, University of Ljubljana (Graduated 1988)
·
Author has 791 answers and 235.8K answer views
·
11mo
Related
What is the significance of prime factors in finding the greatest common divisor (GCD)?
If we know prime factors of two natural numbers m and n, then it is quite easy to find their GCD. It is just necessary to multiply the lower powers of prime factors that appear in both numbers. For example, let m=55125 and n=51450. Now, if we know that m=53⋅32⋅72, and n=73⋅3⋅52⋅2, then
GCD(m,n)=52⋅3⋅72=3675
If the natural numbers m and n do not share any common prime factors, then their GCD is of course 1.
Remember, however, that finding the prime factors of a natural number n is time consuming, because we must try all possible factors
If we know prime factors of two natural numbers m and n, then it is quite easy to find their GCD. It is just necessary to multiply the lower powers of prime factors that appear in both numbers. For example, let m=55125 and n=51450. Now, if we know that m=53⋅32⋅72, and n=73⋅3⋅52⋅2, then
GCD(m,n)=52⋅3⋅72=3675
If the natural numbers m and n do not share any common prime factors, then their GCD is of course 1.
Remember, however, that finding the prime factors of a natural number n is time consuming, because we must try all possible factors that are less or equal than √n. So, prime factorization is not an efficient method to find the GCD; the Euclidean algorithm should be used instead.
Sameer Lal
Studied at Cornell University College of Engineering
·
Updated 8y
Related
Why does the monomial with the lowest power, not including the constant, dominate near the origin?
This is a good question.
Suppose we have the function: f(x)=a0xn+a1xn−1+⋯+an−1x+an
Now consider a neighborhood centered at the origin with an infinitesimally small radius ϵ…what x values are contained in this neighborhood? If we choose a really small radius, this neighborhood will contain x values such that |x|<ϵ which also satisfies x<1. Now for values x<1 we have the following relation:
xk>xk+1
Now it’s easy to show that the an−1x term dominates this expression. The constant does not make a difference near the origin because, after all, it’s constant.
Luca Kenny
BSc in Mathematics, University of Warwick (Graduated 2022)
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Author has 775 answers and 586.1K answer views
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4y
Related
Why can I express the same greatest common divisor for different number pairs? For example: GCD(3101,311) = GCD(311,302) = GCD(302,9) = … = GCD(1,0) =1
By basic arithmetic:
a
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The whole set of common divisors is identical -
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Related questions
Why do we only choose all the same factors in determining the GCD while in LCM we choose all of the factors?
Can you give an example of finding the lowest common multiple (LCM), highest common factor (HCF), and greatest common divisor (GCD)?
What is the process for determining the numbers included in a lowest common multiple or greatest common factor equation?
What is the difference between the highest common factor (HCF), lowest common multiple (LCM), and greatest common divisor (GCD)?
How do you prove that if
gcd
(
a
,
b
)
=
1
then
gcd
(
a
c
,
b
)
=
gcd
(
c
,
b
)
(elementary number theory, gcd and lcm, math)?
What is the difference between the lowest common factor (LCF) and the highest common factor (HCF)?
What is the full form of GCD?
What is the GCD of (120, 3)?
Is there a number that does not have a greatest common divisor (GCD) or highest common factor (HCF), but still has a lowest common multiple (LCM)?
How do you find the lowest common multiple (LCM), highest common factor (HCF), and greatest common divisor (GCD) of three numbers?
What is the lowest common factor (12, 30)?
What is the greatest common factor of 18 and 81?
What did you do to find the GCF given the remaining factors?
How can you find the LCM and GCD without prime factorization?
Is GCD (a,b) equal to GCD (a,-b)?
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50 Hours in Minutes
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Conversion Table
| How Many Minutes Are In? | Answer (rounded) |
--- |
| 35 Hours | 2,100 |
| 36 Hours | 2,160 |
| 37 Hours | 2,220 |
| 38 Hours | 2,280 |
| 39 Hours | 2,340 |
| 40 Hours | 2,400 |
| 41 Hours | 2,460 |
| 42 Hours | 2,520 |
| 43 Hours | 2,580 |
| 44 Hours | 2,640 |
| 45 Hours | 2,700 |
| 46 Hours | 2,760 |
| 47 Hours | 2,820 |
| 48 Hours | 2,880 |
| 49 Hours | 2,940 |
| 50 Hours | 3,000 |
| 51 Hours | 3,060 |
| 52 Hours | 3,120 |
| 53 Hours | 3,180 |
| 54 Hours | 3,240 |
| 55 Hours | 3,300 |
| 56 Hours | 3,360 |
| 57 Hours | 3,420 |
| 58 Hours | 3,480 |
| 59 Hours | 3,540 |
| 60 Hours | 3,600 |
| 61 Hours | 3,660 |
| 62 Hours | 3,720 |
| 63 Hours | 3,780 |
| 64 Hours | 3,840 |
About "Convert date units" Calculator
An online date units converter is a handy tool that helps you quickly and accurately convert time durations from one unit to another. Whether you need to convert seconds, minutes, hours, days, weeks, months, or years, this tool simplifies the process. With this converter, you can easily and quickly convert time periods to a different unit of measurement. For example, it can help you find out what is 50 Hours in Minutes? (The answer is: 3,000).
To use the online date units converter, simply select the unit you want to convert from (e.g., 'Minutes'), enter the quantity you want to convert (e.g., '50'), and choose the target unit you want to convert to (e.g., 'Hours'). Then hit the 'Convert' button to get the results.
For example, if you want to know What is 50 Hours in Minutes, simply select 'Minutes' as the starting unit, enter '50' as the quantity, and select 'Hours' as the target unit. The converter will then display the converted result, which in this case would be 3,000.
This converter can help you with a wide range of time-related calculations, such as calculating the number of seconds in a given number of minutes or the number of days in a particular number of months. It is a practical tool for anyone who needs to work with time durations in different units and wants to save time and avoid errors in their calculations.
Whether you're a student, a researcher, a programmer, or simply someone who wants to know how long it will take to complete a particular task, this online date units converter is a quick and easy way to get the answers you need.
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188465 | https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Microwave_and_RF_Design_II_-_Transmission_Lines_(Steer)/02%3A_Transmission_Lines/2.03%3A_The_Terminated_Lossless_Line | 2.3: The Terminated Lossless Line - Engineering LibreTexts
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2: Transmission Lines
Microwave and RF Design II - Transmission Lines (Steer)
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2.3: The Terminated Lossless Line
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2.3: The Terminated Lossless Line
Last updated May 22, 2022
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2.2: Transmission Line Theory
2.4: Special Cases of Lossless Terminated Lines
Page ID 41014
Michael Steer
North Carolina State University
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. 2.3.1 Total Voltage and Current on the Line
1. Example 2.3.1: Forward- and Backward- Traveling Waves at an Open Circuit
2. Example 2.3.2: Current Reflection Coefficient
2.3.2 Forward- and Backward-Traveling Pulses
2.3.3 Input Reflection Coefficient of a Lossless Line
2.3.4 Input Impedance of a Lossless Line
2.3.5 Standing Waves and Voltage Standing Wave Ratio
Example 2.3.3: Standing Wave Ratio
Example 2.3.4: Standing Waves
VSWR Measurement
Example 2.3.5: Slotted Line Measurement of Impedance
2.3.6 Summary
Microwave engineers want to work with total voltage and current when possible and the art of design synthesis usually requires relating the total voltage and current world of a lumped element circuit to the traveling voltage world of transmission lines. This section develops the important abstractions that enable the total voltage and current view of the world to be used with transmission lines. The first step in this process is in Section 2.3.1 where total voltages and currents are related to forward- and backward-
Figure 2.3.1: A terminated transmission line.
traveling voltages and currents. Insight into traveling waves and reflections is presented in Section 2.3.2. Important abstractions are presented first for the input reflection coefficient of a terminated lossless line in Section 2.3.3 and then for the input impedance of the line in Section 2.3.4. The last section, Section 2.3.5, presents a view of the total voltage on the transmission line and describes the voltage standing wave concept.
2.3.1 Total Voltage and Current on the Line
Consider the terminated line shown in Figure 2.3.1(a). Assume an incident or forward-traveling wave, with traveling voltage V 0+e−ȷβz and current I 0+e−ȷβz propagating toward the load Z L at z=0. The characteristic impedance of the transmission line is the ratio of the voltage and current traveling waves so that
(2.3.1)V 0+(z)I 0+(z)=V 0+e−ȷβz I 0+e−ȷβz=V 0+(0)I 0+(0)=V 0+I 0+=Z 0
The reflected wave has a similar relationship (but note the sign change):
(2.3.2)V 0−e ȷβz−I 0−e ȷβz=V 0−−I 0−=Z 0
The load Z L imposes an additional constraint on the relationship of the total voltage and current at z=0:
(2.3.3)V L I L=V(z=0)I(z=0)=Z L
When Z L≠Z 0 there must be a reflected wave with appropriate amplitude to satisfy the above equations. Now the total voltage
(2.3.4)V(z)=V 0+e−ȷβz+V 0−e ȷβz
and the total current, I(z), is related to the traveling current waves by
(2.3.5)I(z)=V 0+Z 0e−ȷβz−V 0−Z 0e ȷβz=I 0+e−ȷβz+I 0−e ȷβz
Thus at the termination of the line (z=0),
V(0)I(0)=Z L=Z 0V 0++V 0−V 0+−V 0−
This can be rearranged as the ratio of the reflected voltage to the incident voltage:
V 0−V 0+=Z L−Z 0 Z L+Z 0
This ratio is defined as the voltage reflection coefficientat the load,
(2.3.6)Γ L=Γ L V=V 0−(0)V 0+(0)=V 0−V 0+=Z L−Z 0 Z L+Z 0
That is, at the load
(2.3.7)V 0−=Γ LV 0+
The relationship of the traveling waves on the line can also be described using the transmission coefficient T (this is the capital Greek letter tau which looks the same as the English letter ‘T’.)
The voltage transmission coefficient from a port at position z to a port at position 0 is (for the transmission line)
(2.3.8)T=T V=V 0+(at end of line)V 0+(at start of line)=V 0+(0)V 0+(z)=V 0+V 0+e−ȷβz=e ȷβz
The relationship in Equation (2.3.6) can be rewritten so that the input load impedance can be obtained from the reflection coefficient:
(2.3.9)Z L=Z 01+Γ V 1−Γ V
Similarly, the current reflection coefficient can be written as
(2.3.10)Γ I=I 0−I 0+=−Z L+Z 0 Z L+Z 0=−Γ V
The voltage reflection coefficient is used most of the time, so the reflection coefficient, Γ, on its own refers to the voltage reflection coefficient, Γ V=Γ.
There are several special cases that are noteworthy. The most important of these is the case when there is no reflected wave and Γ=0. To obtain Γ=0, the value of load impedance, Z L, is equal to Z 0, the characteristic impedance of the transmission line as seen in Equation (2.3.6).
The total voltage and current waves on the line can be written as
(2.3.11)V(z)=V 0+[e−ȷβz+Γe ȷβz]
(2.3.12)I(z)=V 0+Z 0[e−ȷβz−Γe ȷβz]
From Equations (2.3.11) and (2.3.12) it can be seen that the total voltage and current on the line consist of superpositions of incident and reflected waves.
Example 2.3.1: Forward- and Backward- Traveling Waves at an Open Circuit
A lossless transmission line is terminated in an open circuit. What is the relationship between the forward- and backward-traveling voltage waves at the end of the line?
Solution
At the end of the line the total current is zero, so that I++I−=0 and so
(2.3.13)I−=−I+
The forward- and backward traveling voltages and currents are related to the characteristic impedance by
(2.3.14)Z 0=V+/I+=−V−/I−
Note the change in sign, as a result of the direction of propagation changing but the positive reference for current is in the same direction. Substituting for I− at the termination,
(2.3.15)V+=−V−I+/I−=−V−I+/(−I+)=V−
Thus the total voltage at the end of the line, V TOTAL, is V++V−=2V+. Note that the total voltage at the end of the line is twice the incident (forward-traveling) voltage.
Example 2.3.2: Current Reflection Coefficient
A load consists of a shunt connection of a capacitor of 10 pF and a resistor of 60 Ω. The load terminates a lossless 50 Ω transmission line. The operating frequency is 5 GHz.
What is the impedance of the load?
What is the normalized impedance of the load (normalized to Z 0 of the line)?
What is the reflection coefficient of the load?
What is the current reflection coefficient of the load?
Solution
C=10⋅10−12 F;R=60 Ω;f=5⋅10 9 Hz;ω=2πf;Z 0=50 Ω
Z L=R||C=(1/R+ȷωC)−1=0.168−ȷ3.174 Ω
2. z L=Z L/Z 0=3.368⋅10−3−ȷ0.063.
3. This is the voltage reflection coefficient. Γ L=(z L−1)/(z L+1)=−0.985−ȷ0.126=0.993∠187.3∘.
4. Γ L I=−Γ L=0.985+ȷ0.126=0.993∠(187.3−180)∘=0.993∠7.3∘.
2.3.2 Forward- and Backward-Traveling Pulses
Reflections at the end of a line produce a backward-traveling signal. Forward- and backward-traveling pulses are shown in Figure 2.3.2(a) for the situation where the resistance at the end of the line is lower than the characteristic impedance of the line (Z L<Z 0). The voltage source is a step voltage that is zero for time t<0. At time t=0, the step is applied to the line and it begins traveling down the line, as shown at time t=1. This voltage step moving from left to right is called the forward-traveling voltage wave.
At time t=2, the leading edge of the step reaches the load, and as the load has lower resistance than the characteristic impedance of the line, the total voltage across the load drops below the level of the forward-traveling voltage step. The reflected wave is called the backward-traveling wave and it must be negative, as it adds to the forward-traveling wave to yield the total voltage. Thus the voltage reflection coefficient, Γ, is negative and the total voltage on the line, which is all that can be directly observed, drops. A reflected, smaller, and opposite step signal travels in the backward direction and adds to the forward-traveling step to produce the waveform shown at t=3. The impedance of the source matches the transmission line impedance so that the reflection at the source is zero. The signal on the line at time t=4, the time for round-trip propagation on the line, therefore remains at the lower value. The easiest way to remember the polarity of the reflected pulse is to consider the situation with a short-circuit at the load. Then the total voltage on the line at the load must be zero. The only way this can occur when a signal is incident is if the reflected signal is equal in magnitude but opposite in sign, in this case Γ=−1. So whenever |Z L|<|Z 0|, the reflected pulse will tend to subtract from the incident pulse.
The opposite situation occurs when the resistance at the load is higher than the characteristic impedance of the line (Figure 2.3.2(b)). In this case the reflected pulse has the same polarity as the incident signal. Again, to remember this, think of the open-circuited case. The voltage across the load doubles, as the reflected pulse has the same sign as well as magnitude as that of the incident signal, in this case Γ=+1. This is required so that the total current is zero.
A more illustrative situation is shown in Figure 2.3.3, where a more
Figure 2.3.2: Reflection of a voltage pulse from a load: (a) when the resistance of the load, R L is lower than the characteristic impedance of the line, Z 0; and (b) when R L is greater than Z 0.
Figure 2.3.3: Reflection of a pulse on an interconnect showing forward- and backward-traveling pulses. Z L>Z 0.
complicated signal is incident on a load that has a resistance higher than that of the characteristic impedance of the line. The peaking of the voltage that results at the load is typically the design objective in many long digital interconnects, as less overall signal energy needs to be transmitted down the line, or equivalently a lower current drive capability of the source is required to achieve first incidence switching. This is at the price of having reflected signals on the interconnects, but these are dissipated through a combination of line loss and absorption of the reflected signal at the driver.
2.3.3 Input Reflection Coefficient of a Lossless Line
The reflection coefficient looking into a line varies with position along the line as the forward- and backward-traveling waves change in relative phase. Referring to Figure 2.3.4, at a distance ℓ from the load (i.e., z=−ℓ), the input
Figure 2.3.4: Terminated transmission line: (a) a transmission line terminated in a load impedance, Z L, with an input impedance of Z in; and (b) a transmission line with source impedance Z G and load Z L.
Figure 2.3.5: The forward-traveling wave v+(t,z)=|V+|cos(ωt−βz)=|V+|cos(ωt+ϕ(z)) and the backward-traveling wave v−(t,z)=|V+|cos(ωt+βz)=|V+|cos[ωt+ϕ(z)]. The phase, ϕ, of the forward-traveling wave becomes increasingly negative along the line as z increases, and when reflected the phase ϕ of the backward-traveling wave becomes increasingly negative as the wave moves away from the load (i.e. as z decreases).
reflection looking into a terminated lossless line is
(2.3.16)Γ in|z=−ℓ=V−(z=−ℓ)V+(z=−ℓ)=V−(z=0)e−ȷβℓ V+(z=0)e+ȷβℓ=V−(z=0)V+(z=0)e−ȷβℓ e+ȷβℓ=Γ Le−ȷ2βℓ
Note that Γ in has the same magnitude as Γ L but rotates in the clockwise direction (becomes increasingly negative) at twice the rate of increase of the electrical length βℓ.
It is important to graphical concepts introduced later that there be a full appreciation for the angle of Γ in becoming increasingly negative at twice the rate at which the electrical length of the line increases. Figure 2.3.5 is a way of visualizing this. The transmission line here is λ/4 long with an electrical length of 90∘ and is terminated in a load with reflection coefficient Γ L=+1. At position z=0 the forward-traveling voltage wave is v+(t,0)=|V+|cos(ωt), and this then propagates down the line in the +z direction. The forward-traveling voltage at point z=λ/8 at t=0 will be the same as the voltage at z=0 at a time one-eighth of a period in the past. The voltage at z=λ/8 is v+(t,λ/8)=|V+|cos(ωt−2π/8), i.e. there is a phase rotation of −45∘. Then at z=λ/4, v+(t,λ/4)=|V+|cos(ωt−2π/4), i.e. at time t=0 there is a phase rotation of −90∘ relative to v+(0,0), and this is the negative of the electrical length of the line. The voltage wave reflects at the load and becomes a backward-traveling wave. Here Γ L=+1 and so, at the load, the phase of the backward- and forward-traveling waves are the same. The backward-traveling wave continues to travel in the −z direction and its phase at t=0 becomes increasingly negative as z gets closer to the input of the line. The phase of the backward-traveling wave at z=0 is rotated −90∘ with respect to the backward-traveling wave at the load, and has rotated −180∘ relative to the forward-traveling wave at z=0. For a lossless line, in general, the angle of Γ in=[phase of V−(z=0) relative to the phase of V+(z=0)]+(the phase of Γ L)=−2(electrical length of the line)+(the phase of Γ L).
2.3.4 Input Impedance of a Lossless Line
The impedance looking into a lossless line varies with position, as the forward- and backward-traveling waves combine to yield position-dependent total voltage and current. At a distance ℓ from the load (i.e., z=−ℓ), the input impedance seen looking toward the load is
(2.3.17)Z in|z=−ℓ=V(z=−ℓ)I(z=−ℓ)=Z 01+|Γ|e ȷ(Θ−2βℓ)1−|Γ|e ȷ(Θ−2βℓ)=Z 01+Γ Le ȷ(−2βℓ)1−Γ Le ȷ(−2βℓ)
Another form is obtained by substituting Equation (2.3.6) in Equation (2.3.17):
Z in=Z 0(Z L+Z 0)e ȷβℓ+(Z L−Z 0)e−ȷβℓ(Z L+Z 0)e ȷβℓ−(Z L−Z 0)e−ȷβℓ=Z 0Z Lcos(βℓ)+ȷZ 0cos(βℓ)Z 0cos(βℓ)+ȷZ Lcos(βℓ)(2.3.18)=Z 0Z L+ȷZ 0tanβℓ Z 0+ȷZ Ltanβℓ
This is the lossless telegrapher’s equation. The electrical length, βℓ, is in radians when used in calculations.
2.3.5 Standing Waves and Voltage Standing Wave Ratio
The total voltage on a terminated line is the sum of forward- and backward-traveling waves. This sum produces what is called a standing wave. Figure 2.3.6 shows the total and traveling waveforms on a line terminated in a reactance and evaluated at times equal to multiples of an eighth of a period. Here the traveling waves have the same amplitude indicating that the termination of the line is reactive, |Γ|=1. The interesting property here is that the total voltage appears as a standing wave with fixed points called nodes where the total voltage is always zero. This is more easily seen in Figure 2.3.7(a), where the total voltage is overlaid for many times. If the termination has resistance, then the magnitude of the backward-traveling wave will be less than that of the forward-traveling wave and the overlaid total voltage is as shown in Figure 2.3.7(b). This is still a standing wave, but the minima are now not zero. The envelope of this standing wave is shown in Figure 2.3.7(c), where there is a maximum amplitude V max and a minimum amplitude V min.
Now this situation will be examined mathematically to relate the standing wave to the reflection coefficient. If Γ=0, then the magnitude of the total voltage on the line, |V(z)|, is equal to |V 0+| anywhere on the line. For this reason, such a line is said to be “flat.” If there is reflection the magnitude of the total voltage on the line is not constant (see Figure 2.3.7(b)). Thus from Equations (2.3.11) and (2.3.12):
(2.3.19)|V(z)|=|V 0+||1+Γe 2ȷβz|=|V 0+||1+Γe−2ȷβℓ|
Figure 2.3.6: Evolution of a standing wave with a reactive load as the sum of forward- and backward-traveling waves (to the right and left, respectively) of equal amplitude evaluated at times t equal to eighths of the period T. At t=T/8 and t=5T/8 the total voltage everywhere on the line is zero.
Figure 2.3.7: Standing waves as an overlay of waveforms at many times: (a) when the forward-and backward-traveling waves have the same amplitude; (b) when the waves have different amplitudes; and (c) the envelope of the standing wave. N is a node (a minimum) and AN is an antinode (a maximum). Nodes, N, are separated by λ/2. Antinodes, AN, are separated by λ/2.
where z=−ℓ is the positive distance measured from the load at z=0 toward the generator. Or, setting Γ=|Γ|e ȷΘ,
(2.3.20)|V(z)|=|V 0+||1+|Γ|e ȷ(Θ−2βℓ)|
where Θ is the phase of the reflection coefficient (Γ=|Γ|e ȷΘ) at the load. This result shows that the voltage magnitude oscillates with position z along the line. The maximum value occurs when e ȷ(Θ−2βℓ)=1 and is given by
(2.3.21)V max=|V 0+|(1+|Γ|)
Similarly the minimum value of the total voltage magnitude occurs when the phase term is e ȷ(Θ−2βℓ)=−1, and is given by
(2.3.22)V min=|V 0+|(1−|Γ|)
A mismatch can be defined by the voltage standing wave ratio (VSWR):
(2.3.23)VSWR=V max V min=(1+|Γ|)(1−|Γ|)
Also
(2.3.24)|Γ|=VSWR−1 VSWR+1
Notice that in general Γ is complex, but VSWR is necessarily always real and 1≤VSWR≤∞. For the matched condition, Γ=0 and VSWR=1, and the closer VSWR is to 1, the closer the load is to being matched to the line and the more power is delivered to the load. The magnitude of the reflection coefficient on a line with a short-circuit or open-circuit load is 1, and in both cases the VSWR is infinite.
To determine the position of the standing wave maximum, ℓ max, consider Equation (2.3.20) and note that at the maximum
(2.3.25)Θ−2βℓ max=2nπ,n=0,1,2,…
Here Θ is the angle of the reflection coefficient at the load:
(2.3.26)Θ−2nπ=22π λ gℓ max
Thus the position of the voltage maxima, ℓ max, normalized to wavelength is
(2.3.27)ℓ max λ g=1 2(Θ 2π−n),n=0,−1,−2,…
Similarly the position of the voltage minima is (using Equation (2.3.20))
(2.3.28)Θ−2βℓ min=(2n+1)π
After rearranging the terms,
(2.3.29)ℓ min λ g=1 2(Θ 2π−n+1 2),n=0,−1,−2,…
Summarizing from Equations (2.3.27) and (2.3.29):
The distance between two successive maxima is λ g/2.
The distance between two successive minima is λ g/2.
The distance between a maximum and an adjacent minimum is λ g/4.
From the measured VSWR the magnitude of the reflection coefficient |Γ| can be found. From the measured ℓ max the angle Θ of Γ can be found. Then from Γ the load impedance can be found.
In a similar manner to that above, the magnitude of the total current on the line is
(2.3.30)|I(ℓ)|=|V 0+|Z 0|1−|Γ|e ȷ(Θ−2βℓ)|
Hence the standing wave current is maximum where the standing-wave voltage amplitude is minimum, and minimum where the standing-wave voltage amplitude is maximum.
Z in in Equation (2.3.18) is a periodic function of length with period λ/2 and it varies between Z max and Z min, where
(2.3.31)Z max=V max I min=Z 0×VSWR and Z min=V min I max=Z 0 VSWR
Example 2.3.3: Standing Wave Ratio
In Example 2.3.2 the load consisted of a capacitor of 10 pF in shunt with a resistor of 60 Ω. The load terminated a lossless 50 Ω transmission line. The operating frequency is 5 GHz.
What is the SWR?
What is the current standing wave ratio (ISWR)? (When SWR is used on its own it is assumed to refer to VSWR.)
Solution
From Example 2.3.2 Γ L=0.993∠187.3∘ and so
VSWR=1+|Γ L|1−|Γ L|=1+0.993 1−0.993=285
2. ISWR=VSWR=285
Example 2.3.4: Standing Waves
A load has an impedance Z L=45+ȷ75 Ω and the system reference impedance, Z 0, is 100 Ω.
What is the reflection coefficient?
What is the current reflection coefficient?
What is the SWR?
What is the ISWR?
The power available from a source with a 100 Ω Thevenin equivalent impedance is 1 mW. The source is connected directly to the load, Z L. Use the reflection coefficient to calculate the power delivered to Z L.
What is the total power absorbed by the Thevenin equivalent source impedance?
Discuss the effect on power flow of inserting a lossless 100 Ω transmission line between the source and the load.
Solution
The voltage reflection coefficient is
Γ L=(Z L−Z 0)/(Z L+Z 0)=(45+ȷ75−100)/(45+ȷ75+100)=(93.0∠(2.204 rads))/(163.2∠(0.4773 rads))(2.3.32)=0.570∠(1.726 rads)=0.570∠98.9∘=−0.0881+ȷ0.563=Γ V
2. The current reflection coefficient is
(2.3.33)Γ I=−Γ V=0.0881−ȷ0.563=0.570∠(98.9∘−180∘)=0.570∠81.1∘
3. The SWR is the VSWR, so
(2.3.34)SWR=VSWR=V max V min=1+|Γ V|1−|Γ V|=1+0.570 1−0.570=3.65
4. The current SWR is ISWR=VSWR.
5. To determine the reflection coefficient of the load, begin by developing the Thevenin equivalent circuit of the load. The power available from the source is P A=1 mW, so the Thevenin equivalent circuit is
Figure 2.3.8
The power reflected by the load is
P R=P A|Γ L 2|=1 mW⋅(0.570)2=0.325 mW
and the power delivered to the load is
P D=P A(1−|Γ L 2|)=0.675 mW
6. It is tempting to think that the power dissipated in R TH is just P R. However, this is not correct. Instead, the current in R TH must be determined and then the power dissipated in R TH found. Let the current through R TH cdot be I, and this is composed of forward-and backward-traveling components:
I=I++I−=(1+Γ I)I+
where I+ is the forward-traveling current wave. Thus
P A=1 2|I+|2R TH=1 2|I+|2×100=1 mW=10−3 W
so I+=4.47 mA, and
I=(1+Γ I)I+=(1+0.0881−ȷ0.563)×4.47×10−3 A,|I|=5.48 mA
The power dissipated in R TH is
(2.3.35)P TH=1 2|I|2R TH=1 2(5.48×10−3)2R TH=1.50 mW
The circuit is that shown in part (e) and so the current in R TH is the same as the current in Z L. Thus the power delivered to the load Z L is due to the real part of Z L:
(2.3.36)P D=1 2|I|2ℜ(Z L)=1 2(5.48×10−3)2×45=0.676 mW
7. Inserting a transmission line with the same characteristic impedance as the Thevenin equivalent impedance will have no effect on power flow.
VSWR Measurement
The measurement of standing waves can be used to calculate the impedance of a load. The device that does this measurement, called a slotted line, is shown in Figure 2.3.9(a). A probe is inserted a small distance into the transmission line to measure the electric field. The RF electric field produces an RF voltage on the probe that is rectified by the diode detector. The DC voltage at the output of the detector is proportional to the total voltage on the line. The probe can be moved along the line and the ratio of V max to V min determined. This is just the VSWR. To find the complex load impedance it is also necessary to determine the position of the node of the standing wave. From the measured VSWR the magnitude of the reflection coefficient |Γ| can be found. From the measured ℓmax the angle Θ of γ can be found. From γ the load impedance can be found. This is demonstrated in the next example.
Figure 2.3.9: Measurement of standing waves: (a) coaxial slotted line; (b) schematic of slotted line; (c) measured standing wave.
Example 2.3.5: Slotted Line Measurement of Impedance
A slotted line is used to determine the properties of the standing wave on a terminated 50 Ω line see Figure 2.3.7(c). V max=5 V and V min=2 V, and the first minimum is 2 cm from the load. The guide wavelength is 10 cm. What is the load impedance Z L?
Solution
Now VSWR=V max/V min=5/2=2.5. So from Equation (2.3.24)
(2.3.37)|Γ|=|Γ L|=VSWR−1 VSWR+1=2.5−1 2.5+1=0.428
Equation (2.3.29) and the position of the first node can be used to determine the angle of Γ L. For the first node (minimum), n=0 and
(2.3.38)ℓ min λ g=1 2(Θ 2π+1 2)
Rearranging,
(2.3.39)Θ=2π(2ℓ min λ g−1 2)radians
Now ℓ min=2 cm and λ g=10 cm. So, in degrees,
(2.3.40)Θ=360(2ℓ min λ g−1 2)=360(2 2 10−1 10)=−36∘
Thus Γ L=0.428∠(−36∘)=0.3463−ȷ0.2516, so the load impedance is (where Z 0=50 Ω)
(2.3.41)Z L=Z 0(1+Γ L 1−Γ L)=83.2−ȷ51.3 Ω
2.3.6 Summary
This section related the physics of traveling voltage and current waves on lossless transmission lines to the total voltage and current view. First the input reflection coefficient of a terminated lossless line was developed and from this the input impedance, which is the ratio of total voltage and total current, derived. At any point along a line the amplitude of total voltage varies sinusoidally, tracing out a standing wave pattern along the line and yielding the VSWR metric which is the ratio of the maximum amplitude of the total voltage to the minimum amplitude of that voltage. This is an important metric that is often used to provide an indication of how good a match, i.e. how small the reflection is, with a VSWR=1 indicating no reflection and a VSWR=∞ indicating total reflection, i.e. a reflection coefficient magnitude of 1.
This page titled 2.3: The Terminated Lossless Line is shared under a CC BY-NC license and was authored, remixed, and/or curated by Michael Steer.
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2.2: Transmission Line Theory
2.4: Special Cases of Lossless Terminated Lines
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Catherine R.
asked • 09/22/19
How is a termination sequence different from a stop codon
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A terminator sequence is a nucleic acid sequence that causes RNA polymerase to stop its activity. It marks the end of a gene or operon, therefore stopping transcription. A stop codon is an mRNA nucleotide triplet that signals the end of the newly formed polypeptide chain during translation. There are three stop codons: UAG, UAA, and UGA. They do not code for an amino acid.
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Complement of a Set: Definition, Properties, Examples, Facts
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Complement of a Set – Definition, Properties, Examples, Facts, FAQs
Home » Math Vocabulary » Complement of a Set – Definition, Properties, Examples, Facts, FAQs
What Is the Complement of a Set?
Definition of Complement of a Set
Properties of the Complement of a Set
Solved Examples on the Complement of a Set
Practice Problems on Complement of a Set
Frequently Asked Questions about the Complement of a Set
What Is the Complement of a Set?
In set theory, the complement of a set is the set of all elements that belong to the universal set but not to the original set.
The complement of a set A is denoted by A’ or Ac.
Example:
Universal set \=U\= Set of all integers
U\={…,−3,−2,−1,0,1,2,3,…}
Let A be the set of even integers.
A\={…,−2,0,2,4,…}
Here, the complement of A is the set of odd integers.
Complement of set A\=A′\={…,−3,−1,1,3,…}
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Definition of Complement of a Set
The difference between the universal set U and set A is called the complement of that set A.
If U is a universal set and A be any subset of U, then the complement of A is the set of all members of the universal set U, which are not the elements of A.
Mathematically, the complement of a set A with respect to a universal set U is defined as
A′\={x∈U:x∈A}
The complement of A is the set of all elements x in U that are not in A.
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Complement of a Set Symbol
The complement of the set A is denoted as A’ or Ac . The notation of the complement of a set uses an apostrophe (‘) or a superscript c after the name of the set.
Also, as we know that the complement of A is the difference between Universal set and the set A, we can write
A′\=U−A
Complement of a Set: Venn Diagram
A universal set is typically represented using a rectangular box. Subsets of the universal set are generally represented by a circle. The complement of a set is the region of the universal set outside the set A.
The Venn diagram of the complement of set A is shown below. Here, the shaded portion in yellow shows the complement of set A.
If we have two sets that intersect each other, the complements of sets can be represented as follows:
Cardinality of the Complement of a Set
Let the cardinality of the set A be n(A), the cardinality of the universal set U be n(U), then the cardinality of the complement of A, which is represented by n(A’) or |A’| is given by
n(A′)\=n(U)−n(A)
|A′|\=|U|−|A|
Properties of the Complement of a Set
Let’s discuss the properties of the complement of a set.
Complement Laws
The union of a set A and its complement A’ is equal to the universal set U.
A∪A′\=U
The intersection of set A and A’ is the empty set
A∩A′\=∅
Law of Double Complementation
The complement of the complement of a set A is equal to A itself.
(A′)′\=A
Law of Empty Set and Universal Set
An empty set or null set (∅) is the complement of the universal set. The universal set is the complement of the empty set.
∅′\=U
U′\=∅
De Morgan’s Laws
De Morgan’s laws are a set of two fundamental laws in a set theory that relate the complement of set operations. They are named after the mathematician Augustus De Morgan, a British mathematician .
The complement of the union of two sets A and B is equal to the intersection of their complements.
(A∪B)′\=A′∩B′
The complement of the intersection of two sets A and B is equal to the union of their complements.
(A∩B)′\=A′∪B′
How to Find the Complement of a Set
To find the complement of a set, follow the steps mentioned below:
Step 1: Write the elements of the universal set. Write down all the elements in the original set.
Step 2: Find the difference between U and A. In other words, identify all the elements in the Universal set that do not belong to the original set.
Step 3: The elements identified in step 3 form the complement of the set.
Example: U\={1,2,3,4,5,6,7,8,9,10}
A\={1,3,5,7,9}
A′\=U−A\={2,4,6,8,10}
Facts about the Complement of a Set
The complement of an empty set is the universal set.
The complement of a universal set is the empty set.
The set and its complement are disjoint sets.
The complement of a set is unique.
The cardinality of a complement of a set is the difference between the cardinality of the universal set and the cardinality of the given set.
Conclusion
In this article, we have discussed the complement of a set, the notation, the Venn diagram, the properties of the complement of a set, and the method to find the complement. Let’s solve a few examples and practice problems.
Solved Examples on the Complement of a Set
Let set B\={Monday, Tuesday, Wednesday, Friday}. Find the complement of B.
Solution:
B\={Monday, Tuesday, Wednesday, Friday}.
Here, U\= Set of all days in a week
U\={Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
Let’s find the complement of B.
B′\=U−B
B′\={x|xis a day of the week and x is not in B}
B′\={Thursday, Saturday, Sunday}.
Let Y\={x|xis a positive even integer}. Find the complement of Y, if U is the set of positive integers.
Solution:
Y\={x|xis a positive even integer}
Here, U is the set of positive integers.
Therefore, Y′\={x|xis a positive integer and x is odd}
Y′\={1,3,5,7,9,…}.
If U\={1,2,3,4,5,6,7,8,9,10} and A\={2,3,5,7}. Find the complement of A.
Solution:
U\={1,2,3,4,5,6,7,8,9,10}
A\={2,3,5,7}
∴A′\=U−A\={1,4,6,8,9,10}
If B\={x:xis a vowel in English alphabets}, then find B’.
Solution:
B\={x:xis a vowel in English alphabets.}
B\={a,e,i,o,u}
B′\={b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
B′\={x:xis not a vowel in English alphabets.}
B′\={x:xis a consonant in English alphabets.}
Let U\={1,2,3,4,5,6,7,8,9}. If A\={1,2,3} and B\={4,5,6} and C\={7,8,9}, then find (A U B U C)’.
Solution:
A\={1,2,3} and B\={4,5,6} and C\={7,8,9}
AUBUC\={1,2,3,4,5,6,7,8,9}\=U
∴(AUBUC)′\=U′\=∅
If U\=1,2,3,4,5 and C\={1,3,5} then find (C’)’?
Solution:
(C′)′\=C
U\={1,2,3,4,5} and C\={1,3,5}
C′\={2,4}
Again taking the complement of C’, we get
∴(C′)′\={2,4}′\={1,3,5}\=C
Practice Problems on Complement of a Set
Complement of a Set - Definition, Properties, Examples, Facts, FAQs
Attend this quiz & Test your knowledge.
1
A∪A′\=
A
A'
U
∅
CorrectIncorrect
Correct answer is: U
A∪A′\=U
2
If U\={a,b,c,d,e,f} and B\={a,c,e}, what is the complement of B?
{a,b,d,f}
{a,b,c,d,e,f}
The empty set
{b,d,f}
CorrectIncorrect
Correct answer is: {b,d,f}
The complement of B consists of all elements in the universal set that are not in B. B′\={b,d,f}.
3
U′\=
{0}
∅
{1}
U
CorrectIncorrect
Correct answer is: ∅
U′\=∅
4
If U\={a,b} and the set D\={b}, what is the complement of D?
{a,b}
{a}
{b}
∅
CorrectIncorrect
Correct answer is: {a}
U\={a,b}
D\={b}
D′\=U−D\={a}
5
(A∪B)′\=
A′∪B′
A′∩B′
A∪B
A∩B
CorrectIncorrect
Correct answer is: A′∩B′
By De Morgan’s law, (A∪B)′\=A′∩B′
Frequently Asked Questions about the Complement of a Set
What is the meaning of a complement set?
The meaning of a complement set is the set of all elements in the universal set that are not in the given set.
What is the notation for the complement of a set?
The symbol used to denote the complement of a set is usually an apostrophe (‘) or a superscript c, placed after the set symbol. For example, if A is a set, then its complement can be denoted by A’ or Ac.
What is the formula for the complement of a set?
The formula for the complement of a set A with respect to a universal set U is given byA’ \=U−A\={x∈U|x∈A}
What is the difference between a set and its complement?
A set and its complement have no element in common.Thus, A−A′\=A
What is the complement of the empty set?
The complement of the empty set is the universal set.
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15 Puzzle question - trying to understand reasoning here
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I'm following some problem about the 15 puzzle from the MIT Discrete Maths course. They are talking about this starting position:
1. row A B C D
2. row E F G H
3. row I J K L
4. row M O N
Where there is exactly one inversion (O and N).
I figured out some questions on my own:
a) Can a row move change the order of tiles?
My answer: A row move can change the index of the tile moved only my going left -1 or going right +1. As no other tile moves, the order of tiles overall doesn't change.
b) How many pairs of tiles will have their relative order changed by a column move.
My answer: In a column move we either move the tile down a column, sliding it after the next 3 tiles or up one column, sliding it before the previous 3 tiles. Thus a column move changes the relative order of exactly 3 tiles.
c) What effect does a row move have on the parity of inversions?
My answer: As described in a) the order of tiles doesn't change in a row move, therefore the parity of inversions stays unchanged for a row move.
d) What effect does a column move have on the parity of number of inversions
e) The previous problem part implies that we must make an odd number of column moves in order to exchange just one pair of tiles (N and O, say). But this is problematic, because each column move also knocks the blank square up or down one row. So after an odd number of column moves, the blank can not possibly be back in the last row, where it belongs!
I tried playing around and for d) I claim that: a column move changes the parity of the number of inversions n n by either +3 or -3, always flipping the parity n n to the opposite side (even n n becomes odd, odd n n becomes even after a column move.)
The previous problem part implies that we must make an odd number of column moves in order to exchange just one pair of tiles
~~I don't understand this sentence. Why?~~
EDIT: After reading the problem again, and taking into account they talk about the start state which has an odd number (1) of inversions, it makes more sense now. We need to go to an even number of inversions (0) by making an odd number of column moves, as each column move flips the parity to the oppsite side, see d)
But this is problematic, because each column move also knocks the blank square up or down one row. So after an odd number of column moves, the blank can not possibly be back in the last row, where it belongs!
Hm, why not? I still dont' get what they mean with So after an odd number of column moves, the blank can not possibly be back in the last row, where it belongs.
discrete-mathematics
proof-explanation
alternative-proof
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edited Jul 12, 2017 at 5:34
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asked Jul 11, 2017 at 6:28
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The following article may be of help: The Fifteen Puzzle - A New Approach, Mathematics Magazine (Mathematical Association of America), Vol 90, No 1, February 2017, pp 48-57. –user348749 Commented Jul 11, 2017 at 10:22
For your first question: what is the parity of the number of inversions due to an exchange of a single pair of tiles? –Tob Ernack Commented Jul 11, 2017 at 10:31
@TobErnack If I exchange a single pair of tiles, like switching their position, I either create a new inversion if they have been in order before or I "fix" an inversion if they have been inverted before, no? So exchanging a single pair of tile switches the parity of the number of inversions to the opposite. –BMBM Commented Jul 12, 2017 at 5:20
@TobErnack Oh I get it, if the number of inversions in the start state is odd, we must make an odd number of column moves to get to an even number of inversions. Can you give me a hint how to interpret So after an odd number of column moves, the blank can not pos- sibly be back in the last row, where it belongs!? –BMBM Commented Jul 12, 2017 at 5:26
Ok, I think I get it now, in the desired state the blank square is in the bottom right in row 4 (where it also resides in the starting position). Each column move changes the row number of the blank square +1 or -1. So after an odd number of moves it can't be in an even row number. Sorry, I think I made the question much harder buy not realizing they were taking about this specific starting position and make observations based on that. I was thinking too far ahead. Maybe I should try to come up with an answer for my own question for future reference later. –BMBM Commented Jul 12, 2017 at 5:38
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Theorem. No sequence of moves transforms the board below on the left into the board below on the right.
1. row A B C D A B C D
2. row E F G H E F G H
3. row I J K L I J K L
4. row M O N M N O
a) Can a row move change the order of the tiles?
A row move can change the position of a tile only +1 or -1, but the position of no other tile changes. So a row move can never change the order of tiles.
b) How many pairs of tiles will have their relative order changed by a column move?
A column move changes the order of a tile either -4 or +4. When the tile is moved down, it moves in front of the next 3 tiles. When the tile is moved up, it moves behind the previous 3 tiles. Hence a column move changes the relative order for 3 pairs of tiles.
c) We define an inversion to be a pair of letters L1 and L2 for which L1 precedes L2 in the alphabet, but L1 appears after L2 in the order of the tiles.. What effect does a row move have on the parity of the number of inversions??
As a row move doesn't change the order of tiles, a row move has no effect on the number of inversions, so the parity stays the same.
d) What effect does a column move have on the parity of the number of inversions? Prove your answer.
As a column move changes relative order of 3 tiles, the number of inversions is flipped to the opposite, from even to odd and from odd to even.
e) The previous problem part implies that we must make an odd number of column moves in order to exchange just one pair of tiles (N and O, say). But this is problematic, because each column move also knocks the blank square up or down one row. So after an odd number of column moves, the blank can not possibly be back in the last row, where it belongs! Now we can bundle up all these observations and state an invariant, a property of the puzzle that never changes, no matter how you slide the tiles around.
Lemma: In every configuration reachable from the position shown below, the parity of the number of inversions is different from the parity of the row containing the blank square.
1. row A B C D
2. row E F G H
3. row I J K L
4. row M O N
Proof of the lemma by induction:
Base case:
After 0 moves, there is exactly 1 inversion (odd) and the row number of the blank square is 4 (even).
Inductive step:
If a row move happens, the row number of the blank square doesn't change, neither does the number of inversions change.
If a column move happens, the parity of the number of inversions is flipped (+3 or -3). Also, the parity of the row number of the blank square is flipped (+1 or -1). Given the start state with odd number of inversions and even row number for the blank square, the parity of those two variables will continue to differ after flipping them both to the opposite side.
e) Prove the theorem that we originally set out to prove.
As per the lemma, the parity of the number of inversions will always be different from the parity of the the row number for the blank square. In the desired end state their parity is equal, which will never be possible given the start state where those two variables have different parity.
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edited Jun 12, 2020 at 10:38
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answered Jul 20, 2017 at 6:32
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It appears you first had the question without the solution then found the recitation solution. –sunspots Commented Apr 24, 2024 at 17:55
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188469 | https://goldbook.iupac.org/terms/view/E02142/plain | Title: enthalpy of activation Long Title: IUPAC Gold Book - enthalpy of activation DOI: 10.1351/goldbook.E02142 Status: current Index: quantity Definition The standard enthalpy of activation (\Delta ^{\ddagger}H^{\,\unicode{x26ac}}) is the enthalpy change that appears in the thermodynamic form of the rate equation obtained from conventional transition state theory. This equation is only correct for a first order reaction, for which the rate constant has the dimension reciprocal time. For a second order reaction, for which the rate constant has the dimension (reciprocal time) × (reciprocal concentration), the left hand side should be read as (k\ c^{\,\unicode{x26ac}}), where (c^{\,\unicode{x26ac}}) denotes the standard concentration (usually (\pu{1 mol dm-3})). [k = \frac{k_{\rm{B}}\ T}{h}\ \rm{e}^{\frac{\Delta ^{\ddagger }S^{\,\unicode{x26ac}}}{R}}\ \mathrm{e}^{\frac{-\Delta ^{\ddagger }H^{\,\unicode{x26ac}}}{R\ T}}] The quantity (\Delta ^{\ddagger}S^{\,\unicode{x26ac}}) is the standard entropy of activation, and care must be taken with standard states. In this equation (k_{\rm{B}}) is the Boltzmann constant, (T) the absolute temperature, (h) the Planck constant, and (R) the gas constant. The enthalpy of activation is approximately equal to the activation energy; the conversion of one into the other depends on the molecularity. The enthalpy of activation is always the standard quantity, although the word standard and the superscript (^{\circ}) on the symbol are often omitted. The symbol is frequently (but incorrectly) written (\Delta H^{\ddagger }), where the standard symbol is omitted and the (\ddagger ) is placed after the (H). Related Terms - Boltzmann constant: - Planck constant: - activation: - activation energy: - enthalpy: - entropy of activation: - gas constant: - molecularity: - rate constant: - standard concentration: - transition state theory: Sources - Green Book, 2nd ed., p. 56 ( - PAC, 1993, 65, 2291. 'Nomenclature of kinetic methods of analysis (IUPAC Recommendations 1993)' on page 2294 ( - PAC, 1994, 66, 1077. 'Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)' on page 1113 ( - PAC, 1996, 68, 149. 'A glossary of terms used in chemical kinetics, including reaction dynamics (IUPAC Recommendations 1996)' on page 164 ( Other Outputs - html: - json: - xml: Citation: Citation: 'enthalpy of activation' in IUPAC Compendium of Chemical Terminology, 5th ed. International Union of Pure and Applied Chemistry; 2025. Online version 5.0.0, 2025. 10.1351/goldbook.E02142 License: The IUPAC Gold Book is licensed under Creative Commons Attribution-ShareAlike CC BY-SA 4.0 International ( for individual terms. Collection: If you are interested in licensing the Gold Book for commercial use, please contact the IUPAC Executive Director at executivedirector@iupac.org . Disclaimer: The International Union of Pure and Applied Chemistry (IUPAC) is continuously reviewing and, where needed, updating terms in the Compendium of Chemical Terminology (the IUPAC Gold Book). Users of these terms are encouraged to include the version of a term with its use and to check regularly for updates to term definitions that you are using. Accessed: 2025-09-29T04:07:59+00:00 |
188470 | https://ntrs.nasa.gov/api/citations/19940009325/downloads/19940009325.pdf | NASA Contractor Report 4544 / P Application of Symbolic and Algebraic Manipulation Software in Solving Applied Mechanics Problems Wen-Lang Tsai and Noboru Kikuchi (NASA-CR-4544) APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION SOFTWARE IN SOLVING APPLIED MECHANICS PROBLEMS (Michigan Univ.) 164 p HI/31 N94-13798 Unclas 0187791 CONTRACT NCA2-136 August1993 N/ A National Aeronautics and Space Administration NASA Contractor Report 4544 Application of Symbolic and Algebraic Manipulation Software in Solving Applied Mechanics Problems Wen-Lang Tsai and Noboru Kikuchi University of Michigan Department of Mechanical Engineering Ann Arbor, Michigan Prepared for Ames Research Center CONTRACT NCA2-136 August 1993 NationalAeronautics and Space Administration Ames Research Center Moffett Field, California 94035-1000 TABLE OF CONTENTS SUMMARY CHAPTER I.
HISTORY OF SYMBOLIC AND ALGEBRAIC MANIPULATION ......
1.1 Introduction .........................
2 1.2 History of SAM systems ....................
3 1.3 Conclusion .........................
11 1.4 References .........................
13 II.
SURVEY OF THE LITERATURE ON SYMBOLIC AND ALGEBRAIC MANIPULATION ........................
15 2.1 Introduction .........................
15 2.2 Review of SAM applications in engineering ..............
16 2.3 Conclusion .........................
24 2.4 References .........................
26 III.
CAPABILITIES OF THE SYMBOLIC AND ALGEBRAIC MANIPULATORS 29 3.1 Introduction ........................
29 3.2 What can the symbolic and algebraic manipulators do?
.........
29 3.3 References .........................
51 IV.
APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO AUTOMATIC PROBLEM FORMULATION ..............
52 4.1 Introduction .........................
52 4.2 Derivation of equation of motion by SAM ..............
53 4.3 Autornatic tensor formulation for shell problem ............
57 4.4 Approximation of a function by Fourier series ............
65 4.5 Template for nonlinear numerical analysis ..............
6 4.6 Triangular stiffness and mass matrix construction ...........
72 iii PreCEDING PAGE BLANK NOT FILMED 4.7 Closed form solution of stiffness matrix of four-node element .......
80 4.8 Significance and conclusion ...................
89 4.9 References .........................
91 V.
APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO A MATERIALLY NONLINEAR PROBLEM --- RIGID-PLASTIC RING COMPRESSION ........................
93 5.1 Introdt, ction .........................
93 5.2 Preliminary Ik)rmulation ....................
94 5.3 Matrix method for tile ling compression problern ...........
98 5.4 Finite element analysis ....................
100 5.5 Application of symbolic manipulation ...............
1(13 5.6 Numerical evaluation and result treatment ..............
107 5.7 References ................
.........
113 VI.
APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO THE PLATE PROBLEMS ....................
115 6.1 Introduction ........................
115 6.2 Preliminary formulation ....................
116 6.3 Methodology for solving shell problem by FEM ...........
118 6.4 Symbolic and algebraic manipulation application to plate problems .....
119 6.5 References .........................
14(I VII.
CONCLUSIONS .......................
145 7.1 Introduction ........................
145 7.2 Advantages of symbolic and algebraic rnanipulation ..........
145 7.3 Internal swelling and mathematical limitations ............
146 7.4 Symbolic and algebraic manipulation and education ..........
148 7.5 Contributions of this study ...................
149 7.6 Prospects and continuations of this research .............
15(1 iv 7.7 References .........................
152 APPENDIX A ..........................
153 APPENDIX B ..........................
154 3.1 3.2 4.1 4.2 4.3 4.4 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 LIST OF FIGURES Error between the solutions of REDUCE and Gauss elimination ......
36 Beam under uniform load ...................
41 Dynamic system for demonstration of syrnbolic and algebraic manipulation .53 Pictorially vector notations of shell ................
58 Convergence of Fourier series approximation ............
68 Behavior of optimal location vs. variable x3 .............
71 Configuration of domain and boundary conditions ...........
94 Plot of frictional stress vs. relative velocity .............
95 Arctangent approximation of fi'ictional stress .............
97 Physical configuration for ring compression problem ..........
97 Discretization of ring cross section ................
1()1 Deformed configuration after the 8th stage for m=().5 .........
108 Deformed configuration after the 7th stage for re=l).0 .........
108 Velocity distribution of m=0.5 case ...............
109 Velocity distribution of m=O.I) case ...............
110 Effective stress distribution ..................
111 Effective stress distribution ..................
112 Plate with hole under uniform uniaxial tension load ..........
125 Stress distribution of plate under uniaxial tension ...........
130 Physical configuration of plate bending problem ...........
134 Deformed shape of plate under bending load ............
136 Stress distribution of outer fiber of plate ..............
137 Three different sizes of mesh for testillg the convergence of plate bending .138 Convergence of plate bending solution ..............
138 Deformed shape of clamped plate under unilbl'm transverse load .....
139 vi LIST OF TABLES Table 1.1 3.1 3.2 3.3 List of symbolic and algebraic systems ...............
12 Comparison of solutions for 55 Hilbert matrix ............
35 Comparison of solutions for 66 Hilbert mauix ............
35 Comparison of solutions for 77 Hilbert matrix ...........
35 vii NOMENCLATURE SAM V f N Na_ M M_a K £ X, y, Z U, v, W R L t I q P k 0 T at_ Da D da_ Ea_ Fa Pa_ K_ 0 V IJl symbolic and 'algebraic manipulation strain energy; potential energy; velocity domain of integration stress resultant; shape function stress tensor moment resultant; mass matrix moment tensor stiffness matrix strain smoothing parameter curvature of curve Cartesian coordinates displacement components radius of curve linear operator; Lagrangian; length time imagimu'y symbol load per unit; rotation tensor position vector spring constant angle kinetic energy metric tensor operator of covariant derivative material property matrix; flow matrix curvature tensor strain tensor; Young's modules force components in surface of shell displacement gradient bending tensor determinant of displacement gradient poisson's ratio Jacobian matrix ...
VIII wi si , ij _ 0", C .f U d u A AU d h m g weight parameter area coordinates of triangular element natural coordinates of quadrilateral element Cauchy stress effective su'ess and strain rate friction stress velocity vector Lagrange multiplier die velocity alea difference between U and U I j Generalized displacement vector thickness of shell friction factor yield shear stress ix SUMMARY Compared to numerical analysis, symbolic and algebraic manipulation is unfamiliar to people in the research field.
As its name implies, symbolic and algebraic manipulation can be simply interpreted as a computerized operation which can retain symbols throughout computations and express results in terms of symbolic forms.
For example, the coefficients a, b, and c in the quadratic polynomial equation ax2+bx+c=O do not need to be known in order to find its roots.
The equation itself can be directly input to a computer and the results will be (-b + 4bZ-4ac) , (-b -4b z-4ac) x= 2a and x:.-2a If the numerical values are required, three coefficients can be specified and solutions will be expressed as numbers.
From the example above, at least two unique characteristics of symbolic and algebraic manipulation can be observed.
First unlike numerical analysis, the solutions from symbolic and algebraic manipulation are exact and therefore no round-off error is introduced.
Second, the solutions are the same as those derived by hand.
Therefore the extension of human capability to handle more sophisticated fomulations becomes feasible by computer.
In the first chapter of this report, the history of symbolic and algebraic manipulation is introduced.
The sencond chapter chronologically reviews the literature regarding the application of symbolic and algebraic manipulation in the engineering field.
The capabilities of symbolic and algebraic manipulators are demonstrated in chapter three by selected examples.
Chapters four through six demonstrate applications of symbolic and algebraic manipulation.
Chapter four describes the automatic formulation of applied mechanics problems, chapter five covers the materially nonlinear, rigid-plastic ring compression problem, and chapter six discusses plate problems.
The final chapter summarizes the overall conclusions of this report.
It is well known that there are some difficuilties existing in the symbolic and algebraic field.
The report proposes a remedy to avoid the difficulties and successfully accomplishes the applications.
Due to this breakthrough, the solution of some previously insolvable problems become available.
In addition, one of the advantages found in this research is believed to be crucial for improving the execution efficiency of numerical programs.
CHAPTER I HISTORY OF SYMBOLIC AND ALGEBRAIC MANIPULATION 1.1 Introduction Symbolic and Algebraic Manipulation (abbreviated as SAM) software is one of the new products of modem technology for use with highly developed digital computers.
Traditionalists might say that SAM software is a misuse of modern computers.
This is true if the viewpoint is adopted that a computer is a machine which only counts numbers.
This viewpoint, however, severely limits the emerging artificial intelligence capabilities of computers.
For instance, in addition to numbers, there are many symbols which define appropriate mathematical relations in a calculus book.
Can we ask a computer to do these analytical derivations for us? This is a great question which finally led to the birth of SAM and added "soul" to the computer, to make it think more like a human brain.
This idea, which originated before 1953, has had an impact on a variety of fields, such as science, industry and education.
Therefore, although its history is not as long as that of the classical sciences, its impact has been so large that a record of its history is deserving.
At the initiation of this dissertation effort in 1986, there were already a number of SAM systems available on the market.
Some of them were more than ten years old, such as FORMAC, REDUCE and MACSYMA.
Others were just being developed, such as muMATH and MATHEMATICA.
Ironically, most relevant documents either ignore the history of these systems or just skim over them briefly.
Only one book, written in 1969 by Jean E.
Sammet , includes historical details, however, it is too old to cover recent developments.
Most of the major systems used today have been produced since then.
Therefore, it is necessary to collect, rewrite, and update the history of the SAM systems.
It is hoped that the interested researchers will get a complete picture of the development of SAM systems.
Through an understanding of the history they will be able to grasp the direction of the field and devote themselves towards making a further contribution.
This is the ma.jor purpose of the chapter.
It is much more important than just knowing how to run the SAM systems.
The first ideafor usingcomputers to doSAM canbetracedbackto two mastertheses published in 1953(,). Threeyearslater,whatis believedto bethe earliestSAM system , calledPM, wasdeveloped at IBM . Now therearemanySAM systems on themarketfor variouscomputers.
Someof themaredesigned for generalpurpose usage, while othershave beendeveloped for particularapplications.
Generally speaking, theevolutionof symbolicand algebraic manipulation canbeclassified intothreestages.
Theyare: 1. The first generation(1953-1965)---software to appearin this generationwas PM, ALGY, FORMAC,MATHLAB andALTRAN. Because of thelimitation of hardware capacity, the systemsin this stagewere small in size and immature in content.
Therefore,mostof thembecame obsolete or wererevised.
2. The2nd generation (1966-1975)---software to appear in this generation wasREDUCE and MACSYMA. Thesesystemstook advantage of the improvementof hardware memorycapacity.Theycontainmanybuilt-in functions,arelargeandarefor general purpose usage.
All of themrunon themainframe.
3. The 3rd generation(1976-present)---some representatives of morerecentsystems are muMATH, MATHEMATICA, and DERIVE. Unlike the systemsof the second generation, thesystems in thisgeneration aredesigned to runonmicrocomputers.
This has been possible due not only to the improvement of memory capacity in microcomputers, but also due to the requirement of most users who just need quick checks or moderate manipulations.
Details of the histories of the systems will be described in the following subsections individually.
For the sake of clarification, a summary is also included in Table 1.1.
1.2 History of SAM systems 1.2.1 PM PM is believed to be the earliest computerized algebraic system in the world.
It was developed by George E. Collins at the IBM research center in Yorktown Heights, New York.
Although the first document was published in 1966 its beginning dates back to 1956.
Written in assembly language for the IBM 701 computer, PM contained the subroutines for addition, subtraction and multiplication of multiple-precision integers, and subroutines for performing the same operations on multivariate polynomials with multiple-precision integer coefficients.
Between 1956 and 1966 PM was reprogrammed for the newer IBM computers (e.g.
709, 7090 and 7094), and augmented to include new operations (such as integer god) with various improvements (e.g.
the incorporations of list processing and dynamic storage allocation).
In 1966 Dr.
Collins became a professor of computer science at the University of Wisconsin.
With the aid of graduate students, the PM system was converted to the SAC-I system in 1973.
In spite of its eventual replacement, PM, as the first SAM system, was still very significant in the field.
1.2.2 ALGY Although ALGY has few functions, it was one of the earliest SAM systems in the world.
The developmental work was started at Western Development Lab-Philco Co.
in Paio Alto, California by Bernick, Callender and Sanford, around 1961 .
It was interactive and allowed expressions written in a notation similar to FORTRAN as input, with some deviations.
For example, the $ was used instead of to represent exponentiation, and all natural integers were expressed as fractions, for example, 0 and 1 were denoted by 0/1 and 1/1, respectively.
It only contained a few commands, such as : • OPEN : expanding the expression in the parenthesis • SBST : making substitution • FCTR : factoring a given expression • TRGA : expanding the sin(a+b) into sin(a)cos(b)+sin(b)cos(a) Which is why the authors said that only two hours instruction was enough to use it.
Although ALGY didn't come into extensive use, some of its ideas were succeeded by the FORMAC system, which is still popular today.
1.2.3 FORMAC FORMAC is an acronym of FORmula MAnipulation Compiler.
It was developed by J.
E.
Sammet and Robert G.
Tobey at IBM's Boston Advanced Programming Department in July, 1962 .
Five months later (December, 1962), the first complete draft of language specifications was prepared and implementation design started immediately thereafter.
After 18 months of extensive experiments, the first complete version was successfully running on the IBM 7090/94 computer in April, 1964.
For the sake of obtaining feedback from users to make further improvement to the system, FORMAC was released for public use by the authors themselves (not by IBM) in November, 1964.
This version of FORMAC was written in 4 FORTRAN IV. Three years later (November, 1967), the new version of FORMAC written in PIM was released by the authors for use on IBM/360 systems.
The FORTRAN version of FORMAC kept most commands and notations of FORTRAN IV.
In addition, there were a couple of new commands added to allow it to do algebraic manipulations.
For example, LET assigns symbols to variables instead of numbers in FORTRAN, SUBST makes substitution, EXPAND removes all parentheses in expressions and COEFF obtains the coefficients of variables.
The major PL/I FORMAC capabilities can be divided into the following categories : 1. User control of simplification : EXPAND for expanding the parentheses expression, DIST for applying the distributive law to all products of sums, etc.
2. Substitution : EVAL(expr,a,b) replaces a in expr by b.
3. Differentiation : DERIV performs partial differentiation.
4.
Expression analysis : COEFF(exprl,expr2) returns the coefficient of exprl in expr2.
NUM and DENOM return the numerator and denominator, respectively.
HIGHPOW and LOWPOW return the highest and lowest power.
5. Storage allocation : SAVE(var) for storing the var to secondary storage.
6. Output : PRINT_OUT(expr) to pnnt out the required expressions.
7. Built-in functions : these include trigonometric, logarithm, exponentiation, square root, hyperbolic function, etc.
8. User defined function : the user can define functions as needed.
FORMAC is now one of the most popular SAM systems.
It is the first reasonably general purpose system to receive extensive usage worldwide.
With the advantages of longer history and larger numbers of users, its accumulated contributions to SAM field are remarkable.
In 1977, the new version, called FORMAC 73, was released to replace the old one.
5 1.2.4 MATHLAB MATHLAB l system was developed by C. Engeiman and his employees at MITRE Co.
in 1964 .
Its source language is LISP, but the commands are defined as English words.
For example, PLEASESIMPLIFY(x,y) is the command to simplify x and name it as y. In the fall of 1967, the first version of MATHLAB was replaced by the second version, MATHLAB 68, which operated on a PDP-6 machine with 256 K core memory.
The input and output were through a teletype-like keyboard with a fixed character display scope.
The notations in the second version were more ALGOL-like.
MATHLAB was the first complete on-line system.
1.2.5 ALTRAN ALTRAN is a system developed at the BELL TELEPHONE Laboratory in Murray Hill, New Jersey by W.
S. Brown, M.
D. Mcllroy, D.
C. Leagues and G.
S. Stoller .
It was running in late 1964 on the IBM 7090/7094, 7040/44, etc.
The basic languages which ALTRAN adapted were a mixture of FORTRAN II and FORTRAN IV.
Since it was limited to use in the BELL Lab., its contributions to the SAM field were small.
1.2.6 REDUCE REDUCE was developed by A.
C. Hearn of Rand Corporation, California, in 1963 .
At that time, he met Dr. John McCarthy, an inventor of the LISP language, who suggested the use of LISP for the problems of elementary particle physics.
Since then, Dr.
Hearn, as a theoretical physicist, has worked in the SAM area.
In August 1966, the first publication was issued .
This paper only talked about the specific application of SAM techniques to elementary particle physics.
Two years later (1968), the first paper describing a general algebra system, "REDUCE", was published . The name of REDUCE originated from this paper.
Its name is not an acronym.
According to the description from the author himself, its name was actually intended as a wit.
He said "algebra system then as now_tended to produce very large expressions for many problems, rather then reduce the results to a more manageable form".
The system at this time was called REDUCE for distinction from the new version, REDUCE 2, which appeared in 1970.
The big improvement was that the whole system was written in an ALGOL-like dialect (call RLISP), rather than the parenthesized notation of LISP in which REDUCE was written.
At this time, the REDUCE 2 system was also released to users, making the beginnings of a user community.
Thereafter, REDUCE 2 was implemented successfully on 1 MA'rHLAB is not to be confused with MATLAB. MATLAB is the numerical software for matrix operations, while MATHLAB is another symbolic and algebraic manipulator.
6 theMichiganTerminal System (MTS) of the University of Michigan by Mike Alexander.
After a long silence, REDUCE 3 was distributed in 1983.
Several significantly new packages were added in this version, such as analytic integration, multivariate factorization, arbitrary precision real arithmetic and equation solving.
Following REDUCE 3, upgraded versions were also released.
They were REDUCE 3.1 released in 1984, REDUCE 3.2 in April, 1985, REDUCE 3.3 on July 15, 1987.
Each of them contains bug fixes and additional capabilities.
Instead of implementation on MTS, the REDUCE 3.3 was first implemented on the APOLLO workstation in the Computer Aided Engineering Network (CAEN) of the University of Michigan.
REDUCE 3.3 was also updated once in January 15, 1988.
REDUCE system has become one of the most well-known SAM systems.
Its general purpose design makes it possible to be used in a wide variety of areas.
Its contributions are confirmed by the number of papers published in different fields.
1.2.7 SCHOONSCHIP SCHOONSCHIP was designed by M.
Veltman at CERN, Switzerland in 1964 .
Its major applications are in the field of high energy physics, but it is sufficiently general to be used for other calculations.
It can deal easily with expressions of 104 to 105 terms on the CDC 6000 computer.
It was limited to use within CERN.
1.2.8 ANALITIK ANALITIK was developed at the Institute of Cybernetics in Kiev, Soviet Union, by the direction of the well known Soviet cybernetician and academician V. M. Glushkov .
The first paper discussing the system features was published in 1964.
The language it used was ALGOL-like and close to that of traditional mathematical notation and natural language.
It possessed interactive and batch processing modes.
Since its implementation is highly machine dependent, ANALITIK has only run on the MIR-2 computer.
1.2.9 FLAP FLAP was written in LISP 1.5 by A.
H.
Morris, Jr.
at the U.S.
Naval Weapons Laboratory in Dahlgren, VA.
prior to 1967 [11. Obviously, the FLAP system wasn't released to the public.
7 1.2.10 SAC The SAC system was developed by Dr.
George E.
Collins at the University of Wisconsin, Madison.
The first version, SAC-I, was distributed in 1967 .
This was a highly portable general purpose system, developed to replace one of the very earliest computer algebra system, PM in IBM [see 1.2.1].
SAC-1 was replaced by SAC-2 in July, 1980.
The SAC-2 was programmed in ALDES language, which was designed by Rudiger Loos and G.E.
Collins in 1973 to 1974.
The SAC-2 system also provided the translator from ALDES to standard FORTRAN to maintain its portability.
1.2.11 MACSYMA MACSYMA is an acronym of project MAC's SYmbolic MAnipulator.
It was originally designed by C.
Engelman, W.
Matin, J. Moses for project MAC at M.I.T.
in 1968.
The implementation of it began in July, 1969.
The system has quintupled in size since the first paper describing it appeared in 1971 ( ).
It was made available over the ARPA networks in May, 1972.
MACSYMA has a lot of built-in mathematical functions and graphic facilities which have made it one of the most powerful SAM Systems in the world.
Unfortunately, the University of Michigan didn't have it until September, 1988.
The one implemented on the APOLLO workstation in CAEN of the University of Michigan still doesn't have a graphics package.
1.2.12 SCRATCHPAD Although the name of SCRATCHPAD was chosen in 1970, the initial work on it can be traced back to 1965.
The SCRATCHPAD system was designed principally by James H.
Griesmer, Richard D. Jenks, Fred Blair, David Yun, and their colleagues, at the IBM Thomas J.
Watson Research Center, Yorktown Heights, New York ( 118]).
Unfortunately, the name of SCRATCHPAD was not used for the first paper, presented in Bonn in 1970.
One year later, a revised version by Dick Jenks, called SCRATHPAD/1, was demonstrated at SYMSAM/II in March 1971.
After combining some new features, such as history file (allowing users to backtrack), and system commands, the first completed SCRATCHPAD/1 manual was eventually published in 1975.
After this, there seemed a stagnation in the progress of the SCRATCHPAD system due to personnel changes.
Jim Griesmer left the group to be a manager of education at IBM research and Dick Jenks went to the University of Utah for a sabbatical.
When Dick Jenks returned to Yorktown Heights in the fall of 1977, David Yun agreed to organize the "mode-base" ideas originated by Dick Jenks in 1973.
This led to the 8 NEWSPADwhichthereafter wasrenamed to SCRATCHPAD84at theNew York conference in 1984.However,thenameof SCRATCHPAD84 wasnotquiteappropriatesinceit would takemorethanoneyearto finishthesystem.
Therefore it waschanged intoSCRATCHPADII, which is thenameusednow.It became availablein 1985for testandevaluationto a limited numberof usersfrom anIBM ownedmainframeviatelenet,CSNETandARPANET. As yet, it isnotcommercially available.
1.2.13 CAMAC The CAMAC system was designed by Vera Pless in 1973 at M.I.T.
.
The first version of it ran interactively and was written in FORTRAN with sections in assembler language.
When Vera Pless moved to Chicago Circle in 1975, the CAMAC system was transferred to the Circle's IBM 370-158 by William Pattern.
The name of CAMAC is an acronym of Combinatorial and Algebraic Machine Aided Computation.
As the name implies, it was for a specific application.
1.2.14 SHEEP SHEEP was designed by I. Frick at the University of Stockholm, Sweden in 1975 [(20].
) It was specialized for manipulating components of tensors.
The source language it uses is MACRO-10.
It runs on DEC PDP 10 and PDP 20. The first version of SHEEP, now called SHEEP 1, was written in assembler code for the DEC-10/20 computer.
Unlike the first version, SHEEP 2 is written in standard LISP.
1.2.15 ORTOCARTAN ORTOCARTAN is written in LISP.
It was designed by Andrzej Krasinski in Poland in 1977 .
Its name is an acronym and is due to the specific application to the calculation of Riemann, Ricci, Einstein and Weyl tensors from a given metric tensor using an ORTHonormal set of CARTAN forms.
Although the author said it could be relatively easily extended for other uses, such as inverting matrices of arbitrary rank, it did not come into wide use.
1.2.16 MAPLE The MAPLE system was designed by Bruce Char, Keith Geddes, W.
Morven Gentleman and Gaston Gonnet at University of Waterloo, Canada in December 1980 ( ).
The name "MAPLE" is not an acronym but rather it was simply chosen as a name with a Canadian identity.
There were two goals which oriented MAPLE's design.
The first was to be 9 usedon a time sharingmainframecomputer.
The second wasto runit on a microprocessor-basedworkstation. This was the major difference betweenMAPLE and REDUCE (or MACSYMA).
1.2.17 muMATH wntten in muSIMP (a LISP-like language), muMATH was designed and developed by David R.
Stoutmeyer and Albert Rich at the University of Hawaii in 1977 .
The first version was called muMATH-77 and was experimental.
Two years later, the Software House, Inc., was founded by David Stoutmeyer and the first product, muMATH-79, was distributed to users for the CP/M-80 operation system or Apple II family Z80 machine with 64 K bytes core memory required.
The second product called, muMATH-83, was not released for the IBM personal computer until 1983. The muMATH-83 needs 256 K bytes RAM memory.
Recently, DERIVE has taken over the place of muMATH-83.
The significant improvement is that DERIVE combines the numerical, algebraic and graphical functions together, rather than just algebraic functions of muMATH.
The DERIVE system requires 512 K memory space for normal execution.
1.2.18 MATHLIB & SMP MATHLIB is an interactive general purpose SAM system.
It was originally designed and developed under the auspices of the Department of Mathematics at Harvey Mudd College, California, in 1978.
It was one of the products of Innosoft International Inc., of Claremont, California and became commercially available in 1983.
It can perform numerical and symbolic operations.
In addition, its graphical output is device-independent and allows it to be processed by over 150 different graphics devices.
The PRS subroutine embodied in the algebraic subsystem of MATHLIB has more than 250 built-in functions for manipulation of mathematical expressions.
SMP is another product of Innosoft International Inc..
It was designed at the California Institute of Technology.
It's written in C language and was originally developed to run on VAX/780 under the UNIX operating system.
In addition, there is a special design character in SMP to allow for easy conversion between operating systems.
It needs at least 2.5 megabytes of memory space for typical usage.
10 1.2.19 MATHEMATICA MATHEMATICA is a recent product of Wolfram Research, Inc.
There are several versions of MATHEMATICA for a variety of computers, such as for Apple Macintosh, DEC VAX, IBM ,Cray, and so forth.
It was designed and implemented by Stephen Wolfram, Daniel Grayson, Roman E. Maeder, and their colleagues at the University of Illinois in 1988 .
It integrates the algebraic manipulation, numerical computation, and graphical functions together and allows the resultant expressions to be outputed in Ccode, FORTRAN code, and text form.
Its source language is C. The memory requirement for normal operation is about 3.7 mega bytes.
The MATHEMATICA as well as DERIVE are expected to be two dominant systems in the coming decade.
1.2.20 Mathcad Mathcad is developed by Mathsoft, Inc.
at Cambridge, Massachusetts.
The earlier version appeared on market around 1987.
This system adopted the core functions of MAPLE and extented itself by including the graphic capacity.
It is written in C language.
The latest version 3.1 is available in 1992.
This newest version can run in IBM, Macintosh PCs and unix based machines.
The minimum space requirements are two megabyte RAM and seven megabyte hard disk.
Unlike most of the SAM systems, the command inputs in this system are menu driven.
This allows users to communicate with machine by simply picking and clicking.
This unique feature not only saves users lots of efforts in typing but also reduces human errors which sometimes turn out a unmanageable, hard-to-be-debugged results.
1.3 Conclusion From a history of the SAM system, we can draw the following conclusions : • The SAM systems evolved from small, immature systems to well-designed, multi-function systems, to compact systems which can be used on microcomputers.
• At present, there is no unique best system.
The definition of the best SAM system depends on many variables, such as computer availability, availability of software, familiarization with software, the problem to be solved, software contents, circumference facility, and so forth.
11 system PM i, ALGY FORMAC MATHLAB year 1956 1961 1962 1964 1964 ANALITIK ALTRAN RED UCE 1963 SCHOONCHIP 1964 1964 FLAP SAC MACSYMA SCRATCHPAD CAMAC SHEEP 1967 1967 remarks IBM WDLP Co.
IBM - Boston MITRE Co.
BELL Lab.
Rand Co.
CERN Soviet Unions U.S.
Navy Uni.
of Wisconsin 1968 M.I.T.
1965 IBM-Yorktown Heights 1973 Vera Pless 1975 1977 ORTOCA RTA N MAPLE 1980 muMATH & DERIVE 1977 1977 1977 1988 1992 MATHLIB SMP MATHEMATICA Mathcad 3.1 Sweden Poland Canada Uni.
of Hawaii Harvey Mudd College Caltech Uni.
of Illinois Mathsoft Inc.
Table 1.1 • List of symbolic and algebraic systems 12 1.4 References Jean.
E. Sammet, "Programming Languages • History and Fundamentals", Prentice-Hall inc., 1969.
H.
G.
Kahrimanian, "Analytic differentiation by a digital computer", M.S.
thesis, Temple University, Philadelphia, Pennsylvania, 1953.
J.
Nolan, "Analytic differentiation on a digital computer", S.
M.
thesis, M.I.T., cambridge, Massachusettes, 1953.
George E. Collins, "PM, A system for polynomial manipulation", ACM, Vol.
9, No.
8, Aug.
1966.
M.
D.
Bernick, E.
D. Callender, J.
R. sanford, "ALGY-An Algebraic manipulation program",Proc.
WJCC, voi.
19, pp 389-392, 1961.
C. Engelman, "MATHLAB-A program for on-line machine assistance in symbolic computations", Proc.
FJCC, vol.
27, pt.2, pp.
117-126, Nov., 1965.
Gerhand Rayna, " REDUCE • software for algebraic computation", Springer-Verlag, 1987.
A.
C. Hearn, "computation of Algebraic Properties of Elementary Particle Reactions using a digital computer", comm.
of the ACM, 9, pp 573-577, 1966.
A.
C.
Hearn, "REDUCE-A user oriented interactive system for algebraic simplification", Interactive systems for Experimental Applied Mathematics, pp 79-90, (edited by M. Klerer and J. Reinfelds), Academic Press, New York, 1968.
H.
Strubbe, "presentation of the SCHOONSCHIP system", EURUSAM 74, pp 55-59, 1974.
Nisse Husberg, Jouko Seppanen, "ANALITIK • principal features of the language and its implementation", EUROSAM 74, pp 24-25, 1974.
Joel Moses, "MACSYMA - The fifth year", ACM SIGSAM bulletin EUROSAM'74, Stockholm, pp 105-110, Aug 1974.
"MACSYMA reference Manual",version ten, Mathlab Group, MIT, 1983.
13 Richard Pavelle, " MACSYMA : capacities and applications to problems in engineenng and the sciences", Symbolics Inc., Cambridge, MA,1985.
W.
A Martin and R. J. Fateman, '"Fhe MACSYMA system", proc.
2nd symposium on symbolic and algebraic manipulation, pp 59-75, ACM, March 1971.
Robert S.
Sutor, "The Scratchpad II computer algebra language and system", EUROCAL 85, pp 32-33, voi. 12, 1985.
James H.
Griesmer, "A history of the SCRATCHPAD Project (1965-1977)", IBM newsletter, vol. 1, no.
2, Jan.
15, 1986.
Richard D.
Jenks, "A history of the SCRATCHPAD Project (1977-1986)", IBM newsletter, Vol.
1, no. 3, May 15, 1986.
Jeffrey S. Leon, Vera Pless, "CAMAC 1979", symbolic and algebraic computation EUROSAM 79, pp 249-257, 1979.
Lars Hornfeldt, "A system for automatic generation of tensor algorithms and indicial tensor calculus, including substitution of sums", Symbolic and algebraic computation, EUROSAM 79, pp 279-290, 1979.
I. Frick, '"Fhe computer algebra system SHEEP, What it can and cannot do in general relativity",Inst, of Theoretical physics, University of Stockholm, 1977.
Andrzej Kransinski, "ORTOCARTAN---A program for algebraic calculation in general relativity", ACM SIGSAM, vol.
17, pp 12-18, 1983.
Bruce Char, Keith Geddes, Gaston Gonnet, "The MAPLE symbolic computation system", ACM SIGSAM, vol.
17, pp 31-42, 1983.
Bruce Char, Keith Geddes, W.
Morven Gentleman, Gaston Gonnet, '"/'he design of MAPLE : a Compact portable and powerful computer algebra system", Computer Algebra, proceedings Eurocal'83, pp 101-115, 1983.
C.
Woof f, D.
Hodgkinson, "muMATH : A microcomputer algebra system", Academic press, 1987.
Stephen Wolfram, "MATHEMATICA---a system for doing mathematics by computer", Addison-Wesley, 1988.
14 CHAPTER II SURVEY OF THE LITERATURE ON SYMBOLIC AND ALGEBRAIC MANIPULATION 2.1 Introduction The documents published in the symbolic and algebraic manipulation field are not as plentiful as those in the area of numerical analysis.
However, after a careful classification of the existing documents, one finds that the developmental history is closely related to research directions and content of the publications.
In general, the documents about SAM may be divided into four categories.
They are : 1. About SAM system itself --- More than half of the existing papers belong to this class.
Most of them were published in the period of the first generation.
The contents are focused on the following topics" (a)The introduction of the new SAM system, including the capacities, functions, etc.
.
(b) The technical reports of softwares .
(c) The data structure, language and implementation .
. Applications to science --- This class of publications is the second largest of the existing SAM papers.
One of the major impetuses in developing SAM systems was due to the requirements from scientists, especially in the fields of elementary particle, general relativity and celestial mechanics.
Some of the famous examples were collected in the paper by Hearn .
One of them is the recalculation of Delaunay's moon coordinates by Deprit, Henrard and Rom in 1970 The others are such as Campbell and Hearn's analysis of the Feyman diagram , Rudiger Loos' work about Archimedes' cattle problem , and Roberts' and Boris' on the solution of partial differential equations .
15 .
Applications to engineering ---In fact, most engineering problems are not solvable analytically.
Therefore, numerical approximation usually predominates in the solution of engineering problems.
This is one of the reasons why the engineering applications of SAM has not been as popular as those on science.
However, there are two factors which necessitate the use of symbolic and algebraic manipulation in engineering.
The first is that the accuracy requirement of a solution of numerical approximation becomes more and more strict today.
The second is that the problem to be solved usually involves more sophisticated algebraic manipulation due to the more strict requirements of solutions.
Due to these factors, the applications of symbolic and algebraic manipulation to engineering problems has became more popular recently.
Some of the publications, such as those written by Madson, Smith and Hoff , Levi , Wilkins , Noor and Anderson , Korncoff and Fenven , Steinberg and Roache , will be discussed in more detail in the next paragraph.
. Application in the other fields --- In addition to science and engineering, Symbolic and algebraic manipulation has been applicable in other fields, such as information management , education and business .
As time goes on, more and more applications will be reported in various fields.
This is due to the fact that : 1. The ongoing improvement in the memory space of hardware systems, especially personal computers.
2. The availability of variously sound SAM software systems.
However, in order to see that scratch paper is replaced by the computer screen in all areas, the people in the educational field should assume the responsibility of utilizing this new tool.
As the discussion in the paper, written by Richard Pavelle in 1985, points out , only about 20 percent of people in the related field are aware of the existence of SAM system and less than a quarter of these actually use them.
2.2 Reviews of SAM applications in engineering (1) Computer algorithms for solving non-linear problems A paper [ 16] published in 1965 is believed to be the earliest document which employed computerized symbolic and algebraic manipulation to solve an engineering problem.
The 16 authors,W. A. Madson,L. B. SmithandN. J. Hoff, developed theirown softwareto find the solution for thepost-bucklingbehavior of thin-walledcircularcylindrical shellsunderaxial compression.
Themajorcommands developed bythemwere: SERIESMULT: To expand theexpressions, e.g.(asin(x)+bcos(y)) n.
TRI.GSPAND : To treat non-double trigonometric terms, such as sin2(x), sin(x)sin(2x)cos(y) etc., into double trigonometric terms like cos(y)cos(x).
SEARCHSTORE : To search and collect the coefficients of like trigonometric function, then store them.
NEWTNRAPH : To solve the nonlinear system equations obtained from the calling of the last three commands by using the Newton-Raphson iteration method.
The application of the above commands to the shell post-buckling problem started at the assumption of radial displacement, w=t .Aifcos( ixt/ Ax)cos(jny/ Ay ) Then by the strain-displacement relationship and Hook's law, the stresses could be obtained.
As the stresses (therefore the Airy stress function) were known, the membrane energy, bending strain energy and the potential of axial load could be derived.
The resultant total potential energy was then minimized with respect to the coefficients of radial displacement w.
The system equations obtained after the minimization then could be solved by calling the NEWTNRAPH command.
(2) Symbolic algebra by computer-applications to structural mechanics One of the earliest publications of symbolic manipulation application in the engineering field was in 1971, when only a few SAM systems existed.
Only REDUCE and FORMAC were mentioned in this paper.
At the beginning of the paper by I. M. Levi, he described the story of SAM application in seeking the minimum theoretical post-buckling load for a thin circular cylindrical shell under axial compression.
Starting from 1941, Von Karman and Tsien indicated that a low post-buckling load could be found with only two terms included in the series expression of normal displacement.
This inconsistency in the Von Karman-Tsien's solution was not found until J. Kempner increased the series into three terms and found a further lowering of the post-buckling load in 1954.
Since then, additional investigations were conducted as new terms were added into the series solution.
But finally everybody was limited 17 by their inability to solve the complex algebraic equations without error.
The drive to seek the minimum load did not end until 1965 when Madson, Smith and Hoff of Stanford University wrote the special program in ALGOL to increase the series into 14 terms and found the minimum load approached zero, which revealed the basic fallacies in the application of the Karman-Tsien procedure.
The second topic in the paper talked briefly about the derivation of a stiffness matrix for a compatible triangular plate bending element by symbolic and algebraic manipulation.
Then an example of the calculation of creep strain rate in plate and shell problems was demonstrated using SAM.
This computation started from the x, y components of stress which were expressed as a double trigonometric series, followed by the calculation of equivaleat stress, and ended in the substitution of the above quantities into strain rate equations.
The resultant fortran codes were printed out by REDUCE.
The paper ended with a brief discussion on the REDUCE capacities.
(3) Applications of symbolic algebra manipulation language for composite structures analysis This paper was published in 1973 by Dick J.
Wilkins, Jr.
of General Dynamics/Convair Aerospace Division, 1::olt Worth, Texas.
The author used PL/I FORMAC to calculate the strain energy for an anisotropic shell.
The strain energy can be written as x N Y N M_ My M_ !r "E x Cy g_y K" x Ky K_y dO (2.1) Where • N • stress resultants.
• M • moment resultants.
• e • mid-plane strain.
K • curvalure.
• Q: shell sttrface area.
18 • V" strainenergy.
and the constitutive equations are expressed as follows • Ny N. M My 'A, I A12 A 13 BI!
B12 B13 At2 A 22 A 23 Bt2 B22 B23 A13 A 23 A33 BI3 B23 B33 BII BI2 BI3 Dll Di2 Di3 812 822 823 DI2 D22 D23 Bl3 B23 B33 Dr3 D23 D33 X Ey K" x K'y K. j (2.2) Where • Aij , Bij , Dij are the pertinent constitutive components of Hooke's Law.
The equation (2. l) and (2.2) can be combined into the form V = fa[{e}r[A]{e}+2{e}r[B]{x}+ {r }r [D ]{ r } ]d Q (2.3) Then the displacements are approximated as w - C:.XmY.
(2.4) , _ OX., ,-.,, _ OY.
v -C,,,X,,.-----(2.6) Where the Cij are constants to be determined by Rayleigh-Ritz method.
By Vlasov shell theory, the strain-displacement relations are au E x ---"-dv w e, """+ T 3u Ov oXw _ x " t_X 2 w I( y == R' 3Zw 1 3u 1 o_ r_ =-2 + oxoy ROy R Ox (2.7) (2.8) (2.9) (2.10) (2.11) (2.12) 19 WhereR hereis theradiusof shellcurvature.
With theassumption of asymmetric constitutive matrix,thecalculation of theintegrand of equation(2.3) wasdoneby FORMACby thesubstitutionof equations (2.4) to (2.12)into (2.3).TheRayleigh-Ritzmethodwasthenappliedto takethepartialderivativewith respect to all undeterminedconstantsin the displacementseries.This was also manipulated by @X , @Y 4 FORMAC.
The resultant expressions were the energy variation which is in terms of ----, T and their derivatives.
Since FORMAC was incapable of performing the symbolic integration at that time, the informal "symbolic integration" was done by examining each term in the energy variation for a specific combinations of derivatives.
Each time a certain type was found, it was replaced by a symbol and an appropriate constant to allow for the non-dimensionalization of the integrals.
There were a total of twelve different integrations in the energy variation equation.
The final expressions were then slightly modified into FORTRAN code by adding DO loops and suitably changing the indices by hand.
(4) Computerized Symbolic Manipulation in Structural Mechanics --- Progress and Potential In the beginning of the paper , A. K. Noor and C. M.
Anderson introduced the symbolic and algebraic manipulator MACSYMA.
These included the brief history, basic capacities and special commands, as well as associated packages.
The second part of the paper gives three applications in the structural mechanics field by using MACSYMA.
They are • 1. Generation of characteristic arrays of finite elements for a shear flexible shallow shell element --- There were three types of basic integrals for linear problems and three types of basic integrals for geometrically nonlinear problems.
These were (a) Linear problems A '1 .
,,N'NJdI2 B ,, .. f ,, N tgaN Jd [2 O i - f ,, o_,,N 3aNid12 (2.13) (2.14) (2.15) (b) Geometrically nonlinear problems 20 C qk -f ,.N 3,NJO Nkdg2 a # (2.16) ,jk ' J o_ N kd g'2 D u_t .. f ,,O_,N 0#N r (2.17) qkn i l k Ec,arp " f ,.aoN cgoN cgrN OpN"dl-2 (2.18) The evaluation of integrals in equations (2.13) and (2.14) can be performed analytically by MACSYMA.
However in general the integrand of equations (2.15) to (2.18) cannot be integrated exactly due to the existence of a Jacobian determinant in the denominator of the integrand 2 Therefore the hybrid approach (numerical quadrature plus symbolic manipulation) was proposed.
The number of integrations to be performed can be substantially reduced by the help of permutative and Dihedral symmetries.
2. Evaluation of effective stiffness and mass coefficients of continuum models for repetitive lattice structures --- The symbolic manipulations by MACSYMA included the evaluation of strain components, calculation of strain energy (with the thermoelastic strain energy) and kinetic energy, computation of stiffness and thermal coefficients as well as effective mass coefficients, forming the Lagrangian of the system and finally obtaining the governing differential equations.
The numerical analysis started as soon as the governing equations were obtained.
This numerical analysis was also done in MACSYMA.
The results of mode shapes were then plotted out by MACSYMA's graphic facility.
3. Application of the Rayleigh-Ritz technique to the free vibration analysis of laminated composite elliptic plates --- The tasks done by MACSYMA in this application were (a) Selecting approximation functions for each of the fundamental unknowns displacement amplitude with undetermined coefficients and developing analytic expressions for the specific strain and kinetic energies as quadratic functions of the undetermined coefficients.
(b) Differentiating specific strain and kinetic energies with respect to the undetermined coefficients symbolically.
(c) Evaluating stiffness and mass coefficients by performing integrations over volume.
2This has been done successfully, see the details in chapter three.
21 (d) Simplifying the expressions for the nonzerostiffnessand masscoefficients and developingFORTRANcode.
After stiffnessandmasscoefficients hadbeenevaluated, thevibrationfrequencies and modeshapes could beobtainednumericallyby usinganyscheme for generalized eigenvalue problems.
The lastpart of thepaperdiscussed theproblemswhich limited theapplicability of computerized symbolic manipulation.The major problemsmentionedin the paper were summarized asfollows.
were 1. Production of large expressions during the computation (intermediate expressions swell), 2. Slow speed of symbolic computation.
3. Low portability of large symbolic manipulation systems.
4. Need for analyst interaction during the symbolic computation.
5. Inability to estimate the storage requirements and CPU time for symbolic computations.
6. Problems associated with interface between algebraic and numerical calculations.
In addition, the authors suggested the directions of future research in this field.
They 1. Reduction of a general (tensor) formulation of structural mechanics problem to its computational level.
2. Hybrid computations.
3. Approximate symbolic integration of rational functions.
(5) Symbolic generation of finite element stiffness matrices As the title implies , the authors A. R. Korncoff (Boeing computer service, Seattle WA) and S. J. Fenves (Carnegie-Mellon University) used the symbolic processor MACSYMA to assist in the development of a software to generate the stiffness matrices for finite element analysis.
These included the construction of the strain-displacement matrix, calculation of the 22 determinantof theJacobian, andmultiplicationof relevantmatrices.The integrandwasthen integrated symbolicallyif it wasintegrable (e.g.theconstant straintriangleelement).
Otherwise it would beoutputasthefunctionof theproblemparameters for furthernumericalevaluation (e. g. four-node quadrilateralelement)3. In addition,the softwaregavetwo optionsin the materialpropertymatrix. Onewas"user-supplied" materialproperty.Theotherwas"library supplied"which providesone-dimensional elasticity,plainstress, plain strain,axisymmetric, and3-D linearisotropicelasticity Theexample of constructing theisoparametric formulation for constant straintrianglewasalsoshownin theappendix of thepaper.
(6)Symbolicmanipulation andcomputational fluid dynamics TheauthorsS.Steinberg andP.J.Roache employedthesymbolicandalgebraic manipulatorVAXIMA, a VAX versionof MACSYMA, to transformthephysicaldifferential equationand theboundary conditionsinto therectangular regionandthenconstructed theso calledstencilcoefficientmatrixfor afinite difference scheme.
Themajorideascamefrom the generalelliptic problems. In physicalcoordinates,the linear elliptic equation could be expressed as /1 /1 Lf -O=f + b, + cf + d OX_dX Ox (2.19) Where aij, bi, c, d were given and were the function of coordinates in general.The problem was to find a numerical approximation solution which satisfies equation (2.20) and the given boundary conditions.
Lf=0 (2.20) Since the physical domain is not regular in general, it is necessary to transform the physical coordinates into the rectangular, computational coordinates in which the finite difference scheme could be constructed easily.
This coordinate transformation involved the calculations of the Jacobian matrix, its determinant, and cofactors.
The equation in new coordinates would become I1 #I -, ,a,,&,&, .
ae, (2.21) 3 The integration (2.15) for a four-node isoparametrical quadrilateral element has been obtained exactly and will be discussed in detail in chapter three of this report.
23 Wherethetilde denoted thatthequantities werefunctions of computational coordinates.
After the transformationof equationand boundaryconditions had beendone, the centered differencemethod wasemployed to construct thefinitedifferencescheme asfollows" X C,.j.(et, eve3)g(e l+iAe I,Iq, Iqkl,,3 ee2 +j Ae z, e3 +k Ae3)= R(et,e ve 3) (2.22) Where the coefficient ci,j, k were called the stencil and was constructed by symbolic manipulation.
Taking advantage of the symmetric property, the number to be computed for ci,j, k could be dropped to 10 from 27. The resultant expressions of ci,j, k then could be coded in the FORTRAN language for the next numerical scheme.
2.3 Conclusion After making a survey of the publications on symbolic and algebraic manipulation, the following conclusions are drawn ' . None of the papers applying SAM to engineering problems tried to get closed-form solutions They kept traditional methodology by increasing the terms of the approximation function to get more accurate solutions.
This is due to the difficulty in solving generally partial differential equations or integral equations analytically.
. The papers discussing the application of symbolic and algebraic manipulation on the finite element analysis stop at the step of making a local stiffness matrix, local mass matrix, etc.
The same situation also occurred in finite difference analysis.
This was because (a) The finite element and finite difference methods are themselves approximation methods.
The accuracy of results depends on many factors, not just on round-off error or integration error which can be cured by symbolic and algebraic manipulation.
Although it was also one of the purposes to improve the accuracy of the solution, the major consideration in applying symbolic and algebraic manipulation was to help in the formulation of the tedious mathematical equations.
(b) In general engineering problems, the stiffness matrix in FEA and stencil coefficient in FDA are huge in dimension.
The limitation of memory space makes the execution of FEA's (or FDA) job impossible by symbolic and algebraic manipulation.
Therefore it is necessary to be finished by numerical analysis.
24 (c) Althoughmostof theSAM systems alsopossess thecapacityof numericalanalysis, theexecutionspeed of numericalanalysis in symbolicandalgebraicmanipulatoris slower in comparison to that in pure numerical analysis. The difference of efficienciesbetween themis remarkable whenthejob is big.Thereforeit is bestnot to haveit donecompletely in symbolicandalgebraic manipulation.
As thedocuments showed,nobodydid thewholeFEAor FDAjob in symbolicmodealone.
25 2.4 References A.
C.
Hearn,"REDUCE--A user oriented interactive system for algebraic simplification", interactive system for experimental applied mathematics (edited by M.
Klerer and J. Reinfelds), pp 79 -90, academic press, New York, 1968.
Joses Mose,"I'he fifth year---MACSYMA", ACM, vol.
8, No.
3, pp 105-110, Aug.
1974.
J. A. Van Huizen,"FORMAC today, or what can happen to an orphan", ACM, vol.
8, No.
1, pp 5-7, 1974.
B. F. Caviness, H.
I. Epstein,"A note on the complexity of algebraic differentiation", ACM, vol.
I 1, No.
3, pp 4-6, 1977.
Steven ,/. Harrington,"A symbolic limit evaluation program in REDUCE", ACM, vol.
13, No.
1, pp 27-31, 1979.
Edward W.
Ng,"Symbolic integration of a class of algebraic functions", ACM, vol.
8, No.
3, pp 99-102, 1974.
John Fitch,"A simple method of taking nth roots of integers", ACM, vol.
8, No.
4, 1974.
A.
C. Hearn,"Structure • The key to improve algebraic computation", symbolic and algebraic computation by computers (edited by N. Inada and T.
Soma), pp 215-230, 1985.
,/. P.
Fitch,"Implementing REDUCE on a MICRC-computer", computer algebra (edited by./.
A. Van Hulzen), pp 128-136, voi 162, 1983.
A.
C.
Hearn,"Scientific applications of symbolic computation", computer science and scientific compu.
(edited by ,/. M.
Ortega)., pp 83-108, academic press, New York, 1976.
A.
Deprit, ,/. henrard, A.
Rom,"Lunar ephemeris.
Delaunay's theory revisited", Science 168, pp 1569-1570, 1970.
26 G. H. Harrison,"Acompactmethodfor symboliccomputationof RiemannTensor", J.of Computational Physics, 4, pp594-600,1969.
J. A. Campbell, A. C. Hearn,"Symbolic analysis of Feynman Diagrams by computer,J.of Computational Physics,5, pp280-327,1970.
RudigerLoos,"Amthor'ssolutionof Archimedes'cattle problem", ACM, vol. 11, No.1, pp4-7, 1977.
K. V. Roberts,J. P. Boris,"The solutionof partial differential equationsusing a symbolicstyleof Algol", J.of Computational Physics, 8, pp83-105,1971.
W. A. Madson,L. B. Smith, N. J. Hoff, "Computeralgorithms for solving non-linearproblems",J.of SolidsStructures, Voi. 1, pp. 113-138,1965.
I. M. Levi,"Symbolicalgebraby computer-applications to structural mechanics", AIAA paperNo.71-363,pp 19-21, April 1971.
Dick J. Wilkins, Jr.,"Applicationsof a symbolicalgebramanipulationlanguage for compositestructures analysis, Computer & Structures, Vol. 3, pp801-807,1973.
A. K. Noor, C. M. Anderson,"Computerized symbolic manipulation in structural mechanics---progress and potential",Computer& Structures,Vol. 10,pp 95-118, 1979.
A. R. Korncoff, S. J. Fenves,"Symbolicgenerationof finite element stiffness matrices",Computer& Structure,Vol. 10,pp95-118,1979.
S. Steinberg, P. J. Roache,"Symbolicmanipulation and Computational fluid dynamics",J. of Computational Physics,57, pp251-284,1985.
S. Bandyopadhyay,J. S. Devitt,"The role of symbolic computation in the management of scientificinformation",EUROPCAL,pp460-461,1985.
P. S.Wang,"Implications of symboliccomputation for theteachingof mathematics", ACM, vol. 10,No. 3, pp 15-18,1976.
R. H. Lance,"Symbolic computation and the instruction of engineering undergraduate", ASME, HTD-vol. 105,AMD-vol. 97,pp 21-24, 1988.
27 Richard Pavelle," MACSYMA • capacities and applications to problems in engineeringandthesciences", SymbolicsInc.,Cambridge, MA, 1985.
28 CHAPTER III CAPABILITIES OF THE SYMBOLIC AND ALGEBRAIC MANIPULATORS 3.1 Introduction The capabilities of the symbolic and algebraic manipulator are quite system dependent.
Roughly speaking, a system which is designed for general purpose usage usually possesses the functions of differentiation, integration, matrix operation, polynomial manipulation, pattern match, variable substitution and equation solver.
Some systems, such as MACSYMA and MATHEMATICA, have a lot of built-in mathematical functions which allow the users to get the answers by just calling the appropriate command once.
Others, like REDUCE, may need users to write a short program to get the same answers.
This chapter will demonstrate the fundamental capabilities of the symbolic and algebraic manipulators which are available at hand by solving examples of applied mechanics.
Since REDUCE is the oldest system available at The University of Michigan, most examples will be demonstrated by using REDUCE.
Of course, MACSYMA will be employed to help the demonstration if it is necessary.
Unfortunately, REDUCE doesn't possess the graphic function, and the version of MACSYMA being used in The University of Michigan also doesn't include the graphics package, although it is available on the market.
Therefore, the postprocessing of the results from symbolic and algebraic manipulators will be done by other graphics packages.
3.2 What can the symbolic and algebraic manipulators do ?
In this section, some of the most useful operations in symbolic and algebraic manipulation are demonstrated in detail by examples.
They are differentiation, integration, matrix operation, algebraic equation solving, treatment of trigonometric function, differential equation solving, polynomial and rational operation, fortran code output, number system, substitution and built-in functions.
The strategies and particular techniques are also mentioned at the place where they are necessary.
29 3.2.1 Differentiation All differentiations, without exception, can be done analytically by REDUCE.
If it is necessary, REDUCE knows how to apply the chain rule to solve problems.
The powerful capabilities of this analytical differentiation will probably replace traditional numerical differentiation in many cases, such as the evaluations of the Jacobian and Hessian matrices However, the wrong results may be obtained by careless or naive users.
For instance, in finding the first derivative of x xx with respect to x. Two different solutions may be obtained as follows" 1: on time; Time: 134 ms 2: df(x(xx),x); X X 2 X +X(LOG(X) X + LOG(X)X + 1) ..................................................
X Time: 383 ms 3: df(xxx,x); 2 X X X(2LOG(X) + I) Time: 233 ms Here the first solution is correct.
How to judge the correctness of the results is one of the important tasks in symbolic manipulation.
A sound background knowledge in the SAM and problem-related fields is very helpful in checking them.
dx In some cases, the unevaluated differentiation form of functions, such as "k" is desired to be retained throughout the computation.
This also can be done as follows • 4: depend x,t; Time: 84 ms 5: depend y,t; Time: 83 ms 6: p:=axy; P := AXY Time: 150 ms 7: df(p,t); A(DF(X,T)Y + DF(Y,T)X) Time: 133 ms 30 3.2.2 Integration In REDUCE, all the integrations performed by the command INT are indefinite integrations with the integration constants discarded.
If the function is not integrable by REDUCE (the closed-form solution may exist theoretically), the original form will be displayed on the screen.
The definite integration may be obtained by further substituting the upper and lower limits into the results after the indefinite integration.
%A non-integrable case.
8: int(sqrt(a^2-x^2),x); 2 2 INT(SQRT(A - X ),X) Time: 950 ms %An integrable case.
9: int(1/(aa2+xa2),x); X ATAN(---) A A Time: 466 ms In most cases, the non-integrable integrand will become integrable after appropriate manipulation.
This pre-treatment involves the technique of changing the integrating variables in fundamental calculus.
Sometimes the intelligent users can substantially extend the capabilities of the symbolic and algebraic manipulator by suitably combining human intelligence with the tireless and errorless advantages of computer.
For example, if the x in command 8 is substituted by a'cos(t), then dx=-asin(t)dt and the integration of _ x2 with respect to x will become the integration of -a2sin2(t)dt with respect to t, which is integrable by REDUCE.
After the integration is done, the original variable x may be substituted back to get the desired expression in terms of x. The check may be done by skeptics by differentiating the resultant expression to get the original integrand.
The following three commands demonstrate these procedures.
10: int(-aa2(sin(t))^2,t); 2 A (COS(T)SIN(T) - T) ...............................
2 Time: 2134 ms 31 11: sub(cos(t)=x/a,sin(t)=sqrt(1-x2/a2),t=acos(x/a),ws); X 2 2 2 ACOS(---)A - SQRT(A - X )X A ..........................................
2 Time: 716 ms 12: df(ws,x); 2 2 A-X 2 2 SQRT(A - X ) Time: 450 ms 3.2.3 Matrix operation Matrix operation is one of the powerful capabilities of symbolic and algebraic manipulators.
These include the addition and multiplication of matrices, multiplication of matrices and scalars, inverting matrices, calculating the determinant of a square matrix, finding the trace, computing the eigenvalues and associated eigenvectors exactly if they are available and so forth.
3.2.3.1 Matrix multiplication The three body rigid rotation 1-2-3 in dynamics is a good example of the utility of matrix multiplication.
In robotics, it is necessary to find the analytical form of final orientation from which the rotation angles of each arm can be computed.
The final direction cosine matrix d is obtained from the product of three consecutive direction cosine matrices a, b, c.
%Declaring four matrices.
13: matrix a(3,3),b(3,3),c(3,3),d(3,3); %Inputting matrices.
14: a:=mat(( 1,0,0),(0,cos(q 1),-sin(q 1)),(0,sin(q 1),cos(q 1))); A(1,1) := 1 A(1,2) := 0 A(1,3) := 0 A(2,1) := 0 A(2,2) := COS(Q1) A(2,3) :=-SIN(Q1) A(3,1) := 0 A(3,2) := SIN(Q1) A(3,3) -= COS(Q1) 32 Time: 700ms 15:b:=mat((cos(q2),0,sin(q2)),(0,1,0),(-sin(q2),0,cos(q2)))$ Time:367ms 16:c:=mat((cos(q3),sin(q3),0),(-sin(q3),cos(q3),0),(0,0,1))$ Time: 383ms %multiplication of threematrices.
17:d:=abc; D(1,1):=COS(Q2)COS(Q3) D(1,2):=COS(Q2)SIN(Q3) D(1,3) :=SIN(Q2) D(2,l) :=-(COS(Q 1)SIN(Q3)-COS(Q3)SIN(Q 1)SIN(Q2)) D(2,2):=COS(Q 1)COS(Q3)+SIN(Q 1)SIN(Q2)SIN(Q3) D(2,3):=-COS(Q2)SIN(Q1) D(3,1)"=-(COS(Q 1)COS(Q3)SIN(Q2)+SIN(Q 1)SIN(Q3)) D(3,2) :=-(COS(Q 1)SIN(Q2)SIN(Q3)-COS(Q3)SIN(Q 1)) D(3,3):=COS(Q 1)COS(Q2) Time:617ms Theterminators $ incommand lines15and16prohibittheprintingof resultsandsave almosthalf of thetimecompared tocommand line 14whichuses theotherterminator.
3.2.3.2 Matrix inversion Unlikenumerical analysis in whichthetime-consuming operation of matrixinversionis to beavoided,to find theinverseof amatrixsymbolicallyis oneof thesignificantandsimple tasksin symbolicandalgebraicmanipulation.Oneimportantapplicationof it is in solvinga systemof linearequations.
For example, theproblemof finding acurveto fit thegivensetof databy theleastsquare methodresults in solvingasystem of linearequations.
Thecoefficient matrixhereis theHilbert matrixwhichisusuallyusedto investigate thephenomenon of round-off erroraccumulation.
REDUCEcansolvethisproblemexactly.Thenumericalsolutionand symbolicsolutionaretabulated in Table3.1,3.2, 3.3 for threedifferent Hilbert matrix sizes.
As thetables show,thenumerical solutionis notcapable of producing accurate resultsevenfor the caseof the77 Hilbert matrix. The deviationbetweenbothsolutionsis also plotted in Figure3.1.Thesignificance of symbolicandalgebraic manipulation isevident.
8: matrix h(40,40),x(40,1)$ 9: for i:=1:40do forj:=l:40 doh(i,j):=l/(i+j- 1)$ Time:30917ms 10:for i:=1:40dox(i,1):=i$ Time: 583ms 11:h:=(1/h); 33 H(1,1) :=1600 H(1,2) :=-1279200 H(1,3) :=340267200 H(1,4) :=-45113759600 H(1,5) :=3573009760320 H(1,6) :=-187.583012416800 H(40,39):=-1141149866470104951399125616277120066810096080000 H(40,40):= 58520505972825894943544903398826670092825441X) Time: 1355067 ms 12:x:=hx; X(1,1) := -64000 X(2,1) := 102272040 X(3,1) :=-40781023920 X(4,1) := 7204667408120 X(5,1) :=-712815447183840 X(6,1) := 44879235720719400 X(7,1) := - 1948531047013671120 X(8,1) := 61638279321584754360 X(9,1) := - 1478390066741437724160 X( 10,1) := 27706961874232024704320 X(11,1) := -415343205971360018352000 X(12,1) := 5073605405648309638180800 X(13,1) := -51267503668234803803526400 X( 14,1) := 433831946060133824442926400 X( 15,1) := -3105695134246771063279900800 X(33,1) X(34,1) X(35,1) X(36,1) X(37,1) X(38,1) X(39,1) := -275148980879869194392659362117120 := 129027970745724768036578024940720 := -49525830202172298308155472545440 := 15151287095628987186696245706000 := -3551734684918636715496119886400 := 598923394712817307441549441200 := -64662238360272798660726511200 X(40,1) := 3356375056654755429131776400 Time: 96000 ms 34 x Exact REDUCE solution x(1) 125 Gauss elimination solution 124.7899166408321 x(2) -2880 -2875.997324887610 x(3) 14490 14472.49323423709 x(4) -24640 -24613.29019994230 x(5) 13230 13216.83350607823 Table 3.1 • Comparison of solution for 55 Hilbert matrix x Exact REDUCE solution x(1) -216 x(2) 7350 Gauss elimination solution -204.4675087167038 7027.565254454758 x(3) -57120 -54968.63675793860 x(4) 166320 160780.3224850843 x(5) -201600 - 195532.2818417542 x(6) 85932 83555.641 56991533 Table 3.2 • Comparison of solution for 66 Hilbert matrix x Exact REDUCE solution Gauss elimination solution x(1) x(2) x(3) 343 131.3201346851223 - 16128 -7941.23464123 5595 177660 100320.8278414759 x(4) -772800 -4761 54.4030660979 x (5) 1559250 1020735.514691367 x(6) - 1463616 - 1001760.776649569 x(7) 516516 365761.52976973 52 Table 3.3 • Comparison of solution for 77 Hilbert matrix 35 L m t# L @ 10 27 .
lo2s 107--3: 1021" lo19: lO17 -1015 -1013. -1011 10 9 10 7 10 5 1(13 101 0 ' 3'o ' 10 40 rank ¢_I-Ii]_rt matrix {r) 5O Figure 3. I :Error between the solutions of REDUCE and Gauss elimination 3.3 Eigenvalues and Eigenvectors To find the eigenvalues and eigenvectors for a matrix is another important task in the application of symbolic and algebraic manipulation.
Since to solve the eigenvalue problems analytically for an arbitrary dimensional matrix is theoretically impossible, the following examples only show the exact solutions for a 3 by 3 matrix.
21: matrix s(3,3)$ 22: s:=mat((sxx,sxy,sxz),(sxy,syy,syz),(sxz,sxy,szz))$ 23: mateigen(s,eta); 3 2 2 2 2 {{ETA-ETA SXX-ETA SYY-ETA SZZ+ETASXXSYY+ETASXXSZZ-ETASXY 2 -ETASXYSYZ-ETASXZ +ETASYY SZZ+SXX SXY SYZ-SXXSY ySZZ 2 2 2 -SXY SXZ+SXY SZZ-SXYSXZSYZ+SXZ SYY, 36 ARBCOMPLEX(1)(ETASXZ+SXYSYZ-SXZSyy) MAT(I,1):= ...........................................................................
2 2 ETA -ETASXX-ETASYY+SXXSYY-SXY ARBCOMPLEX(1)(ETASYZ-SXXSYZ+SXYSXZ) MAT(2,1):=...........................................................................
2 2 ETA -ETASX.X-ETASY Y+SXXSYY-SXY MAT(3,1):=ARBCOMPLEX(1)}} Time: 1283ms 24: s:-mat((5,1,0),(1,2,4),(0,4,3))$ Time:284ms 25: mateigen(s,eta); 3 2 {{ETA -10ETA +14ETA+53, , 4ARBCOMPLEX(2) MAT(l,1) := ..............................
2 ETA - 7ETA + 9 4ARBCOMPLEX(2)(ETA - 5) MAT(2,1) := ........................................
2 ETA - 7ETA + 9 MAT(3,1) := ARBCOMPLEX(2)}} Time: 666 ms 26: trace(s); SXX + SYY + SZZ Time: 100 ms As the command lines 23 and 25 show, the solutions from MATEIGEN contain three parts.
They are (a) Characteristic equation.
(b) The number of repetition roots.
It's one in the above examples.
(c) Eigenvectors.
REDUCE by calling 37 If the eigenvaluesare required, the characteristic equation needs to be solved in addition.
The equation solver command SOLVE which will be demonstrated later can meet this requirement.
The ARBCOMPLEX(1) which appeared in the eigenvectors is referred to as "arbitrary complex constant".
Examples 24 and 25 also show results from REDUCE by substituting numbers into matrix S. The trace of the matrix is also evaluated by the command TRACE as shown in example 26.
3.2.4 Equation solver REDUCE can solve the polynomial equation up to order three exactly.
If the solution includes the imaginary part, the "I" will show up to represent the imaginary symbol.
The following example solves the characteristic equation obtained above.
27:solve(ETA_'3 - 10ETA2+ 14ETA+53=0,eta); 2/3 {ETA=-((63SQRT(229) 1-691SQRT(3)) SQRT(3) I+(63" SQRT(229)I-691, 2/3 1/3 1/3 1/6 2/3 SQRT(3)) -20(63SQRT(229)I-691_SQRT(3)) 2 '3 -582 SQRT(3) 1/3 2/3 1/3 1/3 1/3 1/6 • 3 I+58"2 "3 )/(6(63SQRT(229)I - 691SQRT(3)) 2 "3 2/3 ETA=((63SQRT(229)I - 691SQRT(3)) SQRT(3)I - (63SQRT(229)I - 691" 2/3 1/3 1/3 1/6 2/3 SQRT(3)) + 20(63SQRT(229)I-691SQRT(3)) 2 3 -582 SQRT(3) 1/3 2/3 1/3 1/3 1/3 1/6 3 I-58'2 3 )/(6(63SQRT(229)I-691SQRT(3)) 2 3 ), 2/3 1/3 ETA=((63SQRT(229)I - 691SQRT(3)) + 10(63SQRT(229)I - 691SQRT(3)) I/3 1/6 2/3 2 3 + 582 Time: 5717 ms 1/3 1/3 1/3 1/6 3 )/(3(63SQRT(229)I - 691SQRT(3)) 2 3 If the numerical mode NUMVAL, complex switch COMPLEX and FLOAT mode are turned on, the numerical solution of three eigenvalues can be obtained in sixteen digits precision by default.
The imaginary parts in the following example are very small and are due to the round-off errors.
28:solve(ETA3 - 10ETA2 + 14ETA + 53=0,eta); {ETA=4.830750950611553d0 + 2.991124223331416d-7"I, 38 ETA=-(1.6167630306368408d0 + 6.960060167347475d-8"I), ETA=6.786003556862447d0 + (-2.295118206596669d-7) I} Time: 1567ms REDUCEcansolvesystems of linearalgebraicequations exactly.The limitation is determined only by thememorycapacity of thehardware system.
Thefollowing examplefinds theminimization of aquadratic function Q=klyln(2}+k2yl y2+k3y2h{2]+k4y2y3+kSy3_2]_k6,y3 subject to yl +y2=2 29: Q: =kl y 1 2+k2" y 1y2+k3 y2 "2+k4" y2 y3+k5 y3 2-k6' y3+y4 (y 1+y2-2) $ Time: 550 ms 30: a:=df(q,yl); A := 2KIY1 + K2Y2+ Y4 Time: 167 ms 31: b:=df(q,y2); B := K2Y 1 + 2K3Y2 + K4Y3 + Y4 Time: 167 ms 32: c:=df(q,y3); C := K4Y2 + 2K5Y3 - K6 Time: 166 ms 33: d:--df(q,y4); D:=YI+Y2-2 Time: 150 ms 34: soive({a=0,b=0,c=0,d=0},{y 1,y2,y3,y4}); 2 4K2K5 - 8K3K5 + 2'K4 - K4K6 {{YI= .....................................................
, 2 4KlK5 - 4K2K5 + 4K3K5 - K4 8KlK5 - 4K2K5 - K4K6 Y2= ....................................................
, 2 4KlK5 - 4K2K5 + 4K3K5 - K4 2 2 16K lK3K5-4K l'K4 +2KlK4 K6-4' K2 K5-K2K4 K6 Y4= .........................................................................................
}} 2 4KlK5 - 4K2K5 + 4K.3K5 - K4 39 Time: 1400 ms After the substitution of yl to y4 into the quadratic function, the minimum is found as follows • 35: q:=q; 2 2 2 Q:-( 16K 1K.3K5-4K l'K4 +4K 1K4K6 -K 1K6 -4"K2 K5-2K2K4K6+ 2 2 2 K2K6 -K3K6 )/(4KlK5-4K2_K3K5-K4 ) Time: 267 ms 3.2.$ Treatment of the trigonometric function REDUCE doesn't even know an equation as simple as sin2(q)+cos2(q)=l.
However REDUCE does possess the potential to learn it.
Due to this powerful capability, the trigonometric functions can be handled easily by just teaching REDUCE the operation rules.
For example, without teaching the operation rule of trigonometry, the determinant Of direction cosine matrix d in command 17 is as follows : 36: det(d); 2 2 2 2 2 2 2 COS(Q1) COS(Q2) COS(Q3) +COS(Q1) COS(Q2) SIN(Q3) +COS(Q1) 2 2 2 2 9 2 2 COS(Q3) SIN(Q2) +COS(Q1) SIN(Q2) SIN(Q3)'+COS(Q2) COS(Q3) 2 2 2 2 2 2 2 SIN(Q1) +COS(Q2) SIN(Q1) SIN(Q3) +COS(Q3) SIN(Q1) SIN(Q2) + 2 2 2 SIN(Q1) SIN(Q2) SIN(Q3) Time: 483 ms After teaching REDUCE the appropriate operation rules, the solution becomes quite simple.
Note that the time consumption in command 38 is longer than that in command 36.
This is due to the extra work needed for simplification.
It also reveals the phenomenon of internal swells.
37: let cos(ql)2+sin(ql)2=l,cos(q2)2+sin(q2)2=l,cos(q3)2+sin(q3)2=l.
Time: 650 ms 38: det(d); 1 Time: 584 ms 4O 3.2.6 Solving differential equation REDUCE is unable to solve the differential equation directly, while MACSYMA does possess this capability.
The following example is a problem of beam deflection w(x) under uniform load q [Figure 3.2].
The differential equation is of the form dx 2 - sw + rx (x - L) (3. 1) Where s and r ,in general, are the function of Young's modules, moment of inertial as well as boundary conditions.
In the case of small deflection with simple supported on both end, the s becomes zero, and r=q/2EL W q X Figure 3.2 • Beam under uniform load.
The boundaries are not specified to find general solution.
The (Cn) in the following examples is the MACSYMA prompt for inputting the command and (Dn) is the solution given by MACSYMA.
(C 1) depends(w,x); (DI) [W(X)] (C2) diff(w,x,2)-sw-rx(x-l)=0; 2 dW (D2) - RX(X-L) + ..........
S W =0 2 dX (C3) Is S P; ode2(d2,w,x); positive, negative, or zero?
41 SQRT(S)X -SQRT(S) X (D0)W = %K1%E + %K2 %E 2 RSX-LRS X+2R 2 S Where the %K1 , %K2 are constants to be determined by boundary conditions.
The %E here is the symbol of exponential function.
Although the next example (no specific physical problem associated with it) is more complex, it only takes 3.6 milliseconds in MACSYMA.
(C4) depends(y,x); (134) W(X)I (C5) diff(y,x,2)+(2/x)diff(y,x)-(2/x2)y-( 1/x2)sin(log(x))=0; (135) dY 2 2---d Y dX 2Y SIN(LOG(X)) 2 X 2 2 dX X X (C6) ode2(d2,y,x); (D6) 3 SIN(LOG(X)) + COS(LOG(X)) %K2 Y= ..........................................
+%K1X+ .......
10 2 X (C7) time(d3); Time: (I37) [3.6d0] 3.2.7 Polynomial and rational operations Polynomial and rational operations is one of the most important and useful functions in symbolic and algebraic manipulation.
There are two occasions in employing these functions.
First, in most cases the problems to be solved are not as simple as the above demonstrations.
Therefore it is necessary to manipulate the formulae into a machine manageable forms before calling the appropriate REDUCE commands to solve them.
Second, sometimes the solutions are restricted to specific forms for particular usage.
In order to get the appropriate forms, there is no way to avoid employing this package.
The capabilities of this package include the controls of the expansion, factorization, and cancellation of common factors, determining the GCD of two polynomials, obtaining the part of polynomial and rational functions, and so forth.
42 %Turningoff expansion switch.
39:off exp; %Inputtingp andq polynomials.
Thenumerical common factors %will beautomatically factored out.
40: p:=(x-1)(5x-3)2(4x+8)^3(9x-6); 2 3 P := 192"(5"X - 3) (3X - 2)(X + 2) (X - 1) 41: q:=(x-1)(5x-3)(4x+8)(x+4); Q :=4"(5'X - 3)(X + 4)(X + 2)(X - 1) %Gettinggreatest commondividerof pandq.
42: gcd(p,q); 4'(5"X - 3)(X + 2)(X - 1) %Turningonexpansion switchandcheckingp,q.
43:onexp; 44:p; 7 6 5 4 3 2 192"(75"X +235'X -163"X -723'X +392'X +664'X -624"X+144) 45:q; 4 4"(5'X + 22"X 3 2 - 5X - 46"X + 24) %Definingafractionr.
46:r.=p/q; 7 6 5 4 3 2 R :=(48'(75"X +235"X -163"X -723'X +392"X +664'X -624"X+144))/ 4 3 2 (5X +22"X -5X -46"X+24) %Turning on the greatest common divider switch and rechecking r.
%The common factors have been cancelled as shown in command 47.
47: on gcd; 48: r; 4 3 2 48"(15"X + 41"X - IOX - 52'X + 24) ....................................................
X+4 %Getting the denominator and numerator of fraction r.
49: den(r); X+4 43 50: num(r); 4 3 2 48"(15"X + 41X - lOX - 52"X + 24) %Gettingtheleadingdegree of numerator of fractionr.
51"deg(num(r),x); 4 %Thenumerator of r in command 49alsocanbefactored asthemultiplications of each factors.
52:on ifactor; 53:factorize(hum(r)); {2,2,2,2,3,X + 2,X + 2,3"X - 2,5"X - 3} 3.2.8 Fortran code output REDUCE can automatically produce the fortran expression, natural style expression (default), and REDUCE code.
The fortran code can be made as a subroutine and be directly input to the fortran main program.
The natural style expression allows it to be looked as hand-written form, while REDUCE code is useful in making a REDUCE subroutine for input into the REDUCE main program.
Since the results from REDUCE are generally very lengthy, the functions of code-conversion make the switch from symbolic and algebraic manipulation to numerical analysis smoother.
This not only saves effort in symbolic and algebraic manipulation, but also rules out all the possibilities of error introduced by hand typing.
The following examples will give a clearer understanding about these functions.
%Inputting the polynomial.
The output forms are in natural style of human being writing.
54: p:=(a+b-c)7; 7 6 6 52 5 52 43 42 P := A +7A B-7A C+21A B -42'A BC+21A C +35"A B -105A B C+105" 4 2 43 34 33 322 3 3 34 A BC -35"A C +35"A B -140A B C +210A B C -140A BC +35"A C 2 5 24 23 2 2 2 3 2 4 2 5 +21A B -105A B C +210A B C -210A B C +105A BC -21A C +7 6 5 4 2 3 3 2 4 5 6 AB -42AB C+105AB C -140AB C +105AB C -42ABC +7AC 7 6 5 2 4 3 3 4 25 67 +B -7B C+21B C -35"B C +35"B C -21B C +7BC -C Time: 1434 ms %Turning on the fortran-code conversion switch.
55: on fort; 44 %Checking the output of fortran code.
56: p; ANS=A7+7.A6B-7.A6C+21.A5B2.42.A t5B •C+ 21.A5C2+35.A4B3-105.A4B2C+ 105.
A4BC2-35.A4C3+35.A3B4-140.A3B '3"C+2 lO.A3B2C2-140.A3BC3+35.A3 C'4+21.A2B5-105.A2B4C+210.A2 B3C '2-210.A2B2C3+ 105.A2BC4-21.A2C 5+7.AB6-42.AB5C+ 105.AB4C2-140.AB '3"C'3+ 105.AB2C4-42.ABC5+7.AC6+B 7-7.B'6"C+21.B5C2-35.B4C3+35.B3C 4-21.B2C5+7.BC6-C7 Time: 850 ms %Changing the number of continuation line in fortran code.
57: cardno!:=10$ 58: p; ANS 1=7.AB6-42.AB5C+ 105.AB4C2-140.AB • '3"C'3+ 105.AB2C4-42.ABC5+7.AC6+B • 7-7.B'6"C+21.B5C2-35.B4C3+35.B3C • 4-21.B2C5+7.BC6-C7 ANS=A7+7.A6B-7.A6C+21.A5B2-42.A5B • C+21.A5C2+35.A4 B3-105.A4B2C+ 105.
• A4B C'2-35.A'4" C3+35.A3B4-140.A3B • '3"C+210.A3B2C2-140.A3B C.3+35.A.3.
• C'4+21.A2B5-105.A2B4C+210.A2B3C • '2-210.A2B2C3+ 105.A2BC4-21.A2C • 5+ANS1 Time: 1034 ms %Turning off fortran-code conversion switch.
59: off fort; %Turning off the 'natural style' function switch• 60: off nat; %Checking the output of REDUCE code.
61: p:=p; P := A7+7A6B-7A 6C+21A5B2-42A5BC+2 lA5C2+35A4B3-105' A4B2C+105A4BC2-35A4C3+35A3B4-140A 3B3C+210A3B2C2-140A3BC3+35A3C4+ 2 lA2B5-105A2B4C+210A2B3C2-2 IOA2B '2'C'3+ 105A2BC4-2 lA2C5+7AB6-42AB5C + 105AB4C2-140AB3C3+ 105AB2C4-42ABC5 +7AC6+B7-7B6C+21 B5C2-35B4C3+35B3C '4-21" B2C5+7BC6-C75 Time: 816 ms 45 3.2.9 Number system In addition to symbolic manipulation, the symbolic and algebraic manipulators can also do numerical analysis.
There are three ways to treat numbers in REDUCE.
They are (a) Integer --- In general, there is no practical limit on the number of digits.
For example, the value of 2 l°°° gives 255 digits.
It only takes 383 milliseconds.
62: A:=2 1000; A: = 107150860718626732094842504_ 181056140481170553360744375038837 03 5105112493612249319837881569512759467291755314682518714528569231 40435984577598574803934567774824230985421 0746050623711418771821530 464749835819412673 987675591655439460770629145711 Time: 383 ms (b) Fraction number ---Numbers that aren't integers and operated with symbols (or numbers) are represented by default by the quotient of two integers with message(s) telling users the conversion.
63: a:=0.999bc; 0.999 represented by 999/1000 999"BC A :--.............
I000 Time: 200 ms 64: a:=0.999"0.5; 0.999 represented by 999/1000 0.5 represented by 1/2 999 A := ........
2000 Time: 217 ms (c)Real number ---It is also possible to ask REDUCE to work the floating point approximations to numbers with arbitrary precision with specified numbers of digit.
%Turning on the numerical mode.
65: on numval; %Turning on the floating system switch.
66: on float; 67: pi; 3.141592653 589793d0 Time: 150 ms 68: on bigfloat; Domain mode FLOAT changed to BIGFLOAT 46 % Specified50digits.
69: precision 505 70:pi; 3.1415926535897932384626433832795028841971693993751 Time: 217ms 3.2.10 Substitution Therearetwo substitution functionsin REDUCE.Oneis for localsubstitution, andthe otheris for globalsubstitution.
Thedifference canberevealed in the followingexamples.
%Definingafunctionf.
71:f:=6.A2R I2U2-3.ATAN(U 1/A)A2 RIR2U 1-9.ATAN(UI/A)A2RIR2U2+3.ATAN(UI/A)A 2R22U I+3.ATAN(U I/A)A2R22U2-2.ATAN(U 1/ A)RI2U I3+6.ATAN(U I/A)RI2U 1"'2'U25 Time: 1234ms %makinga local substitution and calling it as B.
72: b:=sub(atan(u 1/a)=k,f); 2 2 2 2 2 B :=-(3A KRIR2UI+9A KRIR2U2-3A K'R2 U1-3A K 2 2 2 2 3 2 2 R2 U2-6A R1 U2 + 2KZR1 U1 - 6K'R1 U1 U2) Time: 433 ms %Checking the original function f after the local substitution.
It's unchanged.
73: f; U1 2 U1 2 U1 2 2 -(3ATAN( .... )A RIR2UI+9ATAN( .... )A RIR2U2-3ATAN( .... )A R2 U1 A A A U1 2 2 U1 2 3 U1 2 2 -3ATAN( .... )A R2 U2+2ATAN( .... )R1 UI - 6ATAN( .... )R1 U1 U2 A A A 2 2 - 6A R1 U2) Time: 367 ms %Making global substitution.
74: let atan(ul/a)=g; Time: 133 ms %The original function f has been changed.
75: f; 47 2 2 2 2 2 2 -(3A GRIR2UI+9A GRIR2U2-3A G'R2 U1-3A G'R2 U2 2 2 2 3 2 2 -6A R1 U2 + 2G'R1 U1 -6G'R1 UI U2) Time: 283ms 3.2.11 Built-in functions The built-in functions are quite system dependent.
Since REDUCE is designed for general purpose usage, there are not many built-in functions.
However they can be obtained by suitably combined commands of REDUCE.
On the contrary, MACSYMA has many built-in functions which allow users to simply call commands once to get the solution.
Some of these MACSYMA functions are shown in the following paragraphs.
(a) Limit evaluation ---If it is necessary, the function LIMIT in MACSYMA will automatically apply L'Hospital's rule to evaluate the formulae.
(C8)limit(sin(x)/x,x,0,plus); (D8) 1 (C9) time(d8); Time: (I39) [2. 116d0] (C 10) limit((6x2+3x-4)/(x-1),x, l,plus); (D10) INF (C 11) limit((1-x)(1/x),x,0); -1 (Dll) %E (C 12) time(dl 1); Time: (DI2) [2.75d0] (b) Laplace transformation --- The LAPLACE command in MACSYMA can transform the functions in physical domain, such as EXP, LOG, SIN, COS, SINH, COSH, DELTA and ERF, into the s domain.
In addition, it also can transform a differential equation into algebraic equation.
The command for inverse of Laplace transform is also available.
(C 13) laplace( l/sqrt(t),t,s); SQRT(%PI) (D13) ...............
SQRT(S) 48 (C14)time(dl3); Time: (D14) [0.05d0] (DIS) (C 15)laplace(((c-b)exp(a t)+(a-c)exp(b t)+(b-c)exp(c t))/((a-b) (b-c)(c-a)),t,s); B-C A-C C-B ........
+ ........
+ .......
S-C S-B S-A ........................................
(A-B) (B - C) (C-A) (C 16) time(d9); Time: (D16) [0.384d01 (C 17) laplace(sin(at)-atcos(at),t,s); (D17) 2 A 2S 1 ..........
A( .......................
) 2 2 2 22 2 2 S +A (S +A) S +A (C 18) time(dlT); Time: (D18) [0.233d0] %Inverting the Laplace transform.
(C 19) ilt((s+6)/(s^2+4s+ 12),s,t); -2T (D19) %E (C20) time(d4); Time: (D20) 2 SIN(2 SQRT(2) T) (.........................
+ COS(2 SQRT(2) T)) SQRT(2) [0.933d01 %Transforming a differential equation into algebraic equation.
(C21) laplace(diff(y(x),x,2)-3diff(y(x),x)+2y(x)=0,x,s); d v (D21) ....
(Y(X)) - 3 (S LAPLACE(Y(X), X, S) - Y(0)) dX I !X=0 2 +S LAPLACE(Y(X),X,S)+2 LAPLACE(Y(X),X,S)-Y(0) S=0 (C22) time(d21); Time: 49 (D22) [(3.15dOl (C23) laplace(diff(y(x),x,2)+w^2 y(x)-bsin(wx)=0,x,s); !
d !
2 2 (D23) ....
(Y(X)) ! +W LAPLACE(Y(X),X,S)+S LAPLACE(Y(X),X,S) dX t X=0 BW ..........
Y(0) S - 0 2 2 W+S (C24) Time: (D24) time(d23); [0.217d0] (c) Series expansion --- The MACSYMA version in University of Michigan provides the Taylor series and power series expansion capabilities.
Although the function of Fourier series expansion is available on the market, it is not available here.
(C25) taylor(%e^x,[x,0,7]); (D25)/T/ 2 3 4 5 6 7 X X X X X X I+X+--+---+---+ ....
+ ......
+ .......
+...
2 6 24 120 720 5040 (C26) time(d25); Time: (D26) [0.567d0] (C27) powerseries(%e^x,x,0); INF =--= I1 \ X (D27) > .....
/ Il!
I1-0 (C28) time(d27); Time: (D29) [0.45d0] 5O 3.3 References A.
C. Hearn,"REDUCE users' manual", version 3.3, The RAND Co., Santa Monica, CA 90406-2138, July 1987.
"MACSYMA reference manual", The mathlab group laboratory for computer science, MIT, version 10, January 1983.
Stephen Wolfram,"Mathematica-A system for doing mathematics by computer", Addison-Wesley Publishing Compan, Inc., 1988.
Peter V.
O'Neil,"Advanced Engineering mathematics", Wadsworth Publishing Company, 1983.
Richard L.
Burden, J.
Douglas Faires,"Numerical Analysis", Prindle, Weber & Schmidt publishers, Third edition, 1985.
Gilbert Strang,"Introduction to applied mathematics", wellesley-Cambridge press, 1986.
Thomas R. Kane, Peter W.
Likins and David A.
Levinson,"Spacecraft dynamics", MacGraw-Hill book Company, 1983.
Richard Pavelle," MACSYMA • capacities and applications to problems in engineering and the sciences", Symbolics Inc., Cambridge, MA, 1985.
AI Shenk,"Calculus and Analytic Geometry", Scott, Foresman and Company, 1984.
Murry R.
Spiegel,"mathematical Handbook of formulas and tables", Schaum's outline series in mathematics, McGraw-Hill Book Company, 1968.
51 CHAPTER IV APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO AUTOMATIC PROBLEM FORMULATION 4.1 Introduction One of the most impcx_t advantages gained from symbolic and algebraic manipulation is to automatically formulate lengthy mathematical equations without making any errors.
Modern scientists and engineers are continually being challenged with more and more complicated formulas.
With the aid of symbolic and algebraic manipulators, most problems can be treated easily and correctly.
This chapter will demonstrate six automatic formulation examples done by symbolic and algebraic manipulation.
They are : 1. The derivation of equations of motion in dynamics.
2. Tensor formulation for the shell problem.
3. The approximation to a function by Fourier series.
4. The formulation template for the iteration method in nonlinear numerical analysis.
5.
Finite element stiffness matrix and mass matrix construction for 6-node triangular element in a heat transfer problem.
6. Finite element stiffness matrix construction for 4--node isoparametrically quadrilateral element in a plane elasticity problem.
Of course, the use of symbolic and algebraic manipulators as tools to automatically formulate mathematical equations can be extended to any fields.
Although the methodologies are dependent on the problem to be solved, the basic commands used in programming are similar.
52 4.2 Derivation of equation of motion 4.2.1 Introduction by SAM The derivation of equations of motion by the Lagrange method for the system shown in Figure 4.1 involves finding kinetic energy T, potential energy V, and therefore the Lagrangian L.
M, i Figure 4.1 • Dynamic system for demonstration of symbolic and algebraic manipulation The Lagrangian then is partially differentiated with respect to both generalized coordinates and the rate of generalized coordinates.
The equations of motion will be obtained after taking the time derivatives to appropriate terms and assembling the necessary terms.
Mathematically, the Lagrange equations are expressed in the form of where L is the Lagrangian and is defined as the difference between kinetic energy T and potential energy V.
I .
I 1 .2 T -"iM2(x + tO) 2 + _M(rO) 2 + _IO (4.2) V --Mtgr(l_cosO)_M2g(r_rcosO +xsin 0)+ tk(rO)2 (4.3) 4.2.2 REDUCE program and solution 1: on time; Time: 83 ms 53 %declaring theta and x as a function of time.
2: depend theta, time; Time: 67 ms 3: depend x,time; Time: 50 ms %Calculating the total kinetic energy of system.
%The velocity v of body 2 need be evaluated later.
4: re:=( 1/2)i0(df(theta,time))2+( I/2)m lr2(df(theta, time))2+(I/2)m2,v2; 2 2 2 2 DF'(THETA,TIME) I0 + DF(THETA,TIME) R M1 + M2V TE := ...............................................................................
2 Time: 500 ms %Calculating the position vector of body 2.
5: P:=(rcos(theta)-xsin(theta))j+(rsin(theta)+xcos(theta)),i; P := COS(THETA)_'IX+COS(THETA),J,R+SIN(THETA),I,R.SIN(THETA),.I,X Time: 333 ms %The velocity vectors is then obtained by taking the denvative of %the position vector with respect to time.
6: dp:=df(p,time); DP := COSfTHErA)DF'(THETA,TIME)IR-COS(THETA), DF(THETA, TIME)JX+COS(THE A)DF(X,TIME-')I-DF(THETA ,TIME) SIN(THETA) I X-DF'(THET A,TIME)StN(THET A ) 2R.DF(X,TIME) SIN(THETA)J Time: 234 ms %Getting the magnitude square of velocity of body 2.
7: v 2:=lcof(dp, i)2+tcof(di:Q)2; 2 2 2 2 2 2 V :=COS(THETA) DF(THETA,TIME) R +COS(THETA) DF(THE_A,TIME) 2 2 2 X +2COS(THETA) DF(THETA,TIME)DF(X,TIME)R+CC_(T'I-tETA) • 2 2 2 2 2 DF(X,TIME) +DF(THETA,TIME) SIN(THETA) _R +DF(THETA,TIME) 2 2 2 S1N(THETA) X +2DF(THEI'A,TIME)DF(X,T1ME)SIN(THETA) R+ 2 2 DF(X,TIME) SIN(THE--TA) Time: 650 ms %Teaching REDUCE the tri'gonometric rule for simplification of %formulae.
8: let (cos(them)) 2+(sin(theta))2= 1; Time: 150 ms 54 %Inputtingthe potential energy of system.
9: ve:=-m 1 g'r( 1-cos(theta))-m2 g(r ( 1-cos(theta))+xsin(theta))+( 1/2)k (r theta)2; VE := (2COS(THETA)GRMI+2COS(THETA)GRM2-2SIN(THETA) 2 2 • GM2X-2GRM1 - 2GRM2 + KR THETA )/2 Time: 483 ms %calculating the Lagrangian.
10: la:=te-ve; LA := -(2COS(THETA)GRMI+2COS(THETA)GRM2-DF(THETA, 2 2 2 2 2 TIME) I0 - DF(THETA,TIME) R M1 - DF(THETA,TIME) R 2 2 M2 - DF(THEI'A,TIME) M2X -2DF(THETA,TIME)DF(X,TIME) 2 R'M2 - DF(X,TIME) M2 -2SIN(THETA)GM2X-2GRM1-2 2 2GRM2+KR THETA )/2 Time: 617 ms %Deriving the equation of motion for theta coordinate without %any simplification.
11: e l:=df(df(la, df(theta, time)),time)-df(la,theta); E1 := -(COS(THETA)GM2X+DF(THETA,THETA,TIME)DF(THETA, 2 TIME) I0+DF(THETA,THETA,TIME)DF(THETA,TIME)R M 1 2 +DF(THE'FA,THETA,TIME)DF(THETA,TIME)R M2 +DF(THETA, 2 THETA,TI ME) DF(THETA,TIME) M2 X +DF(THETA,THETA,TI ME) 2 DF(X,TIME) R M2-DF(THETA,TIME,2)I0-DF(THETA,TIME,2)R 2 2 M 1-DF(THETA,TIME,2)R M2-DF(THETA,TIME,2)M2X -2 Time: DF(THETA,TIME) DF(X,TI ME) M2 X- DF(X,TI ME,2) R M2 + 2 SIN(THETA) GRMI+SIN(THETA)GRM2-KR THETA) 783 ms %Deriving the equation of motion for x coordinate without simplification.
12: e2:=df(df(la, df(x,time)),time)-df(la,x)+f; 2 E2 := DF(THETA,TIME,2)RM2-DF(THETA,TIME) M2X-DF(THETA, TIME) DF(X,TIME,X)RM2-DF(X,TIME,X)DF(X,TIME)M2 + DF(X,TIME,2)M2 - SIN(THEq'A)GM2+F Time: 483 ms %Teaching REDUCE the 2nd derivative of theta w/t theta and time.is null, and so does x.
55 13:let df(theta_time_theta)--__df(x_x_time)--__df(theta_theta_time)---__df(x_time_x)--; Time:350 ms %turningoff theautomatic expansion switch.
14:off exp; Time:50ms % %Startingpolynomialmanipulation to simplifytheequations of motion.
% %Gettingthecoefficientof angular acceleration term.
15:a1:=lcof(e 1,df(theta,ti me,2)); 2 2 A 1 := (M1 + M2)R + I0 + M2X Time: 617 ms %Getting the coefficient of sin(theta).
16: a2:=lcof(e 1,sin(theta)); A2 := - (M1 + M2)GR Time: 550 ms %Getting the leftover after taking off the above two terms.
17: a3:=e 1-a ldf(theta, time,2)-a2sin(theta); A3 :=-(COS(THETA)G M2X-2DF(THETA,TIME)DF(X,TIME)M2X 2 -DF(X,TIME,2)R M2-KR THETA) Time: 533 ms %Rearranging the equation of motion for theta coordinate.
%The results are simpler than those of in command 11.
18: e 1:=aldf(theta,time,2)+a2sin(theta)+a3; 2 2 E1 := ((MI+M2)R +I0+M2X )DF(THETA,TIME,2)-(COS(THETA)GM2 2 X-2 DF(THETA,TIME)DF(X,TIME)M2X-DF(X,TIME,2) R M2.KR THETA)-(M I+M2)SIN(THETA) G R Time: 400 ms %Getting the coefficient of x acceleration term.
19: c 1:=lcof(e2,df(x,time,2)); C1 := M2 Time: 350 ms %Getting the coefficient of angular acceleration term.
20: c2:=lcof(e2,df(theta, time,2)); C2 := R'M2 Time: 333 ms %Getting the leftover terms by removing the above two terms.
21: c3:=e2-c ldf(x,time,2)-c2df(theta, time,2); 2 C3 := - (DF(THE/'A,TIME) M2X + SIN(THETA)GM2 - F) 56 Time:350ms %equation of motionfor x coordinate.
22:e2:=cldf(x,time,2)+c2df(theta,time,2)+c3; 2 E2 :=-((DF(THETA,TIME) M2X+SIN(THETA)GM2-F)-DF(THETA,TIME,2)RM2 - DF(X,TIME,2)M2) Time:317ms 23:bye; The examplesshownaboveare thedemonstrationof obtaining the left handside formulaeof equation(4.1).Theequations of motionof systemcanbesimply doneby setting the resultsof command18and22 equalto zero.As the systemis complex, the analytical derivationsof equations of motion by handwill become tediousandproneto error. For the cases of complexsystems, The applicationof symbolicandalgebraicmanipulationwill be moresignificant.
4.3 Automatic tensor formulation for shell problem 4.3.1 Preliminary formulation Tensors are convenient mathematical entities for concisely describing physical situations, that are independent of coordinate transformations.
Although the benefits gained by employing tensor notation are quite significant in the related fields, the expressions of tensor formula often rise to errors.
Fortunately, this difficulty can be avoided by the use of symbolic and algebraic manipulation.
This advantage will be demonstrated by formulating the thin shell problem in tensor form and using SAM to expand the resulting tensor equations.
For the sake of convenience, the Latin indices will be referred for the range 1, 2, 3 and the Greek indices are in the range of 1, 2 in the following paragraph without special stress.
On the formulation of the thin shell problem, the only necessary inputs are three parametric equations fi(u 1,u 2) of the shell middle surface.
Based on parametnc equations, the covariant metric tensor aab and its determinant are calculated fl...
3___ (4.4) aqn "/ix, Ill J ,,. 3u a 3u _ a = det (a) (4.5) The contravariant metric tensor is therefore given by 57 3 X =Z fi 2 ,1 u =const.
2 X X Figure 4.2" Pictorically vector notations of shell a ' - cofactor (a,)/a (4.6) The first and second kind of Christoffel symbols for the surface are defined as follows respectively" [afl,y ] - a,, .p + aa, .a -a_., ) -+a -a ) fly -ia ( a .r ,_ pr ,6 (4.7) (4.8) The covariant differentiations of a covariant vector (the 1st order tensor) and the 2nd order tensor are therefore calculated A p .,, " D ,_A Ij " A ¢ .a - { fl Ya } A r f 'ta { '}a A or ' ' " " A or a -fl a _r y ct t (4.9) (4. 10) When the above formulations are done, the curvature tensor and its determinant are then computed by the formulae of 58 d_ - X f_ (4.11) d -det (d_) (4.12) Where the X i are the unit normal vectors of the shell middle surface [see Figure 4.2] and are equal to X ' -a-l12e j k ,jkf.,f.2 (4.13) Where eij k is the generalized Kronecker delta.
The strain tensor is defined as half of the difference between the deformed metric tensor and undeformed metric tensor.
l , E_ -2(a_ - a_) (4. 14) Where the asterisk superscript is denoted as deformed state.
The aal 3 is defined as a' = a +p, +p +prapar + qaqa (4.15) The generalized two-dimensional displacement gradientp, and rotation q are represented as p, =Dav a -d_w (4. 16) qa - d,_v a + w.a (4.17) Where the vct are the in-plane displacements and w is the out of plane displacement of the middle surface of the shell.
Similarly the bending tensor is equal to the difference between the deformed curvature tensor and undeformed curvature tensor.
K¢ -d - d_ Z The deformed curvature tensor d ctl3 is expressed as a I12_.i r d' ,,. (-a'7) [( +p; + O/a)(d + Daq,, + dtj par) -(qP + e_e r_ qr P_)(D a Pap - d tjpqa)] Where O =det(Pa_ The constitutive equations for isotropically elastic and thin shell are (4.18) 59 N _ , 1---[(1 -v)E "# + wt'E] Eh 3 Y M -12(1-v:) [(1-v)K_ + va'Kr] (4.19) (4.20) E co Where the Young's modules E and strain tensor shouldn't be confused here.The final equilibrium equations are then DaN ¢ + 2dr DoM + M D,,dr + -0 D,,D pM _ -d,_dY#M '¢ -deN '¢ -P - 0 (4.21) (4.22) Where the F # and P are the external forces applied in the in-plane and out-plane directions, respectively.
4.3.2 REDUCE program and resultant expressions The REDUCE program shown below is based on the above formulation methodology.
The explanations of the program are also included to facilitate an understanding where it is necessary.
The resultant expressions are too huge to be included here and are available in reference .
ARRAY X(3),C 1(2,2,2),C2(2,2,2); OPERATOR U,V,W,F; MATRIX A(2,2),CONTRA(2,2),D(2,2); % INPUTTING SURFACE PARAMETRIC FUNCTIONS % X(1):=U(1); X(2):=U(2); X(3):=CONSTANT; % % CALCULATING COVARIANT METRIC TENSOR & ITS DETERMINANT % FOR M:=l:2 DO FOR N:=l:2 DO A(M,N): =FOR I:= 1:3 SUM DF(X(I),U(M))DF(X(I),U(N)); DETA:=DET(A); DEPEND W,U(1),U(2); FOR I:=1:2 DO DEPEND V(I),U(1),U(2); %CALCULATING CONTRAVARIANT METRIC TENSOR COMPONENT 60 FORL:=l:2 DO FORM:=l:2 DO IF L=I andM=I THEN CONTRA(L,M):=A(2,2)/DETA ELSEIF L NEQM THEN CONTRA(L,M):=-A(M,L)/DEI"A ELSECONTRA(L,M):=A(1,1)/DETA; % %CALCULATING THE 1ST CHRISTOLFFEL SYMBOL % FOR L:=l:2 DO FOR M:=l:2 DO FOR N:=l:2 DO C I(L,M,N):=(1/2)(DF(A(L,N),U(M))+DF(A(M,N),U(L))DF(A(L,M),U(N))); %CALCUI,ATING THE 2ND CHRISTOLFFEL SYMBOL (SEE EQ.
4.6) FOR L:=l:2 DO FOR M:=1:2 DO FOR N:=l:2 DO C2(L,M,N):=FOR I:=1:2 SUM A(L,I)CI(M,N,I); % %SUBROUTINE FOR CALCULATING THE COVARIANT DERI VATI VE %FOR THE 1ST ORDER TENSOR % PROCEDURE COVD(L, VAR(M)); DF(VAR(M),U(L))-(FOR I:=1:2 SUM C2(I,L,M)VAR(I)); %SUBROUTINE FOR CALCULATINE THE COVARIANT DERIVATIVE %FOR THE 2ND ORDER TENSOR PROCEDURE COVD2(L,FUN(M,N)); DF(FUN(M,N),U(L))-(FOR I:=1:2 SUM C2(I,L,M)FUN(I,N)) -(FOR I:=1:2 SUM C2(I,L,N)FUN(M,I)); MATRIX E(2,2),K(2,2),P(2,2),FF(2,3),XX( 1,3); % %CALCUL,ATING REL.ATIVE ALTERNATE TENSOR %ALSO CALLED AS GENERALIZED KRONECKER DELTA % PROCEDURE EE(I,J,K); IF I=J OR J=K OR K=I THEN 0 ELSE IF (I=l AND J=2 AND K=3) OR (I=3 AND J=l AND K=2) OR (I=2 AND J=3 AND K=I) THEN 1 ELSE - 1; %CALCULATING TIlE CURVATURE TENSOR & ITS DETERMINANT % 61 FORI:=1:2DO FORJ:=1:3DO FF(I,J):=DF(X(J),U(I)); FORI:=1:3DO XX(1,I):=( 1/SQRT(DETA))(FOR J:=l:3 SUM (FORK:=1:3SUM EE(I,J,K)FF(1,J)FF(2,K))); FORI:=1:2 DO FOR 3:=1:2 DO D(I,J):=FOR 11:=1:3 SUM XX(1,I 1)DF(FF(I,I 1),U(J)); DETD:=DET(D); ======_ %CALCULATING THE GENERIZED DISPLACEMENT TENSOR % FOR L:=l:2 DO FOR M:=I:2 DO P(L,M):=DF(V(M),U(L))-(FOR I:=1:2 SUM C2(I,L,M)V(I))-D(L,M)W; % %CALCULATING THE ROTATION TENSOR (SEE EQ. 4.17) % ARRAY Q(2); FOR M:=l:2 DO Q(M):=DF(W,U(M))+(FOR I:=1:2 SUM D(M,I)(FOR J:=l:2 SUM CONTRA(I,J)V(J))); i_---='====." %DERIVING THE STRAIN TENSOR O' ==== = --.,, FOR I:=1:2 DO FOR J:=l:2 DO E(I ,J): =( 1/2) (P(I,J)+P(J,I) +(FOR I1:=1:2 SUM (FOR Jl:=l:2 SUM CONTRA(I 1,J1)P(I,J1)) P(J,I 1))+Q(I)Q(J)); OFF PERIOD; ON FORT; OFF PERIOD; FOR I:=1:2 DO FOR J:=1:2 DO WRITE " E(",I,",",J,")=",E,(I,J); % %CALCULATING THE ABSOLUTE 2-D ALTERNATE TENSOR % PROCEDURE EPS(L,M); IF L=I AND M=2 THEN SQRT(DETA) ELSE IF I_,=2 AND M=I THEN -SQRT(DETA) ELSE 0; % %CALCULATE CONTRAVARIANT ABSOLUTE 2-D ALTERNATE TENSOR % PROCEDURE CEPS(L,M); IF L=I AND M=2 THEN I/SQRT(DETA) ELSE IF L=2 AND M=I THEN -I/SQRT(DETA) 62 ELSEO; % %CALCULATING THE BENDINGTENSOR O_=-'===========-FOR I:=1:2 DO FOR J:=l:2 DO <-.+ a cost_) + b_sin (---if-) (4.23) n-I where the Fourier coefficients are evaluated as follows 65 1 L a 0-fL f(x)dx 1 L a"" (' I L b ,, -_f_L f (x )sin (-----)dx (4.24) (4.25) (4.26) 4.4.2 REDUCE program for generating Fourier series The REDUCE program to generate the Fourier series and output a fortran subroutine for hat function is shown in the follow assuming L=I.0.
This program can be used to generate Fourier series for an arbitrary function by simply changing the input function.
% % Inputting the given function and informations.
% M : number of piecewise bounded interval.
% f(M): the function in the Mth interval.
% c(n) : the upper and lower limit of finite integration.
% 1 : half length of interval.
% k : number of Fourier series terms needed.
O_ -'-----------,, ._ M:=2; K:=50; ARRAY F(M),C(M+I); f(1):=l+x; %c(1) =< x =< c(2) f(2):=l-x; %c(2) =< x < c(3) c( 1):=- 1;c(2):=0;c(3): = 1 ; 1:=(c(m+ 1)-c( 1))/2; % % Obtaining the Fourier series coefficients % a0:=for i:=l:m sum (sub(x=c(i+l),int(f(i),x))-sub(x=c(i),int(f(i),x)))/l; an:=for i:=l:m sum (sub(x=c(i+ 1),int(f(i)cos(npi x/l),x)) -sub(x=c(i),int(f(i)cos(npix/i),x)))/l; bn:=for i:=l:m sum (sub(x=c(i+ 1),int(f(i) sin(npix/l),x)) -sub(x=c(i),int(f(i)sin(npi rdl),x)))/l; on rat; on div; % Generating the Fourier series.
66 fs:=a0/2+fori:=l:k sum sub(n=i,an)cos(ipix/l)+sub(n=i,bn)sin(ipix/l); off echo; onfort; cardno!:=10; % % Outputtingthefortransubroutine of Fourierseries.
% out "fourier.ftn"; subroutine fourier(x,fs)"; implicit real8(a-h,o-z)"; pi=3.141592654"; write " write " write " fs:=fs; write " return"; write " end"; shut"fourier.ftn"; bye; 4.4.3 Resultant fortran subroutine from REDUCE The following results are produced automatically from the above REDUCE program for fifty terms Fourier series case.
This subroutine can be directly input into fortran main program without any troubles• subroutine fourier(x,fs) implicit real8(a-h,o-z) pi=3.141592654 ANS 1=4./625. COS(25.PIX) PI(-2)+4./529.COS(23. . PI X) PI (-2)+4./441.
COS(21.
PI X) Pl (-2)+4./ • 361.COS(19.PIX)PI(-2)+4./289.COS(17.PIX)PI • (-2)+4./225.COS(15.PIX)PI(-2)+4./169.COS( 13.PIX)PI(-2)+4./121.COS( 11. PIX)PI(-2)+4./ . 81.COS(9.PIX)PI(-2)+4./49.COS(7.PIX)PI(-2 • )+4./25.COS(5. PIX) PI(-2)+4./9. COS(3. PI X) PI • (-2)+1./2• FS=4.COS(PIX)PI(-2)+4./2401.COS(49. PIX) PI( . -2)+4./2209.COS(47.PIX)PI(-2)+4./2025.COS(45.
. PIX)PI(-2)+4./1849.COS(43.PIX)PI(-2)+4./ 1681 .COS(41.PIX)PI(-2)+4./1521.COS(39. PIX) • PI(-2)+4./1369.COS(37.PIX)PI(-2)+4./1225. • COS(35.PIX) PI(-2)+4./1089.COS(33.PIX)PI( • -2)+4./961.COS(31.PIX)PI(-2)+4./841.COS(29. • PIX) PI(-2)+4./729.COS(27.PIX)PI(-2)+ANS 1 return end 67 Fourcurvesare plotted in the same figure for three-term, five-term, seven-term and fifty-term cases.
As the figure shows, the hat function can be actually simula_d by Fourier series.
When fifty terms are used, the difference between hat function and its Fourier series is just invisible although they are in different spaces.
This is one of the advantages of the application of symbolic and algelzraie manipulation.
O°O I.o o.8 o o o._l 1.0 Figure 4.3 : Convergence of Fourier series approxlmatlorl 4.$ Template for nonlinear numerical analysis 4.5.1 Introduction In nonlinear numerical analysis, it is necessary to evaluate the Jacobian matrix and to solve the system of equations at each iteration.
Symbolically, the Newton-Raphson iteration can be expressed as {X }<k,.
{X }'-t'-[J]-{F} (4.27) Since the evaluation of the inverse of the Jacobian matrix is quite time-consuming, equation (4.27) is traditionally changed into the following form and then is solved as a linear system of equations at each iteration stage.
[s ]{AX}''--{F} (4.28) Although the avoidance in evaluation of the inverse of the Jacobian does expedite the execution, it still takes time to solve a system of equations, especially for the ease of a large number of unknowns.
With the help of symbolic and algebraic manipulation, the efficiency can be improved further.
68 Instead of transforming equation (4.27)into equation (4.28), it is rearranged into {X }("-{X }(-"-{AX }('--[J ]-{F} .(4.29) The application of symbolic and algebraic manipulation to this problem is to make a template form of the righ_handlside of equation (4.29).
This includes the symbolic evaluation of the Jacobian matrix, its inverse, and the multiplication of the matrix by load vector.
The results are then converted into fortran code for numerical analysis.
The iterations are done by simply substituting the current solutions into the template formulae toget the residuals.
This substitution is much faster than solving the system of equations.
This is a new way to improve the efficiency of program execution.
4.5.2 Preliminary formulation The above idea is demonstrated by solving the Fermat-Weber location problem for the case of a two-dimensional Euclidean space.
Given n fixed points ,(xi, yi) , i=l,..,n, in the plane, the object of the Fermat-Weber problem is to find the best location to minimize the total length of the distances among the optimal location and each point.
Mathematically, it is ): 2 Minimize (x - x, + ( Y - Y ,) (4.30) i-I or rewritten in the following form n _l -ArY Minimize , , t-I I1: (4.31) where (x,y) is the coordinate of the optimal location to be found, and (x,) C i -y, A,-[10 Y .
(xy) 0] (4.32) (4.33) (4.34) Intuitively, the minimization will be accomplished by differentiating equation (4.31) with respect to Y and setting the results to zero.
But since the objective function in equation (4.31) is not differentiable at the exact (xi, Yi) points, a smoothing parameter e is introduced into the object function to avoid the difficulty.
69 The perturbedobjective function now becomes differentiable everywhereand is strictly convex.Then ,n, "---" = 0 will resultin afixedpointiteration y(k.u [_w<) A A r]- _wlk)A,c = , ,, ( ,) , k - 0, 1, 2 ....
(4.36) Where the weight wi(k) is defined as (k) 1 w' " ] t : U: " (4.37, ,-A Y +e 2 4.5.3 REDUCE program Consider the specific problem which is to find the optimal location among three points (0,1), (0,-1) and (x,0) where x varies in the range 0, oo). The REDUCE program is based on equation (4.36).
% waa • WAA in equation (4.36) % wac • WAC in equation (4.36) matrix waa(2,2),wac(2,1); waa( 1,1):= 1/sqrt(r 1)+ 1/sqrt(r2)+ 1/sqrt(r3); waa(2,2): =waa( 1,1); wac(1,1):=xl/sqrt(rl)+x2/sqrt(r'2)+x3/sqrt(r3); wac(2,1):=y 1/sqrt(r 1)+y2/sqrt(r2)+y3/sqrt(r3); wac:=( 1/waa)wac; on fort; off echo; off period; cardno!:=10; out "cssa.ftn"; write " subroutine cssa(xl,yl,x2,y2,x3,y3,pu,x0,y0)"; wnte" implicit real8(a-h,o-z)"; write" r l=(x-x l)2+(y-y 1)2+pu2"; wnte" r2=(x-x2)2+(y-y2)2+pu2"; wnte" r3=(x-x3)2+(y-y3)2+pu2"; wnte" x=",wac(1,1); write " y=",wac(2,1); write " retum", write " end"; shut "cssa.ftn"; bye; 70 4.5.4 Fortran subroutine from REDUCE The fortran subroutine produced here is applicable when the number of points is three.
The coordinates of three points are arbitrary.
The perturbation parameter is an arbitrarily small number except zero.
The resultant optimal locations are plotted in Figure 4.4.
with respect to variable x3.
The execution time for this problem on Apollo workstation Domain 4000 is too small to be measured.
The difference in execution time will be more significant when the size of the problem is increased.
subroutine cssa(x 1,y 1,x2,y2,x3,y3,pu,x,y) implicit real8(a.h,o-z) r l=(x-x 1)2+(y-y 1)2+pu2 r2=(x-x2)2+(y_y2)2+pu2 r3=(x-x3)2+(y-y3)2+pu2 x=(SQRT(R2)SQRT(RI),R3,X I+SQRT(R2)SQRT(R1),R3,X2+ • SQRT(R3)SQRT(R1)R2X 1+SQRT(R3)SQRT(R 1),R2,X3+SQRT • (R3)SQRT(R2)R 1 X2+SQRT(R3)SQRT(R.2)RIX3+R1,R2,X3 • +R lR3X2+R2 R3X 1)/(2SQRT(R2)SQRT(R 1),R3+2,SQRT( • R3)SQRT(R 1)R2+2SQRT(R3)SQRT(R2)R I+R 1,R2+R 1,R3+ • R2R3) Y=(SQRT(R2)SQRT(R1)R3Y 1+SQRT(R2)SQRT(R 1) R3Y2+ • SQRT(R3)SQRT(R1)R2Y I+SQRT(R3)SQRT(R1)R2Y3+S (R3)SQRT(R2)R1 Y2+SQRT(R3)SQRT(R2)R 1Y3+R1 R2QyR3 T • +RIR.3Y2+R2R3Y 1)/(2SQRT(R2)SQRT(R1)R3+2SQRT( • R3)SQRT(R 1) R2+2SQRT(R3)SQRT(R2)R I+R1,R2+R1,R3+ • R2R3) return end I.O | 0.8 O.II 0,4 ' 0.2 0.0 0 o.'= o.', o.'6 ,.'8 IS leeatllelb Figure 4.4 : Behavior of optimal location vs. variable x3 71 4.6 Triangular stiffness and mass matrix construction 4.6.1 Preliminary formulation It is well known that the finite element method has become a powerful tool in solving general engineering problems.
Usually the finite element method is applied in three steps: 1. Domain discretization (pre-processing).
2.
Constructing the stiffness, mass matrices and load vectors etc., then prescribing the boundary conditions and solving them.
3. Results treatment (post-processing).
While computing stiffness and mass matrices, partial differentiation, matrix multiplication, matrix inversion and integration are required.
For a higher order interpolation, the formulation is always very tedious and prone to introduce errors.
With the aid of symbolic and algebraic manipulation, all these troubles can be alleviated.
The example shown in this section will demonstrate the application of symbolic and algebraic manipulation to the automatic construction of stiffness and mass matrix of six-node triangular element for a heat transfer problem.
The stiffness K and mass M matrices are defined as follows assuming unit thickness.
K - fffr • D BdA (4.38) M -fN r N p cpdA (4.39) where the shape functions are N4" 4si s2' N s -4 s, s3' N6" 4$3 $1' 1 t m i N l -s t N 2 -s 2 - -(N4+Ns)" N3-s 3 (4.40) The B matrix is evaluated by the formulae of 72 n T m ON l ON l aN t as I as z as 3 aN 2 ON 2 aN 2 as I as 2 as 3 aN 3 aN 3 aN 3 as--V as--: 7723 ON 4 ON 4 ON 4 aS I aS 2 aS 3 aN s ON 5 aN s as I as 2 as 3 ON 6 ON 6 ON 6 OsI Os2 Os3 bl cl ] b2 c 2 b 3 C 3 The material property matrix D is assumed to be symmetric The b i, ci here are expressed by global nodal coordinates.
1 1 b,-J--(y, -Y3), b2 " ")-'(Y 3-Y,), b3" 7 "(y, -Y2) 1 1 1 c,--y-(x3 -x2), c 2- 7-(x,-x3), c_- 7-(x,-x,) The Jacobian J is equal to (4.41) (4.42) (4.43) J " (xl - x3)(Y2 -Y3) -(Yl -Y3)(x 2 -x3) " 2" area 4.6.2 REDUCE program for stiffness matrix MATRIX NS(6,3),NS 1(2,6),NS2(2,6),NS3(2,6),BB(2,6); ARRAY N(6),B(3),C(3); ..................................
%Inputting the shape functions ..................................
N(4):=4'S 1"$2; N(5):=4"$2"$3; N(6):--4"$3'S1; N( 1):=S 1-( 1/2) (N(6)+N(4)); N(2):=$2-( I/2) (N(4)+N(5)); N(3):=$3-( 1/2)(N(5)+N(6)); (4.44) 73 %C.alculate the Jacobian, area and b(i),c(i) defined in Eq.(4.43)& (4.44) JAC:=(X l-X3) (Y2-Y3)-(X2-X3) (y 1 -Y3); AREA:---JAC/2; B( 1):=(Y2-Y3)/JAC; B(2):=(Y3-Y 1)/JAC; C( 1):=(X3-X2)/JAC; C(2): =(X 1-X3)/JAC; B(3):=-B(1)-B(2); C(3):=-C(1)-C(2); %Calculating B matrix defined in Eq. (4.41).
FOR M:=l:2 DO << FOR I:=1:6 DO << NS(I, 1):=DF(N(I),S 1) ;NS(I,2):=DF(N(1),S2) ;NS(I,3):=DF(N(I),S3); IF M=I THEN BB(M,I):=FOR J:=l:3 SUM (B(J)NS(I,J)) ELSE BB(M,I):=FOR J:=l:3 SUM (C(J)NS(I,J))>>>>; FOR I:=1:2 DO <<FOR J:=l:6 DO <>>>; %Given the symmetric material matrix.
MATRIX D(2,2); D: =MAT((K 11 ,K 12),(K 12,K22)); MATRIX LO(6,6),NU(6,6),CC(6,6),SKE,(6,6); O'_ ...................................
%Obtaining the stiffness matrix.
SKE:=(AREA/3)(TP(NS1)D, NS I+TP(NS2)DNS2+TP(NS3),D,NS3)$ %Making the appropriate substitution to simplify the final expression.
COB:=-(X3Y2-X3Y 1-X2Y3+X2Y I+X 1Y3-XIY2); FOR I:=1:6 DO FOR J:=l:6 DO <>; %Outputting the resultant fortran subroutine.
ON FORT; OFF ECHO; OFF PERIOD; OUT "sym"; WRITE " subroutine stiff(xl,y 1,x2,y2,x3,y3,dl 1,d12,d22,ske)"; WRITE" implicit real8(a-h,o-z)"; WRITE" dimension ske(6,6)"; WRITE" DJ=-(X3Y2-X3Y 1-X2Y3+X2Y I+X IY3-X l,Y2),; FOR I:=1:6 DO FOR J:=l:6 DO IF J>=I THEN WRITE "SKE(",I,",",J,")=",SKF_.(I,.I) ELSE WRITE "SKE( ",I,",",J,")=SKE(.,J,,,,.,I ,")"; 74 WRITE " return"; WRITE" end"; SHUT "sym"; OFFFORT; BYE; 4.6.3 Resultant stiffness matrix made by REDUCE subroutine stiff(x 1,y 1,x2,y2,x3,y3,d 11 ,d 12,d22,ske) implicit real8(a-h,o-z) dimension ske(6,6) DJ=-(X3Y2-X3Y 1-X2Y3+X2Y I+X IY3-X l'Y2) SKE( 1,1)=(D 11 Y22-2D 11Y2Y3+D11 Y32-2D 12"X2' Y2 • +2D12X2Y3+2D12X3Y2-2D12X3Y3+D22X22_2D22 • X2X3+D22X32)/(2DJ) SKE( 1,2)=(D11Y IY2-D 11Y IY3-D11Y2Y3+D I lY32-D12 . X IY2+D12XIY3-D12X2Y I+D12X2Y3+D12X3Y l+D12 • X3Y2-2D 12X3Y3+D22X 1X2-D22X 1X3-D22X2X3+D22 • X32)/(6DJ) SKE(1,3)=-(D11Y IY2-D11Y IY3-D1 lY22+D1 lY2Y3-. D12XIY2+D12XIY3-D12X2Y l+2D12X2Y2-D12X2Y3+ • D 12X3Y 1-D 12X3Y2+D22X lX2-D22X lX3-D22X22+ • D22X2X3)/(6DJ) SKE(1,4)=-(2(D11Y IY2-D11Y IY3-D11Y2Y3+D11.Y3..2 • -D12XIY2+D 12X IY3-D12X2Y I+D12X2Y3+D12X3Y I+ . D 12X3Y2-2D 12X3Y3+D22X lX2-D22XlX3-D22X2X3+ • D22X32))/(3DJ) SKE(1,5)=0 SKE( 1,6)=(2"(D 11' Y 1Y2-D 11Y IY3-D 11 Y22+D 11' Y2Y3-• D12XIY2+D12XIY3-D12X2Y l+2D12X2Y2-D12X2Y3+ • D12X.3Y 1-D12X3Y2+D22XlX2-D22XlX3-D22X22+ • D22X2X3))/(3DJ) SKE(2,1)=SKE(1,2) SKE(2,2)=(D 11Y 1"'2-2"D 11Y IY3+D1 lY32-2D12XlY 1 • +2D 12X 1 Y3+2D 12X3Y 1-2' D 12'X3' Y3+D22X 1'2-2" D22 • X 1X3+D22X32)/(2DJ) SKE(2,3)=(D 11Y l2-D 11Y IY2-D11Y IY3+D1 lY2Y3-2 • D 12X 1Y 1+D 12X 1Y2+D 12 X 1Y3+D 12 X2 Y 1-D 12X2Y3+ • D 12X3Y 1-D 12"X3" Y2+D22X 12-D22X 1 X2-D22 X 1X3+ • D22X2X3)/(6DJ) SKE(2,4)=-(2(D 11Y IY2-D11Y IY3-D11Y2Y3+D 11"Y3"'2 • -D 12X IY2+D 12X IY3-D 12X2Y I+D 12X2Y3+D 12X3Y 1+ • D12X3Y2-2D12X3Y3+D22.X lX2-D22XlX3-D22X2.X3+ • D22X32))/(3DJ) SKE(2,5)=-(2(D 11 Y 12-D 1 IY 1Y2-D 11 Y I Y3+D 11 Y2Y3 • -2D12XIY I+D12XIY2+D12XIY3+D12X2Y 1-D12X2Y3 • +D 12X3Y 1-D 12X3Y2+D22X 12-D22X 1X2-D22X 1 X3+ • D22X2X3))/(3DJ) SKE(2,6)=0 SKE(3,1)=SKE(I,3) SKE(3,2)=SKE(2,3) SKE(3,3)=(D 11Y 1 2-2'D 11 Y l Y2+D 11 Y22-2D 12 X 1 Y 1 • +2D 12X 1Y2+2D12X2Y 1-2' D12X2Y2+D22X 1'2-2" D22 75 • X lX2+D22X22)/(2DJ) SKE(3,4)=0 SKE(3,5)=-(2" (D 11Y 12-D 11Y IY2-D 11Y 1Y3+D 11 Y2Y3 • -2D12XIY I+D12XIY2+D12XIY3+D12X2Y 1-D12X2Y3 • +D12X3Y 1-D12X3Y2+D22Xl2-D22X lX2-D22X1,X3+ • D22X2X.3))/(3DJ) SKE(3,6)=(2(D 11 Y 1Y2-D 11 Y 1Y3-D 11 Y22+D 1 lY2Y3-• D12XIY2+DI2X IY3-D12X2Y l+2D12X2Y2-DI2,X2,Y3+ • D 12X3Y 1- D 12X3Y 2+D22X 1X2-D22X 1 X3-D22X22+ • D22X2X3))/(3DJ) SKE(4,1)=SKE( 1,4) SKE(4,2)=SKE(2,4) SKE(4,3)=SKE(3,4) SKE(4,4)=(4(D 11Y l2-D 11Y IY2-D 11Y IY3+D11"Y2"'2-. DI IY2Y3+D11Y32-2D12XIYI+D12XIY2+D12XI,Y3+ . D 12X2Y 1-2D 12X2Y2+D12X2Y3+D12X3Y l+DI2X3Y2-• 2D 12"X3" Y3+D22X l2-D22X 1X2-D22X 1X3+D22X22-• D22X2X3+D22X32))/(3DJ) SKE(4,5)=(4(D1 IY IY2-D11Y IY3-D 1lY22+D 1 lY2Y3-• D12XIY2+D12XIY3-D12X2Y I+2D12X2Y2-DI2X2Y3+ • D12X3Y 1-D12X3Y2+D22XlX2-D22XlX3-D22,X22+ • D22X2X3))/(3DJ) SKE(4,6)=-(4(D 11Y l2-D11Y IY2-D11Y IY3+D1 lY2Y3 • -2D 12X IY 1+D 12X 1 Y2+D 12X 1Y3+D 12X2Y 1-D 12X2Y3 • +D 12"X3" Y 1-D 12X3Y2+D22X 12-D22X 1X2-D22X l'X3+ • D22X2X3))/(3DJ) SKE(5,1)=SKE(1,5) SKE(5,2)=SKE(2,5) SKE(5,3)=SKE(3,5) SKE(5,4)=SKE(4,5) SKE(5,5)=(4(D11Y l2-D11Y IY2-D11Y IY3+D11"Y2"'2-• DI IY2Y3+D1 lY32-2D12XlY I+D12XIY2+D12XIY3+ • D 12X2Y I-2D 12X2Y2+D 12X2Y3+D 12X3Y l+DI2X3Y2-• 2 D 12X3Y3+D22X 1" 2-D22X 1X2-D22X 1X3+D22X22-• D22X2X3+D22X32))/(3DJ) SKE(5,6)=-(4(D 11Y IY2-D11Y IY3-DI IY2Y3+D11"Y3"'2 • -D 12X IY2+D 12X IY3-D 12X2Y I+D 12X2Y3+D 12X3Y 1+ • D 12X3Y2-2D 12X3Y3+D22X lX2-D22XlX3.D22,X2,X3+ • D22X32))/(3DJ) SKE(6, I)=SKE(1,6) SKE(6,2)=SKE(2,6) SKE(6,3)=SKE(3,6) SKE(6,4)=SKE(4,6) SKE(6,5)=SKE(5,6) SKE(6,6)=(4(D11Y l2-DI IY IY2-D11Y IY3+DI 1"Y2"'2-• D11Y2Y3+D11Y32-2D12XlY I+D12XIY2+DI2XIY3+ • D 12X2Y 1-2D 12X2Y2+D 12X2Y3+D 12X3Y l+DI2X3_y2-• 2D 12X3Y3+D22X l2-D22X lX2-D22X lX3+D22,X22.
• D22X2X3 +D22" X32))/(3DJ) return end 76 4.6.4 REDUCE program for mass matrix MATRIX N( 1,6),SQN(6,6),NS(6,6),DMASS(6,6),SDIF(6,6); N:=MAT((N1 ,N2,N3,N4,N5,N6)); % Calculating the integmnd.
SQN:=TP(N)N; % Inputting shape functions N4:=4"S 1"$2; N5: =4 $2" $3; N6: =4" $3" S 1; N I:=S 1-(1/2)(N6+N4); N2:=S2-(1/2)(N4+N5); N3:=$3-( 1/2)(N5+N6); % Evaluating the integration.
LET $3=1-S1-$2; FOR I:=1:6 DO FOR 3:=1:6 DO IF I<=.I THEN << A: =I NT(SQN(1,3),$2);B:=S UB(S2= 1-S 1,A)- SUB (S2=0,A); C:=I NT(B,S 1);NS(I,J):=SUB(S 1= 1 ,C)-S UB(S 1=0,C)>> ELSE NS(I,J):=NS(J,I); % Outputting the results.
ON FORT; OFF ECHO; OFF PERI OD; OUT "mass.ftn"; WRITE " SUBROUTINE MASS(X 1,Y 1,X2,Y2,X3,Y3,RO,CP, DMASS)"; WRITE" IMPLICIT REAL8(A-H,O-Z)"; WRITE" DIMENSION DMASS(6,6)"; WRITE" AREA=-(X3Y2-X3Y 1-X2Y3+X2Y 1+X IY3-X 1,Y2)/2,; DMASS: =ROCP 2 AREA NS; WRITE " RETURN"; WRITE" END"; SHUT "mass.ftn"; % checking the correctness of results.
..........................................
OFF FORT; OUT "CHECK"; R:=FOR I:=1:6 SUM <>; SDIF:=DMASS-TP(DMASS); SHUT "CHECK"; BYE; 4.6.5 Fortran results of mass matrix SUBROUTINE MASS(X 1,Y 1,X2,Y2,X3,Y3,RO,CP, DMASS) IMPLICIT REAL8(A-H,O-Z) 77 DIMENSION DMASS(6,6) AREA=-(X3Y2-X3Y 1-X2Y3+X2Y l+X 1Y3-X 1Y2)/2 DMASS(1,1)=(AREA CPRO)/30 DMASS(1,2)=-(AREACPRO)/180 DMASS(1,3)=-(AREA CPRO)/180 DMASS( 1,4)=0 DMASS(1,:5)=-(ARF_CP RO)/45 DMASS( 1,6)=0 DMASS(2,1)=-(AREACPRO)/180 DMASS(2,2)=(AREACP RO)/30 DMASS(2,3)=-(AREACPRO)/180 DMASS(2,4)=0 DMASS(2,5)=0 DMASS(2,6)=-(AREACP RO)/45 DMASS(3,1)=-(AREA CPRO)/180 DMASS(3,2)=-(AREACPRO)/180 DMASS(3,3)=(AREA CPRO)/30 DMASS(3,4)=-(AREACPRO)/45 DMASS(3,5)=0 DMASS(3,6)=0 DMASS(4,1)=0 DMASS(4,2)=0 DMASS(4,3)=-(AREACPRO)/45 DMASS(4,4)=(8AREACPRO)/45 DMASS(4,5)=(4AREACP RO)/45 DMASS(4,6)=(4AREACP RO)/45 DMASS(5, I)=-(AREACPRO)145 DMASS(5,2)=0 DMASS(5.3)=0 DMASS(5,4)=(4AREACPRO)145 DMASS(5,5)=(8AREACPRO)/45 DMASS(5,6)=(4AREACPRO)/45 DMASS(6,1)=0 DMASS(6,2)=-(AREACP RO)/45 DMASS(6,3)=0 DMASS(6,4)=(4AREACPRO)/45 DMASS(6,5)=(4AREACPRO)/45 DMASS(6,6)=(8AREACPRO)/45 RETURN END 4.6.6 Checking correctness of results In some cases, the results from REDUCE are lengthy as the stiffness matrix shown above.
It is very important to find a way to check their correctness.
In general, a small problem with a known solution is used to test the correctness of a symbolic program before application to actual problems.
In addition, some specific checking procedures are also available in each individual field.
They require a knowledge of the specific field.
Takin$ the mass matrix problem above as an example, the summation of the entries of the mass matrix should be equal to unit multiplied by the accessory constants.
The symmetry of mass matrix is proven by 78 subtractingthe massmatrix from its transpose to get zerosfor eachentries.The REDUCE programto checkcorrectness is appended in the programshownin subsection4.6.4. The following resultsinclude two parts.The first R is the summationof eachentriesof mass matrix. The secondSDIF(i.j) are the differenceof massmatrix and its transpose.
These checkingresultsconfirmthecorrectness of REDUCEprogram.
0" ...............................................................................
% Checking the correctness of mass matrix by summing each entries % of mass matrix to make an unit multiplied by accessory constants i ...............................................................................
R := AREACPRO I_ .............................................................................
% Checking the symmetry of mass matrix by finding the difference % between mass matrix and its transpose.
O_ .............................................................................
SDIF(1,1) := 0 SDIF(1,2) := 0 SDIF(1,3) := 0 SDIF(I,4) := 0 SDIF(1,5) := 0 SDIF(1,6) := 0 SDIF(2,1) := 0 SDIF(2,2) := 0 SDIF(2,3) := 0 SDIF(2,4) := 0 SDIF(2,5) := 0 SDIF(2,6) := 0 SDIF(3,1) := 0 SDIF(3,2) := 0 SDIF(3,3) := 0 J SDIF(3,4) := 0 SDIF(3,5) := 0 SDIF(3,6) := 0 SDIF(4,1) := 0 SDIF(4,2) := 0 SDIF(4,3) := 0 SDIF(4,4) := 0 SDIF(4,5) := 0 SDIF(4,6) := 0 SDIF(5,1) := 0 SDIF(5,2) := 0 SDIF(5,3) := 0 SDIF(5,4) := 0 SDIF(5,5) := 0 SDIF(5,6) := 0 SDIF(6,1) := 0 SDIF(6,2) := 0 SDIF(6,3) := 0 SDIF(6,4) := 0 SDIF(6,5) := 0 SDIF(6,6) := 0 79 4.7 Closed form solution of stiffness matrix of four-node element 4.7.1 Introduction Although the methodology to construct the stiffness matrix of four-node isoparametric quadrilateral element for the plane elasticity problem is the same as that of the triangular element shown in last section, the techniques are quite different from each other.
In the triangular element, the Jacobian determinant is constant and therefore the integration is straightforward.
However the same advantage can't be gained for the isoparametric quadrilateral element.
In general, the determinant of the Jacobian is a function of the natural coordinates.
Having the of Jacobian determinant in the denominator of the integrand due to the coordinate transformation produces a tremendous difficulty in performing integration analytically, therefore the numerical Gauss quadrature rule is usually introduced to solve this problem.
The discussions of this difficulty and the introduction of Gauss quadrature rule can be found in most relevant literatures, such as Zienkiewicz , Becker & Carey & Oden , Cook , Reddy , Huebner , Weaver & Johnson .
The inability to perform analytic integration introduces the integration error in the finite element results.
The following paragraphs will show that this difficulty has been overcome and the exact closed-form solution has been obtained by appropriate application of REDUCE .
4.7.2 Preliminary formulation The local stiffness matrix for a 2-D isoparametric quadrilateral element is formulated by 1 1 g -• E • B,t, Isl dO (4.45) Vdhere • K" local stiffness matrix.
• E' material property matrix.
• IJI • determinant of Jacobian matrix.
• , r I • natural coordinates.
• t" thickness of element.
8O • B ' strain-displacement matrix.
In general, each entry of strain-displacement matrix B is a function of , Vl, IJIand can be expressed as : b,(,'O, lJI)-c° +c,: +c2?
+ c3_' Igl (4.46) Where ° bij" are denoted by the entry in ith row and'jth column of matrix B • ci : are constants.
For simplicity, the material property matrix E and the thickness t are assumed to be independent of natural coordinates.
The integrand in equation (4.45) therefore will be function of , rI and IJI, too.
The entry of integrand can be expressed specifically as : g,j(,n,IJI) = a°+a' +a,o +a3_'+ a, +ash" +a6_20 +a,_o +a,'o" (4.47) Igl Where d 1, d2,. .... d8 are constants, too.
The Jacobian J is [; 1I 1 J, _ m JIt Jr: " Jzl Jz2 (4.48) And the global coordinate variables x, y can be transformed to local coordinate by the same shape functions as those of field variables.
This is an intrinsic property of an isoparametric element.
4 4 x -, y-2N, y, i -1 _ -1 (4.49) Where • xi' Yi " are global node coordinates.
81 • N i • shape functions.
Specifically, the individual shape functions are N, - 42-(1-)(1 - 1,/) N 2-(1 + )(1-r/) N 3 -4L(l + )(1 + r/) N 4-(1 - )(1 + r/) (4.50) (4.51) (4.52) (4.53) and the entries of Jacobian are derived from equations (4.48) to (4.53).
_X r/ Jll " _ -'(x03 _ i -x2 + x3 -x4)+ (-xl+x2+x3-x4)-a,,r I +b, J_2" d_ ''J'(Yl Y2+Y3-Y4)+-YI+Y2+y-y4)..ayr I +by ox J,l" "-" (xm-x2+ Xn-X4)+';(-x,-x2+x3+x4),,a, _ +c,, J,2" O"'' "4-'(Y,-Ya + Y,-Y4) + t4-'( -Y,-Y, + Y3 + Y4)-a_, +cy (4.54) (4.55) (4.56) (4.57) Therefore, the determinant of the Jacobian will be I J I-J,,J22 - J,2J2, •- (b,,ay - axby) + (axCy -ayc,,)r I + (bxcy + b -U_ + V rI + W yc,) (4.58) where U, V, W are independent of natural coordinates.
Obviously, the determinant of the Jacobian is only a linear function of natural coordinates.
This linearity allows the exact integration to be performed and the logarithm function is expected in the solution.
As the first integration with respect to x is done, there is no longer an integration variables h in the denominator.
Therefore the second integration with respect to h can also be performed analytically.
However, the algebraic operations to finish these two integrations axe too tedious to be handled by hand.
Fortunately, with the help of the symbolic and algebraic manipulator REDUCE, these mathematical operations can be done by computer.
In addition, the solutions can be organized in a systematic way and converted into a FORTRAN-code subroutine to be called by the main program.
All of these procedures and parts of solutions are demonstrated in the next paragraphs by an example of linear elasticity.
The explanations of commands and the time consumed in each individual command are also 82 commentedfor reference.
The total timeconsumed in thisexecutionby REDUCE is around two and half hours.The resultantfortranexpressions for just anelementof stiffnessmatrix occupy almostsixteenpages.
Thesehugeexpressions are the reasonwhy the closedform solutionwasnotavailablebefore.
4.7.3 REDUCE program and explanation O' .......................................
% Turning on the CPU elapse time.
O' .......................................
1: on time; Time: 133 ms I_ .....................................
% Inputting 4 shape functions.
% p and q are natural coordinates.
%p:xi %q: eta i_ .....................................
2: s 1:=( l-p)( l-q)/4; P'Q-P-Q+ 1 S1 := ....................
4 Time: 600 ms 3: s2:=(l+p)(1-q)/4; P'Q-p + Q - 1 S2 := ........................
4 Time: 167 ms 4: s3:=( l+p)( l+q)/4; PQ+P+Q+ 1 $3 := ........................
4 Time: 167 ms 5: s4:=(1-p)(l+q)/4; PQ+P-Q-1 S4 := .......................
4 Time: 166 ms % Expressing x & y by shape function % and global node coordinates.
1_ ...................................
...
.....
83 6: x:=slx l+s2x2+s3x3+s4x4; X := (PQX 1-PQX2+PQX3-PQX4-PXI+PX2+PX3-PX4-QX 1-QzX2+QX3+QX4+X1+X2+X3+X4)/4 Time: 400ms 7: y:=sly l+s2y2+s3y3+s4y4; Y := (PQY 1-PQY2+PQY3-PQY4-PY I+PY2+PY3-PY4-QY 1-QY2+QY3+QY4+yl+Y2+Y3+Y4)/4 Time: 267ms % DeclaringandInputtingmatrix elements to calculate the strain-displacement matrix B.
% c : coefficient matrix % jac : combination of Jacobian matrix % sd : matrix containing the derivative of shape function.
% b : strain-displacement matrix.
% detj : determinant of Jacobian.
% j 11 ,j 12,j21,j22 : element of Jacobian matrix.
I_ .........................................................................
.
.............................
8: matrix c(3,4),jac(4,4),sd(4,8),b(3,8)$ Time: 550 ms 9: c: =mat(( 1,0,0,0),(0,0,0,1),(0,1,1,0))$ Time: 183 ms 10: jac: =mat((j22,-j 12,0,0),(-j21,j 11,0,0), (0,0,j22,-j 12),(0,0,-j21,j I 1))$ Time: 284 ms 11: sd:=mat((df(s 1,p),0,df(s2,p),0,df(s3,p),0,df(s4,p),0), (df(sl,q),0,df(s2,q),0,df(s3,q),0,df(s4,q),0), (0,df(sl,p),0,df(s2,p),0,df(s3,p),0,df(s4,p)), (0,df(sl,q),0,df(s2,q),0,df(s3,q),0,df(s4,q)))$ Time: 833 ms 12: b:=cjacsd/detj$ Time: 417 ms % Inputting material matrix D and calculating integrand.
% D : material matrix (assumed symmetric).
% Ga : integrand.
% th : thickness of element.
O'_ ............................
.
..................................
13: matrix d(3,3),ga(8,8); Time: 316 ms 14: d: =mat((e 11 ,e 12,e 13),(e 12,e22,e23),(e 13,e23,e33))$ Time: 200 ms 15: ga:=tp(b)d b thdetj$ Time: 8067 ms 84 % Evaluating each element of Jacobian.
I_ ............................................
16: on factor; Time: 150 ms 17: on div; Time: 33 ms 18: on rat; Time: 50 ms 19: j 1 l:=df(x,p); 1 Jl 1 := ---((X1 - X2 + X3 - X4)Q - X1 + X2 + X3 - X4) 4 Time: 750 ms 20: j 12:=df(y,p); 1 J12 :=---((Y1 - Y2 + Y3-Y4)Q-Y1 + Y2+Y3-Y4) 4 Time: 550 ms 21: j21:=df(x,q); 1 J21 := ....
((X1 + X2 - X3 - X4) - (X1 - X2 + X3 - X4)P) 4 Time: 667 ms 22: j22:=df(y,q); 1 J22 := ....
((Y 1 + Y2 - Y3 - Y4) - (Y 1 - Y2 + Y3 - Y4)P) 4 Time: 650 ms O'_ ................
'.
......................................
%Making substitution for Jacobian matrix.
% ax=(x 1-x2+x3-x4)/4, bx=(-x 1+x2+x3-x4)/4 % ay=(y 1-y2+y3-y4)/4, by=(-y l+y2+y3-y4)/4 % cx=(-x 1,x2+x3+x4)/4, cy=(-y l-y2+y3+y4)/4 ........................................................
23: j 1 l:=axq+bx; Jll := AXQ+ BX Time: 333 ms 24: j 12:=ayq+by; J12:=AYQ+ BY Time: 117 ms 25: j21:=axp+cx; J21 := AXP+ CX Time: 117 ms 85 26:j22:=ayp+cy; J22:= AYP + CY Time: 116 ms %Calculating the determinant of Jacobian.
27: matrix j(2,2),ske(8,8); Time: 200 ms 28: j :=mat(tj 11,j 12),(j21,j22)); J(1,1) := AXQ+ BX J(1,2) := AYQ + BY J(2,1) := AXP+ CX J(2,2) := AYP + CY Time: 367 ms 29: detj:=det(j); DETJ := - ((AXP + CX)(AYQ + BY) - (AXQ + BX)(AYP + CY)) Time: 333 ms %Making a further substitution and giving the lineality of determinanL 30: let-axcy+aycx=al,axby-aybx=a2,-bxcy+bycx-_.a3; Time: 450 ms 31: detj:--detj; DETJ := - (AIQ + A2P+ A3) Time: 450 ms 32: on exp; Time: 50 ms %Performing the double integration.
%Due to the complication of the expression in the integrand, it is necessary to make the "pre-%treatment" to each element of integrand before integration.
This is a vital step to avoid the %limitation of memory space and finish the job.
33: for i:=1:8 do forj:=l:8 do ifj>=i then <> else ske(i,j):=ske(j,i); Time: 234766 ms %Showing the results for the element in 1st row and 1st column of the local stiffness matrix.
%H l=log(al2-a la2+a l'a3), H2=log(-a l2-a la2+ala3) %H3=log(a 1+a2+a3) , H4=log(-a l+a2+a3) %HS=log(al-a2+a3) , H6=log(-a 1-a2+a3) ...........................................................................................................
34: ske:=ske; 1 -1 1 -1 SKE(1,1) := HITH(---A1 A2E13 ....
A1 8 8 -1 2 A2 A3 El3 + 1 -1 -1 2 1 -1 -1 .... AI A2 A3BX E33 ....
A1 A2 A3BXBYE13 16 8 1 -2 2 1 ...........................
+ .... A2 A3 El3+ ....
El3) 16 16 1 2 -2 1 2 -3 + H6TH( .... AI A2 El3 + .... AI A2 16 48 87 1 1 -1 1 -I .................
+ .... EI3)+TH( ....
AIA2 EI3+---AI A2E13+ 16 4 4 1 -2 1 -2 2 ....
A2 A3BXBYEI3+---A2 A3BY Ell) 3 6 Time: 79317 ms 4.7.4 Fortran subroutine from REDUCE %Converting the results into FORTRAN code subrou ti ne(x 1,y I ,x2,y2,x3 ,y3 ,x4,y4,e I ,e2,e3 ,e4,e5,e6,th,ske) implicit real8(a-h,o-z) dimension ske(8,8) ax=(x 1-x2+x3-x4)/4.
bx=(-x 1+x2+x3-x4)/4.
ay=(y 1-y2+y3-y4)/4, by=( - y 1+ y2+y3-y4)/4.
cx=(-x 1-x2+x3+x4)/4.
cy=(-yl-y2+y3+y4)14.
al=-axcy+aycx a2=axby-aybx a3=-bxcy+bycx h 1--log(a I 2-a I a2+a I a3) h2=log(-al 2-a la2+a l'a3) h3=log(al+a2+a3) h4=log(-al+a2+a3) hS=log(a l-a2+a3) h6=log(-al-a2+a3) ANS 14= I/4A 1(-3)A32AyCXE 13- ltSA 1"(-3)A3"2 • AYCYEI 1-1/16A 1(-3)A32CX2E33+ I/8A 1_¢(-3 • )A32CXCYEI3-1/16A I(-3)A32CY2tE1 l-lJ16 • A2(-2)A32E13+ 1/16E 13 ANS I=H ITHANS2 ANS28=-I/4A 1(-3)A32AYCXEI3+ I/8A I z (-3)A3" • 2AYCYE11+ 1/16A I(-3)A32CX2E33-l/8A 1"( • -3)A32CXCYE13+I/16A 1(-3)A32CY2E 11+ 1/ 16A2(-2)A32E13-1/16'E13 ANS 15=H2THANS 16 A NS48= 1/8A2(-3)A32BXBYE 13- I116A2(-3)A3 • 2BY2E11 ANS29=-H3THANS30 A NS68=-1116A2(-3)A32BX2E33+ 1/8A2(-3)A3 88 2BXBYE13-1/16A2(-3)A32BY2E11 Al'_S49=-H4THANS50 ANS75=1/16A2(-3)A32AX2E33-1/8"A2" (-3)A3" • 2AXAYE13+ 1/8A2(-3)A32AXBXE33+ 1/16"A2"( • -3)A32AY2E11-1/4A2(-3)A32AYBXE13+l/8 • A2(-3)A32AYBYE1 l+I/16A2(-3)A32BX2 • E33-1/8A2(-3)A32BXBYE13+I/16A2(-3)A32 • BY2E11-1/16'E13 Al'_S69=H5THANS70 ANS82=-3/16A2(-2)A3BYCYE11+1/16A2(-3)A32 • AX2E33-1/8A2(-3)A32AXAYE 13+ 1/8"A2"(-3) • A32AXBXE33+ 1/16A2(-3)A32AY2E 11-1/4" • A2(-3)A32AYBXE13+ 1/8A2(-3)A32AYBY • E1l+l/16A2(-3)A32BX2E,33-1/8A2(-3)A32 • BXBYE13+I/16A2(-3)A32BY2E11+1/16"E13 AI'_S83=THA NS84 ske(1,1)=ANS I+ANS 15+ANS29+A NS49+ANS69+ANS76+A NS83 return end 4.8 Significance and conclusion Some conclusions are drawn and the significance of automatic problem formulation is discussed as follows • 1. Improving on-line efficiency --- the closed-form solution of the local stiffness matrix allows us to get a numerical value by simply substituting the global nodal coordinates into a FORTRAN-code subroutine.
This procedure is done in just one step.
Of course, this is faster than the Gauss quadrature rule which usually needs more than one integration point to get a reasonable solution 4 . The symbolic template in the nonlinear numerical analysis also plays the same role.
In the case of a large number of elements (or large dimension size in matrix), the significance in improving on-line efficiency will be greater• 2. Increasing the accuracy of solution --- the closed-form solution is an exact solution.
There is no integration error introduced into the evaluation of the stiffness matrix.
The 4 Reduce integration is an exception and sometimes results in Hourglass drawback.
This special case is ruled out here.
89 accuracy of the Fourier series approximation can be increased as high as the user requires.
Therefore, the results from using symbolic and algebraic manipulation will be more precise than those of pure numerical analysis.
3. Free from hand-calculation and typing error --- As the results in the tensor formulation, derivation of equations of motion and stiffness matrix constructions show, the algebraic expressions are too lengthy to be formulated by hand.
Even supposing that they could be done by hand, it would be so tedious that nobody could guarantee that no mistakes would be made when trying to key them into the computer.
With the use of symbolic and algebraic manipulation, both difficulties are completely solved.
As long as the user inputs the correct commands, there will not be any question about the correctness of the results.
4.
Simplifying FORTRAN programming ---the numerical values of the local stiffness matrix can be obtained by simply using the CALL" command once.
This isn't true when Gauss quadrature integration is employed in the finite element method to evaluate integration.
It is necessary to have a "DO" loop, "CALL _ command and multiplications of various weight coefficients for different integration points.
These will complicate the program and therefore will produce more error sources.
5. Further analysis of symbolic results becomes available --- Sometimes the pre-analysis of the expressions produced from symbolic and algebraic manipulation will lead to a dramatic improvement in the incoming numerical analysis.
The closed form solution makes this analysis feasible.
For example, suppose that the diagonal terms of global stiffness matrix need to be more dominant to improve the ill-condition, this can be achieved by appropriately relocating nodes so that the off-diagonal terms of local stiffness matrix will be smaller or even vanished.
This is the unique advantage that the numerical method does not possess.
9O 4.9 References T.
R. Kane and D.
A.
Levinson,"Dynamics : Theory and Applications", McGraw-Hill, 1985.
L. Meirovitch,"Methods of analytical dynamics",1970.
F. I. Niordson,"shell theory",North-Holland, 1985.
I. S. Sokolnikoff,"Tensor analysis", 2nd edition, John Wiley & Sons lnc, 1964.
W.
H.
Yang,"AM519 course note", Mechanical engineering ,The University of Michigan, Ann Arbor, 1989.
P. V. O'Neil,"Advanced Engineering Mathematics",Wadsworth, 1983.
N.
Kikuchi,"finite element methods in mechanics",Cambridge University press, 1986.
A.
R.
Korncoff and S. J. Fenves,"Symbolic generation of finite element stiffness matrices", Computers & Structures, vol.
10, pp.
119-124, 1979.
R. W.
Luft, Jose M. Roesset and J. J. Connor,"Automatic generation of finite element matrices", J. of the structural division, proc.
of ASCE, pp 349-362, Jan.
1971.
O. C. Zienkiewicz,"The finite element method", 3rd edition, McGraw-Hill, pp.
191, 1977.
E. B. Becker, G.
F. Carey and J. T. Oden,"Finite elements an introduction", vol.
1, Prentice-Hall, 1981 Robert D.
Cook,"Concepts and applications of finite element analysis", 2nd edition, pp.
119, John Wiley & Sons Inc., 1974.
J. N. Reddy,"An introduction to the finite element method", pp. 154, McGraw-Hill, 1984.
K.
H.
Huebner,"The finite element method for engineers", pp. 190, John Wiley & Sons, 1975.
91 W.
Weaver, Jr and Paul R. Johnson,"Finite elements for structural analysis, pp.
111, Prentice-Hall Inc., 1984.
A.
C. Hearn," REDUCE user's manual", Version 3.3, July 1987.
[ 17] W. L. Tsai, "Applications of symbolic and algebraic manipulation software in solving applied mechanics problems", Ph.D.
thesis, Department of mechanical engineering and applied mechanics, The University of Michigan, Ann Arbor, Dec.
1989.
92 CHAPTER V APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO MATERIALLY NONLINEAR PROBLEM ---RIGID-PLASTIC RING COMPRESSION A 5.1 Introduction The application of the finite element method to the rigid plastic problem was originally devised by Lee and Kobayashi (, ), and became popular in the last decade.
This method allowed the deformation behavior of metal to be revealed on the computer screen stage by stage during the process of metal forming.
As a consequence, the design techniques of die, and the manufacturing process were improved.
This contribution to the industrial manufacturing field is recognized to be very significant.
Starting from the principle of virtual work and associating with the normality condition of plasticity, the theoretical analysis of this method leads to an inequality objective function with an equality constraint.
This equality constrained problem is then changed into an unconstrained problem by introducing Lagrange multipliers.
As the stationary requirement is reached, the total unconstrained problem can be solved incrementally by the finite element method and the upper bound solution will satisfy the equilibrium equations, constitutive equations, compatibility equations, incompressibility constraint, and boundary conditions.
Despite the success of the finite element application to the problem, the formulation of it is very tedious.
Especially when the friction boundary condition is considered, the performance of integration along the interface surface and the evaluation of the first and second derivatives with respect to velocity fields is always difficult to obtain by hand.
Therefore, unlike the traditional hand derivation, this chapter will utilize the symbolic and algebraic manipulator REDUCE to do the job of formulation.
The quantities involved in the formulae then can be made into subroutines symbolically at the the element level to facilitate the global assemblage.
As a result, otherwise intractable tasks become possible and free of errors.
93 5.2 Preliminary formulation Consider a body of volume V with the essential boundary condition, velocity /7, prescribed on surface S u and two separated natural boundary conditions, traction F and frictional stress .f', prescribed on S F and Sc, respectively [Figure 5.1].
The actual stress and velocity fields will satisfy the following relationships : 1. Equilibrium conditions" crj,j ,, 0 (neglecting body force ) (5.1) s ( sF(F) V S Figure 5.1: Configuration of domain and boundary conditions Where 2. Compatibility and incompressibility conditions: i j - (u, .i + uj., ) (5.2) e,, .- 0 (5.3) 3. Stress-strain rate relationship (]1 "t • 'J • deviatoric stress.
• ' and E • effective stress and strain-rate, respectively.
4. Boundary conditions : 94 cr o n, .. F 1 on S r (5.5) u, - U on S, (5.6) o jn, =fj (frr) on Sc (5.7) Where the relative velocity between die and deforming body is defined as follows • fr = _u..a . Vrt~ The t-here is the unit base vector along die and working piece interface.
With an admissible velocity field u_', the virtual work principle gives (5.8) cr,ji,aV = F.
'dS + f-.
(fr; + J)dS + (_r n,)U dS (5.9) ¥ e M The frictional stress f" is defined as follows and is plotted in Figure 5.2.
f (V) r -mg mg "-V r Figure 5.2 • Plot of frictional stress vs. relative velocity Wr f0z, --,,, •g, where the friction factor m is in the range of 0 < m < 1 and g is the yield shear stress.
By considering the normality condition of yield surface (cr j -o j)i_j :- 0 (5.1 I) 95 and inequality equationS The equation (5.9) becomes =-0 (5.12) (5.13) All conditions will be met when the left hand side of equation (5.13) reaches the minimum values with respect to U " and satisfies the incompressibility condition of equation (5.3).
Since there is no ambiguity in omitting the asterisk, for the sake of simplicity the admissible field '° will be denoted as u for the following discussions without any special note.
By introducing the Lagrange multiplier !, the equality constraint equation can be included into the objective function.
The stationary value problem for finite element formulation is therefore ----[ agdV -g.
¢dS _ f (9' ). 9, dS + Xi, dV] 0 (5.14) ¥ r: • where gt represents the functional inside the square {]:ket.
Since the frictional stress is not differentiable at the point I7 r , 0 [see Figure 5.2], _ does not exist and the convergent solution will not be available for equation (5.14).
In order to overcome this numerical difficulty, the frictional stress is approximated in terms of arctangent function [see Figure 5.3] r(i7,.) = - m • g • [( 2---)inn -,(a.)]f-' (5.
J5) where the arbitrary constant a is several orders less than the die velocity.
Its function is to exaggerate the argument of arctangent to reduce the error between equation (5.15) and equation (5.10).
The equation (5.15) is absorbed into functional by the following inequality6 5 See appendix A for proof.
6 See appendix B for proof.
96 !
-100.
-EO.
-0.5 --1.
-1.5 -I -Tan (x) !
50.
X !
100.
Figure 5.3 • Arctangent approximation of frictional stress The final form of equation (5.13) is velocity specified :0 Figure 5.4" Physical configuration for nng compression problem 11,0 (5.16) (5.17) 97 5.3 Matrix method for the ring compression problem If a specific problem of ring compression is considered [see Figure 5.4], the surface traction is due to friction force only.
Therefore the term for SF integration in equation (5.17) can be dropped.
By the substitution of the following equations to equation (5.17), firt E' 0 m ,J_c',3 " £ £ I,t - B U (5.18) (5.19) the stationary requirement of the functional results in the following nonlinear equation.
( )K _'dV + U r K U -_dV +ZQ together with an incompressibility constraint equation where • K -B r ° O -fB r D B • C dV ° C" proper matrix such that • D" flow matrix.
U r Q -0 implies the C r d incompressibility condition.
•U: vector of velocity.
It is represented as u" previously.
The derivative part in the second term of equation (5.20) is equal to (5.20) (5.21) 98 a(-) 1 ao _ at Y at 3U " "=---= -'-2 OU (5.22) e 3U _ 3U e The first term of (5.22) is dropped for simplicity due to the assumption that _ is constant for the infinitesimal strain change [see References 2]. Physically, _3 is the current yield strength (denoted as Y) of the material.
AU The equations (5.20) and (5.21) are then perturbed by introducing a small velocity into the velocity vector.
This gives the following equations ' [ 2_a + .b._O(____)au] • K • (U + AU )dr + X • Q • K ,(U + AU)[ ( )+ ('-_-)AUIdV 3U e " - [TO-- + au] (5.23) and where (U +AU) r ,Q =0 (5.24) - -(foo "J, e dVr )dS is contributed by the friction traction.
After neglecting the second and higher order terms, the equations (5.23) and (5.24) can be combined into the following forms • r + OU 2 =-Qr U f 99 9 1 9 I 9 I (5.25) where Ps.s=2_ {1K +[M -(1)ZN]} Hs. s =' "_'-K U -"--E g (5.26) (5.27) 99 2 4 .2 .2 .2 3 .2 E" - " E, + e_ + ee -E',d, -E'_c" 8 -d, do + _y,, (5.28) Es.l " Br A (5.29) fvn T Qs.l = C dV (5.30) [2d, -d -do} . rag 1 =j2 , -dr A .l ""'c" 4., " "¢'!2d0 -.fg,,3 e,.
d_ (5.31) Ks.
s -B r D • B (5.32) Ms.
8-( ) [-_--E -K UIE r (5.33) I T _. -"_C c" (5.34) Ns. s E • U r - K (5.35) k -U r K U floo "lm () - (a .---7)dV, ldS _=-r [.is g , 2 tan i V, (5.36) (5.37) m, a g are constants for a specific material.
5.4 Finite element analysis The domain discretization for ring cross section is shown in Figure 5.5.
The actual domain used by finite element analysis is only the upper half of that given in Figure 5.5 due to the symmetric geometry.
The finite element model contains 96 four-node quadrilateral elements with 117 nodes in total.
For an isoparametric element, the shape functions are 4_-(1 - s)(1 - t ) qz 1 ql" , -(l+s)(l-t) q3-14-'(l+s)(l+t) , q4-1(l-s)(l+t) and 100 4.
4 u (s ,t ) -q, , i -I z -I 4 4 • r+ , z(s,t) = Eq , z r(s,t )-.q, -1 t -I Where • s, t are natural coordinates.
• r, z are physical coordinates.
• u, v are the velocity components in the r, z direction, respectively.
The subscripts in r, z, u, v represent the nodal indices.
The strain rate in axisymmetric case can be expressed in a matrix form as follows : ' I 0. O I.tl 3.0 I 4.0 Figure 5.5" Discretization of ring cross section.
Due to the symmetry of geometry, only the upper half of this cross section is used for finite element analysis.
101 Where o_ 0 i t _ II d-de & 0 vU)-B U (5.38) n mt Ii ° 0 0 0 0 1 1 u T -{u, ., u, u.
u_ u, u.
u,} Iilifo 1 0 ql 0 qa 0 q_ 0 q4 0 ql 0 qa 0 q3 0 q4 ÷ -!
_ -% o o Os c_s 5?
"gr o o 0 0 dr & 77" 77" 0 0 _ & (5.39) qt., 0 q:., 0 qs., 0 q4., 0 ] 1.t 0 q: 0 qs, 0 q4, 0 I qt., 0 q2., 0 qs., 0 q,., ql, 0 q 2, 0 q3 0 q 4t j (5.40) The Jacobian is" J / (5.41) The flow matrix is ; "2 0 D-O 0 C r - (1 0 0 O" 2 -0 0 2 0 -0 1 0 0 T l 1 0 (5.42) (5.43) 102 Sincebilinearvelocity distributionsareassumed for four-nodequadrilateralelement, therelativevelocities alongtheinterface areinterpolated asfollows: r - r z r I - • V, . _ul + r-rzU2 (5.44) If die velocity is specified as uJdl= 1, then the frictional stress will be if" = - {m g • (,-)tan-(--'-)}_ (5.45) and ,Vr m g ()tan-(--.ff-)dVr]dS (5.46) 5.5 Application of symbolic manipulation The difficulty of formulating the equations shown in the last paragraph can be eased by the employment of symbolic and algebraic manipulation.
Based on equation (5.25), the original goal is to make a template form in element level for global assemblage.
However, due to the limitation of memory capacity in hardware systems, the goal is modified to make a template form for individual entries of equation (5.25) only.
There are three kinds of forms produced by REDUCE : . Integrable form --- This is the form which results from the fact that integration can be performed analytically by REDUCE.
The evaluation of equations (5.30) and (5.37) belongs to this class.
There are no natural coordinates in the resultant expressions.
, Non-integrable form --- The equations (5.26), (5.27) and (5.33), for example, are not integrable due to the existence of ', its square and even cubic in the denominator of the integrands.
Moreover, since the resultant expression of " occupies more than sixteen pages, the complete integrands of equations (5.26) and (5.27) are not available.
Therefore, the individual form for quantities, such as K, M, N, k and E, are obtained symbolically, and the summation as well as their integration are carried out numerically , Miscellaneous forms --- Other equations which have nothing to do with integration, such as (5.28), (5.29), (5.31), (5.32), (5.35) and (5.36), are obtained symbolically by matrix operations.
103 a4 The evaluation of _ in equation (5.25) requires a number of delicate pre-treatments.
It is necessary to discuss this subject independently here so that the fundamental nature of symbolic and algebraic manipulation will be revealed.
Intuitively, there are three steps to evaluate _ They are • 1. Evaluation of the integral with respect to relative velocity Vr in (5.37).
2. Evaluation of the integral with respect to interface domain S in (5.37).
3. The results then are differentiated with respect to velocity fields.
Theoretically, there is nothing wrong with the methodology given above.
In fact, REDUCE can only do the first evaluation.
The other two are not feasible.
Therefore, some pre-a,a treatments are necessary to make REDUCE work to evaluate the 70". This will force us to deviate from the formal methodology given above as follows.
--()dV, ]dS - m g v - - m g (T) tan - (-a-)dV, 1"£7dS 2)fs . v, ,v, - - m g ('r [ tan -'(.a-)-_"-]dS I e (5.47) The term 7"0" can be evaluated from equation (5.44). Equation (5.47) becomes 2 /2s t. r,tan a+ , (+)Jo J,, ,+, r-,, ,,^ ---- (-a-)rarau Ou -m g t -4.m , 'fr 2[r2 I v = ;7" tan-(+) - r rz tan -t(v) ]dr I (5.48) Rewrite equation (5.44) into UI--I¢ | V, -, --:_hr + r I _ l--" |t¢ I fl--ti -Yr + Z (5.49) and substitute it in equation (5.48).
0_ -+..s /"'[r 2 t _ -i r -,'- j, tart- (x-r + z) _ • rzta n (rr + Z)]dr (5.50) Since REDUCE is still unable to handle equation (5.50), further treatments are required.
Let t04 Y Z F I W -Tr +7--Yr +Z (5.51) then dr -dW r' (5.52) and equation (5.50) becomes 2 09 -4.
f_'[_ _ -z, 1 Y' Y' r2 tan-l W ]dW (5.53) Note that the upper and lower limits of integration are changed.
Starting from this point, REDUCE can proceed by itself.
The tasks it performs in this specific problem include the integration and back substitution of the variables.
The other quantity, such as _ etc., can be calculated in the similar manner.
The second derivatives are simply computed by differentiation of the results of the first derivatives.
The REDUCE program and a part of fortran solution are presented as follows.
.........................................................................
%REDUCE program to calculate friction part and its denvatives .........................................................................
OFF EXP; A 1:=INT(ATAN(W)(W-ZP)2/YP-ATAN(W)(W-ZP) R2,W)4TMTK/((R1-R2)YP2); A2:=INT(ATAN(W) R 1 (W-ZP)-ATA N(W)(W-ZP) 2/Y P,W)4TMTK/((R1 -R2)YP2); LET YP=(U 1-U2)/(A(R 1-R2)),ZP=-(R IU2-R2U 1)/(A(R l-R2)); LET LOG((A2+U22)/A2)-LOG((A 2+U 1"'2)/A'2) =LOG((A2+U22)/(A2+U 1"'2)); OFF EXP; ON FORT; OFF ECHO; CARDNO!:=10; OUT "look.ftn"; WRITE " SUBROUTINE FRIC(U1,U2,A,TK,TM,R1,R2,B 1,B2,B 11,B 12,B22)"; WRITE" IMPLICIT REAL8(A-H,O-Z)"; B 1:=SUB (W=U2/A ,A 1)-SUB(W=U 1/A ,A 1); B2:=SUB(W=U2/A,A2)-SUB(W=U 1/A,A2); B 11: =DF(B 1,U 1); B22: =DF(B2,U2); B 12:=DF(B 1,U2); WRITE" RETURN"; WRITE" END"; OFF FORT; SHUT "look.ftn"; BYE; 105 C Thefortransubroutine made by REDUCE for frictional part.
SUBROUTINE FRIC(U I,U2,A,TK,TM,R 1,R2,B 1,B2,B 11,B 12,B22) IMPLICIT REAL8(A-H,O-Z) ANS4=LOG((A2÷U22)/(A2+U I2))A3R22.3.LOG • ((A2+U22)/(A2+UI2))ARI2U22+3.LOG((A • 2+U22)/(A2+U I2))AR IR2U IU2+3.LOG((A2+ • U22)/(Az2+UI2))ARIR2U22.3.LOG((A2+U2 , 2)/(A2+U l2))AR22U 1' U2+ARI2U l2-6.AR1 • 2U IU2+5.AR!2U22+AR IR2U I2+6.ARIR2 • U IU2-7.AR lR2U22-2.AR22U I2+2.AR22 , U2"'2 ANS3=-6.ATAN(U2/A)A2R 1"2'U2+3.
ATAN(U2/A)A2 R I R2 U I+9. AT AN(U2/ A ) A 2R I R2U2.3, A T AN(U2/A ) A 2R22U 1-3.ATAN(U2/A)A2R22 U2+2.ATAN(U2/ A )R I 2U23-3. A TAN(U2/A )SRI R2U I U22-A T AN(U2/ A)RIR2U23+3.ATAN(U2/A)R22U IU22-ATAN(U2/ A)R22U23+LOG((A2+U22)/(A2+U 1"'2))A'3' R I2-2.LOG((A2+U22)/(A2+U l2))A3RIR2+ ANS4 ANS2=6. ATAN(U1/A) A 2RI 2U2.3. ATAN(U1/A) A 2 R I R2U1-9. A T AN(U1/A ) A 2RI R2 U2+3, A TAN(U1/A ) A 2R22U I+3.ATAN(U 1/A)A2R22U2-2.ATAN(U1/ A )RI 2U I 3+6.W ATAN(U1/A )iRI 2U I t2U2-6. A TAN( U 1/A)RI2UIU22+ATAN(U 1/A)R IR2UI3-3.ATAN( U 1/A)R l'R2 U I2U2+6.ATAN(U 1/A)RIR2UIU22+ A T AN(U I/A )R22U I3-3. A T AN(U1/A )R22U I 2U2+ ANS3 A NS 1=2. A NS2TKT M 13I=ANS 1/(3.(U I-U2)'3) ANS I=2.ANS2TKiTM B 12=A NS 1/(U 1-U2) 4 RETURN END 106 5.6 Numerical evaluation and result treatment The numerical scheme employed is the Newton-Raphson iteration with a displacement increment of 0.01 in each stage.
The initial guesses are slightly modified from the solution of the elastic ring compression problem.
Since the existence of zero velocities in r-direction at interface nodes will overflow the subroutine FRIC, the problem is modified by assigning a different small number to the r-component of every relevant node.
The question that arises is how small they should be. According to the experiments, only the numbers which are smaller than 10 "12 will achieve convergence with this scheme.
The convergence criterion used here is -ll[I-: 0.00005.
The boundary conditions are specified at two parts of the boundary • 1. Symmetric boundary condition --- The velocities in z direction [see Figure 5.5] are specified to be zeros along the z--O boundary.
2. Die velocity boundary condition --- The velocities at the interface surface between the die and the working piece are specified as unit per second in the negative z direction.
Numerical integration of non-integrable terms is performed by the 4-point Guass quadrature rule.
The assembly of a global matrix is also done numerically.
The equation solver is the Gauss elimination method, from the IMSL subroutine library.
The deformed configurations for friction factor for m=0.5 and m=0.0 are shown in Figures 5.6 and 5.7, respectively.
As the figures show, the deformed shapes are completely different for low and high frictional factors.
The velocity distributions in the deformed states are also plotted in Figures 5.8 and 5.9, respectively.
The neutral lines in both cases are visible from pictures.
Figures 5. 10 and 5.11 also show the effective stress 7 distributions for two cases.
As the shapes of elements are distorted, the error increases and the convergence of the scheme becomes harder.
In order to continue the execution, the technique of adaptive mesh needs to be introduced.
7 The effective stress is defined as I -(r, cr_ - o_cr_ + 3_ + 3z_z + 3rx2_ ) 2 107 < Z 8 5 !
0.0 LU 2.0 3.8 4. U R COORDINATE ON) Figure 5.6 : Deformed configuration after the 8th stage for mffi0.5 £ 3. t, 4.U R OOORI_CATE ON) Flgure 5.7 : Deformed eonflguraUon after the 7th 8tt_e for m-0.0 108 [] 109 110 III Q .g -q °..i .g r_ °° w . r,.,,.
112 5.7 References C.
C.
Chen and S.
Kobayashi,"Rigid plastic finite element analysis of ring compression", ASME AMD-vol.
28, pp 163-- 174,1978.
C.
H.
Lee and S.
Kobayashi,"New solutions to rigid-plastic deformations using a matrix method", Trans.
ASME, J. of Engrg.
for Ind., vol.
95, pp 865-873, 1973.
C. H. Lee and S. Kobayashi,"Analysis of axisymmetric upsetting and plane-strain side pressing of solid cylinders by the finite element method, J. of engineering for industry, pp 445-454, May 197 1.
S.
N.
Shah and S. Kobayashi,"Rigid-plastic analysis of cold heading by the matrix method", proceeding, of the 15th int.
machine tool design & research conf., Birmingham, England, pp 603-610, 1974.
R. Hill,"I'he mathematical theory of plasticity", Oxford, pp 40, 1983.
A.
C. Hearn,"REDUCE user's manual", The Rand corp., Santa Monica, CA 90406, version 3.2, April 1985.
K.
Washizu,"Variational methods in elasticity and plasticity", 3rd edition, Pergamon press, 1982.
L.
M. Kachanov,"Fundamentals of the theory of plasticity", Mir publisher, Moscow, 1974.
D.
R.
J.
Owen, E.
Hinton,"Finite elements in plasticity • theory and practice", Pineridge Press Limited, Swansea, U. K., 1980.
Edited by J.
F.
T.
Pittman, R.
D.
Wood, J.
M.
Alexander, O.
C.
Zienkiewicz,"Numerical methods in industrial forming progresses", Pineridge Press, Swansea, U.
K., 1982.
W.
Szczepinski,"Introduction to the mechanics of plastic forming of metals", Sijthoff & Noordhoff, 1979.
P.
polukhin, S.
Gorelik and V.
Vorontsov,"Physical Principles, of plastic Deformation", Mir Publishers, Moscow, 1982.
113 L E.
Malvern,"lntroduction to the mechanics of a continuous medium", Prentice-Hall, Inc., 1969.
G.
E. Mase,"rheory and problem of continuum mechanics", Schaum's outline series in engineering, MacGraw-Hill Book company, I970.
D.
G.
Luenberger,"Linear and nonlinear programming", 2nd edition, Addison-Wesley, t984.
114 CHAPTER VI APPLICATION OF SYMBOLIC AND ALGEBRAIC MANIPULATION TO THE PLATE PROBLEM 6.1 Introduction Although many shell theories exist in literature, there are only two distinct concepts from which these theories are derived.
One takes the three-dimensional body as a starting point and tries by various means to reduce the problem to the form that can be expressed in a two-dimensional manifold.
This class of theories are called derived theories.
The works by W. T.
Koiter, and E.
Reissner are in this class.
The theory adopted by this study belongs to this class.
The other class of theories consider a shell as a bounded region of some deformable two-dimensional manifold, and are supplemented by one or more fields of vectors over this manifold.
This class of theories are called direct theories.
A. E. Green, P. M. Naghdi, and W.
L. Wainwright worked on this class of shell theories.
Since the real shell is a three-dimensional body, the direct approach has to rely upon some a priori statements.
Despite a large amount of publications using the finite element method to solve plate and shell problems, none deals with the problems by employing the tool of symbolic and algebraic manipulation.
This is not only due to the late availability of software, but also due to the capacity limitations existing in the symbolic and algebraic software 8 . This chapter outlines the simple and universal methodology to solve the plate and shell problems, then presents examples which apply symbolic and algebraic manipulation to them, and finally switches to a numerical method at the point where the symbolic manipulation is stuck by its limitations.
As a consequence of this work, the analytic study of plate and shell problems by computer are pushed a step further.
8 See next chapter for a detail discussion on the capacity limitation of symbolic and algebraic manipulation.
115 6.2 WheFe Preliminary formulation Stating from the equations of equilibrium in three-dimensional state, D GV(u=,z)+p'(ua,z),O 1 (6.]) •Dj" operator of covariant derivative in 3-D space.
• oiJ: 3-D state of stress ( i,j=l, 2, 3).
•pi.
external volume force.
• u a" Gaussian coordinates of surface ( a=l, 2).
• z" normal coordinate of surface.
and applying the following Kirchhoff-Love hypothesis to virtual displacements 6va(u',z ) - (a_ - zd_) • (6v, - z6q,) (6.2) 3(ua, z ) = 6w (6.3) where Y • aa " component of mixed metric tensor.
° da " component of mixed curvature tensor.
Here the rotation &/r is defined as 6q,, -de/iv # + (Sw.,, (6.4) Then, following the lengthy derivations by F. I. Niordson in his Shell Theory , the principle of virtual work gives ffot_v _ ,Se +M ' ,sK . jaa - ffo[F°_vo +p_w IdA + ¢ [T"6vo + M_'&a + Q6w lds (6.5) Where 116 • N ctl3" effective membrane stress tensor.
• M ctl_ • effective moment tensor.
• dF_,o.[3: virtual strain tensor.
• dKctl3 : virtual bending tensor.
• F °t, p : effective load components in surface and normal directions, respectively.
• Tct : membrane force vector acting on the boundary.
• M °t : moment vector acting on the boundary.
• Q : supplemented shear force on the boundary.
• dva, dw : virtual displacements in plane and transverse directions, respectively.
• dret=ctl]dql3 • ec_13: alternate tensor.
Mathematically, the strain tensor Ect[3 and bending tensor Kct[5 are defined as 1 ,, E¢ - 2(a_ - a_ ) K =d" -d (6.6) (6.7) where the superscript asterisks indicate the deformed state and can be derived as follows • a _ a e +Po_ +Pt_ +Pr,,Pt3y + q"qtJ (6.8) 1 Y d_ - ()'[(l+p + P /a)(do_ + Dtjq,, + dcp_) -(qP + eme r_qy p_ )(D ¢ p_ - d opq_,)] (6.9) The generalized two-dimensional displacement gradient p and its determinant in equation (6.9) are expressed as • (6.10) (6.11) 117 After linearization, the strain and bendingtensorscan be expressedin terms of displacements asfollows • E# =,(Dav # + D#v,) - d_w (6.12) K ,DaD#w +d_Dav r +darDavr +vrD d -d r # #rdaw (6.13) be • Considering the isotropic thin elastic shell for simplicity.
The constitutive equations will N "¢ Eh I -" __'77r[( v )E "# + va _ E_ ] M "# Eh r'l --r = j_""_bt( v )K ' + va_ Kr ] (6. 14) (6.15) The substitution of the last four equations into equation (6.5) will result in the weak form which is expressed in terms of displacement vectors.
Before applying the finite element method to solve equation (6.5), a universal methodology is outlined in the next paragraph.
6.3 Methodology for solving shell problem by FEM After the preparations of mathematical formulation, it is necessary to discretize equation (6.5) to solve shell problems by FEM.
However, as equations (6.12) to (6. 15) show, the calculations of constitutive equations and strain-displacement relations involve the evaluations of metric, curvature tensors, and covariant derivatives.
Moreover, in general, the calculations of covariant derivatives require the computations of Christoffel symbols.
If a given geometric domain is complex, these calculations will be tedious.
With the help of symbolic and algebraic manipulation, these tough tasks can be performed by simply giving the parametric equations of the surface.
Here the outline of methodology to solve shell problems is presented as follows : 1 Finding the parametric equations of the middle surface for a given shell.
2 Calculating the Christoffel symbols, the metric and curvature tensors based on the parametric equations.
If the parametric equations are chosen correctly, the resulting metric and curvature tensor should obey the integrability condition.
In other words, they should not violate the Coddazzi-Mainardi equations, Gaussian equations and the regular condition.
3.
Substituting the metric, curvature tensors, and Christoffel symbols into constitutive equations and strain-displacement equations.
118 4 Discretizing the variational form of equation (6.5) and solving it by finite element method.
The above methodology is universal for any shell problems.
Different shell geometries can be solved in the same way by simply feeding appropriate parametric equations into the symbolic and algebraic manipulator REDUCE.
Based on this methodology, an example of plate problem is shown in the following paragraph.
6.4 Symbolic and algebraic manipulation application to plate problems 6.4.1 Methodology 1 Three parametric equations are chosen as follows" fl=ul, f2=u2 ' f3=constant 2 Calculate surface metric, curvature tensors, and Christoffei symbols symbolically.
These calculations are based on their fundamental definitions, given by: • metric tensor" I a _ -- f'+,f,o (6.16) • curvature tensor • i !
d_ - X f., (6.17) where f.'l , f,+z and X '/are surface Gaussian and normal coordinates.
They are shown in Figure 5.2 pictorially, and X/ is defined by • I X + -re fJfk •- a ,jk,l_,2 (6.18) • The 2nd kind of Christoffel symbols is defined as: } I _ Oatj p Oarp 8a a .a_[fly,p].Ta [a- + ,, fl Y a,' a, r] (6.19) The REDUCE program for calculating equations (6.16), (6.17), (6.18) and (6.19) is presented as follows.
The results follow the program.
119 O_OZZZ_Z_Z_ZZZZZZZZZZ_ZZZZZZZZZZZZZ_ZZZZZZZZZZZZZmZZZSS_S_8 % REDUCE PROGRAM FOR CALCULATING METRIC, CURVATURE % TENSORS & CHRISTOFFEL SYMBOLS _OZZZZZZZZZZZZ_ZZ_ZZZZ_ZZZZ_ZZZ_ZZ_Z_ZZZZZZ_ZZ_ZZZZ_ZZZZZ C ............................................................
%INPUTTING THE PARAMETRIC EQUATIONS O'_ ............................................................
MATRIX A(2,2),CA(2,2),F(2,3),D(2,2); ARRAY X(3),C 1(2,2,2),C2(2,2,2),N(3),E(3,3,3),U(2); U(1):=S; U(2):=P; OFF PERIOD; X(1):=U(1); X(2):=U(2); X(3):=CONSTANT; FOR I:=1:2 DO FOR J:=l:3 DO F(I ,J): =DF(X(J),U(I)); FOR ALL T1 LET COS(T1)2+SIN(T1)2= 1; %CALCULATING COVARIANT METRIC TENSOR ...............................................................
FOR M:=l:2 DO FOR N:=l:2 DO A(M,N):=FOR I:=1:3 SUM DF(X(1),U(M))DF(X(I),U(N)); A:=A; DETA:=DET(A); %CALCULATING CONTRAVARIANT METRIC TENSOR {qD .......................................................................
FOR L:=l:2 DO FOR M:=I:2 DO IF L=I AND M=I THEN CA(L,M):=A(2,2)/DETA ELSE IF L NEQ M THEN CA(L,M):=-A(M,L)/DETA ELSE CA(L,M):=A( 1,I)/DETA; O' .........................
7 ..................................................
%CALCULATING THE 1ST & 2ND CHRISTOFFEL SYMBOL WRITE "THE 2ND CHRISTOFFEL SYMBOL'; FOR L:=l:2 DO FOR M:=l:2 DO FOR N:=I:2 DO C 1(L,M,N):=(I/2)(DF(A(L,N),U(M))+DF(A(M,N),U(L)) -DF(A(L,M),U(N))); FOR I_.:=1:2 DO FOR M:=l:2 DO FOR N:=l:2 DO <>; O_ ...............................................
%CALCUI_.ATING NORMAL VECTOR ...............................................
FOR I:=1:3 DO N(I):=(FOR J:=l:3 SUM FOR K:=l:3 SUM E(I,J,K)F( 1,J) F(2,K))/SQRT(DETA); 7o ....................................................
%CALCULATING CURVATURE TENSOR ' ....................................................
FOR J:=l:2 DO FOR K:=l:2 DO D(J,K):=FOR I:=1:3 SUM N(I)DF(F(J,I),U(K)); FOR ALL T1 CLEAR COS(T 1)2+SIN(T I)'2; FOR ALL T1 LET COS(T 1)'2-I=-SIN(T1)2; %OUTPUTTI NG RESULTS 7 .................................
OFF PERI OD; OFF ECHO; OUT "SURFACE"; WRITE " ........................................................
".
THE COMPONENTS OF METRIC TENSOR "" WRITE" WRITE " A:=A; WRITE " WRITE " WRITE " D:=D; WRITE " WRITE " WRITE " THE COMPONENTS OF CURVATURE TENSOR"; THE CHRISTOFFEL SYMBOLS"; .........................................
".
FOR I:=1:2 DO FOR J:=l:2 DO FOR K:=I:2 DO WRITE "CHRIS(" ,I,"," ,J," ," ,K,")=" ,C2(I ,J,K); SHUT "SURFACE"; BYE; The outputs of REDUCE are presented as follows • .......................................................
THE COMPONENTS OF METRIC TENSOR .......................................................
A(1,1) := 1 A(1,2) := 0 A(2,1) := 0 A(2,2) := 1 121 D(1,2) := 0 D(2,1) := 0 D(2,2) := 0 THE CHRISTOFFEL SYMBOLS ...................
..
................
.__.
CHRIS(I, 1,1)-0 CHRIS( 1,1,2)-0 CHRIS( 1,2, I)=0 CHRIS( 1,2,2)=0 CHRIS(2,1,1)-0 CHRI S(2,1,2)-0 CHRIS(2,2,1)=0 CHRIS(2,2,2)=0 As the output from REDUCE shows, all of the Christoffel symbols vanish in the plate case, According to the above solutions, the metric and curvature tensors in matrix form are [aqo ] = (10 , [dqs ]." (0 0 (6.20) 3. After substituting the calculated metric and curvature tensors, the constitutive equations are as follows • For membrane" I!t f: 1 0 12 Eh "77 1-v 22 0 ,'lf "l 0/E2_/ • For bending • io 0], f " lXl-,,"---'- 1 -V /M!
o lie_ The relationships between linearized strain-displacement and are from equations (6.12) and (6.13) • I E ,, -" (D ,,v p + D v a ) 1 -_vp.,, + v,,p) K ¢ - D,,D c w .
w.a _ (6.21) (6.22) bending curvature tensors (6.23) (6.24) 122 where the disappearance of Christoffel symbols in this case eliminates the distinction between covariant differentiation and ordinary partial differentiation.
In addition, the unity of the metric tensors remove the distinction between contravariant and covariant tensors.
4. Finite element method starts (see the following paragraph).
6.4.2 Finite element discretization The element chosen for this topic is the combination of the constant strain triangle (CST) element with the Cheung, King, Zienkiewicz (CKZ) triangle element.
The considerations of selecting this element will be discussed later.
The shape functions for this element are Lot _ _ 2 1 2 1 N,, = 2A[cr(_a_t _ + 7,2_3) - ca(,.r + 7,2_3)] 2 1 2 1 Nay = 2A[b a (_a_r + 7,2:3) - br(,, o + 5",_2,)] (6.25) Where (ct,13,¥) is the permutation of (1,2,3) and no summation convention is applied in equation (6.25).
The constants bi, ci, and A are defined as follows • Y l-Y3 v -y 2 = _ b ..
Y J-Yl b, 2a , z 2---, b 3 = X 3-X 2 Xl-X 3 X2-x I Cl _ 2A , C2 =" 2A ' C3 == 2A 2A --x2y 3 -x3Y 2 + x3y I xjy 3 + x,y z -Xzy I (6.26) Then the deflections u, v, w in local coordinates x, y, z direction can be interpolated by 3 u = Xu L t -I 3 v =_v L I-I o I w =waN _ +-.7.
N _ + .2..l_Nay (6.27) The substitution of equations (6.21) to (6.27) into (6.5) gives the discretization form fL [B: OtBl+ B_D 2B2]dA "d = f fD rr dA + fs r ds + f M r ds (6.28) 123 Where l--v DI = F2" 0 D 2 " 12(- l (6.29) (6.30) 2 0 ---0 'g2 Ii R b 2 b 3 0 0 0 1 a Bz" , c2 c3 bl b2 b3 _ 0 0 a , 0 0 c, c 2 c 3 0 -- a_ 2 0 oooo oooo oooo 1 LI 0 0 0 0 L2 0 0 0 0 L3 0 0 2_ 0 a o Jilt B2 I c2 c3 bl b2 b3 _ 0 bl b2 b3 0 _ C 2 C 3 0 0 c I c: c3 _ o _ t_,J M -[M' M2][b '| c2b z C3]b, , 1_' (6.31) (6.32) (6.33) 124 0 N 1 N, Nly 0 0 N 2 Nzx N2y 0 V 1 W t W 1.x W 1,y 1l 2 V 2 W W W 2 2,x 2 .y 1"13 m N3 N3x N3y ] 3 W3 W3, x W3,y ] (6.34) (6.35) (6.36) t 0 0 0 0 L 0 0 0 0 Ls 0 0 0 0 l L t 0 0 0 0 L 2 0 0 0 0 L 3 0 0 0 0 N, Nix Nly 0 0 N 2 N2, N_ 0 0 N 3 N3x N3y (6.37) 6.4.3 Numerical results and post-process Three boundary conditions are applied to the test problem.
They are • 1. In-plane uniaxial tension [see Figure 6.1] --- In this case, only the membrane component contribute to the stiffness matrix.
The consistent load vector is due to the boundary traction S only.
The REDUCE programs to the stiffness and the consistent load vector are presented as follows.
In addition, the stress distribution can also be calculated symbolically.
_---10cm FEM domain v v v Figure 6.1: Plate with hole under uniform uniaxiai tension load 100 N/cm2 125 % %Symbolicalprogramfor makingstiffness matrixfor platetension problem matrix bc(3,6) ,In(6,15),d(3,3),b(3,15) ,s(15,15); arrayn(3); n(1):=s 1;n(2): =s2;n(3):=s3; In(1,1):=I;ln(2,6):=1;In(3,11):= 1;1n(4,2):= 1; 1n(5,7):= 1;In(6,12):= 1; bc:=mat((bl,b2,b3,0,O,O),(c 1,c2,c3,bl,b2,b3),(O,O,O,c 1,c2,c3)); d:=mat((1,O,v),(O,(1-v)/2,0),(v,O, 1)); b:=bcln; s:=tp(b)dbpjyeh/(2(I-v2)); bl:=(y2-y3)/pj;b2: =(y3-y1)/pj;b3: =(y1-y2)/pj; cl:=(x3-x2)/pj;c2:=(x1-x3)/pj;c3:=(x2-x 1 )/pj; for i:=l: 15do<>; for i:=l: 15do<>; onfort; off period; off echo; out "plane.
ftn"; write " subroutine ske(xl,yl,x2,y2,x3,y3,s)"; write " dimension s(15,15)'; write" implicit real8(a-h,o-z)"; write " pj=(x 1-x3)(y2-y3)-(y l-y3) (x2-x3)"; write ' ye=2.1" 1.Oe07'; write " v=0.29"; write " h=0.2"; for i:=l: 15 do for j:=l: 15 do ifj>=i then write " s(",i,",",j,")=',s(i,j) else write " s(",i,",",j,")=s(",j,",",i,")"; write " return"; write " end"; shut "plane.ftn"; bye; C_Z_ZZZ_ZZZ__ZZZZZ_:_ZZZZZZZZZ_ZSZZZ_ZZSZZZZS_IISZZZZSZ % Symbolic program to calculate the load vector for plate tension problem.
_ZZZZZZZZZZZgZZ_cZZZZZZZZZZZZZIZZZZZ,ZZZZIZZ_,Z_ZZZZZZ matrix f( 1,2),sn(2,15),fn( 15,1); army n(3),ff(15); n( l):=s 1;n(2):=s2;n(3):=s3; f:=mat((fl,f2)); for i:=l step 5 until 11 do <>; fn:=tp(fsn); s2:=O; s3:=l-sl; for i:=l: 15do <>; for i:=1;15 do ff(i):=fn(i,1); on fort; off echo; off period; 126 out "151oad.ftn"; write " subroutineIoad(x1,y1,x2,y2,x3,y3,f 1,f2,f3,fe)"; write " implicit real8(a-h,o-z)"; write " dimensionfe(15)"; write " rl=sqrt((x1-x3)2+(y1-y3)'2)"; write " h=0.2"; for i:=l: 15do write " fe(",i,")=",ff(i); write " return"; write " end"; shut"151oad.ftn"; bye; % REDUCEprogramto construct subroutine to compute % stress distributionfor platetension problem.
matrix d(3,3 ), bc(3,6),fn(6,1 ),stre(3,1 ); array n(3); n( 1):=s 1;n(2):=s2;n(3):=s3; bc: =mat((b 1, b2,b3,0,0,0),(c 1,c2,c3 ,bl, b2,b3),(0,0,0,c 1,c2,c3)); d:=mat(( 1,0,v),(0,( 1-v)/2,0),(v,0,1)); operator u; for i:=1:6 do fn(i,l):=u(i); stre:=d bc fn ye/( 1-v 2); b 1:=(y2-y3)/pj ;b2: =(y3-y 1)/pj ;b3: =(y 1-y2)/pj; c 1:=(x3-x2)/pj ;c2:=(x 1-x3)/pj ;c3 :=(x2- x 1)/pj; on fort; off echo; off period; out "st.ftn"; write " write " write " write " write " write " for i:=1:3 do write " write " return"; write " end"; shut "st.ftn"; bye; subroutine stress(x 1,y 1,x2,y2,x3,y3,u,st)"; implicit real8(a-h,o-z)"; dimension u(6),st(3)"; v=0.29"; pj=(x 1-x3) (y2-y3)-(y 1-y3) (x2-x3)"; ye=2.1 1.0e07"; st(",i,")=",stre(i, 1); The resultant fortran subroutines obtained from the above programs to compute stiffness matrix, load vector and stress distribution are presented as follows.
Since the expression of stiffness matrix is quite lengthy, only a part of it is showed.
The interested researchers may refer to the PH.D.
thesis of Wen-Lang Tsai for details.
SUBROUTINE SKE(X 1,Y 1,X2,Y2,X3,Y3,S) IMPLICIT REAL8(a-h,o-z) DIMENSION S( 15,15) pj=(x 1-:,c3)(y2-y3)-(y l-y3) (x2-_) YE=2.1' 1.0E07 127 V=0.29 H=0.2 s(1,1)=(HYE(V X2"'2-2" VX2X3+VX32-X22+2X2 • X3-X.32-2Y22+4Y2Y3-2Y32))/(4PJ(V2-1)) s(1,2)=(HYE(VX2Y2-VX2Y3-VX3Y2+VX3Y3+X2,Y2_ • X2Y3-X3Y2+X3Y3))/(4PJ(V2-1)) s(1,3)=0 s(1,4)=0 s(1,5)=0 s(1,6)=-(HYE(VX 1X2-VX 1X3-VX2X3+VX32-X 1X2 • +X 1X3+X2X3-X32+2Y2Y3-2Y2y 1-2Y32+2Y3Y 1) • )/(4PJ(V2-1)) s(1,7)=-(H YE(2 VX 1Y2-2' VX 1Y3+VX2 Y3-VX2Y 1-2 • VX3Y2+VX3Y3+VX3Y 1-X2Y3+X2YI+X3Y3-X3Y 1)) •/(4PJ(V2-1)) s(1,8)=0 s(1,9)=0 s(1,1 O) =0 s(1,11)=(HYE(VX IX2-VX 1X3-VX22+VX2X3-X l'X2 • +X 1X3+X22-X2X3+2Y22-2Y2Y3-2Y2y l+2Y3Y 1) • )/(4PJ(V2-1)) s(1,12)=(HYE(2' VX 1Y2-2"VX 1Y3-VX2Y 2+2' VX2Y3 • -VX2Y 1-VX3Y2+VX.3Y1-X2Y2+X2YI+X3Y2-X3Y I)) •/(4PJ(V2-1)) s(1,13)=0 s(1,14)=0 s(1,15)=0 s(2,1)=s(1,2) s(2,2)=(HYE(VY22-2VY2Y3+VY32-2X22+4X2 • X3-2X32-Y22+2Y2Y3-Y32))/(4PJ(V2-1)) s(2,3)=0 s(2,4)=0 s(2,5)=0 s(2,6)=(H YE(VX 1Y2-VX 1Y3+2"V'X2 Y3-2"VX2Y 1-V • X3Y2-VX3Y3+2VX3Y 1-XIY2+XIY3+X3Y2-X3Y3)) •/(4PJ(V2-1)) s(2,7)=(HY E(VY2Y3-VY2Y 1-VY32+VY3Y l+2X 1" • X2-2X lX3-2X2X3+2X32-Y2Y3+Y2Y l+Y32-Y3Y 1)) •/(4PJ(V2-1)) s(2,8)=0 s(2,9)=0 s(2,10)=0 s(2,11)=-(HYE(VX 1Y2-VX 1Y3+VX2 Y2+VX2Y3-2 V • X2Y 1-2VX3Y2+2VX3Y 1-XIY2+XIY3+X2Y2-X2Y3)) •/(4PJ(V2-1)) s(2,12)=-(HYE(VY22-VY2Y3-VY2Y 1 +VY3Y l+2X 1 • X2-2X lX3-2X22+2X2X3-Y22+Y2Y3+Y2y I-Y3Y 1 • ))/(4PJ(V2-1)) s(15,15)=0 return end 128 subroutine load(x 1,y I ,x2,y2,x3,y3,f 1,f2,fe) IMPLICIT REAL8(a-h,o-z) dimension fe(15) rl=sqrt((x3-x2)2+(y3-y2)2) h=0.2 fe( 1)=0 fe(2)=O fe(3)=0 fe(4)=0 fe(5)=0 fe(6)=(H F 1 RL)/2 fe(7)=(H F2RL)/2 fe(8)=O fe(9)=0 fe(10)=0 fe( 11)=(H F1 RL)/2 fe( 12)=(H F2 RL)/2 fe( 13)=0 fe(14)=O fe(15)=0 return end subroutine stress(x 1,y 1,x2,y2,x3,y3,u,st) implicit real8(a-h,o-z) dimension u(6),st(3) v=0.29 pj=(x l-x3) (y2-y3)-(y l-y3) (x2-x3) YE=2.1 1.0E07 st( 1)=(Y E(U(6) VX l-U(6) VX2-U(5) VX l+U(5) VX3+U • (4)VX2-U(4)VX3+U(3)Y2-U(3)Y 1-U(2)Y3+U(2)Y I-• U( 1)Y2+U( 1)Y3))/(PJ(V2-1)) st(2)=-(Y E(U(6)VY2-U(6)VY 1-U(6)Y2+U(6)Y l-U(5) • VY3+U(5)VY I+U(5)Y3-U(5)Y 1-U(4)VY2+U(4)VY3 • +U(4)Y2-U(4)Y3+U(3)VX l-U(3) VX2-U(3)X l+U(3) • X2-U(2) V_'X l+U(2) VX3+U(2)X 1-U(2)X3+U( 1) VX2-U( 1) VX3-U( 1)X2+U( 1)X3))/(2" PJ(V2-1)) st(3)=(Y E(U(6)X I-U(6) X2-U(5)X 1+U(5)X3+U(4) X2-U • (4)X3+U(3)VY2-U(3)VY 1-U(2)VY3+U(2) V_'y I-U(1) • VY2+U(1)VY3))/(PJ(V2-1)) return end The stress distribution of plate under tension is plotted by PATRAN in Figure 6.2.
The stress concentration is visible in the top of the hole.
One interesting phenomenon that should be mentioned here is that the stress at the middle top of the plate is less than the applied load.
This is contributed by the bending effect which produces the compression in the top fiber of the plate.
129 2. Four-edge simply supported platebendingwith uniform transverse load [seeFigure 6.3] --- In this case,the membrane componentof stiffnessmatrix is neglected.The REDUCE programto makethe bendingcomponentof stiffnessmatrix is shownas follows • %REDUCEprogramfor constructing platebending stiffness matrix arrayn(9); matrix sn(3,9),bc(2,3),bn(2,9),ssn(6,9),bcbc(3,6),bssn(3,9),d(3,3) ,ske(9,9),s(9,9); n(1):=s1+s1s2(sl-s2)+s1s3(s1 -s3); n(2):=pj(c3(sI2"s2+s 1s2s3/2)-c2" (s12"s3+s1s2s3/2)); n(3):=pj(b2(s1. 2' s3+s 1s2s3/2)-b3(s1 2' s2+s 1s2s3/2)); n(4):=s2+s2 s3(s2-s3)+s2 s1(s2-s1 ); n(5):=pj(c1(s22' s3+s 1s2s3/2)-c3" (s22s 1 +sIs2s3/2)); n(6):=pj(b3(s22 s1+sIs2s3/2)-b1(s22s3+s Is2s3/2)); n(7):=s3 +s3s1(s3-s1)+s3s2(s3-s2); n(8):=pj(c2" (s3 2s1+s1s2s3/2)-cI(s32"s2+s 1s2s3/2)); n(9):=PJ(b l(s32s2+s ls2s3/2)-b2(s32s1+sl.s2,s3/2)); for i:= 1:9do<>; bc: =mat((b1,b2,b3),(c I,c2,c3)); bn:=bcsn; for i:=1:9do <>; s3:=l-sl-s2; bcbc:=mat((bl,b2,b3,0,0,0),(c 1,c2,c3,b 1,b2,b3),(0,0,0,c 1,c2,c3)); bssn:=bcbcssn; D:=MAT((1,0,V),(0,(I_V)/2,0),(V,0,1)); ske:=tp(bssn)d bssn pj$ for i:=1:9doforj:=l:9 do ifj>=i then<> elses(i,j)=s(j,i); b1: =(y2-y3)/pj ;b2: =(y3-y1 )/pj;b3:--(y 1 -y2)/pj; c1: =(x3-x2)/pj ;c2:=(x1-x3)/pj;c3 :=(x2-x 1 )/pj; onfort; off echo; off period; CARDNO!:=IO; out "z.ftn"; WRITE " WRITE " WRITE " WRITE " WRITE " WRITE " WRITE " SUBROUTINE SKE1(X1,Y1,X2,Y2,X3,Y3,S)"; IMPLICIT REAL8(A-H,O-Z)"; DIMENSIONS(2,9)"; pj=(x1 -x3)(y2-y3)-(y1 -y3)(x2-x3)"; YE=2.1' 1.0E07"; V=0.29"; H=0.2"; FORI:=1:2DO FORJ:=l:9 DO 131 IF j>=i THEN WRITE " S(",I,",",J,")=",S(I,J) ELSE WRITE " S(",I ,"," ,J,")=S(",J," ,",[ ,")" , WRITE " RETURN"; WRITE " END"; WRITE " SUBROUTINE SKE2(X1,Y 1,X2,Y2,X3,Y3,S)"; WRITE" IMPLICIT REAL8(A-H,O-Z)"; WRITE " DIMENSION S(4,9),s 1(2,9)"; WRITE " pj=(x 1-x3)(y2-y3)-(y 1-y3)(x2-x_3)"; WRITE " YE=2.1' 1.0E07"; WRITE" V=0.29"; WRITE" H=0.2"; WRITE " CALL SKEI(XI,Y 1,X2,Y2,X3,Y3,S1)"; WRITE " DO 20 I=1,2"; WRITE" DO 20 J= 1,9"; WRITE" 20 S(I,J)=SI(I,J)"; FOR I:=3:4 DO FOR J:=l:9 DO IFj>=i THEN WRITE" S(",I,",",J,")=",S(I,J) ELSE WRITE " S(",I,",",J,")=S(",J,",",I,")"; WRITE " RETURN"; WRITE" END"; WRITE " SUBROUTINE SKE3(X1,Y I,X2,Y2,X3,Y3,S)"; WRITE" IMPLICIT REAL8(A-H,O-Z)"; WRITE " DIMENSION S(6,9),s2(4,9)"; WRITE " pj=(x 1-x3)(y2-y3)-(y 1-y3)(x2-x3)"; WRITE " YE=2.1" 1.0E07"; WRITE" V=0.29"; WRITE " H=0.2"; WRITE " CALL SKE2(X1,Y1,X2,Y2,X3,Y3,S2)"; WRITE " DO 20 I=1,4"; WRITE " DO 20 J=l,9"; WRITE " 20 S(I,J)=S2(I,J)"; FOR I:=5:6 DO FOR J:=l:9 DO IFj>=i THEN WRITE " S(",I,",",J,")=",S(I,J) ELSE WRITE " S(",I,",",J,")=S(",J,",",I,")"; WRITE " RETURN"; WRITE " END"; WRITE " SUBROUTINE SKE(XI,YI,X2,Y2,X3,Y3,S)"; WRITE" IMPLICIT REAL8(A-H,O-Z)"; WRITE " DIMENSION S(9,9),s3(6,9)"; WRITE" pj=(x 1-x3)(y2-y3)-(y 1-y3)(x2-x3)"; WRITE" YE=2.1" 1.0E07"; WRITE" V=0.29"; WRITE" H=0.2"; WRITE" CALL SKE3(X 1,Y I,X2,Y2,X3,Y3,S3)"; WRITE" DO 20 I=1,6"; WRITE" DO 20 J= 1,9"; WRITE" 20 S(I,J)=S3(I,J)"; FOR I:=7:9 DO FOR J:=l:9 DO IF j>=i THEN WRITE" S(",I,",",J,")=",S(I,J) ELSE WRITE " S(",I,",",J,")=S(",J,",",I,")"; WRITE" RETURN"; WRITE" END"; SHUT "z.ftn"; bye; 132 The resultant fortran subroutineis too large (around 135 pages)to be completelyhere.
Thefollowingis onlyasmallportionof it.
SUBROUTINESKE1 (X 1,Y1,X2,Y2,X3,Y3,S) IMPLICIT REAL8(A-H,O-Z) DIMENSION S(1,9) pj=(x1-x3)(y2-y3)-(yl-y3) (x2-x3) YE=2.1' 1.0E07 V=0.29 H=0.2 ANS5=-4Y3Y 1"'3+2"Y 1"'4 ANS4=-4X2"2' Y2Y 1 +8'X2' 2"Y3'2+2' X22Y 12-16 X2X.3"3-16X2X3Y22+32X2X3Y2 y3.16X2X3 Y3 • 2+5X34+8X32Y 2"'2-16X32Y2Y3+ 10"X3"'2" Y32-4X32Y3y i+2X32Y 1"2+5'Y2"4-16'Y2"'3" Y3-4Y23Y l+24Y22Y32+6Y22y 1"'2-16Y2Y3 • 3-4Y2Y l3+5Y34-4Y33y l+6Y32Y 1"'2+ ANS5 ANS3=2X1"'4-4"X l3X2-4X l3X3+6X 1"'2"X2"'2+6" X 1 2"X3 2+2"X 1'2' Y2" 2-4"X 1"2' Y2Y 1 +2X 1'2'Y3 • 2-4' X 12Y3Y 1 +4X 12Y 12-4'X 1 X2"'3-4" X 1X2 Y22+8'X 1X2Y2' Y 1-4X 1 X2Y 12-4"X 1X3 "3-4" X 1 X3Y3 2+8"X1X3Y3 Y 1-4X1X3Y 12+5"X2"'4-16"X2 • 3X3+24X2 2"X3"2+ 10X22Y22-16X22Y2Y3 +ANS4 ANS2=H3YEANS3 ANSI=ANS2/(18PJ3(V2-1)) S(1,1)=-ANS1 presented 133 Y FEM domain X 10cm /'-L2x Z2x Figure 6.3 • Physical configuration of plate bending problem The load vector in this case is for a uniform transverse pressure only.
The REDUCE program and its output are shown as follows" % REDUCE program to construct the load vector % for plate bending problem.
army n(9),fe(9); n(1):=s l+sls2(s 1-s2)+s ls3(sl-s3); n(2):=pj(c3(sl2s2+s ls2s3/2)-c2(sl2s3+sls2,s3/2)); n(3):=pj(b2(s l2s3+s ls2s3/2)-b3(s l2s2+s ls2s3/2)); n(4):=s2+s2 s3(s2-s3)+s2s l(s2-s 1); n(5):=Pj(c l(s22s3+s ls2s3/2)-c3(s22s l+s I s2s3/2)); n(6):=Pj(b3(s22s l+s ls2s3/2)-b l(s22s3+s I s2s3/2)); n(7):=s3+s3sl (s3-s 1)+s3s2 (s3-s2); n(8):=Pj(c2(s32s l+s ls2s3/2)-c l(s32s2+s ls2s3/2)); n(9):=pj(bl(s32s2+s ls2s3/2)-b2(s32s l+s ls2s3/2)); s3:=l-sl-s2; for i:= 1:9 do <>; b 1:=(y2-y3)/pj ;b2: =(y3-y 1)/pj ;b3: =(y 1-y2)/pj; c 1 :=(x3-x2)/pj ;c2: =(x 1-x3)/pj ;c3:=(x2-x 1)/pj; off period; 134 off echo; ON FORT; out "zload.ftn"; write " SUBROUTINE LOAD(X1,Y 1,X2,Y2,X3,Y3,F3,FE),,.
write " IMPLICIT REAL8(A-H,O-Z),,; WRITE " DIMENSION FE(9)"; write " PJ=(x 1-x3)(y2-y3)-(y 1-y3)(x2_x3),,; N Pt tl ft FOR I:=1:9 DO WRITE FE( ,I, )= ,FE(I); WRITE " RETURN"; write " end"; shut "zload.ftn"; bye; SUBROUTINE LOAD(X 1,Y 1,X2,Y2,X3,Y3,F3,FE) IMPLICIT REAL8(A-H,O-Z) DIMENSION FE(9) pj=(x 1-x3)(y2-y3)-(y 1-y3) (x2- x3) FE(1)=(PJF3)/6 FE(2)=-(PJ F3(2X 1-X2-X3))/48 FE(3)=(PJF3(Y2+Y3_2,y 1))/48 FE(4)=(PJ F3)/6 FE(5)=(PJ F3 (X 1-2X2+X3))/48 FE(6)=-(PJF3(2Y2_Y3_y 1))/48 FE(7)=(PJF3)/6 FE(8)=(PJ F3(X 1+X2-2X3))/48 FE(9)=(PJF3(Y2-2Y3+y I))/48 RETURN end The deformed shape is shown in Figure 6.4 and the stress distribution pattern is in Figure 6.5.
In addition, three different sizes of mesh are tested to investigate the convergence of solution.
They are shown in Figure 6.6.
The convergence trend is presented in Figure 6.7 which is the plot of error vs. element mesh size.
3. Four-edge clamped plate bending under uniform load --- the only difference between this case and the last case is the boundary constraints.
The extra slope constraints are enforced in the edges in this case.
Therefore it is expected that the solution will be stiffer than that of the simply-supported case.
The deformed shape is shown in Figure 6.8.
The stiffer phenomenon is visible by comparing Figures 6.4 and 6.8.
135 I / ° |') 0"!
• 136 I ffl 137 32 element 64 element 98 element Figure 6.6" Three different sizes of mesh for testing the convergence of solution of plate bending $ 0 3 n_m_ • Figure 6.7 : Convergence of plate bending solution 4 138 !
0'!
£',,I E_o ("q I= c7' °," ,c:: .
o I N L L--0 °.
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5, pp 691-705, 1986.
141 Kolbein Bell, "A refined triangular plate bending finite element", Int.
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Smith,"A finite element analysis for 'moderated-thick' rectangular plates in bending", Int.
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Sci., Vol.
10, pp 563-570, 1968.
I. M. Smith and W. Duncan, 'The effectiveness of excessive nodal continuities in the finite element analysis of thin rectangular and skew plates in bending", Int.
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for numerical methods in engi., Vol.
2, pp 253-257, 1970.
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De Veubeke, "A conforming finite element for plate bending", Int.
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Isaac Fried, "Shear in C O and C 1 bending finite elements", Int. J. Solids Structures, Voi.
9, pp 449-460, 1973.
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142 F.
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Fox, L. A.
Schmit, Jr., "The generation of inter-element-compatible stiffness and mass matrices by the use of interpolation formulas", pp 397-423, AFFDL-TR-66-80.
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Weaver & P. R. Johnston, "Finite elements for structural analysis", prentice-Hall, 1984.
S.
Klein, "A study of the matrix displacement method as applied to shells of revolution", pp 275-298, AFFDL-TR-66-80.
P. K. Mishra and S. S. Dey, "discrete energy method for the analysis of cylindrical shells", Computer & Structures, vol 27, No.6, pp 753-762,1987.
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Argyris, De, Fraes, "Matrix displacement analysis of anisotropic shells by triangular elements", J. of the royal aeronautical society, Vol.
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D. J. Dawe, "Rigid-body motions and strain-displacement equations of curved shell finite elements", Int. J. mech.
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of Michigan, Aug., 1987 R. S. Millman, G.
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143 W.L.Tsai, " Applications of symbohc and algebraic manipulation software in solving applied mechanics problems", Ph.D.
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L44 CHAPTER VII CONCLUSIONS 7.1 Introduction The topics discussed in this chapter include the advantages of using symbolic and algebraic manipulation, the difficulties existing in running symbolic and algebraic software, the SAM in education, contributions, and the prospect of future development and application.
Simple examples will be presented to illustrate the points where necessary.
7.2 Advantages of symbolic and algebraic manipulation There are many advantages of application of symbolic and algebraic manipulation.
They can be classified as follows • 1.
Tireless power---Together with human intelligence, the tireless capability in manipulating symbols and numbers has made SAM an indispensable tool in modern computational community.
It has created the potential to challenge both previously intractable problems and new sophisticated formulae.
Due to this advantage, the analytical work is pushed forward.
2. Accuracy --- A solution obtained by symbolic and algebraic manipulation is always exact.
There is no round-off error accumulation.
. Reliability --- The resultant expressions obtained by symbolic and algebraic manipulation will be correct if the input information is right.
In addition, the capability of automatic code generation eliminates any typographic errors and substantially reduces the time in debugging the programs.
= Efficiency --- This is a new advantage found in this research.
The symbolic template in nonlinear numerical analysis can significantly improve the efficiency of program execution.
This advantage is believed to be a crucial solution in the future for fields in 145 whichthedevelopment time playsanimportantrole.Forexample,therealtimecontrol will beoneof them.
7.3 Internal swelling and mathematical limitations As mentioned in the last chapters, there are some difficulties existing in the symbolic and algebraic manipulation.
The followings are the detail discussions : . Memory capacity limitation --- A large amount of memory space is required for symbolic and algebraic manipulation.
This is one of its fundamental limitations.
The amount of memory space needed for running a symbolic and algebraic manipulator varies a lot from one system (hardware and software) to another.
Different software systems need different sizes of memory space, and different hardware systems may provide different amounts of memory space for the same software.
Even the same software running in the same hardware system sometimes may need different memory spaces depending on whether external packages are connected or not.
For example, 834560 bytes RAM (about 815 K bytes) are currently provided (Fa11,1989) for running REDUCE in the Michigan Terminal System (MTS) when it is invoked.
If integration performance is involved in the computation, the external integration package should be manually included and the memory space will be extended to 1048560 bytes (about 1 mega bytes).
The total memory space that MTS can provide during the computation is up to seven mega bytes.
On the other hand, three mega bytes are provided to run REDUCE in an Apollo Domain workstation at the Computer Aided Engineering Network (CAEN) of The University of Michigan.
Unlike MTS, this space can be automatically extended up to six mega bytes during execution if necessary.
When the space requirement is beyond the provisions of hardware, execution will be aborted automatically.
Therefore it is recommended that the symbolic and algebraic manipulator be implemented on a machine with at least one mega bytes RAM capacity to guarantee a safe execution.
To demonstrate the mechanism of internal swell in symbolic and algebraic manipulation, an example is given to calculate the factorial of a number.
Mathematically, a factorial is deft ned as : 1 n if n _.0; n!-(n -1)!
otherwise.
(7.1) The LISP function for this problem is as follows : 146 (defunfactorial(n) (if (= nO) 1 ( n (factorial(1- n))))) Whenthefunctionis calledto calculatethefactorialof four, thebuilding process will occurfirst andthenthecollapsing process follows 9.
(factorial4) -> ( 4 (factorial 3)) -> ( 4 ( 3 (factorial2))) -> ( 4 ( 3 ( 2(factorial1)))) -> ( ,4 ( 3 ( 2 ( 1 (factorial 0))))) -> ('4('3('2( 11)))) -> ('4('3('21))) -> ('4('32)) -> ( 4 6) -> 24 The internal swelling phenomenon occurs during the process of building.
It is unquestionable that this phenomenon will become more serious if a larger number is given.
Moreover if input number is negative, the recursive process will theoretically continue infinitely.
Of course, the execution will be aborted when the provided space is used up.
2. Mathematical limitation --- Strictly speaking, a mathematical limitation should not be completely classified as limitation of symbolic and algebraic manipulation.
For example, the analytical solution for the general 5th polynomial is proven to be non-existent.
9 In some cases, the collapsing process may be impossible and the swelling phenomenon will last to the end of execution if it is not beyond the capacity of the hardware system.
147 Thereforeit is alsoimpossiblefor symbolicandalgebraicmanipulationto solve this problem.However,in additionto theabovementioned example,difficulties in solving mathematical equations whoseanalytical solutions existaresometimes encountered.
The occurrenceof this phenomenon is quite systemdependent.
In general,mostof such occurrences areencountered during integrationand equationsolving (algebraicand differentialequation).
7.4 Symbolic and algebraic mafiipulation and education The impact of SAM to science and engineering is significant.
There are many publications of its application on celestial mechanics, relativity theory, and fluid mechanics.
Compared to such successful applications, the response from educational community is far behind.
So far, schools which officially include symbolic and algebraic manipulation in the content of courses include Cornell University, The University of Pennsylvania, and The University of Michigan.
At Cornell University, MACSYMA was the first system introduced into the graduate-level course, in the winter of 1983.
It was not until the fall of 1984 that Professor Richard Rand introduced the muMATH system into the sophomore engineering mathematics course.
Unlike the MACSYMA system which ran on the mainframe, the muMATH system was implemented on IBM-XT and AT.
At The University of Pennsylvania, Professor H. H. Bau employed MACSYMA in the instruction of approximate analyses.
At The University of Michigan, Professor Noboru Kikuchi has used REDUCE to facilitate courses of finite element methods and applied mathematics since 1985.
The other system, MATHEMATICA, was also implemented into the Macintosh II in the computational laboratory by Professor Kikuchi around 1988.
Others such as Professor P. Papalambros and Professor R. Scott also used REDUCE in the courses on optimal design and finite element method.
The introduction of symbolic and algebraic manipulation into the education field should certainly be encouraged.
So far, some critic views have been reported by students at The University of Michigan.
They are : 1. Since there is no introductory course in symbolic and algebraic manipulation, students always struggle in learning the symbolic and algebraic manipulator itself rather than its application to the subject.
2.
Most of manuals of symbolic and algebraic manipulators, such as REDUCE and MACSYMA, are unfriendly to the users.
It is difficult for new users to understand the new terminologies in such a short time.
148 , There is no appropriate textbook to facilitate to teach symbolic and algebraic manipulation and its application l0 . Unlike numerical analysis, the amounts of symbolic and algebraic manipulation results are not predictable.
Therefore if assignment is not carefully designed, it could turn out just as simple as a symbol (say '0') or several hundred pages of outputs or nothing at all due to the internal swelling problem.
4. Qualified instructors are not easily found.
. The software may not be fully operational.
For instance, the MACSYMA system at The University of Michigan has just a half of its full capabilities.
It is not easy to use because some functions cannot be found even when they are listed in the manual.
The MATHEMATICA system is only implemented in some specific offices and is not yet available for public use.
In order to overcome these problems, some proposals are suggested as follows • , The education of symbolic and algebraic manipulation should start from the early undergraduate period.
It is recommended that the existing "Numerical analysis" course be revised into "Numerical analysis and symbolic manipulation".
The concept of symbolic manipulation, the use of available software, and the complementarity between symbolic manipulation and numerical analysis should be taught in the course.
2. It is urgent to design a textbook for such a new course.
The existing manuals need be revised for easy accessibility.
3. The software systems should be rechecked and made available to the public.
7.5 Contributions of this study The study presented in this report is believed to have made three original contributions to applied mechanics and symbolic manipulation.
They are • A). To applied mechanics" 1. Before this study, all of the advantages from the applications of symbolic and algebraic manipulation were either in handling lengthy formulae, or in increasing the accuracy of solution.
In addition, this report points out for the first time a new advantage in 10 The one written by Gerhand Rayna is rather an experimental book of REDUCE than an application of REDUCE in applied mechanics.
149 improvingtheefficiencyof theexecution of anumerical program.
It isbelievedthatthis advantagewill becrucial in suchapplicationsthat the developmenttime plays an importantrole.Forinstance, if thetemplate is prepared in symbolicform beforehand and implementedin a chip, the datareceivedby a heat-seeking missile can simply be substituted into thetemplate.
Theresponse timewill besubstantially shortened.
1 The closed-form solution of a stiffness matrix of a 4-node quadrilateral isoparametric element was not available before.
This dissertation presents the first analytical solutions of it. The contributions to the finite element analysis by this breakthrough are multifolds.
First, the integration error is eliminated and the solutions are more accurate.
Secondly, the closed-form solution can be automatically coded into a fortran subroutine.
This allows the element stiffness matrix to be obtained by simple substitution of nodal coordinates.
Of course, the fortran programming is simplified and the assemblage of the global stiffness matrix is expedited.
B). To Symbolic and algebraic manipulation • . Although the difficulties of the internal swelling problem and mathematical limitation were well known in the SAM field, no one has given the remedy for it. This report proposes a sysmatic pre-treatment method to avoid these difficulties and then successfully applies it to solve the problems.
7.6 Prospects and continuations of this research As the criterion of the quality of results (in both industry and academia) becomes more and more strict, it is expected that more and more sophisticated formulations will be produced.
Some expectations of future trends are as follows • o The design trend of symbolic and algebraic systems will continuously go towards smaller, more convenient packages for personal computers or even calculators.
However, the mainframe SAM system will still co-exist to process large-expressions.
. Applications of SAM in industry are scare at this time.
However, this situation will change gradually after the teaching of symbolic and algebraic manipulation is actually implemented in schools.
150 , The relationship between numerical analysis and symbolic manipulation will be smoother in the future.
The switch from symbolic manipulation to numerical analysis (or vice versa) is expected to be automatic eventually 4. The gap between theoretical analysis and computational experiments will be smaller and smaller.
, The reconsideration of every problem, equation, and formulation will become necessary.
Regardless of whether they have already been solved or not.
The solved problem can be used to check the correctness of solutions by SAM.
The unsolved problem might then become solvable with the employment of SAM.
6. The inclusion of higher order terms for applied mechanics problems will become popular due to the availability of SAM systems.
, To debug the symbolic program and to check the correctness of results are the important associated tasks of symbolic and algebraic manipulation.
It is expected that the self debugging function of symbolic manipulators will be developed soon.
A technique for the sysmatic checking of results from symbolic and algebraic manipulation should be available in the future.
The following three topics are closely relative to this study, and should be continued in future research.
They are : 1. Extension of methodology in constructing a stiffness matrix for 2-D isoparametrical quadrilateral element to a 3-D problem.
!
a o --'7_ ---) 2. Inclusion of the higher derivative term of 0t, r in Equation (6.23) to Equation (6.25).
This was neglected in the original formulation by Lee and Kobayashi and in this thesis.
3. Extension of methodology presented in section 7.3 to solve the general shell problem.
151 7.7 References P. H.
Winston, B. K. P. Horn, "LISP", 2nd edition, Addison Wesley, 1984.
Robert R Kessler, "LISP, objects and symbolic programming", Scott, Foresman and Company, 1988.
Editor H.
H.
Bau, T.
Herbert, "Symbolic computation in fluid mechanics and heat transfer", ASME HTD-Vol.
105, AMD-Vol.
97, 1988.
P. H. Winston, "Artificial intelligence", 2nd edition, Addison Wesley, 1984.
152 APPENDIX A Proof of Equation (5.12) V Given: ff(17,)--mg t-Prove: ff(17 ).
17;-ff(17).
17_ >0 Proof : Let mg =1 for simplicity, /_ mF(17r)V'--ff(17;) ° _1¢; vr v;t--v;t ) There are three cases for discussions • Case 1 • when V _ > 0 ,equation (A. 1) will be Z0' when V, _0 0, when V, >0 Case 2" when V _ - 0, equality is hold.
Case 3 • when V _ < 0, equation (A. 1) will be V, {>0, when V, aO F -V'(-1-]--), 0, when V, <0 Therefore, ff(fr ).
17;-ff(17_).
17_ a O is proven (A.
1) (A.
2) (A .3) 153 APPENDIX B Proof of Equation (5.16) Given : f(17) .m g [_t _ Prove" /:lf..
dfr - ,y.
d r 0 0 :r(e,) .( ',- ',) (B. _) Proof : Let mg =1 for simplicity, 0 0 /v, >o-• ',--Iv:l, If V, <0= " • df r, -V f0 g t~ t" -- sign ( V , )dV , -_dV , • :1 'r.
',--Iv,I fl_'lf •d_',- Iv,I There are four cases for discussions" 1. V_ >O,V, >0 case" Left side of (B.1)=-Iv:l +Iv,I --V;+V V, - - ]-V7 - V,)=fight side of (B.1) 2. V > 0,V, < 0 case.
(B .2) (B.3) (B .4) Leftsideof (B.1)= IV;l_ Iv,I- v; +v, •¢ -]( V," - V, ) =right side of (B. 1) 3.
V 0 case • Left side°f(B'l)=-Iv:l + Iv,I- v: +v 154 V • .:-V_' + V, --](V -V,)=right side of(B.1) 4.
V_ <0.V, <0 case" Left side of Iv:I-Iv,I-- v: +v V, --(V_ - V,) =-( - rw--r)(V" - V_) IV,I =-1(right side of (B. 1)) (B.5) case 4.
Equation (B.5) implies that equality is hold and both sides of (B. 1) are zero for Therefore equation (B. 1) is proven.
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FUNDING NUMBERS Application of Symbolic and Algebraic Manipulation Software in Solving Applied Mechanics Problems 6. AUTHOR(S) NCA2-136 Wen-Lang Tsai and Noboru Kikuchi 7.
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PERFORMING ORGANIZATION REPORT NUMBER University of Michigan Department of Mechanical Engineering A-93112 Ann Arbor, MI 48109 9.
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ABSTRACT (Maximum 200 words) As its name implies, symbolic and algebraic manipulation is an operational tool which not only can retain symbols throughout computations but also can express results in terms of symbols.
This report starts with a history of symbolic and algebraic manipulators and a review of the literatures.
With the help of selected examples, the capabilities of symbolic and algebraic manipulators are demonstrated.
These applications to problems of applied mechanics are then presented.
They are the application of automatic formulation to applied mechanics problems, application to a materially nonlinear problem (rigid-plastic ring compression) by finite element method (FEM) and application to plate problems by FEM. At the end of the report, the advantages and difficulties, contributions, education and perspectives of symbolic and algebraic manipulation are discussed.
It is well known that there exist some fundamental difficulties in symbolic and algebraic manipulation, such as internal swelling and mathematical limitation.
This report proposes a remedy for these difficulties, and successfully solves the three applications mentioned above.
For example, the closed from solution of stiffness matrix of four-node isoparametrical quadrilateral element for 2-D elasticity problem was not available before.
Due to the work presented here, the automatic construction of it becomes feasible.
In addition, a new advantage of the application of symbolic and algebraic manipulation found in this study is believed to be crucial in improving the efficiency of program execution in the future.
This will substantially shorten the response time of a system.
It is very significant for certain systems, such as missile and high speed aircraft systems, in which time plays an important role.
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solution, in chemistry, a homogenous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible. Air, for example, is a solution consisting chiefly of oxygen and nitrogen with trace amounts of several other gases, and brass is a solution composed of copper and zinc.
A brief treatment of solutions follows. For full treatment, see liquid: Solutions and solubilities.
Life processes depend in large part on solutions. Oxygen from the lungs goes into solution in the blood plasma, unites chemically with the hemoglobin in the red blood cells, and is released to the body tissues. The products of digestion also are carried in solution to the different parts of the body. The ability of liquids to dissolve other fluids or solids has many practical applications. Chemists take advantage of differences in solubility to separate and purify materials and to carry out chemical analysis. Most chemical reactions occur in solution and are influenced by the solubilities of the reagents. Materials for chemical manufacturing equipment are selected to resist the solvent action of their contents.
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liquid: Solutions and solubilities
The liquid in a solution is customarily designated the solvent, and the substance added is called the solute. If both components are liquids, the distinction loses significance; the one present in smaller concentration is likely to be called the solute. The concentration of any component in a solution may be expressed in units of weight or volume or in moles. These may be mixed—e.g., moles per litre and moles per kilogram.
Crystals of some salts contain lattices of ions—i.e., atoms or groups of atoms with alternating positive and negative charges. When such a crystal is to be dissolved, the attraction of the oppositely charged ions, which are largely responsible for cohesion in the crystal, must be overcome by electric charges in the solvent. These may be provided by the ions of a fused salt or by electric dipoles in the molecules of the solvent. Such solvents include water, methyl alcohol, liquid ammonia, and hydrogen fluoride. The ions of the solute, surrounded by dipolar molecules of the solvent, are detached from each other and are free to migrate to charged electrodes. Such a solution can conduct electricity, and the solute is called an electrolyte.
The potential energy of attraction between simple, nonpolar molecules (nonelectrolytes) is of very short range; it decreases approximately as the seventh power of the distance between them. For electrolytes the energy of attraction and repulsion of charged ions drops only as the first power of the distance. Accordingly, their solutions have very different properties from those of nonelectrolytes.
It is generally presumed that all gases are completely miscible (mutually soluble in all proportions), but this is true only at normal pressures. At high pressures, pairs of chemically dissimilar gases may very well exhibit only limited miscibility. Many different metals are miscible in the liquid state, occasionally forming recognizable compounds. Some are sufficiently alike to form solid solutions (see alloy).
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| Howkins & Bourne: Shaw's Textbook of Gynaecology, 18th Edition - E-Book Sunesh Kumar, V. G. Padubidri, Shirish N Daftary Elsevier Health Sciences, Sep 1, 2022 - Medical - 596 pages - Content Organized in a logical sequence that aids learning and enables the students to build sound knowledge of the subject. - Book covers the entire course curriculum in a narrative manner that helps build concepts and makes it easy to retain and reproduce. - Colored Illustrations, pathological images and slides, and supporting sonographs have been included extensively to enhance understanding of various diseases. - Key points at the end of chapter for quick revision. - Self-assessment questions at the end of each chapter help in preparing for expected/frequently asked questions in the examination. - Updated and revised as per new CBME curriculum. - Vertical and horizontal integration of the topics has been done. - Keeping in view wide variation in practice and opinion in the latest suggestions made by WHO and Govt of India, new guidelines pertaining to Indian perspective have been included in chapter Diagnosis of Female Genital Tuberculosis'. - The book is updated with recent guidelines and staging. - Latest FIGO classification Ca Cervix has been included. - Clinical cases included at the end of most of the chapters to provide the students a detailed workup for commonly encountered conditions. Preview this book » |
Selected pages
Table of Contents
Index
Contents
| | |
--- |
| 1 Approach to a gynaecological patient OG 333 OG 351 AN 488 | 1 |
| |
| SECTION 1 ANATOMY PHYSIOLOGY AND DEVELOPMENT OF FEMALE REPRODUCTIVE ORGANSSECTION | 12 |
| |
| SECTION 2 DISORDERS OF MENSTRUATION | 118 |
| |
| SECTION 3 COMMON CONDITIONS IN GYNAECOLOGY | 204 |
| |
| SECTION 4 BENIGN CONDITIONS IN GYNAECOLOGY | 290 |
| |
| SECTION 5 INFECTIONS IN GYNAECOLOGY | 344 |
| |
| | |
--- |
| SECTION 6 URINARY AND INTESTINAL TRACT IN GYNAECOLOGY | 381 |
| |
| SECTION 7 GYNAECOLOGICAL MALIGNANCIES | 419 |
| |
| SECTION 8 IMAGING MODALITIES ENDOSCOPIC PROCEDURES AND MAJOR AND MINOR OPERATIONS IN GYNAECOLOGY | 525 |
| |
| Index | 579 |
| |
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| Title | Howkins & Bourne: Shaw's Textbook of Gynaecology, 18th Edition - E-Book |
| Editors | Sunesh Kumar, V. G. Padubidri, Shirish N Daftary |
| Edition | 18 |
| Publisher | Elsevier Health Sciences, 2022 |
| ISBN | 8131266311, 9788131266311 |
| Length | 596 pages |
| Subjects | › Gynecology & Obstetrics Medical / Gynecology & Obstetrics |
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| Export Citation | BiBTeX EndNote RefMan |
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188473 | http://educ.jmu.edu/~kohnpd/236/TKsection0_4.pdf | 0.4 Exponential and Trigonometric Functions ▷Definitions and properties of exponential and logarithmic functions ▷Definitions and properties of trigonometric and inverse trigonometric functions ▷Graphs and equations involving transcendental functions Exponential Functions Functions that are not algebraic are called transcendental functions. In this book we will investigate four basic types of transcendental functions: exponential, logarithmic, trigono-metric, and inverse trigonometric functions. Exponential functions are similar to power func-tions, but with the roles of constant and variable reversed in the base and exponent: Definition 0.21 Exponential Functions An exponential function is a function that can be written in the form f(x) = Abx for some real numbers A and b such that A ̸= 0, b > 0, and b ̸= 1.
There is an important technical problem with this definition: we know what it means to raise a number to a rational power by using integer roots and powers, but we don’t know what it means to raise a number to an irrational power. We need to be able to raise to irrational powers to talk about exponential functions; for example, if f(x) = 2x then we need to be able to compute f(π) = 2π. One way to think of bx where x is irrational is as a limit: bx = lim r→x r rational br.
The “lim” notation will be explored more in Chapter 1. For now you can just imagine that if x is rational we can approximate bx by looking at quantities br for various rational numbers r that get closer and close to the irrational number x.For example, 2π can be approximated by 2r for rational numbers r that are close to π: 2π ≈23.14 = 2 314 100 = 100 √ 2314.
As we consider rational numbers r that are closer and closer to π, the expression 2r will get closer and closer to 2π; see Exercise 4. In Chapter xxx we will give a more rigorous definition of exponential functions as the inverses of certain accumulation integrals.
Interestingly, the most natural base b to use for an exponential function isn’t a simple inte-ger, like b = 2 or b = 3. Instead, for reasons that will become clear when we study derivatives, the most natural base is the irrational number known as e, and the function ex is therefore called the natural exponential function. The first 75 decimal places of the number e are: 2.71828182845904523536028747135266249775724709369995957496696762772407663035....
Of course, since e is an irrational number, we cannot define e just by writing an approximation of e in decimal notation; we will define e properly once we cover limits in Chapter xxx.
In Exercise 88 you will prove that every exponential function can be written so that its base is the natural number e: 0.4 Exponential and Trigonometric Functions 48 Theorem 0.22 Natural Exponential Functions Every exponential function can be written in the form f(x) = Aekx for some real number A and some nonzero real number k.
Ev-ery exponential function has a graph similar to either the exponential growth graph below left or the exponential decay graph below right, depending on the value of k or b. Of course, if the coefficient A is negative, then the graph of f(x) = Aekx or f(x) = Abx will be an upside-down version of one of these two graphs.
f(x) = ekx with k > 0, f(x) = bx with b > 1 f(x) = ekx with k < 0, f(x) = bx with 0 < b < 1 1 1 Logarithmic Functions Since every exponential function bx is one-to-one, every exponential function has an inverse.
These inverses are what we call the logarithmic functions: Definition 0.23 Logarithmic Functions as Inverses of Exponential Functions The inverse of the exponential function f(x) = bx is the logarithmic function g(x) = logb x.
As a special case, the inverse of the natural exponential function f(x) = ex is the natural logarithmic function g(x) = lnx.
We require that the base b satisfies b > 0 and b ̸= 1, because these are exactly the conditions we must have for y = bx to be an exponential function. In Section xxx we will define logarithms another way, in terms of integrals and accumulation functions.
You should already be familiar with the algebraic rules of logarithms, but we restate them here in case you need a refresher; see Exercises 90–94 for proofs.
0.4 Exponential and Trigonometric Functions 49 Theorem 0.24 Algebraic Rules for Logarithmic Functions For all values of x, y, b and a for which these expressions are defined, we have: (a) logb x = y if and only if by = x (b) logb(bx) = x (c) blogb x = x (d) logb(xa) = alogb x (e) logb(xy) = logb x + logb y (f) logb( 1 x) = −logb x (g) logb( x y ) = logb x −logb y (h) logb x = loga x loga b The first three properties follow from properties of inverse functions, and tell us that logb x is the exponent to which you have to raise b in order to get x. For example, log2 8 is the power to which you have to raise 2 to get 8; since 23 = 8 we have log2 8 = 3. All of these rules also apply to the natural exponental function, since lnx is just logb x with base b = e.
Properties (d) and (e) follow from the algebraic rules of exponents, and properties (f) and (g) are their immediate consequences. The final property in Theorem 0.24 is called the base conversion formula, because it allows us to translate from one logarithmic base to another.
The base conversion formula is especially helpful for converting to base e or base 10 so that we can calculate logarithms on a calculator. For example, log7 2 is equal to ln 7 ln 2, which we can approximate using the built-in ln key on a calculator.
The graphs of logarithmic functions can be obtained easily from the graphs of exponential functions by reflection over the line y = x, as shown below.
g(x) = logb x with b > 1 g(x) = logb x with 0 < b < 1 1 1 Trigonometric Functions There are six trigonometric functions defined as ratios of side lengths of right triangles, or more generally, as ratios of coordinate lengths on the unit circle. We now provide a quick review of the definitions of these functions and their graphical and algebraic properties.
Throughout most of this book we will be using radian measure for angles (not degrees).
Given any angle θ in standard position, the terminal edge of θ intersects the unit circle at some point (x,y) in the xy-plane. We will define the height y of that point to be the sine of θ, while the cosine of θ will be defined as the x-coordinate of that point.
0.4 Exponential and Trigonometric Functions 50 Definition 0.25 Trigonometric Functions for Any Angle Given any angle θ measured in radians in standard position, let (x,y) be the point where the terminal edge of θ intersects the unit circle. The six trigonometric functions of an angle θ are the six possible ratios of the coordinates x and y for θ: θ (x,y) (cos θ, sin θ) x y sinθ = y cosθ = x tanθ = y x cscθ = 1 y secθ = 1 x cotθ = x y Notice that the sine and cosine functions determine the remaining four trigonometric func-tions, since, tanθ is the ratio sin θ cosθ, and the last three trigonometric functions are the reciprocals of the first three.
You should already be familiar with the basic trigonometric identities, but they are re-peated below for your review; see Exercises 95–100 for proofs. The first Pythagorean iden-tity, the even/odd identities, and the shift identities follow easily from the definitions of the trigonometric functions. The sum identities follow from a geometric argument that we will not get into here. The remaining identities can all be proved from the previous identities. In these identities we are using the notation sin2 x as shorthand for (sinx)2.
Theorem 0.26 Basic Trigonometric Identities Pythagorean Identities Even/Odd Identities Shift Identities sin2 θ + cos2 θ = 1 sin(−θ) = −sinθ cos(θ −π 2 ) = sinθ tan2 θ + 1 = sec2 θ cos(−θ) = cosθ sin(θ + π 2 ) = cosθ 1 + cot2 θ = csc2 θ tan(−θ) = −tanθ sin(θ + 2π) = sinθ cos(θ + 2π) = cosθ Sum Identities Difference Identities sin(α + β) = sinαcosβ + sinβ cosα sin(α −β) = sinαcosβ −sinβ cosα cos(α + β) = cosαcosβ −sinαsinβ cos(α −β) = cosαcosβ + sinαsinβ Double Angle Identities Alternate Forms Alternate Forms sin2θ = 2 sinθ cosθ cos2θ = 1 −2sin2 θ sin2 θ = 1−cos 2θ 2 cos2θ = cos2 θ −sin2 θ cos2θ = 2cos2 θ −1 cos2 θ = 1+cos 2θ 2 The graphs of the six trigonometric functions are recorded below. Each of the graphs in the second row is the reciprocal of the graph immediately above it. Remember that you can use the graph of a function f to sketch the graph of its reciprocal 1 f . In particular, the zeros of f will be vertical asymptotes of 1 f , large heights on the graph of f will become small heights on the graph of 1 f , and vice-versa.
0.4 Exponential and Trigonometric Functions 51 y = sin x y = cosx y = tanx π 2π 3π −π −2π −3π −2 −1 2 1 π 2π 3π −π −2π −3π −2 −1 2 1 π 2π 3π −π −2π −3π −3 3 −2 −1 2 1 y = cscx y = secx y = cotx π 2π 3π −π −2π −3π −3 3 −2 −1 2 1 π 2π 3π −π −2π −3π −3 3 −2 −1 2 1 π 2π 3π −π −2π −3π −3 3 −2 −1 2 1 Inverse Trigonometric Functions None of the six trigonometric functions are one-to-one, but after restricting domains we can construct the so-called inverse trigonometric functions. In this section we will focus on the inverses of only three of the six inverse trigonometric functions, those for sine, tangent, and secant. There are many different restricted domains that we could use to obtain partial in-verses to these three functions. We need to pick one restricted domain for each function and stick with it. In this text we will use the restricted domains shown below.
y = sin x restricted to the domain [−π 2 , π 2 ] y = tanx restricted to the domain (−π 2 , π 2 ) y = secx restricted to the domain [0, π 2 ) ∪( π 2 ,π] -1 1 !" !" 2 " 2 " !" !" 2 " 2 " -1 1 -1 1 !" !" 2 " 2 " Each of the restricted functions shown above is one-to-one, and thus invertible. The inverses of these restricted functions are the inverse sine, inverse tangent, and inverse secant functions.
Definition 0.27 The Inverse Trigonometric Functions (a) The inverse sine function sin−1 x is the inverse of the restriction of the function sinx to [−π 2 , π 2 ].
(b) The inverse tangent function tan−1 x is the inverse of the restriction of the function tanx to (−π 2 , π 2 ).
(c) The inverse secant function sec−1 x is the inverse of the restriction of the func-tion secx to [0, π 2 ) ∪( π 2 ,π].
0.4 Exponential and Trigonometric Functions 52 Notice that since the inputs to the trigonometric functions are angles, it is the outputs of the inverse trigonometric functions that are angles. We will interchangeably use the alternative notations arcsinx, arctanx, and arcsec x for these inverse trigonometric functions.
All of the properties of sin−1 x, tan−1 x, and sec−1 x come from the fact that they are the inverses of the restricted functions sinx, tanx, and secx. For example, we can graph the inverse trigonometric functions simply by reflecting the graphs of the restricted trigonometric functions over the line y = x, as shown below.
y = sin−1 x y = tan−1 x y = sec−1 x -1 1 !" !" 2 " 2 " -1 1 -!" !" 2 " 2 " -1 1 !" !" 2 " 2 " Although sin−1 x and (restricted) sinx are transcendental functions, their composition sin−1(sinx) = x is algebraic. This is obvious because these functions are inverses of each other. However, something more general and surprising is true: the composition of any inverse trigonometric function with any trigonometric function is algebraic; see Example 4.
Examples and Explorations Example 1 Finding values of transcendental functions by hand Calculate each of the following by hand, without a calculator.
(a) log6 3 + log6 12 (b) cos 5π 6 (c) sin−1 1 2 Solution.
(a) log6 3 is the exponent to which we would have to raise 6 to get 2; think 6? = 3. It is not immediately apparent what this exponent is. Similarly, it is not clear how to calculate log6 12 without a calculator. However, using the additive property of logs we can write log6 3 + log6 12 = log6(3 · 12) = log6 36 = 2.
The final equality above holds since 62 = 36.
(b) The diagram below left shows where the angle 5π 6 lies on the unit circle. If we draw a line from the point (x,y) where the angle meets the unit circle to the x-axis, we obtain a triangle whose reference angle is 30◦. Using the known side lengths of a 30–60–90 triangle with hypotenuse of length one, we can label the side lengths of our reference triangle, as shown below middle. This in turn means that we know the coordinates (x,y) = (− √ 3 2 , 1 2) of the point at which the terminal edge of θ intersects the unit circle. Therefore cos 5π 6 = − √ 3 2 .
0.4 Exponential and Trigonometric Functions 53 Angle θ = 5π 6 has reference angle 30◦ Side lengths of a 30-60-90 triangle with hypotenuse 1 π 6 is the angle in [−π 2 , π 2 ] whose sine is equal to 1 2 x y 5π 6 q = 5π 6 30° 1 1 2 x y q = 5π 6 30° √3 2 1 2 √3 2 − , ) ( 30° π 6 1 2 (c) If θ = sin−1 1 2, then we must have sinθ = 1 2. There are infinitely many angles whose sine is 1 2, but only one of those angles is in the restricted domain [−π 2 , π 2 ] of sine. Thus θ = sin−1( 1 2) is the unique angle in [−π 2 , π 2 ] whose sine is 1 2, as shown above right. Notice that the triangle must be a 30–60–90 triangle (since its height is 1 2), and therefore the angle θ we are looking for must be 30◦, i.e., π 6 radians. Therefore sin−1 1 2 = π 6 .
Example 2 Solving equations that involve transcendental functions Solve each of the following equations: (a) 3.25(1.72)x = 1000 (b) sinθ = cosθ (c) sec−1 x = π 3 Solution.
(a) To solve for x we will isolate the expression (1.72)x and then apply the natural logarithm so that we can get x out of the exponent: 3.25(1.72)x = 1000 = ⇒ln((1.72)x) = ln ! 1000 3.25 " = ⇒xln(1.72) = ln ! 1000 3.25 " .
It is now a simple matter to solve for x = ln “ 1000 3.25 ” ln(1.72) ≈10.564.
(b) If sinθ = cosθ, then θ is an angle whose terminal edge intersects the unit circle at a point (x,y) with x = y. The only such points on the unit circle are ( √ 2 2 , √ 2 2 ) and (− √ 2 2 ,− √ 2 2 ), as shown below left. The angles that end at these points are all of the form θ = π 4 + πk for some integer k. Thus the solution set for the equation is {...,−3π 4 , π 4 , 5π 4 , 9π 4 ,...}..
Diagram to solve sin θ = cosθ Diagram to solve sec−1 x = π 3 x y 45 π 4 2 2 √ 2 2 √ 45 2 2 √ 2 2 √ 3π 4 x y 30 π 3 3 2 √ 1 2 (c) If sec−1 x = π 3 , then x = sec π 3 = 1 cos π 3 = 1 1 2 = 2. The angle π 3 and the reference triangle we used for this calculation are shown above right.
0.4 Exponential and Trigonometric Functions 54 Example 3 Domains and graphs of transcendental functions Find the domain of each of the following functions. Then use transformations to sketch careful graphs of each function by hand, without a graphing utility.
(a) f(x) = 5 −3e1.7x (b) g(x) = 1 ln(x−2) (c) h(x) = 3sec2x Solution.
(a) The domain of f(x) = 5 −3e1.7x is all of R, and its graph is a transformation of the expo-nential growth function e1.7x shown below left. y = −3e1.7x can be obtained by reflecting the leftmost graph over the y-axis and stretching vertically by a factor of three, as shown below middle. The graph of f(x) = 5 −3e1.7x can now be obtained by shifting the middle graph up five units, as shown below right.
y = e1.7x y = −3e1.7x y = 5 −3e1.7x 1 -3 2 5 (b) For the function g(x) = 1 ln(x−2) to be defined at a value x, we must have x −2 > 0, and thus x > 2. We must also have ln(x −2) ̸= 0, which means that x −2 ̸= 1, and thus x ̸= 3.
Therefore the domain of g(x) is (2,3) ∪(3,∞). To sketch the graph of g(x) = 1 ln(x−2) we start with the graph of y = lnx shown below right, translate to the right two units as shown below middle, and then sketch the reciprocal as shown below right.
y = ln x y = ln(x −2) y = 1 ln(x −2) 1 2 3 4 5 -3 -2 -1 1 2 3 1 2 3 4 5 -3 -2 -1 1 2 3 1 2 3 4 5 -3 -2 -1 1 2 3 (c) The function h(x) = 3sec2x = 3 cos 2x is defined when cos2x ̸= 0. This happens when 2x is not a multiple of π 2 , and thus when x is not a multiple of π 4 . Thus the domain of h(x) is x ̸= π 4 k for positive integers k. To sketch the graph of h(x), we start with the graph of y = secx below left, stretch vertically by a factor of 3 as shown below middle, and then compress horizontally by a factor of 2 as shown below right.
0.4 Exponential and Trigonometric Functions 55 y = secx y = 3sec x y = 3sec2x !Π ! Π 2 Π 2 Π -9 -6 -3 3 6 9 !Π ! Π 2 Π 2 Π -9 -6 -3 3 6 9 !Π ! Π 2 Π 2 Π -9 -6 -3 3 6 9 Example 4 Simplifying compositions of inverse trigonometric and trigonometric functions Write cos(sin−1 x) as an algebraic function, that is, a function that involves only arithmetic operations and rational powers.
Solution.
If we define θ = sin−1 x, then sinθ = x and θ must be in the interval [−π 2 , π 2 ]. Let’s first consider the case where θ is in the first quadrant [0, π 2 ]; the reference triangle for such a θ is shown below left. If we wish θ to have a sine of x then the length of the vertical leg of the triangle must be x. The hypotenuse of the triangle is length 1, since we are on the unit circle. We could also have considered that the sine of θ is “opposite over hypotenuse”; thus one triangle involving our angle θ could have an opposite side of length x and a hypotenuse of length 1. Using the Pythagorean theorem, we find that the length of the remaining leg of the triangle is √ 1 −x2, as shown below right.
Reference triangle for an angle θ in [0, π 2 ] Use Pythagorean Theorem to determine length of remaining leg # # x 1 1!x2 $ Now cosθ is the horizontal coordinate of the point on the unit circle corresponding to θ, or in terms of “adjacent over hypotenuse,” we have: cosθ = √ 1 −x2 1 = # 1 −x2.
The case where θ is in the fourth quadrant, i.e., where θ ∈[−π 2 ,0], is similar and also shows that cosθ = √ 1 −x2. Therefore we have shown that cos(sin−1 x) is equal to the algebraic function √ 1 −x2.
Checking the Answer.
To verify the strange fact that cos(sin−1 x) = √ 1 −x2, try evaluating both sides at some simple x-values. While looking at just a few x-values will not prove that the two expressions are equal for all x, it will at least give us some evidence that the equality is reasonable. For example, at x = 0 we have cos(sin−1 0) = cos0 = 1 and # 1 −02 = √ 1 = 1, 0.4 Exponential and Trigonometric Functions 56 and at x = 1 we have cos(sin−1 1) = cos(π 2 ) = 0 and # 1 −12 = √ 0 = 0.
As a less trivial example, consider x = 1 2. At this value we have cos(sin−1( 1 2)) = cos( π 6 ) = √ 3 2 and $ 1 −( 1 2)2 = $ 1 −1 4 = $ 3 4 = √ 3 2 .
?
Questions. Test your understanding of the reading by answering these questions: ▷Why do we require that A ̸= 0 and b > 0, b ̸= 1 in the definition of exponential functions?
What would the graphs look like when A = 0, when b < 0, b = 0, or b = 1?
▷In the reading we calculated log7 2 by finding ln 7 ln 2 with a calculator. Would we get the same answer if we computed log10 7 log10 2?
▷How do you convert from radians to degrees, or vice-versa?
▷How is the graph of the reciprocal of a function related to the graph of that function?
How can that information be useful for remembering the graphs of y = cscx, y = secx, and y = cotx?
▷How are the unit circle definitions of the trigonometric functions related to the right-triangle definitions of trigonometric functions?
Exercises 0.4 Thinking Back Algebra with exponents: Write each of the following expres-sions in the form Abx for some real numbers A and b.
▷ 32x+1 ▷ 5x23−x ▷ (23x−5)4 ▷ 1 2(3x−4) ▷ 4(3x)2 2x ▷ ( 1 8)x 3(23x+1) Inverse functions: Suppose f and g are inverses of each other.
▷What can you say about f(g(x)) and g(f(x))?
▷If f has a horizontal asymptote at y = 0, what can you say about g?
▷If f has a y-intercept at y = 1, what can you say about g?
Famous triangles, degrees, and radians: The following exer-cises will help you review and recall basic trigonometry.
▷Suppose a right triangle has angles 30◦, 60◦, and 90◦ and a hypotenuse of length 1. What are the lengths of the remaining legs of the triangle?
▷Suppose a right triangle has angles 45◦, 45◦, and 90◦ and a hypotenuse of length 1. What are the lengths of the remaining legs of the triangle?
▷What is a radian? Is it larger or smaller than a de-gree? Compare an angle of one degree with an angle of one radian in standard position.
▷Show each of the following angles in standard posi-tion on the unit circle, in radians: ▷ 3π 4 ▷−4π 3 ▷ 17π 6 ▷21π Concepts 0.
Problem Zero: Read the section and make your own summary of the material.
1.
True/False: Determine whether each of the following statements is true or false. If a statement is true, ex-plain why. If a statement is false, provide a coun-terexample.
(a) True or False: The function f(x) = 3e0.5x −2 is an exponential function.
(b) True or False: Every exponential function f(x) = Aekx has a horizontal asymptote at y=0.
(c) True or False: For all x > 0, ln(x3) = 3lnx.
(d) True or False: For all x > 0, log2 x log2 3 = log6 x log6 3 .
(e) True or False: If (x,y) is the point on the unit circle corresponding to the angle −7π 3 , then x is positive and y is negative.
(f) True or False: The sine of an angle θ is always equal to the sine of the reference angle for θ.
(g) True or False: For any x, 1 −cos2(5x3) = sin2(5x3).
(h) True or False: sec−1 x = 1 cos−1 x.
2.
Examples: Give examples of each of the following. Try to find examples that are different than any in the reading.
(a) Two exponential functions and their inverses.
(b) Two x-values at which tanx is not defined.
(c) Two x-values at which sec−1 x is not defined.
3.
What is the definition of an exponential function, and how is it different from a power function? Is the func-tion f(x) = xx a power function, an exponential func-tion, or neither, and why?
4.
In this exercise we will examine two ways to think of ab when b is an irrational number, and in particular what the quantity 2π represents.
(a) One way to define 2π is to think of it as a limit.
If we take a sequence a1,a2,a3,... of rational numbers that approaches π, then the sequence 2a1,2a2,2a3,... should approach 2π.
Said in terms of limits, this means that: 2π = lim a→π 2a, where each a is assumed to be a rational num-ber.
Can you think of a sequence of rational numbers that get closer and closer to π? (Hint: Think about the decimal expansion of π.) (b) Another way to consider 2π is to write it as an infinite product: 2π = 23 2 1 10 2 4 100 2 1 1000 2 5 10000 2 9 100000 ··· .
What will the next term in the product be? How could 2π equal the product of infinitely many numbers?
Wouldn’t that make 2π infinitely large? Calculate some of the later terms in the product (for example, 2 5 10000 or 2 9 100000 ) and use these calculations to argue that even though 2π can be written as a product of infinitely many numbers, it is not necessarily infinitely large.
5.
Approximate 2 √ 3 by calculating 2r for rational val-ues r that get closer and closer to √ 3. (Hint: You can use the decimal expansion of √ 3 to get a sequence of ratio-nal numbers that approaches √ 3.) 6.
Why can’t we define the number e just by writing it down in decimal notation to lots of decimal places?
0.4 Exponential and Trigonometric Functions 58 7.
Write the exponential function f(x) = 3e−2x in the form Abx for some real numbers A and b. Then write the exponential function g(x) = −2(3x) in the form Aekx for some real numbers A and k.
8.
Fill in each blank with an interval of real numbers.
(a) An exponential function f(x) = Abx represents exponential growth if b ∈ , and exponential decay if b ∈ .
(b) An exponential function f(x) = Aekx repre-sents exponential growth if k ∈ , and ex-ponential decay if k ∈ .
(c) Suppose that ekx = bx for some real numbers k and b. Then k ∈(0,∞) if and only if b ∈ .
(d) Suppose that ekx = bx for some real numbers k and b. Then k ∈(−∞,0) if and only if b ∈ .
9.
In the definition of the logarithmic function logb x, what are the allowable values for the base b, and why?
10.
Fill in the blanks in each statement below.
(a) For all x ∈ , log2 x = y if and only if x = .
(b) For all x ∈ , 3log3 x = .
(c) For all x ∈ , log4(4x) = .
(d) log2 3 is the exponent to which you have to raise to get .
11.
The graphs of y = log2 x and y = log4 x are shown be-low. Determine which graph is which, without using a calculator. (Hint: Think about the graphs y = 2x and y = 4x, and then reflect those graphs over the line y = x.) y = log2 x and y = log4 x 1 2 3 4 -2 -1 1 2 12.
State the algebraic properties of the natural logarithm function that correspond to the eight properties of logarithmic functions in Theorem 0.24.
13.
Use algebraic properties of logarithms, the graph of y = ln x, and your knowledge of transformations to sketch graphs of f(x) = ln(x2) and g(x) = ln( 1 x).
14.
Solve the inequality ln( x+1 x−1) ≥0.
15.
Give a mathematical definition of sin θ for any angle θ. Your definition should include the words “unit circle,” “standard position,” “terminal,” and “coor-dinate.” 16.
Give a mathematical definition of tanθ for any an-gle θ. Your definition should include the words “unit circle,” “standard position,” “terminal,” and “coordi-nate.” 17.
Use the definition of the sine function to explain why sin( π 4 ) is equal to sin( 9π 4 ) and sin(−7π 4 ).
18.
Fill in each blank with an interval of real numbers.
(a) The function f(x) = cos x has domain and range .
(b) The function f(x) = csc x has domain and range .
(c) The restricted tangent function has domain and range .
(d) The function f(x) = sec−1 x has domain and range .
19.
Suppose θ is an angle in standard position whose ter-minal edge intersects the unit circle at the point (x,y).
If y = −1 3, what are the possible values of cosθ? If you know that the terminal edge of θ is in the third quadrant, what can you say about cosθ? What if the terminal edge of θ is in the fourth quadrant? Could the terminal edge of θ be in the first or second quad-rant?
20.
Show that − √ 3 is in the range of tangent by finding an angle θ for which tanθ = − √ 3.
21.
Describe restricted domains for sin x, tanx, and sec x on which each function is invertible. Then describe the corresponding domains and ranges for arcsin x, arctanx, and arcsec x.
22.
Fill in the blanks: (a) sin−1 x is the angle in the interval whose is x.
(b) y = arcsin x if and only if siny = , for all x ∈ and y ∈ .
(c) If tan−1 x = θ and tanθ is positive, then θ is in the quadrant.
(d) If arctanx = θ and sin θ = 1 3, then cos θ = .
23.
Which of the following expressions are defined?
Why or why not?
(a) sin−1(−1 25) (b) sin−1 3 2 (c) tan−1 100 (d) sec−1 π 4 24.
Sketch a graph of the restricted cosine function on the domain [0,π] and argue that this restricted function is one-to-one. Then sketch a graph of cos−1 x, and list the domains and ranges of the inverse cos−1 x of this restricted function.
25.
Without calculating the exact or approximate values of the following expressions, use the unit circle to de-termine whether each of the following quantities is positive or negative.
(a) sin−1(−1 5) (b) sin−1(−2 3) (c) tan−1 2 (d) sec−1(−5) 26.
Find all angles whose secant is 2, and then find sec−1(2).
0.4 Exponential and Trigonometric Functions 59 Skills Find the domains of the functions in Exercises 27–32.
27. f(x) = ln(x + 1) ln(x −2) 28. f(x) = 1 ex −e2x 29. f(x) = 1 p ln(x −1) 30. f(x) = 1 1 −tanθ 31. f(x) = √ secθ 32. f(x) = 2sin−1(x −3) Find the exact values of each of the quantities in Exer-cises 33–44. Do not use a calculator.
33. ln( 1 e2 ) 34. log 1 2 4 35. 4log2 6 −2log2 9 36.
log7 9 log7 1 3 + log3 1 37. tan(−π 4 ) 38. cos( 48π 3 ) 39. csc(−5π 4 ) 40. sin(201π) 41. cos−1(−1) 42. sin−1(−1) 43. arcsec (−2 √ 2) 44. arctan(−1 √ 3) Solve the equations in Exercises 45–50 by hand. When you are finished, check your answers either by testing your so-lutions or by graphing an appropriate function.
45. 2x = 3x−1 46. 2 = 10(1 + 0.19 12 )12x 47. log2( x−1 x+1) = 4 48. sinx = 1 2 49. cos 2x = 1 50. sec−1 x = π Suppose that cos(θ) = 1 6, sin(θ) > 0, sin(φ) = 3 5, and cos(φ) < 0.
Use trigonometric identities to identify the quantities in Exercises 51–56.
51. sin(θ) 52. sin(−φ) 53. cos(2θ) 54. sin(θ + π 2 ) 55. the sign of cos(θ + φ) 56. the sign of tan(θ + π) Write each of the expressions Exercises 57–60 as an alge-braic expression that does not involve trigonometric or in-verse trigonometric functions.
57. sin(cos−1 x) 58. tan(tan−1 2x) 59. sec2(tan−1 x) 60. sin2(tan−1 x) 61. sin(sec−1 x 3 ) 62. csc(2tan−1 x) 63. cos(2sin−1 5x) 64. tan2(2sec−1 x 3 ) Sketch graphs of the functions in Exercises 65–72 by hand, without using a calculator or graphing utility. Indicate any roots, intercepts, and asymptotes on your graphs.
65. f(x) = −( 1 2)x + 10 66. f(x) = −0.25(3x−2) 67. f(x) = 20 −5e−2x 68. f(x) = log 1 2 x 69. f(x) = −log2(x −3) 70. f(x) = sin(2x) + 4 71. f(x) = 2cos(x −π 4 ) 72. f(x) = tan−1(x−2) + π For each graph in Exercises 73–78, find a function whose graph looks like the one shown. When you are finished, use a graphing utility to check that your function f has the properties and features of the given graph.
73.
-4 -3 -2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 74.
-4 -3 -2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 75.
-1 1 2 -5 5 10 15 76.
-5 -4 -3 -2 -1 1 2 3 -4 -3 -2 -1 1 2 77.
-1 1 !" !" 2 " 2 " 78.
-2 -1 1 2 3 4 5 6 !2" !" " 2" Applications 0.4 Exponential and Trigonometric Functions 60 79.
Ten years ago, Jenny deposited $10,000 into an in-vestment account.
Her investment account now holds $22,609.80. Her accountant tells her that her investment account balance I(t) is an exponential function.
(a) Find an exponential function of the form I(t) = Aekt to model Jenny’s investment account bal-ance.
(b) Find an exponential function of the form I(t) = Abt to model Jenny’s investment account bal-ance.
80.
Suppose there were 500 rats on a certain island in 1973, and 1697 rats on the same island ten years later.
Assume that the number R(t) of rats on the island t years after 1973 is an exponential function.
(a) Find an equation for the exponential function R(t) that describes the number of rats on the is-land. Let t = 0 represent the year 1973.
(b) According to your function R(t), how many rats will be on the island in 2020?
(c) How long did it take for the population of rats to double from its 1973 amount? How long did it take for it to double again? And again?
81.
Suppose a rock sample initially contains 250 grams of the radioactive substance unobtainium, and that the amount of unobtainium after t years is given by an exponential function of the form S(t) = Aekt. The half-life of unobtainium is 29 years, which means that it takes 29 years for the amount of the substance to decrease by half.
(a) Find an equation for the exponential function S(t).
(b) What percentage of unobtainium decays each year?
(c) How long will it be before the rock sample con-tains only 6 grams of unobtainium?
82.
Again considering the rock sample described in Ex-ercise 81, answer the following questions: (a) At one point the rock sample contained 900 grams of unobtainium; how long ago?
(b) What percentage of the unobtainium will be left in 300 years?
(c) How long will it be before 95% of the unob-tainium has decayed?
83.
Alina is flying a kite, and has managed to get her kite so high in the air that she has let out 400 feet of kite string. If the angle made by the ground and the line of kite string is 32 degrees, how high is the kite?
84.
Suppose two stars are each 60 light years away from Earth. The angle between the line of sight to the first star and the line of sight to the second star is two de-grees. In other words, if you look at the first star, then turn your head to look at the second star, your head will move through an angle of two degrees. How far apart are the stars?
Proofs 85.
Prove by contradiction that every exponential func-tion f(x) = Abx has the property that f(x) is never zero.
86.
Use the definition of a one-to-one function to prove that every exponential function f(x) = Abx is one-to-one.
87.
Use the base conversion formula for logarithms to prove that the function f(x) = log2 x is equal to the function g(x) = log3 x only when x = 1.
88.
Use logarithms to prove that every exponential func-tion of the form f(x) = Abx can be written in the form f(x) = Aekx, and vice-versa.
89.
Use the definition of a logarithmic function y = logb x to prove that for any b > 0 with b ̸= 1, the quantity logb 1 is equal to zero.
In Exercises 90–94, assume that x, y, a, and b are values which make sense in the expressions involved.
90.
Use the fact that logarithmic functions are the in-verses of exponential functions to prove that: (a) logb x = y if and only if by = x (b) logb(bx) = x (c) blogb x = x 91.
Prove that logb(xa) = a logb x.
(Hint: Start with logb(xa) and replace x with blogb x.) 92.
Prove that logb(xy) = logb x + logb y.
(Hint: Show this is equivalent to the statement xy = blogb x+logb y, and prove this new statement instead.) 93.
Use the results of the two problems above to prove that: (a) logb( 1 x) = −logb x (b) logb( x y ) = logb x −logb y 94.
Prove the base conversion formula logb x = loga x loga b .
(Hint: Set y = logb x and then show that by = x.) 0.4 Exponential and Trigonometric Functions 61 95.
Use the unit circle definitions of sine and cosine to prove the identity sin2 θ + cos2 θ = 1.
96.
Use the first Pythagorean identity sin2 θ + cos2 θ = 1 to prove the second and third Pythagorean identities listed in Theorem 0.26. (Hint: To prove the second iden-tity, divide both sides of the first identity by cos2 x. A similar trick will prove the third identity.) 97.
Use the unit circle definitions of the trigonometric functions to prove the even/odd identities and the shift identities listed in Theorem 0.26.
98.
Use the sum identities and the even/odd identities to prove the difference identities listed in Theorem 0.26.
99.
Use the sum identities to prove the double angle identities listed in Theorem 0.26. (Hint: Note that 2θ is equal to θ + θ.) 100. The four identities listed as alternate forms in Theo-rem 0.26 are alternate ways of writing the double an-gle identity cos 2θ = cos2 θ −sin2 θ. Use this double angle identity, algebra, and the Pythagorean identi-ties to prove these four alternate forms.
Thinking Forward ▷A special exponential limit: Use a calculator to approx-imate eh−1 h for the following values of h: (a) h = 0.1; (b) h = 0.01; (c) h = 0.001. As h gets closer to zero, what number do your approximations seem to ap-proach?
▷Logarithms with absolute values: Sketch a graph of the function f(x) = ln |x|. What is the domain of this function? Is this function even, odd, or neither, and why?
▷Rewriting trigonometric expressions: Use the double an-gle identity sin2 x = 1−cos 2x 2 to rewrite the expres-sion sin4 x cos2 x in terms of a sum of expressions of the form A coskx. (Note: You’ll have to multiply out some expressions, and use the double angle identity more than once.) Appendix A Answers To Odd Problems 83. If f(x) = Ax3 + lower-degree terms and g(x) = Bx3 + lower-degree terms, then f(x)g(x) = ABx6 + lower-degree terms.
Since f and g are cubic we know that A and B are nonzero. Thus AB must also be nonzero, and therefore fg is of degree 6.
85. (a) If f(x) = k for all k, then f(x) = k = 0x + k is also a linear function. (b) If f(x) = mx + b is a linear function with m ̸= 0, then f is a polynomial of degree 1 with coefficients a1 = m and a0 = b. If f(x) = mx + b with m = 0 then f is a polynomial of degree zero with sole coefficient a0 = b.
86. The domain of a quotient f(x) = p(x) q(x) of functions is {x | x ∈Domain(p(x)) ∩ Domain(q(x)) and q(x) ̸= 0}.
Since p(x) and q(x) are polynomials, they are defined on (−∞,∞); thus the domain of f is {x | q(x) ̸= 0}.
Section 0.4 1.
F, T, T, T, T, F, T, F.
3.
A function is exponential if it can be written in the form f(x) = Abx; the variable is in the exponent and a constant is in the base. For a power function, the situation is reversed. xx is neither a power nor an exponential func-tion because a variable appears in both the base and the exponent.
5.
√ 3 ≈1.73205.
We have 21.7 ≈3.2490, 21.73 ≈3.3173, 21.732 ≈3.3219, 21.7320 ≈ 3.3219, 21.73205 ≈3.3220, and so on. Each of these approximations gets closer to the value of 2 √ 3.
7.
f(x) = 3(e−2)x ≈3(0.135)x, g(x) = −2e(ln 3)x ≈−2e1.0986x.
9.
We must have b > 0 and b ̸= 1, since those conditions are necessary for bx to be an ex-ponential function.
11. The graph that passes through (2,1) is log2 x; the graph that passes through (2, 1 2) is log4 x.
13. f(x) = 2ln x is the graph of ln x stretched vertically by a factor of 2; g(x) = −lnx is the graph of ln x reflected over the x-axis.
15. If θ is an angle in standard position, then sin θ the vertical coordinate y of the point (x,y) where the terminal edge of θ intersects the unit circle.
17. The terminal edges of the angles π 4 , 9π 4 , and −7π 4 all meet the unit circle at the same point (and in particular, at the same y-coordinate).
19. cosθ = x is − √ 8 3 if θ is in the third quadrant; cosθ = √ 8 3 if θ is in the fourth quadrant; θ cannot be in the first or second quadrant.
21. See Definition 0.27 for the restricted do-mains of the trigonometric functions, and thus the ranges of the inverse trigonometric functions. The domain of arcsin x is [−1,1], the domain of arctanx is all of R, and the domain of arcsec x is (−∞,−1] ∪[1,∞).
Their ranges are the restricted domains of sin x, tanx, and secx, respectively.
23. Only (a) and (c) are defined.
25. (a) negative; (b) negative; (c) positive; (d) positive 27. (2,3) ∪(3,∞) 29. (2,∞) 31. ... ∪(−5π 2 ,−3π 2 ) ∪(−π 2 , π 2 ) ∪( 5π 2 , 7π 2 ) ∪...
33. −2 35. 4 37. −1 39.
√ 2 41. π 43.
3π 4 45. x = ln 3 ln( 3 2 ) ≈2.70951 47. x = −17 15 49. x = πk, where k is any integer 51.
√ 35 6 53. −17 18 55. Negative 57.
√ 1 −x2 59. x2 + 1 61.
p 1 −( x 3 )2 63. 1 −2(5x)2 65. Start with the graph of y = ( 1 2)x, then reflect over the x-axis and shift up by 10 units.
67.
-1 1 2 15 20 69.
1 2 3 4 5 6 7 8 -3 -2 -1 1 2 3 4 5 6 71.
-2 -1 1 2 !3" 4 !" 4 " 4 3" 4 !5" 4 5" 4 73. f(x) = 2e−x −3 75. f(x) = −5e−x + 10 77. f(x) = −cos2x 79. (a) I(t) ≈10,000e0.08t; (b) I(t) = 10,000(1.085)t.
81. (a) S(t) = 250e−ln 2 29 t ≈250e−0.0239t, or equivalently, S(t) = 250(0.97638)t; (b) 2.39%; (c) 156 years 83. 211.97 feet 85. Seeking a contradiction, suppose that A and b are nonzero real numbers with Abx = 0 for some real number x. Since A ̸= 0 we know bx = 0, and therefore that x(lnb) = ln(bx) = ln0. But this is a contradiction because ln0 is undefined, so the product x(lnb) of real numbers cannot equal ln 0.
87. log2 x = log3 x ⇔ ln x ln 2 = ln x ln 3 ⇔ (ln3)(lnx) = (ln2)(lnx) ⇔ (ln x)(ln3 − ln2) = 0 ⇔lnx = 0 ⇔x = 1.
89. y = logb x if and only if by = x. Since the only solution to by = 1 is y = 0 (if b ̸= 0), we know that logb 1 = 0.
91. Since x = blogb x, we have logb(xa) = logb((blogb x)a) = logb(b(logb x)a = (logb x)a = a logb x.
93. (a) logb( 1 x) = logb(x−1) = −logb x; (b) logb( x y ) = logb(xy−1) = logb x −logb y.
95. (a) For any angle θ, (cosθ,sin θ) are the coor-dinates of the point where the terminal edge of θ meets the unit circle. Since the equa-tion of the unit circle is x2 + y2 = 1 ad we have x = cos θ and y = sin θ, we must have sin2 θ + cos2 θ = 1.
97. For any angle θ, sin θ is the y-coordinate where the terminal edge of θ meets the unit circle. The angle −θ is the angle of the same magnitude as θ but opening in the clock-wise direction from the x-axis, and therefore its terminal edge will be the same as the ter-minal edge of θ except flipped over the x-axis. Therefore the y-coordinates of these two terminal edges have the same magni-tude but opposite signs; in other words, sin(−θ) = −sinθ. The remaining even/odd identities can be proved in a similar fashion.
99. sin 2θ = sin(θ + θ) = sin θ cos θ + sinθ cos θ = 2sin θ cos θ.
The identity for cos 2θ is proved similarly, and the alternate forms follow from the first two forms and the Pythagorean identity.
Section 0.5 1.
F, T, T, F, T, T, F, F.
3.
If C is true, then D must be true. If C is false, then D may or may not be true.
5.
“For all x > 0, we have x > −2.” and “If x > 0, then x > −2.” 7.
The original statement is true. The converse is “Every rectangle is a square,” which is false. The contrapositive is “Everything that is not a rectangle is not a square,” which is true.
9.
The contrapositive is “Not(Q)⇒Not(P),” which is logically equivalent to P ⇒Q.
11. It is always better to switch! Can you explain why?
13. All integers greater than or equal to 4.
15. True. The negation is “For all real numbers x, x ≤2 and x ≥3.” 17. True. The negation is “There exists a real number that is both rational and irrational.” 19. True. The negation is “There exists x such that, for all y, y ̸= x2.” (In other words, “There exists x for which there is no y with y = x2.)” 21. True. The negation is “There is some integer x greater than 1 for which x < 2.” 23. False. One counterexample is x = 1.35.
25. False.
27. False. One counterexample is x = −1.
29. True. One example is x = 3.
31. True. The negation is “There exist real num-bers a and b such that a < b but 3a + 1 ≥ 3b + 1.” 33. True.
35. False.
37. True.
39. True. One example is x = −1, since for all y we have |y| > −1.
41. True.
43. False. The only counterexample where the two sides of the double implication are not equivalent is x = 0, y = 0.
45. (a) B ⇒(Not A); (b) (Not B) ⇒A 47. (a) (Not A) ⇒(Not B); (b) A ⇒B 49. (a) C ⇒(A and B); (b) Not(C) ⇒(Not A) or (Not B) 51. (a) (B and C) ⇒A; (b) ((Not B) or (Not C)) ⇒(Not A) 53. (a) The converse is “If x is rational, then x is a real number.” (b) The contrapositive is “If x is irrational, then x is not a real number.” (c) x = π is a counterexample to the original and the contrapositive.
55. (a) The converse is “If x ≥3, then x > 2.” (b) The contrapositive is “If x < 3, then x ≤2.” which is false. (c) x = 2.5 is a counterexam-ple to both the original and the contraposi-tive.
57. (a) The converse is “If √x is not a real num-ber, then x is negative.” (b) The contraposi-tive is “If √x is a real number, then x is non-negative. (c) No possible counterexamples for any of the statements.
59. (a) The converse is “If |x| = −x, then x ≤0.” (b) The contrapositive is “If |x| ̸= −x, then x > 0.” (c) No possible counterexamples for any of the statements. |
188474 | https://structnotes.com/asm-example-2/ | ASM Example 2 – StructNotes
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ASM Example 2
<< Back to Axial, Shear, & Moment DiagramsStep 1: Find the support reaction forces/moments.
We should always begin by drawing the free body diagram (FBD). Knowing all the externally applied forces and moments, we can determine the reactions at the supports by summing the forces and moments.The hinge acts like a pin, so in this case where there are only externally applied vertical forces, it just resists vertical forces. To understand how to represent these forces in the FBD, let us temporarily remove the hinge and see the beam as two members. The right member is pushed down by a 5k load, and without the hinge to support it, this right member is going to tilt down. So, for the beam to be stable, we know the hinge serves to push up the right member. Since the summation of forces at the hinge has to be zero, when put back in place, the hinge has to push down on the left member.
Additionally, as stated in the list of basic rules, the left fixed support has a moment reaction, M A.
For consistency, the up forces and counterclockwise moments are positive.Note that R A and R B are positive values, meaning that they are pointing upwards, according to our sign convention. They also match the direction of the arrows in our current FBD. However, if we selected the reaction forces to point downwards in the FBD, both values would be negative, meaning that the actual reaction forces are pointing upwards. Either way gives us the same results. Therefore, the main takeaway is that the positive and negative signs have to be consistent with the FBD and chosen sign convention.
Step 2: Determine axial/shear forces.
Right off the bat, we know that there are no external axial forces, so there are no internal axial forces. Therefore, there is no need to draw an axial force diagram (AFD).
To determine the internal shear forces (V), we are going to look at three segments. The first segment represents the beam member to the left of the hinge. The second segment represents the part of the right beam member between the hinge and the point load. The third segment includes the part of the right beam member to the right of the point load. Generally, we can find the shear force along each segment of the beam by summing all forces.
Mathematically, we can say the shear force at distance x from the left is:Step 3: Draw axial/shear force diagrams.
Once we have the shear equation for each segment, we can plot the shear force diagram (SFD). Note that the shear force at the supports is equal to the reaction. Also, as stated in the list of basic rules, the SFD does not have to be continuous. However, it is continuous at the hinge, because, as we will see in a bit, the moment is zero. Therefore, shear, being the rate of change of moment, is unchanged.Step 4:Determine bending moment.
Similarly, to determine the bending moments (M z), we are going to look at the same three segments. Generally, we can find M z along each segment of the beam by summing all the moments.
Mathematically, we can say the bending moment at distance x from the left is:Step 5:Draw bending moment diagram.
Once we have the bending moment equation for each segment, we can plot the bending moment diagram (BMD). Note that the bending moment is zero at the supports and the hinge. This is true for pinned and roller end supports and hinges without externally applied moments. Also, as stated in the list of basic rules, the BMD is continuous.Recall that shear is the rate of change or derivative or slope of moment. From calculus, we know that there is a local maximum or minimum value when the slope of a function is zero. Therefore, when there is a zero shear value (zero slope), there has to be a corresponding local maximum or minimum moment value. Because shear is the derivative of moment, or inversely, moment is the anti-derivative of shear, we can determine the critical points in the BMD by finding the area under the SFD up to that point.
To the right of the hinge, the maximum bending moment is (2.5k)(2.5ft) = 6.25 k/ft. To the left of the hinge, the SFD never crosses zero, so the maximum bending moment has to be at the ends. In this case, it is at the fixed end, M A = 37.5 k/ft.
TL;DR
There are two main takeaways from this example:
Shear is continuous at hinges.
Bending moment is zero at hinges.
<< Back to Axial, Shear, & Moment Diagrams
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188475 | https://artofproblemsolving.com/wiki/index.php/Heptagon?srsltid=AfmBOoqA0_01X4_g9zs01hmz0tybLiuAvSu50SOFKVhO3bB02rUn24CZ | Art of Problem Solving
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Heptagon
A heptagon is a polygon with seven sides. The sum of the interior angles of a heptagon is Therefore, the measure on an interior angle in a regular heptagon is A heptagon will have 14 diagnols and can be split into 5 triangles.
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188476 | https://www.purplemath.com/modules/rtnladd.htm | Select a Course Below
Intro to Adding (and Subtracting) Rational Expressions
How to Add Rational ExpressionsHow to Subtract & ExamplesMore Examples
Purplemath
Addition and subtraction are the hardest things you'll be doing with rational expressions because, just like with regular fractions, you'll have to convert these polynomial fractions to their lowest common denominators. Everything you hated about adding fractions, you're gonna to hate worse with rational expressions. But stick with me; you can do this!
Content Continues Below
MathHelp.com
Adding Rational Expressions
Let's refresh by looking at an example with regular (that is, strictly numerical) fractions:
To find the common denominator, I first need to find the least common multiple (LCM) of the three denominators. (For old folks like me, whenever you see "LCM", think "LCD", or "lowest common denominator". In this context, they're pretty much the same thing.) There are at least a couple ways of doing this. I could use the "listing" method, where I list the multiples of the three denominators until I find a number that is in all three lists, like this:
5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55,...
25: 25, 50, 75, 100, 125, 150, 175, 200,...
10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100,...
The first multiple to occur in all three lists is 50, so this will be the lowest common denominator.
Another method I could use (and this is the method I prefer) for finding the common denominator is the factor method. It works by finding the prime factorization of each denominator, and then using a chart to find the factors needed for the common denominator. It looks like this:
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In either case, the common denominator will be 50.
To convert each fraction to the common denominator, I will multiply each denominator by what it needs in order to turn it into 50. For instance, in the 2/5, the denominator needs to be multiplied by 10, since 10 × 5 = 50. To keep things fair, I multiply the top by 10 as well. This is because 10 ÷ 10 = 1, and multiplying things by 1 doesn't actually change them. So I get:
Converting the other fractions to the same denominator, I get:
This doesn't simplify or reduce, so my final answer is:
Content Continues Below
The process works similarly for rational expressions.
How do you add rational expressions?
What does this look like, in practice? Let's dive in:
To find the common denominator, I need to find the least common multiple of x, x2, and 2x.
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In the strictly-numerical example above, I used the "listing" method to find the common denominator: all I had to do was multiply each denominator by 1, then 2, then 3, then 4,... and so on, until I found a match.
But in this case, I've got numbers and variables, so just multiplying by numbers is not going to work.
Clearly, the "listing" method won't work for rational expressions. I'll have to use the factor method instead. Here's what I get:
So my common denominator for these polynomial fractions will be 2x2. To convert the 2/x to the common denominator, I will need to multiply by 2x/2x. Why? Because the denominator already has one copy of x, but it needs a 2 and a second copy of x:
Similarly, for the 3/x2, I will multiply by 2/2; and for the 1/2x, I will multiply by x/x. This gives me:
Then the answer is:
This expression cannot be further simplified!
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The x's cannot cancel off, and the 2 cannot cancel into the 6. Why not? Because you can only cancel factors, not terms. You cannot reach inside the understood parentheses around the "5x + 6" factor and rip off arms and legs in an effort to beat the denominator into submission. Don't even try!
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188477 | https://simple.wikipedia.org/wiki/Hardy%E2%80%93Weinberg_law | Hardy–Weinberg law - Simple English Wikipedia, the free encyclopedia
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Beginning
1 Generalisation
2 References
Hardy–Weinberg law
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From Simple English Wikipedia, the free encyclopedia
The Hardy–Weinberg law was developed independently by an Englishmathematician, G.H. Hardy, and a Germandoctor, Wilhelm Weinberg. This concept is also known as the Hardy–Weinberg equilibrium, Hardy–Weinberg theorem or Hardy–Weinberg principle. Sometimes Weinberg's name is placed first.
The law is a foundation of population genetics, and it is still taught to students today. It states that the proportions of alleles of all genes in any population will remain the same unless perturbed (disturbed). That applies to all loci on all chromosomes in the population. Possible perturbations are:
♦ gene mutation♦ natural selection♦ small population size where random effects like genetic drift and inbreeding may occur. H/W populations are assumed to be infinite in size.♦ assortative mating instead of random mating. In effect, this would split the population down into small groups, see item above.♦ migration into or out of the population under study.
It follows that any systematic change in the frequency of alleles in a population must be due to the effect of one or more of these causes. Of course, like all aspects of Mendelian inheritance, the expected proportions of alleles are probabilities. It was for that reason that statisticaltests of significance, such as standard errors, were developed.
Although all changes must be due to perturbations, not all perturbations lead to changes. The classic case is balancing selection, such as heterozygote advantage: "Heterosis: the heterozygote at a locus is fitter than either homozygote". Balancing selection leads to an equilibrium population with Hardy–Weinberg proportions
Generalisation
[change | change source]
The law permits the prediction of genotype frequencies from a knowledge of gene frequencies.
If alleles A and a are in the proportions p and q, the three zygotic types AA, Aa, and aa are in the proportions p 2: 2pq: q 2.
Thus, equations can be written in terms of allele frequencies, and hypotheses about how phenotypes are inherited can be tested from population data.
References
[change | change source]
↑Hardy G.H. (1908). "Mendelian proportions in a mixed population". Science. 28 (706): 49–50. Bibcode:1908Sci....28...49H. doi:10.1126/science.28.706.49. ISSN0036-8075. PMID17779291.
↑Weinberg W. 1908. Über den Nachweis der Vererbung beim Menschen. Jahreshefte des Vereins für vaterländische Naturkunde in Württemberg64: 368–382.
↑Gillespe J.H. 2004. Population genetics: a concise guide. Johns Hopkins, Baltimore MD.
↑Ford E.B. 1975. Ecological genetics, 4th ed. London: Chapman & Hall.
↑Maynard Smith J. 1998. Evolutionary genetics. Oxford. p65
↑Ford E.B. 1965. Genetic polymorphism, p26, Heterozygous advantage. MIT Press 1965.
↑Crow J.F. 1988. Eighty years ago: the beginnings of population genetics. Genetics119: 473–476
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08 Dec 2024
Tags: MathematicsMathematicsCoordinateGeometryDistanceFormula
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Distance Formula in Coordinate Geometry
This calculator determines the distance between two points in a 2D plane using the distance formula.
Explanation
Distance Formula: The distance formula is a fundamental concept in coordinate geometry. It allows us to calculate the distance between any two points in a plane. The formula is derived from the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Related Questions
Q: How is the distance formula derived?
A: The distance formula is derived from the Pythagorean theorem. Consider two points P1(x1, y1) and P2(x2, y2). We can form a right-angled triangle with these points and the point (x2, y1). The distance between P1 and P2 is the hypotenuse of this triangle. Using the Pythagorean theorem, we get (distance)^2 = (x2 - x1)^2 + (y2 - y1)^2. Taking the square root of both sides gives us the distance formula.
Q: What are some applications of the distance formula?
A: The distance formula has numerous applications in various fields, including:
Geometry: Calculating the length of line segments, finding the perimeter of shapes, and determining the distance between points.
Physics: Calculating the distance traveled by an object, finding the displacement of an object, and determining the magnitude of vectors.
Computer Science: Developing algorithms for pathfinding, navigation, and calculating distances in geographical information systems (GIS).
Variables
| Symbol | Name | Unit |
---
| x1 | X-coordinate of P1 | |
| y1 | Y-coordinate of P1 | |
| x2 | X-coordinate of P2 | |
| y2 | Y-coordinate of P2 | |
Calculation Expression
Distance Formula: The distance between two points P1(x1,y1) and P2(x2,y2) is calculated using the distance formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
sqrt((x2 - x1)^2 + (y2 - y1)^2)
Calculator
X-coordinate of P1 ():
Y-coordinate of P1 ():
X-coordinate of P2 ():
Y-coordinate of P2 ():
Solve
Calculated values
Considering these as variable values: y1=0.0, x1=0.0, y2=0.0, x2=0.0, the calculated value(s) are given in table below
| Derived Variable | Value |
--- |
| Distance Formula | 0.0 |
Sensitivity Analysis Graphs
Distance Formula: The distance between two points P1(x1,y1) and P2(x2,y2) is calculated using the distance formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Impact of null on Distance Formula
Impact of null on Distance Formula
Impact of null on Distance Formula
Impact of null on Distance Formula
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What is the general formula used to calculate the distance between two points P1(x1
y1) and P2(x2%20and%20P2(x2)
y2)??)
Is the formula dependent on the type of coordinate system being used (e.g.
Cartesian
polar
etc.)??)
Can the same formula be applied to calculate the distance between two points in three-dimensional space?
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188479 | https://testbook.com/question-answer/a-square-waveguide-carries-te11-mode-whose-axial-m--5dcd72cdf60d5d0970a86867 | [Solved] A square waveguide carries TE11 mode whose axial magnetic fi
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Electromagnetic Theory
Waveguides and Guided Waves
Rectangular Waveguide
Cutoff Frequency
Question
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A square waveguide carries TE 11 mode whose axial magnetic field is given by H z=H 0 cos(π x 8)cos(π y 8)A/m, where wave guide dimensions are in cm. What is the cut-off frequency of the mode?
This question was previously asked in
ISRO Scientist ECE Dec 2017 Official Paper
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5.5 GHz
6.5 GHz
7.5 GHz
8.5 GHz
Answer (Detailed Solution Below)
Option 3 : 7.5 GHz
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Detailed Solution
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Concept:
For a TE MN mode of Electromagnetic wave;
H z=H z 0 cos(m π a.x)cos(n π b.y)e−γ z.e j ω z.a^z
Calculation:
Comparing the given equation with the standard equation;
m = 1, and n = 1.
H z=H 0 cos(π x 8)cos(π y 8)A/m
Given a = b = √8 cm.
f c=c 2(1 a)2+(1 b)2=c 2×1 a×2
f C=3×10 10 2×1 8×2
f C=3×10 10×2 2×2 2=0.75×10 10 c m
f c = 7.5 GHz.
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More Rectangular Waveguide Questions
Q1.When does the propagation occur for any mode of propagation in a rectangular waveguide?
Q2.If we take E=1 2 m 0 v 2 and p=m 0 v then the phase velocity of the corresponding wave packet is
Q3.The cut off frequency of TEM wave is
Q4.Which of the following is True for signal travelling through a Wave-Guide?
Q5.A rectangular waveguide has dimensions 1 cm × 0.5 cm. Its cut off frequency is :
Q6.A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m. If the line is distortionless, the attenuation constant is
Q7.If in a rectangular waveguide for which a = 2b, the cut-off frequency for TE02 mode is 12 GHz, the cut-off frequency for TM11 mode is -
Q8.A pitot static tube is used for measuring the velocity of a gas flowing in a duct. The velocity is proportional to
Q9.A device, if terminated to radiate energy, is called a ______.
Q10.Which type of waveguide are shown in the figure.
More Waveguides and Guided Waves Questions
Q1.Which of the following is NOT possible in a circular wave guide?
Q2.When does the propagation occur for any mode of propagation in a rectangular waveguide?
Q3.If we take E=1 2 m 0 v 2 and p=m 0 v then the phase velocity of the corresponding wave packet is
Q4.The cut off frequency of TEM wave is
Q5.Which of the following is True for signal travelling through a Wave-Guide?
Q6.A rectangular waveguide has dimensions 1 cm × 0.5 cm. Its cut off frequency is :
Q7.____ number of 3-dB couplers are needed for a 64 × 64 bi-directional star coupler.
Q8.The core refractive index and a relative refractive index difference of a multimode step-index fiber are specified as 1.5 and 2%, respectively. At operating wavelength of 1300 nm, the approximate number of propagating modes is 1000. The diameter of the fiber core is :
Q9.The maximum digital transmission rates for unipolar return-to-zero data transmissions over an optical fiber 10-km long with specified pulse-spreading constant of 10 ns/km is :
Q10.With the reduction in actual permittivity, the relative permittivity ______.
More Electromagnetic Theory Questions
Q1.According to Coulomb's law the force of repulsion existing between two point charges is _.
Q2.______ allow partial current to flow through them.
Q3.Gauss's law is an important result of _.
Q4.According to Maxwell's equation which of the following is correct?
Q5.What is the expression for the Poynting vector of an isotropic point source at a distance 'R' from the source?
Q6.The 3 dB beam width of the antenna is called
Q7.Which of the following is NOT possible in a circular wave guide?
Q8.Wave propagation occurs without attenuation as in free space (for σ = 0) in case of a/an
Q9.The characteristic impedance of transmission lines is also known as
Q10.A radiator which radiates uniformly in all directions is called an
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188480 | https://archive.nptel.ac.in/content/storage2/courses/112103016/module1/lec1/5.html | ===============
Module 1 : Classical Thermodynamics
Lecture 1 : Review of Thermodynamics
1
2
3
4
5
6
Exceptions:
Table : 1.2 Symbols for Exceptional Properties
ParameterPropertyTraditional symbol
Temperature Intensive property T
Mass Extensive property m
Number of moles Extensive property n
Specific extensive properties, i.e., extensive properties per unit mass are intensive properties. For example: specific volume, specific energy, density, etc.
It may be worth mentioning here that cyclic integral of the differential of any thermodynamic property must be zero. For example,
However, change in volume between state points 1 and 2 is
Path and process
An operation in which one or more of the properties of a system changes is called a change of state. The sucession of states passed through during a change of state is called the path of the change of state.
The path of thermodynamic states that a system passes through as it goes from an initial state to a final state is known as the thermodynamic process.
Different thermodynamic processes are given in Table 1.3
Table : 1.3 Different Thermodynamic Processes
Sl NoName of the processParameter Held ConstantRemarks
1Constant pressure(Isobaric)p = constantv = (mR/P)T
2Constant volume(isochoric) processV = constantp = (mR/V)T
3Constant temperature (isothermal) processT = constant pV = constant
4Polytropic processn pV n = constant
5Adiabatic processno heat flow across the systemboundary pv γ = constant
6Isoenthapic processh = constant h = constant
1
2
3
4
5
6 |
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Adding with Linking Cubes! Addition Worksheets. Interlocking Counting Blocks
Created by
Teaching with Faith and Joy
These 12 Worksheets are perfect to use when you are teaching your students to add with Unifix Cubes!
PreK - 2 nd
Basic Operations, Math, Other (Math)
$2.50
Original Price $2.50
Rated 4.79 out of 5, based on 86 reviews
4.8(86)
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Cube Addition Task Cards - Addition to 10 with Unifix Cubes or Snap Cubes
Created by
Mrs Learning Bee
This fun and interactive low-prep Kindergarten math game will help students to consolidate their addition to 10 skills with unifix cubes. This game is designed to help children understand the concept of adding two groups together to find a total.How to use:Students build and solve addition problems to 10 with unifix cubes or snap cubes. An optional recording page is also included - students can colour in the cubes and record the answer. This resource includes:96 different task cards for addit
K - 1 st
Math, Numbers
CCSS
K.CC.A.3
, K.CC.B.4
, K.CC.B.4a
+4
$3.50
Original Price $3.50
Rated 5 out of 5, based on 10 reviews
5.0(10)
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Addition with Connecting Cubes/Fact Families/Turn Around Facts/Centers K-2
Created by
Teach K to 2
This is an easy prep math center for children to practice addition facts using connecting cubes. Students can use actual cubes to make trains, or just use crayons or colored pencils to color in the cube trains. These worksheets are a great way to begin teaching Fact Families or Turn Around Facts. There are 3 sheets for children to make a fact and write it 2 ways with a cube train in 2 colors. Then there are 6 sheets for children to make 5 different addition sentences for numbers 5-10. I hope t
K - 2 nd
Basic Operations, Math
CCSS
K.OA.A.1
, 1.OA.C.6
, 2.OA.B.2
$3.25
Original Price $3.25
Rated 4.79 out of 5, based on 38 reviews
4.8(38)
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Addition to 20 With Linking Cubes | Math Cards For Kindergarten
Created by
The Teachers Pet - Lindsay Simpson
These simple addition to 10 cards are a great way to introduce adding to 10 and ways to make ten, to your Kindergarten students. Students will choose a card and using a dry-erase marker to complete the addition sentence. What's included: 28 cards total3 pages of addition to 5 (12 cards total)4 pages of addition beyond 5 to 10 (16 cards)How to use:Cut cards (4 per page) and laminate.Provide students with dry-erase markers and snap cubes for a hands-on learning activityBe sure to follow me on I
PreK - K
Math
$1.50
Original Price $1.50
Rated 5 out of 5, based on 12 reviews
5.0(12)
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Addition With Snap Cubes, 36 Cards, Counting, Math Centers, Connecting Cubes
Created by
Let's go to learn
You will receive 12 PDF pages with 36 addition with snap cubes cards. There are only a few items needed to set up an engaging math center for practicing patterns: ► Snap cubes in a variety of colors. ► Our addition cards :) ► Trays and small containers to group cards with the right materials (optional). Print the addition cards and grab a set of colored connecting cubes. Set them out and let your little one/student solve the additions! ⭐⭐⭐ Digital download only. Blocks are NOT included. Listing
PreK - K
Numbers, Other (Math)
Also included in:SNAP CUBES MATS Growing Bundle, Connecting Cubes Task Cards, Fine Motor Skills
$3.65 Original Price $3.65
$3.29
Price $3.29
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Lessons with Llama: Solving Addition Word Problems with the CUBES method PPT
Created by
Miss Romanovich's Resources
This is a 53 slide lesson to teach how to solve addition word problems. This PowerPoint teaches what word problems are, reviews addition vocabulary, introduces the CUBES method, and shows an example word problem. There are 8 fun word problem game question slides to keep students engaged. A matching word problem worksheet is included for students to do as they participate in the lesson!It is narrated by the adorable Larry the Llama. For more adorably educational PPT Lessons with Llama, follow th
2 nd - 3 rd
Math, Other (Math)
CCSS
2.OA.A.1
$4.00
Original Price $4.00
Rated 4.86 out of 5, based on 7 reviews
4.9(7)
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Grades K - 1 Addition and Subtraction with Unifix Cubes: Activity Book BUNDLE
Created by
Luminous Learning
Illuminating Math in the Minds of Struggling Learners 81 pages of printable addition and subtraction games, worksheets, and task cards Help children master their addition and subtraction facts with our specialized supports:visual aids such as unifix cubes, clear math examples, step-by-step instructions, and color-coded hints. Use the kindergarten and 1st grade addition and subtraction games and activities for independent practice, small group work, HW assignments, or assessment tools. Grades
K - 1 st
Arithmetic, Basic Operations, Numbers
CCSS
K.OA.A.1
, K.OA.A.4
, K.OA.A.5
+3
$10.50
Original Price $10.50
Rated 4.88 out of 5, based on 8 reviews
4.9(8)
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Multi-Step Word Problems with CUBES Strategy (Addition& Subtraction)- 3rd Grade
Created by
Miss Hurley's Special Education Strategies
This product includes daily multi-step word problems involving addition and subtraction that give students the opportunity to apply their learning in real world math situations. These multi-step word problem activities also include the CUBES strategy and scaffolds built within the pages for students who may have difficulties with breaking down word problems and identifying the correct operation to use. These multi-step word problems require students to solve multi-digit addition and subtraction
3 rd - 4 th
Applied Math, Math, Other (Math)
CCSS
2.OA.A.1
, 3.OA.A.3
, 3.OA.D.8
+1
$6.99
Original Price $6.99
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Addition Task Cards: Building with Snap cubes
Created by
Fortunate Firstie
Use these addition task cards for math centers, intervention, early finishers, and more! These task cards are great for hands on practice adding numbers between 1 & 10 as well as 11 & 20! Two task cards PER PAGE! Over 175 addition problems TOTAL!!How to use:Using snap cubes, build the first number. Then add the designated amount of unifix cubes. Write the addition number sentence and find the sum!
PreK - 2 nd
Basic Operations, Math, Mental Math
CCSS
K.NBT.A.1
, 1.NBT.B.2b
$3.00
Original Price $3.00
Rated 5 out of 5, based on 4 reviews
5.0(4)
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Addition Worksheets With Ten frames, dominoes, cubes and more manipulatives!
Created by
GroovyKinders
Addition Worksheets With Ten frames, dominoes, cubes, and more manipulatives! Look at the preview! Preview includes all the pages so you can see them ahead of time. Math Manipulatives Included- Cubes, Dice, Dominoes, Pictures, and Ten Frames. Addition Problems: The worksheet will present addition problems using numbers or simple equations. Presentation: Alongside each addition problem, there is a visual representation of the manipulatives to help students connect the abstract concept of addition
PreK - 1 st
Math, Other (Math)
CCSS
K.CC.A.2
, K.CC.A.3
, K.CC.B.4
+8
Also included in:Kindergarten Math Bundle!! Counting, Tracing, Comparing, Adding and Subtracting
$1.50
Original Price $1.50
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Adding to 10 with Cubes
Created by
Kindergarten Swag
This product contains cards for addition to 10 using cubes. Students can put physical cubes on their card mats and write the addition sentences below. Laminate these, provide cubes, and dry erase markers to use as a small group or center activity.
PreK - 1 st
Math
$3.00
Original Price $3.00
Rated 4.91 out of 5, based on 13 reviews
4.9(13)
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FREE Addition with Linking Cubes up to 10
Created by
Imaginative Teacher
I have quite a few younger year one students who really struggle with basic addition. They also often confuse the concept of addition with subtraction. For them, using hands on manipulatives rather than fingers is an absolute MUST! This is a hands on activity I created to help those students really visualise adding single digit numbers up to ten, record their thinking with coloring in and then writing the corresponding number sum.
K - 1 st
Math, Numbers
FREE
Rated 4.96 out of 5, based on 25 reviews
5.0(25)
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Wish List
Build and Add with Cubes – Addition Activity for Pre-K & K
Created by
Peachy in Pre-K
Make early addition hands-on and visual with this interactive math activity using linking cubes! In this activity, students draw a card that displays two colored cubes with numbers inside—representing an addition problem (e.g., a red 3 + a blue 2). Students then count out the correct number of linking cubes in each color, stack them together, and use a dry erase marker to write the total in the blank space provided. This engaging activity builds number sense, reinforces one-to-one corresponden
PreK - K
Basic Operations, Numbers
CCSS
K.CC.A.3
, K.CC.B.4
, K.CC.B.4a
+2
Also included in:BUNDLE! Linking Cubes Activity Bundle | Fine Motor, Math, and Literacy | Pre-K
$3.00
Original Price $3.00
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Addition Boom Cards to 10 with Unifix Cubes | Hands-On Math Practice
Created by
MightyMinis
Make addition practice engaging and hands-on with these digital Boom Cards! Students use Unifix cube manipulatives to build each addition problem, count the total, and type in the answer. These interactive, self-checking task cards are perfect for Kindergarten and 1st Grade math centers, RTI, small groups, and distance learning. ✨ What’s Included: Boom Cards for addition facts up to 10 Visual prompts with Unifix cubes Interactive “build, count, and type” format Self-checking for immediate fe
PreK - 2 nd
Basic Operations, Numbers, Other (Math)
$2.00
Original Price $2.00
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Second Grade Addition and Subtraction Towers With Unifix Cubes
Created by
MotivatingMath
Students will have fun building towers with unifix cubes and solving math problems that go along with their towers. Students will practice addition and subtraction along with adding and subtracting with a 10. These 17 pages include 5 building activities that are perfect for math centers or it can be used as a whole group activity.
2 nd
Math, Numbers, Other (Math)
$0.98
Original Price $0.98
Rated 4.61 out of 5, based on 16 reviews
4.6(16)
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Addition Task Cards with Unifix/Snap Cubes / Sums to 10/
Created by
Pinar Yesildag Kirik
These Addition Task Cards are a fun and interactive way for students to practice addition up to 10 using Unifix cubes. With 52 task cards in each version, your students will have plenty of opportunities to enhance their math skills! Key Features: Two Versions to Choose From: You can select between two versions of clip art: a playful, fun counting cube clip art or a more traditional, straightforward Unifix cubes clip art. Both versions are designed to engage students and make math practice enjoy
K - 1 st
Mental Math, Numbers
CCSS
K.CC.A.3
, K.CC.B.4
, K.CC.B.4a
+1
$2.00
Original Price $2.00
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Addition Cards with Unifix Cubes for Photo Box Task Cards
Created by
Teaching Tiny Humans with Miss Seely
These Addition Task Cards are the perfect hands-on math tool for young learners, designed to fit perfectly in a plastic photo box for easy storage and accessibility. With two different options to choose from, these task cards make practicing addition fun and engaging while offering versatility to meet different learning needs. Option 1: Addition to 5This set features addition problems that focus on sums up to 5. You can laminate the cards and then use Expo markers to color in and show the dif
K - 2 nd
Basic Operations, Math, Other (Math)
$3.00
Original Price $3.00
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Building Addition with Unifix Cubes and QR
Created by
Sassy Little Teacher
This task card set includes 24 building addition steps with QR codes for easy self-check! The student can write their answer in the box and then check using a QR reader! These facts focus on addition within 20. The facts are printed 4 to a page for easy printing. Thanks for visiting my store! Rate and follow for freebies and sales! Enjoy! :)
K - 2 nd
Arithmetic, Numbers, Other (Math)
$2.00
Original Price $2.00
Rated 4.9 out of 5, based on 24 reviews
4.9(24)
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Addition Worksheets Teen Numbers With Ten frames, cubes and images!
Created by
GroovyKinders
Addition Worksheets Teen Numbers With Ten frames, dominoes, cubes, and more manipulatives! Look at the preview! Preview includes all the pages so you can see them ahead of time. Math Manipulatives Included- Cubes, Pictures, and Ten Frames. Addition Problems: The worksheet will present addition problems using numbers or simple equations. Presentation: Alongside each addition problem, there is a visual representation of the manipulatives to help students connect the abstract concept of addition wi
K - 2 nd
Math, Other (Math)
CCSS
K.CC.A.2
, K.CC.A.3
, K.CC.B.4
+8
Also included in:Teen Number Bundle
$1.00
Original Price $1.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
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48 Addition/Subtraction Math Task Cards with Unifix Cubes
Created by
Teaching While Abroad
This resource has 48 different addition and subtraction questions using unifix cubes. Students will be able to see a visual representation of the equation, as well as read the word problem with targeted math vocabulary. These are perfect for a formative assessment center, or as a write the room activity! Do you want TPT credits to go towards your next purchase?Be sure to leave me a review and let me know how much you love this product, or how I can improve it for you and your classroom needs! O
K - 2 nd
Basic Operations, Mental Math, Numbers
$2.00
Original Price $2.00
Rated 4 out of 5, based on 1 reviews
4.0(1)
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Addition to 10 with Unifix Cubes 48 Addition Problems
Created by
Educate Little Minds
Learn to add with 48 addition problems. - Single digit addition to 10 is included. - Pictures of Unifix Cubes are included on each addition problem. - 8 Addition problem on each page - It can be use with TpT Digital Activities. - It can be used as task cards.
PreK - 2 nd
Arithmetic, Basic Operations, Math
$0.49
Original Price $0.49
Rated 4.8 out of 5, based on 5 reviews
4.8(5)
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Roll , Add and Build with Unifix cubes
Created by
4K LIFE
Engage young learners with "Roll, Add, Build"! Students roll two dice, add the dots, and build a Unifix cube tower to match the sum on the activity sheet. This hands-on activity makes early addition and number recognition fun and concrete. Includes: Activity sheets and simple instructions. Perfect for: Math centers, small groups, and independent practice in 4K & 5K. Reinforces number sense, counting, and basic addition. Just add dice and Unifix cubes!
PreK - 1 st
Basic Operations, Numbers
$1.00
Original Price $1.00
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Adding with Unifix Cubes
Created by
Jhildee K-3
Use these cards to have your students practice adding with the unifix cubes! Have students lay cubes on top or just count the pictures. Record on the recording sheet!
PreK - 2 nd
Basic Operations, Numbers
$2.00
Original Price $2.00
Rated 5 out of 5, based on 6 reviews
5.0(6)
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Unifix Cube Fun! Addition with sums to 10
Created by
Pretty Primary Teaching
This teddy bear math addition page will be bear-y exciting for your littles! The bright colors and large font make this great for preschool, kindergarten, or first grade students who are just learning how to add. It is also a great resource for learning support classrooms who are working on part, part, whole with basic facts. This is a great way to begin number sense and fact fluency from the earliest grades! Why Teachers & Students Love It :✔️ Large and colorful visual math practice ✔️ Great f
PreK - 1 st
Arithmetic, Basic Operations, Numbers
CCSS
K.OA.A.1
, K.OA.A.5
, 1.OA.B.3
+1
$1.00
Original Price $1.00
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188482 | https://www.youtube.com/watch?v=RU3Aj-W_u7k | Intersection of y=x^2 and x=y^2
mroldridge
34800 subscribers
5 likes
Description
978 views
Posted: 26 Jan 2024
The parabola y=x^2 and its INVERSE, which is x=y^2, only meet in two places: (0,0) and (1,1). I solve this Graphically AND Algebraically in this video.
1 comments
Transcript:
we're going to do the intersection of Y = x^2 which is a regular Parabola and X = y^2 which you'll notice is the same equation but with X and Y flipped this is actually the inverse of that now the first thing we're going to do is graph y = x^2 surely you know what that looks like it starts at 0 0 it has points at 1 comma 1 and -1a 1 as well as -2 comma 4 and 2 4 it's a regular Parabola where the y coordinate of each point is the X but squared so -2 2 is 4 -1 2 is 1 0 2ar is 0 Etc now we're going to have to do this for X = y^2 as well now again this is the inverse of that so one option for you is to take the parabola and reflect it along the line yal X that's one strategy but I'm also going to point out that you can take values of Y Square them and that gives you the corresponding X so as an example if I started at -2 here sorry positive2 here that's a y of 2 2^2 is 4 which means the corresponding x coordinate is pos4 this here is 4 comma 2 which again is the inverse of the point on the original Parabola 2 comma 4 now a y value of 1 squared gives you an x coordinate of 1 and you end up with this same Point again that's because when you flip the X and Y coordinates of the point 1 comma 1 you get 1 comma 1 the same point and actually that's the case for 0 comma 0 as well now A Y of -1 would get squared to give you an x coordinate of POS 1 and a y -2 would be squared to give you four take a look at these purple points here you'll note that this is a parabola on its side because it's been reflected along the diagonal y = x anyways the point of intersection between these two points are the points where they meet so one of them is here the point 0 is one of the points of intersection and the only other one is here at 1 1 again you'll note that this these are the points on the original Parabola where the X and Y coordinates are equal to each other very nice now that's how you do it graphically one other option for you is substitution now we know y = x^2 and perhaps you can say well X itself is y^2 so I'm going to replace x with y^2 but don't forget the extra squar here this is y = x² and I'm allowed to make that substitution because this equation says X = y^2 that means we're looking for any of the points where the Y value is the same number raised to the power of 4 now 0 to the power of 4 is 0er that's what gives you this set and 1 to the^ of 4 is also one I challenge you to find another number that satisfies that kind of arrangement there aren't any and this is the graph that proves it but you can work it out algebraically if you're feeling up to it congratulations there are two points of intersection between these two curves and here they are you did it best of luck |
188483 | https://www.youtube.com/watch?v=UNvHo5E9ENM | AI solves Trigonometry Equation Step-by-Step. AI Homework Assistant. 🤯
Pen and Paper Science
8440 subscribers
7 likes
Description
551 views
Posted: 13 May 2023
⭐️ Try for Free: www.ai-tutor.io
📚 In this video we use AI technology to help us solve the trigonometry equation sin(2x) - cos(x) = 0. These equations can be very difficult to solve, but thanks to our trusty AI-tutor we get hints that help us on our way.
⏱ Timestamps:
00:00 start
00:48 First Hint
02:20 Factorisation
04:30 First subproblem
05:13 Second subproblem
06:30 FINAL SOLUTION
🙋♀️ If you have any questions, remarks, or suggestions for future topics, let me know in the comment section below or join the Discord Server!
--
🎥 Subscribe:
👍 Thanks for watching!
aitutor #trigonometry #mathematics
5 comments
Transcript:
start in this video we will solve this trigonometry equation now because these equations can become quite difficult especially if you don't know how to start we will use the AI tutor to help us this is specialized AI technology that is fine-tuned to help students solve exercises and learn and for teachers trying to figure out new exercises you can try it out for free yourself at AI tutor dot IO let's get into it we need to find a specific angle X for which the sine of two times this angle minus the cosine of this angle is equal to zero now let's say that you don't have a clue of how to start this exercise then you can just ask the AI tutor to help you it will respond by stating the equation such that you know that you've put in the right equation and it says that we need to use the First Hint double angle identity for the sign and it asks us if we remember what it is let's say that you indeed remember it and you say that this sine of 2 times x is equal to the sine of x multiplied by the cosine of x so then you answer the AI tutor with what you found and actually it corrects us here because it says that we forgot this factor of 2 in front of this sine of x cosine of x and this is indeed correct sine of two x is equal to 2 times the sine of x multiplied by the cosine of x then it asks us to substitute this into our original equation and if we do this then we get the following the sine of 2x simply becomes 2 times this sine of x multiplied by the cosine of x minus this original cosine of x which doesn't change and this has to be equal to zero so this is already our first step and at this point you can have a go at the rest of the exercise yourself however if you get stuck and you want more help you can simply ask the AI tutor by giving the equation where you are so in this case 2 times the sine of x times the cosine of x minus the cosine of x is equal to 0. it confirms that this is indeed the equation that we have if we substitute our double rule for the sine of X and then it asks us to factor this equation in fact it asks us to find Factorisation the common factor in both of these terms now this is not difficult because we see that in both of these terms we have a cosine of x and then it asks us to factor out this cosine of x however at this point you might have some more questions for instance why do we want to factor the left hand side of this equation so you can just ask this question why do we want to factor this equation and then it answers with an explanation of why we want to factor it namely factoring the left hand side of this equation makes it from a difference of two terms which has to be equal to 0 to a product of two factors which has to be equal to zero basically what we then get is an equation that looks as follows a multiplied by B which is equal to zero now A and B can be anything and still this equation is very easy to solve because this equation is true if a is equal to 0 because then we get 0 times B which is 0 or B is equal to zero because then we get that a multiplied by zero is also equal to zero so basically the problem then simplifies a lot if we factor out this equation and that's why factoring out this cosine of x will help us solve our equation with this valid question answered by the AI tutor we follow its hint and just factor out the cosine we get that the cosine of x the angle that we're looking for multiplied by two times the sine of x minus 1. has to be equal to zero and if we fill this in in the AI tutor to see if we didn't make any mistakes then it indeed says that we didn't make any mistakes and that the problem now becomes two sub problems namely each part of this product the cosine as the first part and these brackets as the second part has to be equal to zero in order for this equation to be true so let me just make some space and put this equation on the top then we can look at the two sub problems First subproblem the first one is a very easy one so the cosine of x has to be equal to zero now we know when the cosine of x is equal to zero if we just draw the coniometric circle then we know that the cosine of x is 0 at pi over 2 which is on top of the circle and minus pi over 2. therefore we find that the cosine of x is zero if x is equal to pi over 2 plus n times pi where n is just a whole number so we've already found our first solution which is also a solution of our original equation but there are more Second subproblem angles for which this original equation is indeed true and this brings us to the second sub problem namely when this factor between brackets is equal to zero so if we write this down in equation we get two times D sine of x minus 1 has to be equal to zero now we can rewrite this equation to isolate the sine of x because that's a function for which we readily know its values then we get that design of X or angle has to be equal to 1 over 2. and if we look at our chronometric Circle then we know by heart or you can look this up that the sine of x is equal to one-half when the angle itself is equal to pi over 6 or on the other side here which is 5 pi over 6. and for each of these Solutions we can add an entire circle so adding 2 pi this means that for this sub problem we get solutions that are the following when X is equal to pi over 6 Plus n times pi and when X is equal to 5 pi over 6 Plus n times pi and this is the second set of solutions FINAL SOLUTION for which our original equation is true and thus we have found all of the angles X for which the sine of 2 times this angle minus the cosine of this angle is equal to zero |
188484 | https://ocw.mit.edu/courses/6-231-dynamic-programming-and-stochastic-control-fall-2015/resources/mit6_231f15_lec15/ | 6.231 DYNAMIC PROGRAMMING LECTURE 15 LECTURE OUTLINE
• Review of basic theory of discounted problems
• Monotonicity and contraction properties
• Contraction mappings in DP
• Discounted problems: Countable state space with unbounded costs
• Generalized discounted DP
• An introduction to abstract DP 1DISCOUNTED PROBLEMS/BOUNDED COST
• Stationary system with arbitrary state space
xk+1 = f (xk , u k, w k), k = 0 , 1, . . .
• Cost of a policy π = {μ0, μ 1, . . . }
Jπ (x0) = lim E
{N −1∑ αkg(xk, μ k(xk), w kN →∞ wkk=0 ,1,... k=0
})
with α < 1, and for some M , we have |g(x, u, w )| ≤
M for all ( x, u, w )
• Shorthand notation for DP mappings (operate on functions of state to produce other functions)
(T J )( x) = min E ,
u∈U(x)
{g(x, u w ) + αJ
w
(f (x, u, w ))} , ∀ x
T J is the optimal cost function for the one-stage problem with stage cost g and terminal cost αJ .
• For any stationary policy μ
(TμJ)( x) = E
w
{g(x, μ (x), w ) + αJ (f (x, μ (x), w ))} , ∀ x2“SHORTHAND” THEORY – A SUMMARY
• Cost function expressions [with J0(x) ≡ 0]
Jπ (x) = lim ( Tμ0 Tμ1 · · · k
Tμk J0)( x), Jμ(x) = lim ( Tμ J0)( x)
k→∞ k→∞
• Bellman’s equation: J∗ = T J ∗, Jμ = TμJμ
• Optimality condition:
μ: optimal <== > TμJ∗ = T J ∗
• Value iteration: For any (bounded) J and all
x:
J∗(x) = lim ( T kJ)( x)
k→∞
• Policy iteration: Given μk,
− Policy evaluation: Find Jμk by solving
Jμk = Tμk Jμk
− Policy improvement: Find μk+1 such that
Tμk+1 Jμk = T J μk3MAJOR PROPERTIES
• Monotonicity property: For any functions J and
J′ on the state space X such that J(x) ≤ J′(x)for all x ∈ X, and any μ
(T J )( x) ≤ (T J ′)( x), ∀ x ∈ X,
(TμJ)( x) ≤ (TμJ′)( x), ∀ x ∈ X.
• Contraction property: For any bounded func-tions J and J′, and any μ,max ∣∣(T J )( x) − (T J ′)( x)
x
∣ ≤ α max
x
′
)
∣ ∣∣J(x) − J′(x)∣,
max ∣∣(TμJ)( x)−(TμJ (x) α max J(x) J′(x)
∣
.
x
∣ ≤ x
∣ − ∣
• Shorthand writing of th
∣
e contrac
∣
tion property
∣
‖T J −T J ′‖ ≤ α‖J−J′‖, ‖T ′ ′
μ
J−TμJ ‖ ≤ α‖J−J ‖,
where for any bounded function J, we denote by
‖J‖ the sup-norm
‖J‖ = max
x∈X
∣∣J(x)∣∣.4CONTRACTION MAPPINGS
• Given a real vector space Y with a norm ‖ · ‖
(see text for definitions).
• A function F : Y 7 → Y is said to be a contraction mapping if for some ρ ∈ (0 , 1), we have
‖F y − F z ‖ ≤ ρ‖y − z‖, for all y, z ∈ Y. ρ is called the modulus of contraction of F .
• Linear case, Y = ℜn: F y = Ay + b is a con-traction (for some norm ‖ · ‖ ) if and only if all eigenvalues of A are strictly within the unit circle.
• For m > 1, we say that F is an m-stage con-traction if F m is a contraction.
• Important example: Let X be a set (e.g., state space in DP), v : X 7 → ℜ be a positive-valued function. Let B(X) be the set of all functions
J : X 7 → ℜ such that J(s)/v (s) is bounded over s.
• The weighted sup-norm on B(X):
‖J‖ = max |J(s)|
s∈X
v(s) .
• Important special case: The discounted prob-lem mappings T and Tμ [for v(s) ≡ 1, ρ = α]. 5A DP-LIKE CONTRACTION MAPPING
• Let X = {1, 2, . . . }, and let F : B(X) 7 → B(X)be a linear mapping of the form (F J )( i) = b(i) + a(i, j ) J(j), i
j
∑
∈X
∀
where b(i) and a(i, j ) are some scalars. Then F is a contraction with modulus ρ if
∑
j∈X
|a(i, j )| v(j)
v(i) ≤ ρ, ∀ i
[Think of the special case where a(i, j ) are the transition probs. of a policy ].
• Let F : B(X) 7 → B(X) be the mapping (F J )( i) = min( FμJ)( i), i
μ∈M
∀
where M is parameter set, and for each μ ∈ M , Fμ
is a contraction from B(X) to B(X) with modulus
ρ. Then F is a contraction with modulus ρ.6CONTRACTION MAPPING FIXED-POINT TH.
• Contraction Mapping Fixed-Point Theorem: If
F : B(X) 7 → B(X) is a contraction with modulus
ρ ∈ (0 , 1), then there exists a unique J∗ ∈ B(X)such that
J∗ = F J ∗.
Furthermore, if J is any function in B(X), then
{F k J} converges to J∗ and we have
‖F k J − J∗‖ ≤ ρk‖J − J∗‖, k = 1 , 2, . . . .
• Similar result if F is an m-stage contraction mapping.
• This is a special case of a general result for contraction mappings F : Y 7 → Y over normed vector spaces Y that are complete : every sequence
{yk} that is Cauchy (satisfies ‖ym − yn‖ → 0 as
m, n → ∞ ) converges.
• The space B(X) is complete [see the text (Sec-tion 1.5) for a proof]. 7GENERAL FORMS OF DISCOUNTED DP
• Monotonicity assumption: If J, J ′ ∈ R(X) and
J ≤ J′, then
H(x, u, J ) ≤ H(x, u, J ′), ∀ x ∈ X, u ∈ U (x)
• Contraction assumption:
− For every J ∈ B(X), the functions TμJ and
T J belong to B(X).
− For some α ∈ (0 , 1) and all J, J ′ ∈ B(X), H
satisfies
∣∣H(x, u, J )−H(x, u, J ′)∣∣ ≤ α max ∣∣J(y) )
y
−J′(y
∈X
∣
for all x
∣
∈ X and u ∈ U (x).
• We can show all the standard analytical and computational results of discounted DP based on these two assumptions (with identical proofs!)
• With just the monotonicity assumption (as in shortest path problem) we can still show various forms of the basic results under appropriate as-sumptions (like in the SSP problem) 8EXAMPLES
• Discounted problems
H(x, u, J ) = E{g(x, u, w ) + αJ (f (x, u, w ))}
• Discounted Semi-Markov Problems
n
H(x, u, J ) = G(x, u ) + ∑ mxy (u)J(y)
y=1
where mxy are “discounted” transition probabili-ties, defined by the transition distributions
• Deterministic Shortest Path Problems
()
{ a JH(x, u, J = xu + u) if u 6 = t,
axt if u = t
where t is the destination
• Minimax Problems
H(x, u, J ) = max [g(x, u, w )+ αJ
w∈W(x,u )
(f (x, u, w ))] 9RESULTS USING CONTRACTION
• The mappings Tμ and T are sup-norm contrac-tion mappings with modulus α over B(X), and have unique fixed points in B(X), denoted Jμ and
J∗, respectively (cf. Bellman’s equation ). Proof :From contraction assumption and fixed point Th.
• For any J ∈ B(X) and μ ∈ M ,lim T kμ J = Jμ, lim T kJ = J∗
k→∞ k→∞
(cf. convergence of value iteration ). Proof : From contraction property of Tμ and T .
• We have TμJ∗ = T J ∗ if and only if Jμ = J∗
(cf. optimality condition ). Proof : TμJ∗ = T J ∗,then T ∗
μ
J = J∗, implying J∗ = Jμ. Conversely, if Jμ = J∗, then TμJ∗ = TμJμ = Jμ = J∗ = T J ∗.
• Useful bound for Jμ: For all J ∈ B(X), μ ∈ M ‖Jμ − J ‖T
‖ ≤ μJ − J‖
1 − α
Proof : Take limit as k → ∞ in the relation
kk
‖T kμ J−J‖ ≤ ∑ T
ℓ=1
‖ ℓμJ−T ℓ−1
μ
J‖ ≤ ‖ TμJ−J‖ ∑ αℓ−1
ℓ=1 10
RESULTS USING MON. AND CONTRACTION I
• Existence of a nearly optimal policy : For every
ǫ > 0, there exists μǫ ∈ M such that
J∗(x) ≤ Jμǫ (x) ≤ J∗(x) + ǫv (x), ∀ x ∈ X
Proof : For all μ ∈ M , we have J∗ = T J ∗ ≤ T J ∗
μ
.By monotonicity, J∗ ≤ T k+1
μ
J∗ ≤ T kμ J∗ for all k.Taking limit as k → ∞ , we obtain J∗ ≤ Jμ.Also, choose μǫ ∈ M such that for all x ∈ X,
‖Tμǫ J∗−J∗‖ = ∥∥(Tμǫ J∗)( x)−(T J ∗)( x)∥∥ ≤ ǫ(1 −α)From the earlier error bound, we have
‖Jμ
‖− ∗ ‖T
‖ ≤ μJ∗ − J∗
J ,
1 − α ∀ μ ∈ M
Combining the preceding two relations,
∣∣Jμǫ (x) − J∗(x)∣∣
v(x) ≤ ǫ(1 − α)1 − α = ǫ, ∀ x ∈ X
• Optimality of J∗ over stationary policies :
J∗(x) = min Jμ(x),
μ∈M
∀ x ∈ X
Proof : Take ǫ ↓ 0 in the preceding result. 11 RESULTS USING MON. AND CONTRACTION II
• Nonstationary policies: Consider the set Π of all sequences π = {μ0, μ 1, . . . } with μk ∈ M for all k, and define for any J ∈ B(X)
Jπ (x) = lim sup( Tμ0 Tμ1 · · · Tμk J)( x), x
k→∞
∀ ∈ X,
(the choice of J does not matter because of the contraction property).
• Optimality of J∗ over nonstationary policies :
J∗(x) = min Jπ (x),
π∈Π
∀ x ∈ X
Proof : Use our earlier existence result to show that for any ǫ > 0, there is μǫ such that
∗
‖Jμǫ −
J ‖ ≤ ǫ(1 − α). We have
J∗(x) = min Jμ(x) ≥ min Jπ (x)
μ∈M π∈Π
Also
T kJ ≤ Tμ0 · · · Tμk−1 J
Take limit as k → ∞ to obtain J ≤ Jπ for all
π ∈ Π. 12 MIT OpenCourseWare
6.231 Dynamic Programming and Stochastic Control
Fall 2015
For information about citing these materials or our Terms of Use, visit: |
188485 | https://math.stackexchange.com/questions/1177613/pie-conflicts-with-complementary-counting | combinatorics - PIE conflicts with Complementary counting - Mathematics Stack Exchange
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PIE conflicts with Complementary counting
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Asked 10 years, 7 months ago
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How many 4 4-digit numbers have at least two consecutive digits that are the same?
I solved this problem in two ways, as are shown below:
PIE: We apply the Principle of Inclusion-Exclusion. This principle tells us that the overall number of desirable 4 4 digit numbers is Number with one pair of two consecutive digits−Number with two pairs of two consecutive digits+Number with 3 pairs of two consecutive digits.Number with one pair of two consecutive digits−Number with two pairs of two consecutive digits+Number with 3 pairs of two consecutive digits.
We solve the third case first, as it is the most simple. If three pairs are the same, all the digits are the same. Thus, there are 9 9 digits possible, as 0000 0000 is not a 4 4 digit number. Thus, there are 9 9 numbers.
Now, we solve the first case. We either have the first two digits the same, or the second or third two. In the first subcase, we have 9 9 choices for the first two digits, as the first cannot be a 0 0, and 10 10 choices for each of the last two digits, as they are unrestricted. Thus, there are 9⋅10 2 9⋅10 2 in the first subcase.
In the second subcase, there are 9 9 choices for the first number, then 10 10 choices for the digit shared by 2 2 of the last 3 3 digits. There are then 10 10 choices for the last digit, which yields 9⋅10 2 9⋅10 2 possibilities for each subcase of the subcase, so 18⋅10 2 18⋅10 2 possibilites overall.
Thus, in the first main case, there are 9⋅10 2+18⋅10 2 9⋅10 2+18⋅10 2 cases overall, which yields 2700 2700 cases.
Finally, we attack the second case, where there are two pairs of equal digits. The two subcases are that the two digits are a pair of three in a row, or are separated in a block of four in a row. This is three subcases overall, which is correct because we are choosing (3 2)(3 2) from the 3 3 possible equal pairs of digits.
In the first subcase, we can have the three digits at the beginning of the 4 4 numbers or at the end. If they are at the beginning, they have 9 9 possibilities as they cannot be 0 0, and the last digit has 10 10. If they are at the end, they have 10 10 possibilities and the remaining digit has 9 9.
In the second subcase, there are 9 9 possibilities, for this is simply the same as the three pairs case.
Thus, we get 90+90+9=189 90+90+9=189 possibilities overall for the second case.
Summing the three cases, we get 189+2700+9=2898 189+2700+9=2898 possibilities.
Complementary counting:
We use complementary counting and subtract the number of 4 4 digit numbers with no digits the same from the total number of 4 4 digit numbers. There are 9⋅10 3=9000 9⋅10 3=9000 4 digit numbers and there are 9⋅9⋅8⋅7 9⋅9⋅8⋅7 4 digit numbers with no digits the same. Thus, there are 9000−9⋅9⋅8⋅7=4464 9000−9⋅9⋅8⋅7=4464 numbers.
Why are these two, supposedly correct, methods giving me different answers? Where did I make my error?
Thank you!
combinatorics
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asked Mar 6, 2015 at 1:36
rk_347rk_347
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You want to test every number between 1000 and 9999 for the condition that either the first digit equals the second, or the second equals the third, or the third equals the fourth.
Your inclusion exclusion principal strategy is incorrect. If implemented correctly, it would return the number of four digit numbers that either have exactly 1 or 3 pairs, but this is not what we want. You also counted wrong here
You complementary counting method is also incorrect. 1231 is not an acceptable number for your count, but your method will include it. This is why your result is too high.
The correct answer is 2439.
To correctly use inclusion exclusion:
First we consider all the four digit numbers whose first two digits are the same. Lets call this set of numbers F 1 F 1. You correctly count the size of this at 900.
Next we consider all the four digit numbers whose second two digits are the same. Lets call this set F 2 F 2. The middle digits might be 00, 11, ..., or 99 -- all in all 10 possibles. The first digit can be 1-9 and the last digit can be 0-9. 10910 gives us another 900 possibles for this set.
Now we need all the four digit numbers whose third two digits are the same. Lets call this set F 3 F 3. There are 91010 or 900 numbers in this set as well.
Unfortunately F 1 F 1, F 2 F 2 and F 3 F 3 are not disjoint sets. Therefore we must use the inclusion exclusion formula:
|F 1∪F 2∪F 3|=|F 1|+|F 2|+|F 3|−|F 1∩F 2|−|F 1∩F 3|−|F 2∩F 3|+|F 1∩F 2∩F 3||F 1∪F 2∪F 3|=|F 1|+|F 2|+|F 3|−|F 1∩F 2|−|F 1∩F 3|−|F 2∩F 3|+|F 1∩F 2∩F 3|
We have some more counting to do.
For F 1∩F 2 F 1∩F 2 we have numbers of the form, 111x, 222x, ... 999x. 90 of these.
For F 1∩F 3 F 1∩F 3 we have numbers of the form XXYY, again 90 of these.
For F 2∩F 3 F 2∩F 3 we have numbers of the form x000, x111, ... x999. Again, 90 of these.
For F 1∩F 2∩F 3 F 1∩F 2∩F 3 we have numbers of the form XXXX, so 9 of these.
Plug it all up and we have 900+900+900-90-90-90+9 or 2439.
A tiny Java program to check your work:
```
public static void main(String[] args) {
int count = 0;
for (int a=1000; a<10000; a++) {
String b = Integer.toString(a);
char[] c = b.toCharArray();
if(c==c || c == c || c == c) {
count++;
}
}
System.out.println(count);
}
```
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answered Mar 6, 2015 at 2:27
JonnyJonny
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We can solve this problem using the Complement Principle.
As you noted, there are 9000 9000 four digit positive integers. We must subtract the number of four digit numbers in which no two consecutive digits are equal. If no two digits are consecutive, there are nine ways of selecting the thousands digit (since we cannot use 0 0), which leaves us nine ways to select the hundreds digit (since it can be any digit other than the thousands digit), nine ways of selecting the tens digit (since it can be any digit other than the hundreds digit), and nine ways of selecting the units digit (since it can be any digit other than the tens digit). Therefore, there are
9000−9 4=9000−6561=2439 9000−9 4=9000−6561=2439
positive integers with four digits in which at least two consecutive digits are the same.
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edited Mar 29, 2015 at 17:43
answered Mar 7, 2015 at 14:17
N. F. TaussigN. F. Taussig
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188486 | https://www.facebook.com/groups/boomersbyjukeboxnostalgia/posts/1582107662384609/ | BABY BOOMERS (1946 - 1964) by Jukebox Nostalgia | Hey how many know why we are called boomers | Facebook
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Why are people born between 1946 and 1964 called baby boomers?
Summarized by AI from the post below
BABY BOOMERS (1946 - 1964) by Jukebox Nostalgia
Christine Mckee
· April 12, 2024 ·
Hey how many know why we are called boomers?? I have more info just have to find it.
Hey how many know why we are called boomers?? I have more info just have to find it.
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Cathy Bingham
Because there was a “baby boom” when the greatest generation came home from WWII.
1y
13
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Heather Slaughter McGrath Lawless
Apparently… Right after World War II, there was an influx or a “boom” of babies being born that following generation. Hence… Baby boomers! And I guess as we aged… They dropped the word “baby”…
1y
Jonne Dmochowski
Thought it had to do with the end of WW11. All the men came home from war. Got married and there. Was a baby boom. I was born in 1948 and this is what I was told.
1y
4
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188487 | https://www.mdpi.com/2227-9717/9/10/1843 | Typesetting math: 100%
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Table of Contents
Abstract
Introduction
Sulfate-Reducing Bacteria in Various Biotopes
Conditions Determining the Viability of Sulfate-Reducing Bacteria
Cultivation and Storage of Sulfate-Reducing Bacteria
Preservation of Intestinal Microorganisms
Summary of the Obtained Information with the Possibility of Application to SRB
Conclusions
Author Contributions
Funding
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Acknowledgments
Conflicts of Interest
References
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Open AccessFeature PaperReview
Distribution of Sulfate-Reducing Bacteria in the Environment: Cryopreservation Techniques and Their Potential Storage Application
by
Ivan Kushkevych
Ivan Kushkevych
SciProfilesScilitPreprints.orgGoogle Scholar
1,,
Aneta Kovářová
Aneta Kovářová
SciProfilesScilitPreprints.orgGoogle Scholar
1,
Dani Dordevic
Dani Dordevic
SciProfilesScilitPreprints.orgGoogle Scholar
2,
Jonah Gaine
Jonah Gaine
SciProfilesScilitPreprints.orgGoogle Scholar
3,
Peter Kollar
Peter Kollar
SciProfilesScilitPreprints.orgGoogle Scholar
4,
Monika Vítězová
Monika Vítězová
SciProfilesScilitPreprints.orgGoogle Scholar
1, and
Simon K.-M. R. Rittmann
Simon K.-M. R. Rittmann
SciProfilesScilitPreprints.orgGoogle Scholar
5,
1
Department of Experimental Biology, Faculty of Science, Masaryk University, Kamenice 753/5, 625 00 Brno, Czech Republic
2
Department of Plant Origin Food Sciences, Faculty of Veterinary Hygiene and Ecology, University of Veterinary Sciences Brno, Palackého tř. 1946/1, 612 42 Brno, Czech Republic
3
Department of Science, Technology Engineering and Mathematics, Faculty of Biological and Pharmaceutical Sciences, Institute of Technology, Clash, Tralee, V92 CX88, Co. Kerry, Ireland
4
Department of Pharmacology and Toxicology, Faculty of Pharmacy, Masaryk University, Palackého třída 1946/1, 612 00 Brno, Czech Republic
5
Department of Functional and Evolutionary Ecology, Archaea Physiology & Biotechnology Group, Universität Wien, 1090 Wien, Austria
Authors to whom correspondence should be addressed.
Processes 2021, 9(10), 1843;
Submission received: 26 August 2021
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Revised: 8 October 2021
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Accepted: 13 October 2021
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Published: 18 October 2021
(This article belongs to the Special Issue Feature Review Papers in Section "Environmental and Green Processes")
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Abstract
Sulfate-reducing bacteria (SRB) are a heterogeneous group of anaerobic microorganisms that play an important role in producing hydrogen sulfide not only in the natural environment, but also in the gastrointestinal tract and oral cavity of animals and humans. The present review was written with the inclusion of 110 references including the time period from 1951 to 2021. The following databases were evaluated: Web of Science, Scopus and Google Scholar. The articles chosen to be included in the review were written mainly in the English and Czech languages. The molecular mechanisms of microbial cryoprotection differ depending on the environment where microorganisms were initially isolated. It was observed that the viability of microorganisms after cryopreservation is dependent on a number of factors, primarily colony age, amount of inoculum, cell size or rate of cooling, and their molecular inventory. Therefore, this paper is devoted to assessing the performance and suitability of various cryopreservation methods of intestinal bacteria, including molecular mechanisms of their protection. In order to successfully complete the cryopreservation process, selecting the correct laboratory equipment and cryopreservation methodology is important. Our analysis revealed that SRB should be stored in glass vials to help mitigate the corrosive nature of hydrogen sulfide, which can affect their physiology on a molecular level. Furthermore, it is recommended that their storage be performed in distilled water or in a suspension with a low salt concentration. From a molecular biological and bioengineering perspective, this contribution emphasizes the need to consider the potential impact associated with SRB in the medical, construction, and environmental sectors.
Keywords:
anaerobic microorganisms; sulfate-reducing bacteria; hydrogen sulfide; toxicity
Graphical Abstract
1. Introduction
Sulfate-reducing bacteria (SRB) are microorganisms that occur in different ecosystems globally [1,2,3,4,5]. They can also be isolated from the gastrointestinal tract and the oral cavity of humans and animals [6,7,8,9,10,11]. The cultivation of SRB is sometimes fastidious, as they require anaerobic conditions, strict temperature regulations and precise pH requirements . Consequently, research on SRB is uncommon, and a method of long-term cryopreservation has not been thoroughly developed. Therefore, in this work we review methods for cryopreservation and their application for preservation of SRB [13,14].
For cryopreservation, it is necessary to choose the right laboratory equipment in which long-term storage can be performed . Another important step in cryopreservation is to select the right cryoprotectant to maximize the viability of the microorganism after freezing . Although the type and concentration of the chosen cryoprotectant is critical, the possibility to combine different preservation compounds to achieve successful cryopreservation may be equally important. The viability of microorganisms is influenced by a number of factors . Colony age, amount of inoculum, cell size or rate of cooling may impact the survival of the culture. Furthermore, viability can also vary between individual species within the same genus .
This review focuses on the following main points: SRB in various biotopes, conditions determining their viability, molecular aspects of cryopreservation, cultivation and storage, preservation of intestinal microbial communities and their viability and comparison of the obtained information with the possibility of application to SRB (Figure 1).
Figure 1.
The scheme shows the information outlined in the review.
The aim of this systematic review was to describe the distribution of SRB in various environments, the individual steps related to the methods of cryopreservation of intestinal microorganisms and the subsequent comparison of the findings as to whether they are applicable to the long-term storage of SRB.
2. Sulfate-Reducing Bacteria in Various Biotopes
In terms of physiology, ecology and function, SRB can be isolated from various biotopes (water, mud, river sediment, sea sediment, human and animal intestinal tracts, etc.) . SRB are mostly classified in the class Deltaproteobacteria and differ from other classes in their characteristic type of metabolism . SRB use sulfate reduction respiration to obtain necessary energy [19,20,21,22]. Another descriptive feature of SRB is their cellular shape. Their cells can be spherical, oval, spiral or vibroid [1,18]. The positive occurrence of SRB is characterized by a strong odor of hydrogen sulfide emission [23,24]. At the moment, the classification of sulfate-reducing microorganisms has been validly revised. However, according to Bergey’s Manual of Determinative Bacteriology (1994), SRB were divided on the basis of 16S rRNA into the following groups [17,23,24]:
Gram-negative mesophilic SRB; these do not form spores and are one of the most widespread SRB in nature (genera Desulfovibrio, Desulfobotulus, Desulfobulbus, Desulfohalobium and Desulfomicrobium);
Gram-positive spore-forming bacteria; these are a typical representative (genus Desulfotomaculum) that can be identified from soil samples (according to the updated classification, these microorganisms are represented and included in order Clostridiales);
Gram-negative thermophilic sulfate-reducing microorganisms (genus Thermodesulfobacterium);
Gram-negative thermophilic archaeal sulfate-reducing microorganisms; these include members of the genus Archaeoglobus that can only be found in anaerobic, underwater, hydrothermal environments because they require salt and high temperature for their growth.
The majority of these groups use sulfate as a terminal electron acceptor during anaerobic respiration. The presence of SRB with high metabolic activity can be identified by the blackening of water and sediments .
SRB are important hydrogen-utilizing organisms that, despite their occurrence in other ecosystems, colonize the digestive tract of mammals [11,25,26,27,28,29]. Previous studies have indicated that SRB play an important role in the development of intestinal bowel disease. SRB are also an important factor with regards to food biotechnology, and they can also play a role in part methylation of mercury. Certainly, the presence of different microorganisms in the gut, and the application of probiotics, can influence the eco-physiology of SRB in the intestinal environment . Moreover, SRB have successfully adapted to almost all ecosystems on Earth .
2.1. Water Environment
SRB are often found in aquatic polluted environments [1,12]. Pollution can be of anthropogenic or of natural origin. The presence of sulfate can lead to a number of microbial processes and sulfide formation. Large microbial pollution due to the growth of SRB was recorded in canals and ports such as Venice or the city of Bruges in Belgium. SRB can also be found in the aqueous phases of oil and gasoline storage tanks .
In marine waters, SRB can be found more in the upper layers of sediment, where low redox conditions are encountered. Their known competitors, methanogenic archaea, are commonly found in the lower parts of sediment . When SRB and methanogenic archaea occur together in marine sediments, they do not compete against each other, but rather complement each other in the degradation of organic matter. For instance, in marine sediment, SRB and methanogenic archaea are often present together, but methanogens degrade non-competitive substrates and produce methane . Samples of Desulfovibrio spp., Desulfotalea and Desulfuromonas have been found in the upper part of the marine sediment (100 cm) [1,31]. At the same time, Desulfosporosinus and Desulfovibrio have been most often isolated from the deeper layers. However, a study conducted by Barton & Hamilton (2007) reported that the amount of SRB was low in comparison to the total microbial population inhabiting saltwater environments. It was confirmed that SRB from deep-sea habitats are much more barotolerant than species from near-surface environments .
A relatively high population density of SRB has been observed in wastewater biofilms. The composition of the microbial community in wastewater depends on the ability of the organisms to adhere to the surface of the biofilm . Six major genera of SRB have been found in wastewater biofilms: Desulfomicrobium, Desulfovibrio, Desulfonema, Desulforegula, Desulfobacterium, and Desulfobulbus . The authors found that Desulfobulbus spp. generated the highest population density of about 108–109 cells per cm−3 from SRB . High sulfur reduction was found in a narrow anaerobic zone, which was located 150–300 µm below the biofilm surface. As a result, the biofilm formed in the wastewater facilitated the growth of anaerobic SRB under aerobic conditions .
2.2. Surfaces of Corrosive Metals
The colonization of surfaces by SRB is one of the issues of the oil and gas industry, as hydrogen sulfide produced by SRB can cause corrosion and contamination of hydrocarbon products . Corrosion of iron and ferrous alloys occurs not only in aquatic but also in terrestrial environments, regardless of nutrient content, temperature, pressure and pH . Microbial corrosion (MIC) is a biological process that damages the surfaces of corrosive materials due to the action of not only SRB, but also other microorganisms such as aerobic and autotrophic bacteria [24,34]. A scheme of iron metal corrosion by SRB is shown in Figure 2.
Figure 2.
Scheme of iron metal corrosion by SRB (modified from Barton and Fauque, 2009 ).
SRB consume H2, and as a result, depolarize the cathode. When Fe2+ is released from the anode, a depression is formed in the metal and insoluble FeS is created. H+ from ionizing water is combined with electrons to form H2 for SRB .
2.3. Corrosion of Concrete, Stone Elements and Masonry
Concrete pipes can also be subject to microbial corrosion . The main cause of corrosion of concrete pipes is the metabolic process of SRB. Bacteria grow in water sediment at the bottom of the pipes and hydrogen sulfide is formed there. Once the hydrogen sulfide is produced by SRB, an aerobic zone occurs, and the sulfate-oxidizing bacteria begin to form sulfuric acid, which gradually dissolves the stone surfaces [32,34].
MIC is the result of a chemical interaction between a metal material and the environment in which the metal is located . The result is a loss of material. Most often, it is an electrochemical process in which electrons from a metal are transported through several redox reactions to a final electron acceptor that is close to the metal surface . There are several mechanisms by which SRB affect corrosion , including biofilm formation and attachment on the anode side. In this process, a set of natural bacteria, including SRB, accumulates on the metal surface; it is assumed that the effect of the so-called “quorum sensing” tunes the oxidation, localizes the bacteria on the metal material and creates a depression in this place . Another mechanism by which SRB accelerate corrosion is depolarization at the cathode, which occurs because SRB consume H2 facilitated by hydrogenases. Corrosion can be prevented by the use of protective materials . Plastic pipes with an uneven inner surface or pipes that are highly alkaline on their walls .
2.4. Gastrointestinal Tract
The large intestine is a complex microbial ecosystem inhabited by a number of different microbial species . The abundance and composition of organisms plays an important role in human metabolism, and also in the health, disease or physiology of the human body. There are about 1011–1012 microbial cells in 1 g of intestinal contents. 143 stool samples were examined for the abundance of SRB, and it was found that 83% of the specimens contained SRB at a concentration of 102–1011 per 1 g of feces . It was also shown that the incidence of SRB influence the number of methanogenic archaea [36,37]. It is well-known that SRB and methanogenic archaea compete for nutrients in the gastrointestinal tract (GIT). A study revealed a negative correlation between the concentration of methane in the breath to the number of SRB in fecal samples .
In the intestinal microbial composition, hydrogen-utilizing microorganisms play an important role in the metabolism of molecular hydrogen (H2) and sulfur . Due to the fact that SRB use H2 as an electron donor, they facilitate fermentation processes [38,39].
Anaerobic bacteria represent an integral component of the human microbiome. While many of them are associated with maintaining optimal health, others are involved in a variety of pathological processes, both in immune-competent and immunocompromised individuals [40,41,42]. The most common SRB species that occur in humans and animals are: Desulfovibrio (64–81%), Desulfobacter (9–16%), Desulfobulbus (5–8%), Desulfomonas (3–10%) and Desulfotomaculum (2%). The genus Desulfovibrio is the most common genus of SRB [43,44]. Desulfovibrio is the most isolated genus of SRB and is found in samples in which inflammatory bowel disease has been confirmed [4,5,45,46,47]. Desulfotomaculum ruminis and D. acetoxidans originate from intestines .
SRB most commonly occur along with Actinobacteria, Firmicutes, and Proteobacteria . SRB significantly affect the pH in the gastrointestinal tract, since they form hydrogen sulfide and acetic acid, and these substances lower pH . The growth conditions of intestinal SRB in the GIT are greatly influenced by the concentration of sulfates, which varies among individuals. However, this depends on the type of diet .
3. Conditions Determining the Viability of Sulfate-Reducing Bacteria
Representatives of SRB occur at all sites that meet anaerobic conditions . SRB are able to tolerate temperatures from –5 °C to 75 °C. They are able to tolerate a large pH range (5–9.5) and a large osmotic pressure range [1,12].
The presence of sulfate and lactate in the human gut contributes significantly to the growth support of SRB [4,8]. This is also related to the subsequent accumulation of their metabolites, acetate and hydrogen sulfide, in the gastrointestinal environment. In a mixed culture, the growth of SRB was supported by increased sulfur availability . Sulfated polysaccharides, such as mucin and chondroitin sulfate, could be used by SRB as electron acceptors. It has also been shown that when sulfate concentrations increased, the growth of Desulfovibrio desulfuricans also increased .
3.1. Physical Conditions
Temperature. SRB strains, which are classified as mesophilic, are known to have a temperature optimum of about 30 °C, but can also tolerate up to 45–48 °C . However, such a large temperature range is more connected with environmental SRB [12,49]. Intestinal SRB species are grown at 37 °C. This temperature corresponds to warm-blooded animals and humans. Most thermophilic SRB were found in geothermal environments and in oil field waters . The optimum growth temperature of thermophilic SRB (Thermodesulfobacterium) is from 54 °C to 70 °C, and the maximum temperature at which the bacteria are still able to grow is 85 °C .
pH. As mentioned above, SRB are able to tolerate a range of pH from 5 to 9.5, but this depends on the environment from which they originally isolated . Although the pH level in the large intestine of humans or animals is limited and depends on a number of different factors (composition and enzymatic activity of intestinal microorganisms, digestion process and consumed food), the pH in the human digestive tract is most often reported to be around 7.6 to 8. However, the pH in the colon is at least one unit lower and lies between 5.7 in the caecum and 6.7 in the rectum . The pH measured in feces is 7 . Microbial growth is said to reach a maximum when the medium has a pH between 7 to 8 . When the medium reached a pH of less than 6, a 26% decrease in microbial growth was recorded when compared to the medium at a pH of 7 to 8 . It was concluded that adults and elderly people (64 to 83 years) supported a higher number of SRB than young people aged around 15 to 20 years .
3.2. Competition and Coexistence with Other Intestinal Microorganisms
Coexistence between hydrogenotrophic microbes (SRB and methanogenic archaea) and hydrogenogenic microbes (Clostridium, Bacteroides and Escherichia) is essential to maintain fermentative processes in the gut . Desulfovibrio, which uses H2 produced by Clostridium and Bacteroides, can serve as an example . Methanobrevibacter competes with Desulfovibrio for H2. The inhibition of SRB by methanogenic archaea (increased methanogenesis) results in the accumulation of short chain fatty acids with succinate and lactate . A concentration of 200 mg L−1 of hydrogen sulfide is regarded as an upper limit and can cause inhibition of methanogenesis . The increased availability of sulfate in the gut can lead to the inhibition of methanogenesis .
When sulfate is present in the large intestine, the occurrence of SRB is promoted . However, when sulfate levels are reduced, methanogenic archaea predominate in the large intestine that can strongly compete with SRB for the availability of important metabolites of the intestinal microbiota [11,54]. Sulfate present in the digestive tract can come from both exogenous and endogenous environments . Exogenous sources most often include drinking water and diet. Specific sulfate concentrations have been measured in more than 200 commonly available foods and beverages . Foods high in sulfate (>10 µmol g−1 or up to 1 mg g−1) include certain types of bread, soy flour, dried fruit and sausages. Beverages which commonly contain sulfates (>2.5 µmol mL−1 or 0.25 mg mL−1) include some beers, ciders and wines. About 95% of the sulfate is absorbed in the gastrointestinal tract, and the remaining 5% can be found in the feces. Sources of sulfates of endogenous origin include sulfate mucins, sulfate-conjugated bile, and also, for example, chondroitin sulfate . Sulfate ions in organic compounds need to lose the sulfate group, and in that case, sulfate becomes available. Sulfate does not only serve to support SRB growth and suppress growth of methanogenic archaea, but during sulfate dissimilation reduction, sulfate is used as the final electron acceptor [56,57,58,59].
If sulfate is present, Desulfovibrio and Desulfobulbus intestinal species are able to use H2 in the intestinal environment. The coexistence of SRB and methanogenic archaea found in the same ecosystem is possible if both groups of microorganisms use different electron donors . Lactate is one of the main electron donors that occurs within the large intestine . Lactate can be produced by lactic acid-producing bacteria such as Lactobacillus, Streptococcus, Bifidobacterium and others. Lactic acid as a final product of metabolism is then used by SRB .
3.3. Biochemical Characteristics of Sulfate-Reducing Bacteria
For intestinal bacteria, polysaccharides, starch and cellulose serve as the main sources of energy and carbon [1,12]. They can also use a certain amount of oligosaccharides and proteins. The main products of intestinal metabolism are acetate, short chain fatty acids, propionate, butyrate, H2 and CO2. The most common electron donors of SRB are H2, lactate and acetate [11,12]. Frequent removal of H2 from the lumen by SRB is essential for maintaining a healthy gut . In contrast, sulfate or sulfite serve as electron acceptors (similar to thiosulfate and sulfur, in some cases) for the formation of hydrogen sulfide. Other possible electron donors for SRB growth are fatty acids, glutamate, serine, alanine, ethanol, and a variety of other organic acids such as succinate and pyruvate . The most frequently used electron donors of intestinal SRB are lactate, pyruvate, acetate and ethanol (Figure 3).
Figure 3.
Different types of electron donors used by intestinal SRB which were isolated from human fecal samples (data from Gibson et al., 1988 ).
Gibson et al., 1988, found that SRB differ with regards to their substrate utilization sprectrum; Desulfovibrio spp. use lactate and H2, Desulfobacter spp. utilize acetate, Desulfobulbus spp. use propionate and H2, Desulfomonas spp. utilize lactate, and lastly, acetate and butyrate are used by Desulfotomaculum spp. .
Inhibitors. It is clear that, for SRB, one of the main inhibitors is molecular oxygen (O2) [12,24]. However, it is mentioned that SRB are capable of some adaptation to the environment where O2 is present . For example, SRB colonizing a drinking water biofilm have been able to survive up to 72 h of exposure to aeration. Of the SRB, the genus Desulfovibrio is the most tolerant to an environment with a certain amount of O2 .
High concentrations of various metals are reported to be inhibitors of SRB growth. The toxicity of individual metals depends on the experimental conditions (amount of inoculum, number of cells, pH, temperature). Molybdates have been found to inhibit SRB growth by inhibiting sulfate reduction, thereby reducing the possibility of sulfate transport into the bacterial cell and, thus, reducing possible energy production .
The following organisms were used to test the degree of toxicity of molybdates: Desulfotomaculum ruminis, Desulfovibrio vulgaris, and Desulfovibrio desulfuricans (two species). Postgate C medium was used in the experiment. Bacterial growth in the medium was inhibited at a concentration ranging from 40–200 µmol L−1 of the molybdate. Simultaneously, selenium inhibition of SRB growth was tested. Selenium inhibited SRB growth at a concentration between 160 and 320 µmol L−1. The presence of 50 µmol L−1 thiosulfate completely suppressed the effect of selenium. A comparison of the effect of selenium and molybdates shows that molybdates have more mechanisms of inhibition than selenium, since the addition of thiosulfate to the molybdate resulted in only partial inhibition .
Hydrogen sulfide. Hydrogen sulfide formation is dependent on SRB growth, which is strongly influenced by environmental pH [45,46,62,63,64]. This compound is highly toxic. If hydrogen sulfide is not effectively removed from the gut, its accumulation can lead to damage of the colon’s epithelial cells. It has been reported that increased levels of hydrogen sulfide in the gut may be associated with inflammatory bowel disease [44,55]. Hydrogen sulfide can be removed in the gut by detoxification with the help of intestinal epithelial cells or, due to ongoing bacterial growth, can be incorporated into cellular material [37,44]. The accumulation of hydrogen sulfide in the intestinal environment can be the result of sulfate metabolism inhibition in mammals and increased SRB activity . Hydrogen sulfide toxicity can even cause DNA damage , leading to the formation of an unstable genome, the accumulation of mutations and, in extreme cases, the outbreak of colorectal cancer .
The extent that hydrogen sulfide influences the human gut is still not fully understood [44,55]. Further knowledge of this could be useful particularly in the medical sector, as hydrogen sulfide production can lead to inflammatory bowel diseases [46,47]. The construction industry is likewise affected by SRB. Here, SRB are responsible for corroding various surfaces, and thereby the structural integrity of these materials is weakened. It is therefore desirable to find suitable protection to combat these issues . Furthermore, SRB are involved in water pollution . Water pollution caused by SRB occurs mostly in canals.
4. Cultivation and Storage of Sulfate-Reducing Bacteria
The cultivation conditions for SRB differ depending on the isolation spot and their specific growth rate . Generally, 1–5 days of cultivation under anaerobic conditions are reported. SRB derived from the environment possess lower specific growth rates than SRB that have been isolated from animals and humans. Some studies have described the differences in the growth of mesophilic and thermophilic SRB . The growth of mesophilic SRB is usually very slow (from a few days to 2 weeks) at 30 °C, and at the same time, they require more anaerobic conditions than most other anaerobic microorganisms . Thermophilic SRB grow much faster (12–18 h) at 55 °C .
Beerens and Romond (1977) isolated SRB in tubes containing liquid Postgate medium . The bacterial suspension was transferred to the medium in test tubes, and 2 mL of paraffin was pipetted over the upper surface to maintain anaerobic conditions. The bacteria could then be cultured up to 37 °C in the thermostat for 7 days. Then, it was possible to observe an increase in bacteria, which was manifested by blackening of the medium, mainly at the bottom of the tubes and slightly along their walls .
Some intestinal SRB are very difficult to cultivate or yet uncultivable , such as, for example, the family Desulfovibrionaceae. Due to the fact that SRB occur together with other microorganisms, such as Bacteroides, Pseudomonas, and Clostridium, they are demanding not only for the mentioned cultivation, but mainly for the necessary isolation of a pure SRB culture . Postgate media and other media that exist are created primarily for species of the genus Desulfovibrio which are native to the natural environment .
4.1. Isolation of Intestinal Sulfate-Reducing Bacteria
The following procedure can be utilized for the isolation of intestinal SRB: samples should be cultured in sterilized Eppendorf tubes, completely filled with liquid medium and incubated at 37 °C . To detect the presence of SRB, it is necessary to add Mohr’s salt to the sterilized medium . Mohr’s salt [(NH4)2Fe(SO4)2·6H2O] readily dissociates into free Fe2+ ions that interact with hydrogen sulfide. 10% Mohr’s salt solution and 1% Na2S·9H2O solution, which must be anaerobically prepared to avoid oxidation of S2-, must be autoclaved separately . Then, both solutions may be added to the sterile medium. After adding the Mohr’s salt solution and 20–30 µL of Na2S·9H2O solution to the medium, a black circle forms in the medium, which confirms the interaction of hydrogen sulfide and the free Fe2+ complex released from the Mohr’s salt. The observable black coloration of the medium is the result of a growing SRB culture due to the FeS complex. SRB require an O2-free environment for growth [1,12]. The establishment of an ideal anoxygenic environment can be achieved in two ways:
The entire tube or Eppendorf is filled to the brim with Postgate medium;
One mL of sterile liquid paraffin is added dropwise to the surface of the medium, thus keeping anaerobic conditions.
It was also reported that the addition of a 3% solution of Na2SO3·7H2O to the medium inhibits other bacteria [3,49].
4.2. Media for Cultivation of Sulfate-Reducing Bacteria
Postgate medium B is reported as a basic medium for the detection and cultivation of Desulfovibrio and Desulfotomaculum; the pH of this medium should be between 7.0–7.5 . Postgate medium C is a clear medium for biomass production of the genus Desulfovibrio at pH 7.5 (Table 1). Postgate medium E is used for the isolation of “pure” cultures of SRB with the recommended pH of 7.6 and also serves for the census of grown bacterial populations .
Table 1.
Composition of different cultivation media.
5. Preservation of Intestinal Microorganisms
One of the main reasons for the development of preservation methods of intestinal microorganisms is the possible production of therapeutic microbiota , such as fecal transplants and probiotics . The microbiome is a unique natural resource that has possible medical applications. The result of the currently occurring human way of life entails the extinction of microbial species due to exposure to adverse environmental conditions such as diet, antibiotic use or stress [67,68,69,70]. Therefore, it is important to find suitable methods for preserving the intestinal microbiome to allow the timely restoration of the intestinal microflora with minimal safety risks. The best methods for long-term storage of microorganisms are cryopreservation and lyophilization [13,14].
Lyophilization, also known as freeze-drying under vacuum, is a method in which the sample is first frozen. The amount of solvent (usually water) is then reduced to a level where biological activities or chemical reactions are no longer promoted. The amount is reduced first by sublimation and then by desorption (as a secondary drying process). However, one of the disadvantages of lyophilization is that it has a low percentage of microorganisms that survive storage at very low temperatures . Mutations, membrane damage, protein denaturation, and water crystallization are also frequently observed in this method [16,71].
5.1. Cryopreservation of Intestinal Microorganisms
This is one of the methods for long-term storage of cultures . Cryopreservation is a process in which viable microbes are preserved by cooling to temperatures below 0 °C . It is the only method that is widely applicable and has been proven to be reliable for the preservation of microorganisms. However, it is necessary that the cryopreservation be tuned for each individual type of microorganism . At such low temperatures, all biological activities, including biochemical processes, are suspended (anabiosis) . When temperatures drop 0 °C, water becomes inaccessible, and thus, dehydrated cells, which have no further access to water, can be stored at temperatures below 0 °C .
The following procedure is used for the cryopreservation of Escherichia coli 451-B [75,76]. Some of the most important conditions of the medium, which serves to restore the growth of the microorganisms after cryopreservation, are osmolarity and pH level . Most cryoprotectants are toxic to cells at normal temperatures. Therefore, it is desirable to remove cryoprotectants from the medium after thawing (by centrifugation) or to reduce their concentrations (by dilution).
5.2. Methods of Cryopreservation
Long-term storage methods are still being developed and tested . It is found that the temperature is not sufficient for the long-term storage of living microorganisms, while the use of extreme freezing boxes and boxes with solid carbon dioxide is much more advantageous (Figure 4). In general, temperatures above −30 °C usually give poor results due to a number of side effects .
Figure 4.
The five most common cryopreservation methods and the stated ideal temperatures inside the boxes in °C; data from Hubálek, 1996 .
Liquid nitrogen (−196 °C) is currently the most widely used liquefied gas used in cryopreservation due to low consumption and its safety in handling .
5.3. Laboratory Equipment, Instruments and Materials
Microorganisms are stored using various laboratory equipment. In cryopreservation, it depends on the material from which the ampoules or vials are made . Glass or plastic ampoules are commonly used. These materials differ in their behavior in heat or cold conduction during cryopreservation and subsequent thawing. In polypropylene bottles, the conduction of heat or cold is slower than in glass ampoules.
Glass ampoules/vials. Borosilicate glass ampoules are commonly used in liquid nitrogen storage [77,78]. Glass ampoules must be sealed to prevent probable suspension leakage or liquid nitrogen contamination. After immersion in liquid nitrogen, the liquid from the LN can penetrate a incompletely closed ampoule. This ampoule then has a high probability of exploding due to the rapid expansion of nitrogen gas. It is therefore necessary to carry out a pre-freeze inspection.
Another vessel can be glass ampoules with a screw cap (10·30 mm) and a capacity of 2 mL [78,79]. These vessels have been successfully used in the cryopreservation of strictly anaerobic phototrophic bacteria. These ampoules contained O2-impermeable butyl rubber stoppers and plastic autoclavable screw caps that allow the samples to be inoculated directly into the center of the hole. Storage in ampoules with screw caps was not only sufficient, but also safe, advantageous and sufficient to create an anaerobic environment without the need for an anaerobic chamber . This method is therefore time-saving and relatively inexpensive. Thanks to these advantages, it could be used in small laboratories or large culture collections.
Plastic bottles. Plastic cryovials are safer and much better to handle than glass ampoules . Therefore, glass vials are being replaced by plastic ones. Polypropylene cryogenic vials (47·12 mm, 2 mL volume) with screw caps and silicone seals are often used. Polypropylene and polycarbonate (1–5 mL) cryovials are excellent for storing samples in liquid nitrogen vapors (around −130 to −170 °C) and in conventional refrigerated boxes, but are less useful for direct loading. Indeed, liquid nitrogen can penetrate vials during storage in the same manner as immersing glass vials in liquid nitrogen [15,81,82,83,84]. A possible solution is to immerse the cryovial in liquid nitrogen only for a short time (<30 min), in order to achieve the effect of deep freezing.
The use of plastic bottles is advantageous as they are more durable in comparison to glass bottles and the risk of an explosive reaction is eliminated. It is less problematic if some liquid nitrogen enters the vials and the surrounding liquid nitrogen becomes contaminated with non-pathogenic organisms than if unsterilized liquid nitrogen were to contaminate a partially closed vial. As a result, special polypropylene resealable packaging for cryovials is available. Although these packages prevent the penetration of liquid nitrogen into ampoules and vials, they take up more space in freezer containers.
Pasteur pipettes. As a possible alternative to plastic cryovials, disposable plastic Pasteur pipettes (Figure 5) are also available . The Figure 5 shows a sealed Pasteur pipette “head” labeled with a felt-tip pen (A), a sealed Paster pipette “head” labeled with a ball-point pen (B), an aluminum stand for holding pipettes during cryopreservation (C), part of the pipette after separation of the tip from the rest of the pipette and subsequent sealing of the end with a 2 mL volume of liquid (D), and an unmodified pipette (E). These plastic pipettes are made of polyethylene and have a 4 mL capacity with dimensions of 4.2·1.4 cm and a tip length of about 11 cm.
Figure 5.
Plastic (polyethylene) Pasteur pipette used for freezing. Taken from the Walter Reed Army Institute of Research; by Jackson et al., 1981 .
Prior to cryopreservation itself, a portion of the tip needs to be separated from the pipette head in which the cell suspension will be stored. The place where the tip separated from the rest of the pipette must be heat sealed. The advantage is that there is no significant heating of the suspension during sealing. After sealing, it is beneficial to verify whether the complete healing of the material has been achieved and there is no leakage of the heterogeneous mixture. If the suspension drips, this can be corrected by resealing . Another advantage is, of course, the cheaper price of pasteurized pipettes compared to plastic cryovials. Working with pipettes and filling their contents with the examined sample is much easier than with plastic or glass ampoules. The author confirms that during the cryopreservation of more than 200 sealed pipettes in LN for one year, no contamination with liquid nitrogen occurred. However, if a frozen pipette is dropped or otherwise mishandled, it may break.
5.4. Cryoprotectants
Different types and combinations of cryoprotectants are an integral part of cryopreservation methods and can highly influence the results of cryopreservation. Their task is to protect cells from possible damage that can be caused by osmotic stress due to the presence of external ice and from the dangerous formation of internal ice crystals . A large number of different mixtures of cryoprotectants can protect microorganisms from drying out and radiation, and can protect proteins from thermal damage [13,85].
The most common cryoprotectants are solutions with glycerol, sucrose, ethylenediaminetetraacetic acid (EDTA) and dimethylsulfoxide (DMSO). Other cryoprotectants are trehalose, methanol, glucose, 1,2-propanediol, proline, glycine, fructose, galactose and lactose. The most commonly used ingredients are glycerol and DMSO . It is recommended that they be tested at a concentration of 10% (v/v) before use .
When using permeable cryoprotectants, proper equilibration must not be ignored . In cases such as glycerol or DMSO, it is recommended to allow the cell suspension to equilibrate for a period of time with a certain temperature before the freezing process takes place. For glycerol, the time is 30–60 min at ambient temperature. For DMSO, it is advisable to leave the suspension for 15–30 min at 4 °C. This is because DMSO takes about 15–30 min for the cryoprotectant to penetrate the cell wall and the cell membrane and balance the intracellular solutions before the cell suspension can be frozen. For slow cryoprotectants such as glycerol, they need more than 30 min for this process. Semipermeable cryoprotectants penetrate only through the cell wall, not through the cell membrane. Examples are monosaccharides, amino acid disaccharides and other low molecular weight polymers. In some methanogenic archaea, betaine, α-glutamate, the beta-amino acids β-glutamine as well as N-ε-acetyl-β-lysine have been described to act as osmoprotective substances. Some of these osmoprotective compounds might also be considered cryoprotective compounds . Recently, a novel cryoprotectant, carboxylated poly-L-lysine, was examined and found to indicate restricted mobility and increased solution viscosity of eukaryotic cells. In addition, intermolecular interactions facilitated the glass transition of carboxylated poly-L-lysine, which prevents intracellular ice formation and osmotic shock during freezing .
High molecular weight polymer cryoprotectants, such as proteins, polysaccharides, and others, are impermeable and only protect the cell from the formation of external ice, and this process does not require equilibration time. However, the possible use of a combination of penetrating and non-penetrating cryoprotectants may be advantageous due to differences in the cell wall structure of individual intestinal microorganisms . The temperature at which the state of the suspension equalizes depends mainly on the degree of toxicity of the cryoprotectant used . An example is DMSO, which is less toxic at 0–5 °C than at higher temperatures. However, the absence of toxic effects is not necessarily an indicator of a good cryoprotectant .
Glycerol is sterilized by autoclaving at 121 °C for 15 min . When bacterial cells are placed in liquid nitrogen, glycerol prevents the internal formation of crystals that form at high low temperatures .
One of the most common methods of cryopreserving anaerobic bacteria fecal specimens (Bifidobacterium, Enterobacter, Enterococcus, and Lactobacillus) is using glycerol as a cryoprotectant . The authors compared the number of bacteria before and after freezing. There were 10.3 log (CFU g–1) anaerobes before the freezing process and a decrease of ≈0.05 % after freezing. Bifidobacterium formed the dominant part of the anaerobic sample in the intestinal microflora. Before freezing, 9.7 log (CFU g–1) was present; however, after cryopreservation a significant decrease (9.0 log (CFU g–1) was recorded. Enterococcus, the sample of which was less than 8 logs (CGU g–1) before storage, was not significantly altered by cryopreservation. Enterobacterium behaved similarly to Enterococcus; thus, no significant decrease was observed .
To study the effect of glycerol as a cryoprotectant for the protection of E. coli, two different concentrations were used [76,91]. The 3% glycerol concentration resulted in a complete protection, whereas the 1% concentration had only partial protection. A comparison was still made if no amount of glycerol, i.e., no cryoprotectant, was used. The 3% glycerol concentration resulted in an approximately 95% survival rate. However, it was demonstrated that the rate of freezing influenced viability when using 1% glycerol concentration. Viability values ranged from 53% to approximately 86%. Without the use of any cryoprotectant, viability decreased to 20% during freezing and was greatly affected by the cooling rate.
DMSO could be sterilized by filtration through glass filters . DMSO bottles which have been open for an extended period of time should be avoided, as oxidation and degradation of products can arise as a result. DMSO is solid at 18 °C.
Sugars: Sucrose. Sugars are another classic group of cryoprotectants used in cryopreservation . Disaccharides such as maltose, sucrose and trehalose can protect the cell from freezing by causing it to shrink, thereby reducing the internal formation of ice crystals. It is necessary to use a 5% sucrose concentration when rapid freezing and cooling occurs to completely protect E. coli (>90% viability). As with glycerol, the authors performed an experiment with half the concentration (2.5%) of sucrose. The test showed that a 5% concentration of sucrose almost completely protected E. coli. At 2.5% concentration, the cooling rate depended. If the cooling rate was below 5 °C min−1, the 2.5% sucrose concentration was almost as effective as 5%. However, if the cooling rate was above 5 °C min−1, 2.5% gave only moderate protection .
Tween 80. In Lactobacillus bulgaricus, the positive effect of the addition of classical cryoprotectants (10% glycerol, DMSO, combinations thereof etc.) on viability was not known . Therefore, it was proposed to observe changes in viability that occurred after the addition of polysorbate 80 (Tween 80) to the medium in different genera of Lactobacillus bulgaricus. Tween 80 is a surfactant that should protect the cell surface . After the addition of Tween 80 to the strain L. bulgaricus NCS4, the strain appeared to be the most stable (Table 2) . The increase in bacterial growth after the addition of Tween 80 was not as significant in NCS2 and NCS3 as in the strain NCS1. However, L. bulgaricus of the strain NCS1 did not have complete protection against damage.
Table 2.
Addition of Tween 80 to the medium and its effect on viability (data from Smittle et al., 1972) .
Tween 80 was used also to store E. coli . The protective effect of Tween 80 was related to the cooling rate of the suspension. Viability was observed at 1% and 0.5% Tween 80 concentrations. The results show that at 1% concentration, the viability of E. coli was highest when the cooling rate was above 10 °C min−1. When the cooling rate was reduced below 10 °C min−1, the protective effect of Tween 80 decreased rapidly. Even at a cooling rate of 1 °C min−1, Tween 80 was slightly toxic. 0.5% of the Tween 80 concentration was toxic even at a cooling rate of 10 °C min−1. In summary, Tween 80 is a good cryoprotectant for rapid cooling.
Synergistic effect of cryoprotectants. Some cryoprotectants are good to combine because they can act well or even better on the cell together than if they were to produce an effect on their own [85,94]. The synergistic effect can thus increase the protection of microorganisms during freezing. However, it is not necessary to combine only cryoprotectants with each other. They can also interact with key cell molecules. One component may have a higher effect than the other; however, when added together the results are beneficial to cryopreservation processes .
Effect of glycerol and Tween 80 in E. coli. Cryopreservation of E. coli has been reported to result most commonly in cell membrane and cell wall damage . Glycerol is said to be able, under certain conditions, to protect the membrane and the wall from possible damage. However, Tween 80 fails to protect the cell wall. Tween 80 is only able to prevent membrane damage. Tween 80 also protects against the possible entry of certain substances into the cell. Examples of these substances might include sucrose and glucose-6-phosphate dehydrogenase. Tween 80 thus plays a significant role in promoting viability .
5.5. Viability
Effective storage of cells facilitates preservation or restoration to the initial state in which the cells were prior to freezing. These states may include morphology, metabolism, capsule formation, mucus production, adhesion, gene expression, and overall changes in genetic material, as well as any other alterations that could occur during cryopreservation .
Cell size is also related to the quality of preservation . A smaller cell size gives higher viability values after cryopreservation than larger cells. Viability is usually measured by CFU counting . Thus, it is the number of cells that survived the storage process at very low temperatures. In this method, Petri dishes with bacteria are cultured in a thermostat at a temperature that corresponds to the growth conditions of the microorganism. To determine viability, the dishes are then compared with experimental and control microorganisms. One of the other methods for determining the preservation quality of microorganisms, or viability, is based on the ability to form colonies or coatings . Another aspect that influences viability is the rate or degree of cooling of a given sample of microorganisms . In some microorganisms, several days to 4 weeks of pre-cultivation and adaptation to low temperatures (around 5 °C) can have a positive effect on cryopreservation . This adaptation allows the cell to alter their proteome and membrane composition .
Differences in viability within the same genus. When cryopreserving different species of the same genus with liquid nitrogen under comparable circumstances and typical conditions, different results may occur . An example is the bacterium Lactobacillus bulgaricus, which is classified as a microorganism that is used as a probiotic and is often used in dairy products . During cryopreservation (at −196 °C), it was evident that within the number of colonies and lactic acid production, viability varied from species to species. L. bulgaricus NCS1 was sensitive to freezing, while NCS4 displayed a tolerance to freezing, and therefore, zero deaths were recorded (Table 3).
Table 3.
Stability of individual species of L. bulgaricus during cryopreservation in liquid nitrogen (data from Smittle et al., 1972 ).
Influence of cell membranes. One of the other variants of good cryoresistance may be the variability or fluidity of cell membranes (cytoplasmic and mitochondrial membranes) . The organization and overall stability of membranes is determined by the strength of the interactions between its individual components . The interactions are sensitive to temperature, pH, ionic strength and the volume of surrounding water. Membrane variability and the strength of interactions are determined by the composition of membrane sterols and phospholipids and especially by the composition of unsaturated fatty acids. The smaller the amount of unsaturated aliphatic acids, the less fluid or variable (more rigid) the membranes are and the lower their cryoresistance .
Concentration, age and physiological state of microorganisms. The age and physiological state of microorganisms are considered to be the main factors determining the ability to survive stress . It is generally accepted that cryopreservation of microbial cultures in the stationary phase has a higher resistance than in the exponential growth phase [95,98]. Péter & Reichart (2001) measured the percentage of survival at different stages of growth after the freezing and subsequent thawing process of Lactobacillus plantarum and E. coli . The survival rate of L. plantarum rate was 78.9% and 89.7% from the exponential and stationary growth phase, respectively. Moreover, E. coli exhibited a survival rate of 47.4% and 54.7% from the exponential and stationary phase, respectively. In addition, it was observed that higher concentrations of E. coli frozen at a temperature of about −30 °C resulted in a higher survival rate after cryopreservation than when a lower concentration of cells had been frozen . The following formula is used to calculate the percentage of recovered E. coli cells:
𝑅=78.97−12.4·1010𝑥
R is the percentage of recovered cells; x is a number of E. coli cells in 1 mL of distilled water before freezing; 78.97 is a coefficient; 12.4·1010 is a constant number of cells that always die during cryopreservation despite the initial concentration of cells.
It should be noted that the coefficient of survival remains constant. A high proportion of the cells in suspension concentrate around the ice crystals during the freezing process. It is also noteworthy to mention that cells can directly disperse into ice crystals. A higher cooling rate may give a higher probability that the cells will be dispersed inside ice crystals. The probability that the cells will be inside the ice crystals may decrease if freezing is done very slowly .
Polysaccharides and proteins. Calcott & MacLeod (1974) mention that there is an association between the amount of protein and the possible increase in survival after the freezing process in E. coli. The authors report that increased cryotolerance is demonstrated in species with lower protein content. It has also been found that storage substances such as glycogen or polyglucose, which are stored inside the E. coli cell, have protected the cell from the strong temperature changes that can occur during cryopreservation [75,76].
5.5.1. Stress Factors
During long-term storage at deep low temperatures, a large number of stress factors act on the cell. Grout et al., (1990) found that the rate of cooling significantly affects the survival of individual cells. If the cooling rate is too fast, intracellular ice crystals form. During slow cooling, the fatal damage is related to the external hypertonic effect of the solution. Freezing and thawing causes high stress for all cells . Therefore, emphasis is given to the procedures, methods and substances used, which determine the extent to which the organism is still able to respond to stress factors to which it is exposed during the entire cryopreservation process. It controls how the cell and cytoplasm respond to cooling, heating and overall changes in the chemical environment. The main factors influencing the viability of samples during freezing are: exposure to low temperatures, the mechanical and physical action of ice crystals, changes in the external and internal character of the environment in solution, damage to membrane structures (especially impacts on semipermeability), cytoplasmic accumulation and shrinkage of cytoskeletal structures . Cryopreserved microorganisms can be divided according to the phase of damage during the process of cryopreservation and subsequent thawing into five categories :
Cooling to temperatures above 0 °C (cold shock);
Further gradual cooling below 0 °C (frost damage);
Storage temperature;
Heating to room temperature (defrost damage);
Recovery.
Related to this is Mazur’s best-known “two-factor” hypothesis, which suggests that upon slow cooling, ice crystals form extracellularly, and cells are damaged due to exposure to concentrated solutions. However, upon rapid cooling, cells are destroyed by the formation of ice crystals directly within the cells [100,101,102].
The cellular response to freezing is schematically shown in Figure 6. During slow cooling (A), the cell is able to maintain osmotic balance by the outflow of water; the cell shrinks and only external ice crystals are formed. In the second case, with optimal cooling (B), the cell is unable to release water fast enough and thus maintain the osmotic balance; therefore, there is only a small amount of shrinkage of the cell and the cell contains a few ice crystals. During rapid cooling (C), the cell does not lose any water. Due to this effect, within the cell, small ice crystals develop.
Figure 6.
Schematic representation of the cellular response to freezing: (A) slow cooling, (B) optimal cooling, and (C) very fast cooling; data from Gao & Critser, 2000 .
Salinity and Temperature. Storing microorganisms in distilled water leads to a higher viability compared to cryopreservation in physiological NaCl solution . This was examined using E. coli 451-B . The organism was frozen in either distilled water or 0.85% NaCl solution. Moreover, ultra-fast freezing to −196 °C followed by slow or rapid heating, or slow cooling to −22 °C and slow heating or rapid heating were examined. The viability of E. coli was up to 78% when distilled water and ultra-fast cooling and rapid thawing were used. Using the same freezing and thawing procedure but using NaCl solution resulted in 12% viability. In contrast, viability decreased to 48% with slow cooling to −22 °C in distilled water and rapid heating, while viability was 41% when using NaCl solution .
Damage to the cell membrane and cell wall. During cryopreservation and subsequent thawing of microorganisms in saline conditions, it was observed that both the cell wall and the cell membrane were damaged [75,76]. Membrane damage leads to increased permeability to small molecules and atoms such as potassium, nucleotides, amino acids. Moreover, sucrose can enter the cell in the same way as water .
Oxidative stress. Another important stress factor in strictly anaerobic intestinal microorganisms is oxidative stress . The O2 tolerance ability varies from species to species. For example, Bacteroides thetaiotamicron is able to create a protective mechanism against O2 by scavenging enzymes that prevent the rapid formation of reactive O2 and perform remediation upon exposure to O2 [103,104]. By adding riboflavin, cysteine and HCl in the preparation of the buffer, an environment can be created that protects the bacteria from possible contact with O2 during cryopreservation . However, highly susceptible strictly anaerobic bacteria may still require a completely O2-free environment.
5.5.2. Osmoregulation and Osmotic Stress
Osmoregulation. Osmoregulation might indicate cryoresistance in microorganisms . Therefore, it is important to determine the optimal cooling rate in order to maintain osmoregulation during cooling.
Osmotic stress. It is suggested to prevent the osmotic stress of microorganisms that occurs during the recovery process after thawing by using hypertonic solutions (1.75% NaCl, 7.5% glycerol or sorbitol) . It has also been suggested that osmotolerance may be strain specific [106,107]. The main consequence of freezing is the exclusion of various molecules from the crystal lattice and their accumulation in the residual external liquid [16,73]. Thus, a hypertonic environment is created around the cell. As cooling continues, osmotic stress begins to act on the unfrozen cell. The cell responds to osmotic stress by losing water and the cell thus becomes dehydrated. If the cell drains the water faster than the heat, the later it will decrease with decreasing temperature. The water in the external environment will freeze and the osmotic stress of the environment will increase. The lower the ambient temperature, the higher the concentration of the unfrozen solution and the higher the leakage of water from the cytoplasm .
6. Summary of the Obtained Information with the Possibility of Application to SRB
Studies concerning the long-term storage of SRB in the form of cryopreservation are uncommon. Because SRB are able to corrode iron and, after a long time, stainless steel, it was initially recommended that glassware be used for longer-term storage . However, this has been refuted in the cryopreservation of intestinal bacteria, and the use of plastic ampoules or vials has been recommended [15,81,82,83,84]. A possible alternative to preserve SRB may be the use of screw-capped glass ampoules [79,80], with which the maintenance of anaerobic conditions was successfully performed and had a number of advantages.
The use of silicone or thin rubber or rubber handles in SRB storage has been shown to be permeable to O2 . The air that penetrates inside the container in which the SRB are stored can impair their growth. Hence, thick rubber or polyvinyl chloride plugs, tubes or other connections may be used. The establishment of anoxygenic conditions during long-term storage could also be performed using paraffin . SRB differ from each other not only within individual genera, but also in terms of species within one genus. The genus Desulfovibrio is considered to be quite resistant to the presence of O2 and is generally much easier to isolate and cultivate than most other SRB genera .
Comparing Desulfovibrio desulfuricans ssp. desulfuricans to Desulfovibrio acrylicus, the optimal pH of D. acrylicus is 7.4, while the optimal pH D. desulfuricans ssp. desulfuricans is 7.2 to 7.8. The optimal growth temperature is almost identical: 30–37 °C for D. acrylicus and 30–36 °C for D. desulfuricans ssp. desulfuricans. However, the optimal NaCl concentration is 0–1 g L−1 for D. desulfuricans ssp. desulfuricans and 18 g L−1 for D. acrylicus [12,59,108,109].
SRB can be found in fresh and salt waters [12,110]. Growth of SRB was observed at 1% NaCl concentration, but when NaCl concentration was 3%, no growth was observed. This reaction to the saline environment has also been reported in E. coli . Storage of intestinal microorganisms in distilled water led to a higher viability than when performing cryopreservation in NaCl solution, depending on the freezing/thawing method that had been applied. Regarding the freezing process of SRB, it has been mentioned that it is best to use glycerol (10%) as a cryoprotectant . Long-term freezing is possible to store SRB at −80 °C or in liquid nitrogen . For shorter storage in the range of 4–6 weeks, SRBs can be stored at 4–6 °C.
7. Conclusions
The aim of this review was to describe the distribution of SRB in various environments, the methods of cryopreservation of intestinal bacteria, and to compare whether the information is applicable for the cryopreservation of SRB. According to the obtained information, it is recommended that SRB are stored in glass vials to prevent corrosion. The storage should be performed in distilled water or in a solution with a low salt concentration. However, differences in viability among the species of the same SRB genus might be observed. According to the present studies and literature information, an unambiguous description of SRB cryopreservation would need more extensive laboratory research. Certainly, vast phylogenetic diversity influences the processes of method development of cryopreservation of SRBs. In particular, the molecular interactions between the cell, the cryoprotective compound and freezing mediator molecules might be worth considering. Moreover, standardization of the methods and determination of the optimal protectant for cryopreservation of SRB for the purpose of later medical and biotechnological applications would be desirable, since no universal method is known and method standardization would certainly have economic impacts in different industries.
Author Contributions
Conceptualization, A.K. and I.K.; writing—original draft preparation, A.K., I.K., D.D. and S.K.-M.R.R.; writing—review and editing, A.K., D.D., J.G., P.K. and S.K.-M.R.R.; visualization, A.K.; supervision, I.K. and M.V.; project administration, I.K. and M.V.; funding acquisition, M.V., P.K. and S.K.-M.R.R. All authors have read and agreed to the published version of the manuscript.
Funding
This study was supported by Grant Agency of the Masaryk University (MUNI/A/1425/2020). Open access funding was provided by the University of Vienna.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Acknowledgments
This study was supported by the Grant Agency of the Masaryk University (MUNI/A/1425/2020).
Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1.
The scheme shows the information outlined in the review.
Figure 2.
Scheme of iron metal corrosion by SRB (modified from Barton and Fauque, 2009 ).
Figure 3.
Different types of electron donors used by intestinal SRB which were isolated from human fecal samples (data from Gibson et al., 1988 ).
Figure 4.
The five most common cryopreservation methods and the stated ideal temperatures inside the boxes in °C; data from Hubálek, 1996 .
Figure 5.
Plastic (polyethylene) Pasteur pipette used for freezing. Taken from the Walter Reed Army Institute of Research; by Jackson et al., 1981 .
Figure 6.
Schematic representation of the cellular response to freezing: (A) slow cooling, (B) optimal cooling, and (C) very fast cooling; data from Gao & Critser, 2000 .
Table 1.
Composition of different cultivation media.
| Salts (g L−1) | Postgate B | Postgate C | Postgate E | Modified by Kovac & Kushkevych, 2017 |
--- ---
| Na2SO4 | – | 4.5 | 1 | 3 |
| KH2PO4 | 0.5 | 0.5 | 0.5 | 0.3 |
| K2HPO4 | – | – | – | 0.5 |
| NH4Cl | 1 | 1 | 1 | 1 |
| CaCl2·6H20 | – | 0.06 | 1 | 0.06 |
| Yeast extract | 1 | 1 | 1 | 1 |
| MgSO4·7H2O | 2 | 0.06 | 2 | 0.1 |
| CaSO4 | 1 | – | – | – |
| Ascorbic acid | 0.1 | – | 0.1 | 0.1 |
| Thioglycolic acid | 0.1 | – | – | – |
| (NH4)2SO4 | – | – | – | 0.2 |
Table 2.
Addition of Tween 80 to the medium and its effect on viability (data from Smittle et al., 1972) .
| Lactobacillus bulgaricus | Medium | Death (%) | Loss of Acid Production (%) |
--- --- |
| L. bulgaricus NCS1 | –T80 | 67 | 5 |
| +T80 | 3 | 1 |
| L. bulgaricus NCS2 | –T80 | 83 | 32 |
| +T80 | 38 | 9 |
| L. bulgaricus NCS3 | –T80 | 65 | 70 |
| +T80 | 49 | 40 |
| L. bulgaricus NCS4 | –T80 | 7 | 7 |
| +T80 | 0 | 5 |
Table 3.
Stability of individual species of L. bulgaricus during cryopreservation in liquid nitrogen (data from Smittle et al., 1972 ).
| Lactobacillus bulgaricus | Storage Time | | | |
--- ---
| 1. Day | | 2. Day | |
| Death (%) | Loss of Acid Production (%) | Death (%) | Loss of Acid Production (%) |
| L. bulgaricus NCS1 | 95 | 73 | 99 | 69 |
| L. bulgaricus NCS3 | 54 | 31 | 72 | 32 |
| L. bulgaricus NCS4 | 0 | 8 | 0 | 8 |
| | |
--- |
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Kushkevych, I.; Kovářová, A.; Dordevic, D.; Gaine, J.; Kollar, P.; Vítězová, M.; Rittmann, S.K.-M.R.
Distribution of Sulfate-Reducing Bacteria in the Environment: Cryopreservation Techniques and Their Potential Storage Application. Processes 2021, 9, 1843.
AMA Style
Kushkevych I, Kovářová A, Dordevic D, Gaine J, Kollar P, Vítězová M, Rittmann SK-MR.
Distribution of Sulfate-Reducing Bacteria in the Environment: Cryopreservation Techniques and Their Potential Storage Application. Processes. 2021; 9(10):1843.
Chicago/Turabian Style
Kushkevych, Ivan, Aneta Kovářová, Dani Dordevic, Jonah Gaine, Peter Kollar, Monika Vítězová, and Simon K.-M. R. Rittmann.
2021. "Distribution of Sulfate-Reducing Bacteria in the Environment: Cryopreservation Techniques and Their Potential Storage Application" Processes 9, no. 10: 1843.
APA Style
Kushkevych, I., Kovářová, A., Dordevic, D., Gaine, J., Kollar, P., Vítězová, M., & Rittmann, S. K.-M. R.
(2021). Distribution of Sulfate-Reducing Bacteria in the Environment: Cryopreservation Techniques and Their Potential Storage Application. Processes, 9(10), 1843.
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2015 AMC 12A Problems/Problem 15
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (for those like myself who couldn't think of something smarter to do)
6 See Also
Problem
What is the minimum number of digits to the right of the decimal point needed to express the fraction as a decimal?
Solution 1
We can rewrite the fraction as . Since the last digit of the numerator is odd, a is added to the right if the numerator is divided by , and this will continuously happen because , itself, is odd. Indeed, this happens twenty-two times since we divide by twenty-two times, so we will need more digits. Hence, the answer is
Solution 2
Multiply the numerator and denominator of the fraction by (which is the same as multiplying by 1) to give . Now, instead of thinking about this as a fraction, think of it as the division calculation . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, is the minimum number of digits to the right of the decimal point needed.
Solution 3
The denominator is . Each adds one digit to the right of the decimal, and each additional adds another digit. The answer is .
Solution 4 (for those like myself who couldn't think of something smarter to do)
First, we can write the denominator as and forget about the (however, we will need to add back to our final answer). Noticing that we divide 123456789 by 2048 using long division. We get 60281.63525390625 as the result (though the process seems intimidating, it actually doesn't take that long, just a couple of minutes). From there, we notice that there are 11 places after the decimal point in the quotient, which means there will be another 11 after we divide this quotient by 2048 again. So, there will be places after the decimal in the final quotient. We add back for a total of
See Also
2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14Followed by
Problem 16
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188489 | https://fiveable.me/key-terms/calc-ii/hyperbolic-functions | Hyperbolic functions - (Calculus II) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Definition
Hyperbolic functions are a set of mathematical functions that are analogs of the ordinary trigonometric functions but are based on hyperbolas instead of circles. They include hyperbolic sine ($$\sinh$$), hyperbolic cosine ($$\cosh$$), and others, which are essential in various calculus applications such as integrals, differential equations, and trigonometric substitution. These functions exhibit properties similar to trigonometric functions but have distinct geometric interpretations related to hyperbolas.
5 Must Know Facts For Your Next Test
Hyperbolic functions can be defined using exponential functions, with $$\sinh(x)$$ and $$\cosh(x)$$ being derived from combinations of the exponential function.
The derivatives of hyperbolic functions mirror those of trigonometric functions: $$\frac{d}{dx} \sinh(x) = \cosh(x)$$ and $$\frac{d}{dx} \cosh(x) = \sinh(x)$$.
The identities for hyperbolic functions are analogous to trigonometric identities; for example, $$\cosh^2(x) - \sinh^2(x) = 1$$ is similar to the Pythagorean identity for sine and cosine.
Hyperbolic functions are useful in solving differential equations, especially in applications related to physics and engineering where hyperbolic trajectories are involved.
They also appear in calculus when evaluating integrals that require substitutions or transformations involving hyperbolic identities.
Review Questions
Compare and contrast hyperbolic functions with trigonometric functions in terms of their definitions and key properties.
Hyperbolic functions like $$\sinh$$ and $$\cosh$$ are defined using exponential functions and exhibit properties similar to trigonometric functions. For instance, both have derivatives that relate them to one another: the derivative of $$\sinh(x)$$ is $$\cosh(x)$$ just like the derivative of sine is cosine. However, while trigonometric functions are based on circular geometry, hyperbolic functions stem from hyperbolic geometry, leading to different identities and values across their ranges.
Explain how hyperbolic functions can be applied in integral calculus and give an example of such an application.
Hyperbolic functions often simplify integrals involving square roots or exponential expressions. For example, when integrating $$\int \sqrt{x^2 + a^2} \, dx$$, a substitution using the hyperbolic function $$x = a \, \sinh(t)$$ can transform the integral into a simpler form. This substitution exploits the relationship between hyperbolas and areas under curves, making it easier to evaluate integrals involving complex algebraic expressions.
Evaluate the significance of understanding hyperbolic functions in solving real-world problems, particularly in physics and engineering contexts.
Understanding hyperbolic functions is crucial in fields like physics and engineering as they describe real-world phenomena involving hyperbolic trajectories, such as in special relativity or certain types of wave equations. They model situations where variables grow exponentially or follow specific paths dictated by the laws of physics. Recognizing these patterns allows engineers and scientists to apply mathematical models effectively, leading to innovations in technology and deeper insights into natural processes.
Related terms
Inverse Hyperbolic Functions:
Functions that provide the angle whose hyperbolic sine, cosine, or tangent gives a specific value; these are denoted as $$\text{arsinh}$$, $$\text{arcosh}$$, and $$\text{artanh}$$.
Exponential Functions:
Functions of the form $$f(x) = e^x$$, which serve as the foundation for defining hyperbolic functions; specifically, $$\sinh(x) = \frac{e^x - e^{-x}}{2}$$ and $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$.
Trigonometric Functions:
Functions like sine, cosine, and tangent that relate angles to side lengths in right triangles; hyperbolic functions can be seen as their counterparts for hyperbolas.
"Hyperbolic functions" also found in:
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188490 | https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_(Arnold)/04%3A_Fractions/4.09%3A_Solving_Equations_with_Fractions | Skip to main content
4.9: Solving Equations with Fractions
Last updated
: Aug 30, 2024
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4.8: Order of Operations with Fractions
5: Decimals
Page ID
: 24084
David Arnold
College of the Redwoods
( \newcommand{\kernel}{\mathrm{null}\,})
Undoing Subtraction
We can still add the same amount to both sides of an equation without changing the solution.
Example 1
Solve for x: x−56=13x−56=13.
Solution
To “undo” subtracting 5/6, add 5/6 to both sides of the equation and simplify.
x−56=13 Original equation.x−56+56=13+56 Add 56 to both sides.x=1⋅23⋅2+56 Equivalent fractions, LCD = 6.x=26+56 Simplify.x=76 Add.
x−56=13 x−56+56=13+56 x=1⋅23⋅2+56 x=26+56 x=76 Original equation. Add 56 to both sides. Equivalent fractions, LCD = 6. Simplify. Add.
It is perfectly acceptable to leave your answer as an improper fraction. If you desire, or if you are instructed to do so, you can change your answer to a mixed fraction (7 divided by 6 is 1 with a remainder of 1). That is x=116x=116.
Checking the Solution
Substitute 7/6 for x in the original equation and simplify.
x−56=13 Original equation.76−56=13 Substitute 7/6 for x.26=13 Subtract.13=13 Reduce.
x−56=13 76−56=13 26=13 13=13 Original equation. Substitute 7/6 for x. Subtract. Reduce.
Because the last statement is true, we conclude that 7/6 is a solution of the equation x − 5/6 = 1/3.
Undoing Addition
You can still subtract the same amount from both sides of an equation without changing the solution.
Example 2
Solve for x: x+23=−35x+23=−35.
Solution
To “undo” adding 2/3, subtract 2/3 from both sides of the equation and simplify.
x+23=−35 Original equation.x+23−23=−35−23 Subtract 23 from both sides.x=−3⋅35⋅3−2⋅53⋅5 Equivalent fractions, LCD = 15.x=−915−1015 Simplify.x=−1915 Subtract.
x+23=−35 x+23−23=−35−23 x=−3⋅35⋅3−2⋅53⋅5 x=−915−1015 x=−1915 Original equation. Subtract 23 from both sides. Equivalent fractions, LCD = 15. Simplify. Subtract.
Readers are encouraged to check this solution in the original equation.
Exercise
Solve for x: x+34=−12x+34=−12
Answer
: −5/4
Undoing Multiplication
We “undo” multiplication by dividing. For example, to solve the equation 2x = 6, we would divide both sides of the equation by 2. In similar fashion, we could divide both sides of the equation
35x=410
35x=410
by 3/5. However, it is more efficient to take advantage of reciprocals. For convenience, we remind readers of the Multiplicative Inverse Property.
Multiplicative Inverse Property
Let a/b be any fraction. The number b/a is called the multiplicative inverse or reciprocal of a/b. The product of reciprocals is 1.
ab⋅ba=1.
ab⋅ba=1.
Let’s put our knowledge of reciprocals to work.
Example 3
Solve for x: 35x=41035x=410.
Solution
To “undo” multiplying by 3/5, multiply both sides by the reciprocal 5/3 and simplify.
35x=410 Original equation.53(35x)=53(410) Multiply both sides by 5/3.(53⋅35)x=2030 On the left, use the associative property to regroup. On the right, multiply.1x=23 On the left, 53⋅35=1. On the right, reduce: 2030=23.x=23 On the left, 1x=x.
35x=410 53(35x)=53(410)(53⋅35)x=2030 1x=23 x=23 Original equation. Multiply both sides by 5/3. On the left, use the associative property to regroup. On the right, multiply. On the left, 53⋅35=1. On the right, reduce: 2030=23. On the left, 1x=x.
Checking the Solution
Substitute 2/3 for x in the original equation and simplify.
35x=410 Original equation.35(23)=410 Substitute 2/3 for x.615=410 Multiply numerators; multiply denominators.25=25 Reduce both sides to lowest terms.
35x=410 35(23)=410 615=410 25=25 Original equation. Substitute 2/3 for x. Multiply numerators; multiply denominators. Reduce both sides to lowest terms.
Because this last statement is true, we conclude that 2/3 is a solution of the equation (3/5)x = 4/10.
Exercise
Solve for y: 23y=4523y=45
Answer
: 6/5
Example 4
Solve for x: −89x=518−89x=518.
Solution
To “undo” multiplying by −8/9, multiply both sides by the reciprocal −9/8 and simplify.
−89x=518 Original equation.−98(−89x)=−98(518) Multiply both sides by −9/8.[−98⋅(−89)]x=−3⋅32⋅2⋅2⋅52⋅3⋅3 On the left, use the associative property to regroup. On the right, prime factor.1x=3⋅32⋅2⋅2⋅52⋅3⋅3 On the left, −98⋅(−89)=1. On the right, cancel common factors.x=−516 On the left, 1x=x. Multiply on the right.
−89x=518 −98(−89x)=−98(518) [−98⋅(−89)]x=−3⋅32⋅2⋅2⋅52⋅3⋅3 x=−516 Original equation. Multiply both sides by −9/8. On the left, use the associative property to regroup. On the right, prime factor.1x=3⋅32⋅2⋅2⋅52⋅3⋅3 On the left, 1x=x. Multiply on the right. On the left, −98⋅(−89)=1. On the right, cancel common factors.
Readers are encouraged to check this solution in the original equation.
Exercise
Solve for z: −27z=421−27z=421
Answer
: −2/3
Clearing Fractions from the Equation
Although the technique demonstrated in the previous examples is a solid mathematical technique, working with fractions in an equation is not always the most efficient use of your time.
Clearing Fractions from the Equation
To clear all fractions from an equation, multiply both sides of the equation by the least common denominator of the fractions that appear in the equation.
Let’s put this idea to work.
Example 5
In Example 1, we were asked to solve the following equation for x:
x−56=13.
Take a moment to review the solution technique in Example 1. We will now solve this equation by first clearing all fractions from the equation.
Solution
Multiply both sides of the equation by the least common denominator for the fractions appearing in the equation.
x−56=13 Original equation.6(x−56)=6(13) Multiply both sides by 6.6x−6(56)=6(13) Distribute the 6.6x−5=2 On each side, multiply first. 6(56)=5 and 6(13)=2.
Note that the equation is now entirely clear of fractions, making it a much simpler equation to solve.
6x−5+5=2+5 Add 5 to both sides.6x=7 Simplify both sides.6x6=76 Divide both sides by 6.x=76 Simplify.
Note that this is the same solution found in Example 1.
Exercise
Solve for t: t−27=−14
Answer
: 1/28
Example 6
In Example 4, we were asked to solve the following equation for x.
−89x=518
Take a moment to review the solution in Example 4. We will now solve this equation by first clearing all fractions from the equation.
Solution
Multiply both sides of the equation by the least common denominator for the fractions that appear in the equation.
−89x=518 Original equation.18(−89x)=18(518) Multiply both sides by 18.−16x=5 On each side, cancel and multiply. 18(−89)=−16 and 18(518)=5.
Note that the equation is now entirely free of fractions. Continuing,
−16x−16=5−16 Divide both sides by −16.x=−516 Simplify.
Note that this is the same as the solution found in Example 4.
Exercise
Solve for u:
−79u=1427
Answer
: −2/3
Example 7
Solve for x: 23x+34=12.
Solution
Multiply both sides of the equation by the least common denominator for the fractions appearing in the equation.
23x+34=12 Original equation.12(23x+34=)=12(12) Multiply both sides by 12.12(23x)+12(34)=12(12) On the left, distribute 12.8x+9=6 Multiply: 12(23x)=8x, 12(34)=9, and 12(12)=6.
Note that the equation is now entirely free of fractions. We need to isolate the terms containing x on one side of the equation.
8x+9−9=6−9 Subtract 9 from both sides.8x=−3 Simplify both sides.8x8=−38 Divide both sides by 8.x=−38 Simplify both sides.
Readers are encouraged to check this solution in the original equation.
Exercise
Solve for r: 34r+23=12
Answer
: −2/9
Example 8
Solve for x: 23−3x4=x2−18.
Solution
Multiply both sides of the equation by the least common denominator for the fractions in the equation.
23−3x4=x2−18 Original equation.24(23−3x4)=24(x2−18) Multiply both sides by 24.24(23)−24(3x4)=24(x2)−24(18) On both sides, distribute 24.16−18x=12x−3 Left: 24(23)=16, 24(3x4)=18x. Right: 24(x2)=12x, 24(18)=3.
Note that the equation is now entirely free of fractions. We need to isolate the terms containing x on one side of the equation.
16−18x−12x=12x−3−12x Subtract 12x from both sides.16−30x=−3 Left: −18x−12x=−30x. Right: 12x−12x=0.16−30x−16=−3−16 Subtract 16 from both sides.−30x=−19 Left: 16−16=0. Right: −3−16=−19.−30x−30=−19−30 Divide both sides by −30.x=1930 Simplify both sides.
Readers are encouraged to check this solution in the original equation.
Exercise
Solve for s: 32−2s5=s3−15.
Answer
: Add texts here. Do not delete this text first.
Applications
Let’s look at some applications that involve equations containing fractions. For convenience, we repeat the Requirements for Word Problem Solutions.
Requirements for Word Problem Solutions
Set up a Variable Dictionary. You must let your readers know what each variable in your problem represents. This can be accomplished in a number of ways:
Statements such as “Let P represent the perimeter of the rectangle.”
Labeling unknown values with variables in a table.
Labeling unknown quantities in a sketch or diagram.
Set up an Equation. Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement.
Solve the Equation. You must always solve the equation set up in the previous step.
Answer the Question. This step is easily overlooked. For example, the problem might ask for Jane’s age, but your equation’s solution gives the age of Jane’s sister Liz. Make sure you answer the original question asked in the problem. Your solution should be written in a sentence with appropriate units. 5. Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it’s possible that your equation incorrectly models the problem’s situation, so you could have a valid solution to an incorrect equation. The important question is: “Does your answer make sense based on the words in the original problem statement.”
Example 9
In the third quarter of a basketball game, announcers informed the crowd that attendance for the game was 12,250. If this is two-thirds of the capacity, find the full seating capacity for the basketball arena.
Solution
We follow the Requirements for Word Problem Solutions.
Set up a Variable Dictionary. Let F represent the full seating capacity. Note: It is much better to use a variable that “sounds like” the quantity that it represents. In this case, letting F represent the full seating capacity is much more descriptive than using x to represent the full seating capacity.
Set up an Equation. Two-thirds of the full seating capacity is 12,250.
Two-thirds of Full Seating Capacity is 12,25023⋅F=12,250
Hence, the equation is
23F=12250.
Solve the Equation. Multiply both sides by 3 to clear fractions, then solve.
23F=12250 Original equation.3(23F)=3(12250) Multiply both sides by 3.2F=36750 Simplify both sides.2F2=367502 Divide both sides by 2.F=18375 Simplify both sides.
Answer the Question. The full seating capacity is 18,375.
Look Back. The words of the problem state that 2/3 of the seating capacity is 12,250. Let’s take two-thirds of our answer and see what we get.
23⋅18375=23⋅183751=23⋅3⋅61251=23⋅3⋅61251=12250
This is the correct attendance, so our solution is correct.
Exercise
Attendance for the Celtics game was 9,510. If this is 3/4 of capacity, what is the capacity of the Celtics’ arena?
Answer
: 12,680
Example 10
The area of a triangle is 20 square inches. If the length of the base is 212 inches, find the height (altitude) of the triangle.
Solution
We follow the Requirements for Word Problem Solutions.
Set up a Variable Dictionary. Our variable dictionary will take the form of a well labeled diagram.
Set up an Equation. The area A of a triangle with base b and height h is
A=12bh.
Substitute A = 20 and b = 212.
20=12(212)h.
Solve the Equation. Change the mixed fraction to an improper fraction, then simplify.
20=12(212)h Original equation.20=12(52)h Mixed to improper: 212=52.20=(12⋅52)h Associative property.20=54h Multiply: 12⋅52=54.
Now, multiply both sides by 4/5 and solve.
45(20)=45(54h) Multiply both sides by 4/5.16=h Simplify: 45(20)=16 and 45⋅54=1.
Answer the Question. The height of the triangle is 16 inches.
Look Back. If the height is 16 inches and the base is 212 inches, then the area is
A=12(212)(16)=12⋅52⋅161=5⋅162⋅2=(5)⋅(2⋅2⋅2⋅2)(2)⋅(2)=5⋅2⋅2⋅2cot22⋅2=20
This is the correct area (20 square inches), so our solution is correct.
Exercise
The area of a triangle is 161 square feet. If the base of the triangle measures 4014 feet, find the height of the triangle.
Answer
: 8 feet
Exercises
Is 1/4 a solution of the equation x+58=58?
Is 1/4 a solution of the equation x+13=512?
Is −8/15 a solution of the equation 14x=−115?
Is −18/7 a solution of the equation −38x=2528?
Is 1/2 a solution of the equation x+49=1718?
Is 1/3 a solution of the equation x+34=1312?
Is 3/8 a solution of the equation x−59=−1372?
Is 1/2 a solution of the equation x−35=−110?
Is 2/7 a solution of the equation x−49=−863?
Is 1/9 a solution of the equation x−47=−3163?
Is 8/5 a solution of the equation 1114x=4435?
Is 16/9 a solution of the equation 1318x=10481?
In Exercises 13-24, solve the equation and simplify your answer.
2x−3=6x+7
9x−8=−9x−3
−7x+4=3x
6x+9=−6x
−2x=9x−4
−6x=−9x+8
−8x=7x−7
−6x=5x+4
−7x+8=2x
−x−7=3x
−9x+4=4x−6
−2x+4=x−7
In Exercises 25-48, solve the equation and simplify your answer.
(x + \frac{3}{2 = \frac{1}{2})
x−34=14
−95x=12
73x=−72
38x=87
−19x=−35
25x=−16
16x=29
−32x=87
−32x=−75
x+34=59
x−19=−32
x−47=78
x+49=−34
x+89=23
x−56=14
x+52=−98
x+12=53
−85x=79
−32x=−59
x−14=−18
x−92=−72
−14x=12
−89x=−83
In Exercises 49-72, solve the equation and simplify your answer.
−73x−23=34x+23
12x−12=32x+34
−72x−54=45
−76x+56=−89
−97x+92=−52
59x−72=14
14x−43=−23
87x+37=53
53x+32=−14
12x−83=−25
−13x+45=−95x−56
−29x−35=45x−32
−49x−89=12x−12
−54x−53=87x+73
12x−18=−18x+57
−32x+83=79x−12
−37x−13=−19
23x+29=−95
−34x+27=87x−13
12x+13=−52x−14
−34x−23=−23x−12
13x−57=32x+43
−52x+95=58
94x+43=−16
At a local soccer game, announcers informed the crowd that attendance for the game was 4,302. If this is 2/9 of the capacity, find the full seating capacity for the soccer stadium.
At a local basketball game, announcers informed the crowd that attendance for the game was 5,394. If this is 2/7 of the capacity, find the full seating capacity for the basketball stadium.
The area of a triangle is 51 square inches. If the length of the base is 812 inches, find the height (altitude) of the triangle.
The area of a triangle is 20 square inches. If the length of the base is 212 inches, find the height (altitude) of the triangle.
The area of a triangle is 18 square inches. If the length of the base is 412 inches, find the height (altitude) of the triangle.
The area of a triangle is 44 square inches. If the length of the base is 512 inches, find the height (altitude) of the triangle.
At a local hockey game, announcers informed the crowd that attendance for the game was 4,536. If this is 2/11 of the capacity, find the full seating capacity for the hockey stadium.
At a local soccer game, announcers informed the crowd that attendance for the game was 6,970. If this is 2/7 of the capacity, find the full seating capacity for the soccer stadium.
Pirates. About one-third of the world’s pirate attacks in 2008 occurred off the Somali coast. If there were 111 pirate attacks off the Somali coast, estimate the number of pirate attacks worldwide in 2008.
Nuclear arsenal. The U.S. and Russia agreed to cut nuclear arsenals of long-range nuclear weapons by about a third, down to 1, 550. How many long-range nuclear weapons are there now? Associated Press-Times-Standard 04/04/10 Nuclear heartland anxious about missile cuts.
Seed vault. The Svalbard Global Seed Vault has amassed half a million seed samples, and now houses at least one-third of the world’s crop seeds. Estimate the total number of world’s crop seeds. Associated Press-Times-Standard 03/15/10 Norway doomsday seed vault hits half-million mark.
Freight train. The three and one-half mile long Union Pacific train is about 2 1 2 times the length of a typical freight train. How long is a typical freight train? Associated Press-Times-Standard 01/13/10 Unusally long train raises safety concerns.
Answers
No
No
Yes
Yes
No
Yes
−52
25
411
715
89
1013
−1
−518
6421
−512
−1621
−736
8156
−29
−298
−3572
18
−2
−1637
−4170
499
83
−2120
−4944
−717
4735
−1427
52159
−2
47100
19,359
12
8
24,948
There were about 333 pirate attacks worldwide.
1,500,000
4.8: Order of Operations with Fractions
5: Decimals |
188491 | https://www.merriam-webster.com/thesaurus/MEANDER?pronunciation&lang=en_us&dir=a&file=appreh04 | Synonyms of meander
as in to wander
as in tangle
meander
1 of 2
verb
verb
Synonyms & Similar Words
meander
noun
Related Words
Synonym Chooser
How does the verb meander contrast with its synonyms?
Some common synonyms of meander are ramble, roam, rove, traipse, and wander. While all these words mean "to go about from place to place usually without a plan or definite purpose," meander implies a winding or intricate course suggestive of aimless or listless wandering.
When would ramble be a good substitute for meander?
The words ramble and meander can be used in similar contexts, but ramble stresses carelessness and indifference to one's course or objective.
When might roam be a better fit than meander?
The meanings of roam and meander largely overlap; however, roam suggests wandering about freely and often far afield.
When can rove be used instead of meander?
While in some cases nearly identical to meander, rove suggests vigorous and sometimes purposeful roaming.
When is traipse a more appropriate choice than meander?
Although the words traipse and meander have much in common, traipse implies a course that is erratic but may sometimes be purposeful.
When is it sensible to use wander instead of meander?
The synonyms wander and meander are sometimes interchangeable, but wander implies an absence of or an indifference to a fixed course.
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188492 | https://www.combinatorics.org/ojs/index.php/eljc/article/download/v28i3p41/pdf/ | Permutation reconstruction from a few large patterns
Maria Jo˜ ao Gouveia
Departamento de Matem´ atica Faculdade de Ciˆ encias Universidade de Lisboa Lisboa, Portugal
mjgouveia@fc.ul.pt
Erkko Lehtonen ∗
Centro de Matem´ atica e Aplica¸ c˜ oes Faculdade de Ciˆ encias e Tecnologia Universidade Nova de Lisboa Caparica, Portugal
e.lehtonen@fct.unl.pt
Submitted: Apr 29, 2021; Accepted: Aug 4, 2021; Published: Aug 27, 2021
© The authors. Released under the CC BY-ND license (International 4.0).
Abstract
Every permutation of rank n > 5 is reconstructible from any dn/ 2e + 2 of its (n − 1)-patterns.
Mathematics Subject Classifications: 05A05
1 Introduction
The theory of permutation patterns and pattern avoidance has been an active field of research in the past decades. The fundamental notion of the theory is the pattern in-volvement relation. Writing a permutation π ∈ Sn as a word π1π2 . . . π n where πt = π(t), a permutation τ ∈ Sis called a pattern of π if τ1τ2 . . . τ is order-isomorphic to some sub-word πt1 πt2 . . . π t` of π. For further information and background on permutation patterns, we refer the reader to B´ ona and Kitaev . A reconstruction problem concerns whether a mathematical structure can be uniquely recovered from a collection of some derived structures that convey partial information on the original structure. The kind of reconstruction problems we discuss in this paper are exemplified by the famous unsolved problem in graph theory that concerns whether or not a finite simple graph is uniquely determined, up to isomorphism, by the collection of its one-vertex-deleted subgraphs. This problem was posed by Kelly and Ulam , and it
∗This work is funded by National Funds through the FCT – Funda¸ c˜ ao para a Ciˆ encia e a Tecnologia, I.P., under the scope of the project UIDB/00297/2020 (Center for Mathematics and Applications) and the project PTDC/MAT-PUR/31174/2017.
the electronic journal of combinatorics 28(3) (2021), #P3.41
is conjectured to hold for every graph with at least three vertices. Analogous reconstruc-tion problems have been formulated and studied for many other kinds of mathematical structures. The problem of reconstructing a permutation from its patterns has been considered by several authors. It is known that, for n > 5, every n-permutation is reconstructible from its multiset of ( n − 1)-patterns (Smith , Raykova ) and from its set of ( n − 1)-patterns (Ginsburg ). Raykova and Smith also considered decks (i.e. multisets) composed of ( n − k)-patterns for k > 1 and, for every k > 1, they proved the existence of natural numbers M such that all permutations of rank n > M are reconstructible from their multisets of ( n − k)-patterns. Although they did not determine Nk, the smallest of those numbers M , they could provide an upper bound and a lower bound for Nk.When a permutation is reconstructible, its deck contains a sufficient amount of in-formation for uniquely determining the permutation. Nonetheless, there may be some redundancy; perhaps some permutations can be reconstructed from only a few cards. Our main goal is to shed some light on how much information is needed for reconstruc-tion, independently of the permutation we may consider. In this way we sharpen the previous results on the reconstructibility of permutations from patterns. In this paper we succeed to answer, for k = 1, one of the open problems posed by Ginsburg in : Can we find a non-trivial function fk : {n ∈ N | n > Nk} → N so that fk(n) is the smallest integer m such that every permutation π ∈ Sn is uniquely determined by any of its par-tial (n − k)-decks of cardinality m? More precisely, here we prove that when k = 1 the function fk is defined by f1(n) = dn/ 2e + 2. This paper is organized as follows. In Section 2, we introduce the basic notions and notation that will be needed throughout the paper. The main questions we are addressing in this paper are formulated precisely in Section 3. In Section 4, we look into certain special cases in which a permutation can be easily reconstructed from a few cards that contain the same monotone segment. Section 5 is devoted to our main result (Theorem 19): every permutation of rank n > 5 is reconstructible from any partial deck of cardinality dn/ 2e+2. Its proof is constructive and yields a reconstruction algorithm. For the sake of illustrating the reconstruction method, several examples are given. We discuss some open problems in the final Section 6.
2 Preliminaries
Let the symbols N and N+ denote the set of all nonnegative integers and the set of positive integers, respectively. For n ∈ N+, denote by [ n] the set {i ∈ N | 1 6 i 6 n}. For a set A
and k ∈ N, we denote by (Ak
) the set of all k-element subsets of A.A finite multiset M over a nonempty set X is a pair ( X, χ M ), where χM : X → N is a map, called a multiplicity function , such that the support or underlying set Supp( M ) :=
{x ∈ X | χM (x) 6 = 0 } is finite. We say that x is an element of M and we write x ∈ M if
x ∈ Supp( M ). For x ∈ X, the number χM (x) is called the multiplicity of x in M . In this paper only finite multisets are discussed and we will refer to them simply as multisets.
For a multiset M , the sum ∑
x∈X
χM (x) is a well-defined natural number, and it is
the electronic journal of combinatorics 28(3) (2021), #P3.41 2
called the cardinality of M and is denoted by |M |.Let M and M ′ be multisets over X. We say that M is a submultiset of M ′ if χM (x) 6
χM ′ (x) for all x ∈ X. The intersection of M and M ′ is the multiset M ∩ M ′ over X given by the multiplicity function χM ∩M ′ (x) = min( χM (x), χ M ′ (x)) for all x ∈ X.We may represent a multiset M as a list enclosed in angle brackets where each element
x ∈ X occurs χM (x) times (the order of elements in the list does not matter), e.g.,
〈0, 0, 1, 1, 1, 1, 2, 3, 4, 4, 4〉. We also write xm to mean m occurrences of x. (This will not create confusion because we will not be dealing with exponentiation in this paper.) Using this shorthand, the above multiset can be written briefly as 〈02, 14, 2, 3, 43〉.If ( ai)i∈I is a finite indexed family of elements of X, then we will write 〈ai | i ∈ I〉 for the multiset in which the multiplicity of each element x ∈ X equals |{ i ∈ I | ai = x}| .For n ∈ N+, the set of all permutations of [ n] is denoted by Sn; these are called permutations of rank n. We may write a permutation π ∈ Sn as a word π1π2 . . . π n where
πp = π(p) for every p ∈ [n]. For a nonempty subset P = {p1, p 2, . . . , p } ⊆ [n] with p1 < p 2 < · · · < p, we denote by πP the permutation τ ∈ Ssuch that the words πp1 πp2 . . . π p and τ1τ2 . . . τ ` are order-isomorphic (with respect to the natural order of integers); when P = [ n] \ { p} we write
π − p instead. If τ is such a permutation πP , then τ is called a pattern of π and π is said to involve τ . We write τ 6 π when τ is a pattern of π.Let π ∈ Sn. The reverse and the complement of π are the permutations πr and πc in
Sn, respectively, given by the rules πr(t) = π(n − t + 1) and πc(t) = n − π(t) + 1, for all
t ∈ [n]. It is well known that the pattern involvement relation is preserved under reverses and complements of permutations, as well as under taking inverses (i.e., if τ 6 π, then
τ r 6 πr, τ c 6 πc, and τ −1 6 π−1). For a permutation π ∈ Sn and k ∈ N+, the ( n − k)-deck of π is the multiset deck n−k(π) := 〈πP | P ∈ ( [n]
n−k
)〉. The elements of deck n−k(π) are called the ( n − k)-cards
of π. If the number n − k is clear from the context, we speak of the deck and the cards
of π. Any submultiset of the deck of π is called a partial deck of π.A permutation π ∈ Sn is reconstructible from its ( n − k)-cards if for all σ ∈ Sn it holds that deck n−k(π) = deck n−k(σ) if and only if π = σ.As we can easily observe from Lemma 6, the ascending (identity ) and the descending
permutations of rank n, i.e., ιn = 12 . . . n and δn = n(n − 1) . . . 1, respectively, are the unique permutations that admit an ( n − 1)-deck with a unique element. Hence, for n > 3,
ιn and δn are examples of permutations that are reconstructible from their ( n − 1)-decks. In Section 3 we present some constructions of permutations that preserve the existence of common cards. Some of those examples are particular cases of direct sums and skew sums of permutations. For σ ∈ Sm and τ ∈ Sn, the direct sum σ ⊕ τ and the skew sum
σ τ are the permutations of rank m + n given by the rules (σ ⊕ τ )( t) =
{
σ(t), if 1 6 t 6 m,
τ (t − m) + m, if m + 1 6 t 6 m + n,(σ τ )( t) =
{
σ(t) + m, if 1 6 t 6 m,
τ (t − m), if m + 1 6 t 6 m + n.
the electronic journal of combinatorics 28(3) (2021), #P3.41 3
Cn,k
k
1 2 3 4
n
2 2∗ — — —3 3∗ 3∗ — —4 4∗ 6∗ 4∗ —5 4† 10 ∗ 10 ∗ 5∗
6 4 14 † 20 ∗ 15 ∗
7 5 18 34 † 35 ∗
8 5 ? ? ?Table 1: Values of Cn,k for small parameters n and k. Asterisks indicate values equal to ( nn−k
), and daggers indicate values equal to ( nn−k
) − 1.
3 Common partial decks
We are now ready to formulate the main question that we are addressing in this paper.
Definition 1. For n, k ∈ N+, with n > k , define Hk(n) to be the smallest integer m such that every permutation π ∈ Sn is uniquely determined by any of its partial ( n − k)-decks of cardinality m.Does the number Hk(n) exist for given n, k ∈ N+? It is clear that Hk(n) exists if and only if every permutation of rank n is reconstructible from its ( n − k)-deck, and that
Hk(n) 6 ( nn−k
) in this case. As explained in the introduction, for every k ∈ N+, there exists a smallest number Nk such that every permutation of rank n > Nk is reconstructible from its ( n−k)-deck (see Smith and Raykova ). Consequently, Hk(n) exists for every
n > Nk, given k ∈ N+. In particular, H1(n) exists for n > 5 and H2(n) exists for n > 6since N1 = 5 (see Smith ) and N2 = 6 (see Raykova ).
Problem 2. Given n, k ∈ N+ such that n > Nk, what is the value of Hk(n)?
Definition 3. For n, k ∈ N with 1 6 k < n , define Cn,k to be the largest number m for which there exist π, σ ∈ Sn such that π 6 = σ but π and σ have m common ( n − k)-cards, i.e., |deck n−k(π) ∩ deck n−k(σ)| = m.Clearly, every permutation of rank n is reconstructible from its ( n − k)-deck if and only if Cn,k < ( nn−k
). It follows immediately from the definitions that Hk(n) = Cn,k + 1 whenever n > Nk.In Table 1 we present the numbers Cn,k for small values of n and k that were discovered by an exhaustive computer search. Values equal to ( nn−k
) are indicated with asterisks – this means that not every permutation of rank n is reconstructible from its ( n−k)-deck. Values equal to ( nn−k
) − 1 are indicated with daggers – this means that there exist permutations of rank n such that all ( n − k)-cards are necessary for unique reconstructibility.
the electronic journal of combinatorics 28(3) (2021), #P3.41 4
Focusing on the case k = 1, the pairs of distinct n-permutations with the maximum number Cn, 1 of common ( n − 1)-cards were determined for 2 6 n 6 8; all such pairs are presented in Table 2, one pair in each row. Note that for n 6 4, the listed permutations are precisely the permutations that are not reconstructible from ( n − 1)-decks. At this point, we would like to draw the reader’s attention to the following facts. On the one hand, the computational evidence seems to suggest that Cn, 1 = dn/ 2e + 1 and hence H1(n) = dn/ 2e + 2 for all n > 5. On the other hand, by noticing that
• for n = 2 k, the two distinct permutations ( ιk−1 1) ⊕ ιk and ( ιk 1) ⊕ ιk−1 have
dn/ 2e + 1 common ( n − 1)-cards, and
• for n = 2 k + 1, the two distinct permutations ιk−1 ⊕ δ2 ⊕ ιk and ιk ⊕ δ2 ⊕ ιk−1 have
dn/ 2e + 1 common ( n − 1)-cards, we may immediately conclude that Cn, 1 > dn/ 2e + 1; consequently, H1(n) > dn/ 2e + 2 for all n > 5. In the next section, we are going to show that this lower bound is actually exact. We end this section by highlighting some constructions of permutations that preserve the existence of common cards. For σ, τ ∈ Sn, the following holds: 1. For every p ∈ [n], σr − p = ( σ − (n − p + 1)) r and σc − p = ( σ − p)c.2. For every i ∈ [n], σ−1 − i = ( σ − σ−1(i)) −1.3. If σ and τ have m common ( n − 1)-cards, then each of the pairs ( σr, τ r), ( σc, τ c), and ( σ−1, τ −1) also has m common ( n − 1)-cards. 4. If σ and τ have m common ( n − 1)-cards, then the following pairs of ( n + 1)-permutations have m common n-cards: (1 ⊕ σ, 1 ⊕ τ ), (1 σ, 1 τ ), ( σ ⊕ 1, τ ⊕ 1), and ( σ 1, τ 1). It is not difficult to verify that the pairs of permutations of Table 2 for n > 5 can be built from permutations of lower rank by using some of these constructions. (Note that for n = 5 we make use of the non-reconstructible pairs of 4-permutations.)
4 Monotone segments and their relevance for reconstructibility
In this section we highlight some cases where the presence of a monotone segment in cards of a permutation is relevant to, and sometimes enough for, its reconstruction. Given words v and w over some alphabet A, we write v v w if v is a consecutive subword of w, i.e., there exist (possibly empty) words x and y over A such that xvy = w.Recall that we may write a permutation π ∈ Sn as a word π1π2 . . . π n over [ n]. We call a nonempty consecutive subword of π1π2 . . . π n a segment of π. We denote the segment
πsπs+1 . . . π t by π[s, t ]. If τ and τ ′ are segments of π and τ v τ ′, then we say that τ is a
subsegment of τ ′ and τ is a proper subsegment of τ ′ if τ v τ ′ and τ 6 = τ ′. A segment π[s, t ]is ascending if πi+1 = πi + 1 for all i ∈ [s, t − 1], and π[s, t ] is descending if πi+1 = πi − 1
the electronic journal of combinatorics 28(3) (2021), #P3.41 5
n = 2 12 21
n = 3 132 213 231 312
n = 4 2413 3142
n = 5 12435 13245 13524 14253 24135 31425 35241 42531 52413 53142 53421 54231
n = 6 123546 124356 123564 124563 123645 126345 124356 132456 124356 132546 124635 125364 135246 142536 146352 153642 163524 164253 164532 165342 214635 215364 231456 234156 235461 243561 241356 314256 241365 314265 312456 412356 352416 425316 352461 425361 365421 465321 463512 536412 463521 536421 524136 531426 534216 542316 543621 546321 562413 563142 612435 613245 613524 614253 624135 631425 635241 642531 645231 653421 645321 653421 651432 654132 652413 653142 653214 654213 653421 654231
n = 7 1235467 1243567 7645321 7653421
n = 8 12346578 12354678 12346785 12356784 12348567 12384567 12354678 12435678 18756432 18764532 23415678 23451678 23465781 23546781 41235678 51234678 48765321 58764321 76453218 76534218 76548321 76584321 81235467 81243567 87564321 87645321 87615432 87651432 87643215 87653214 87645321 87653421 Table 2: Permutations of rank n with Cn, 1 common ( n − 1)-cards, for 2 6 n 6 8.
the electronic journal of combinatorics 28(3) (2021), #P3.41 6
for all i ∈ [s, t − 1]. A segment is monotone if it is either ascending or descending. An ascending (descending) segment is maximal if it is not a proper subsegment of any ascending (descending) segment. The length of a segment π[s, t ] is defined as t − s + 1, that is, the cardinality of the interval [ s, t ]. A segment is nontrivial if its length is greater than 1. The initial and the final values of a segment π[s, t ] are πs and πt, respectively. We sometimes view a segment π[s, t ] as the set {πi | i ∈ [s, t ]} of its elements, and with slight misuse of notation, the meaning of expressions such as i ∈ π[s, t ] or I ⊆ π[s, t ] will be clear. Henceforth we will consider mainly ( n − 1)-cards and (partial) ( n − 1)-decks of an
n-permutation; therefore we will refer to them simply as cards and ( partial ) decks .The following notation will be used to specify ( n − 1)-cards of an n-permutation. For
π ∈ Sn and p ∈ [n], we denote by
• π ↓ i the ( n − 1)-permutation obtained by deleting from π the entry with value i,i.e., if p = π−1(i), then π ↓ i = π − p
and we recall that
• π − p denotes the ( n − 1)-permutation obtained by deleting from π the entry at the
p-th position, i.e., the ( n−1)-permutation order-isomorphic to π1 . . . π p−1πp+1 . . . π n.In other words, π ↓ i = π[n]{ π−1(i)} and π − p = π[n]{ p}. More generally, for a subset
J ⊆ [n], we define π − J := π[n]\J and π ↓ J := π[n]\π−1(J). For a permutation π ∈ Sn and a set I ⊆ [n], we denote by deck I (π) the partial deck 〈π ↓ i | i ∈ I〉.Any permutation π can be recovered when π−1(i) and π ↓ i are known for some i ∈ [n], as is made explicit in the next lemma. For any τ ∈ Sn−1 and p, v ∈ [n], we denote by
τ ↑p v the permutation in Sn obtained by adding 1 to all entries greater than or equal to v and inserting at position p an entry with value v, more precisely, τ ↑p v = a1 . . . a n,where
at =
τ (t), if t < p and τ (t) < v ,
τ (t) + 1 , if t < p and τ (t) > v,
v, if t = p,
τ (t − 1) , if t > p and τ (t) < v ,
τ (t − 1) + 1 , if t > p and τ (t) > v.More generally, if τ = j1j2 . . . j m ∈ Sm, p ∈ [m + 1], and σ = k1k2 . . . k is a word over [n] with n := m +, then we define τ ↑p σ as the permutation a1a2 . . . a n ∈ Sn, where
apap+1 . . . a p+−1 = σ and a1 . . . a p−1ap+ . . . a n is order-isomorphic to j1j2 . . . j m.
Lemma 4. For any π ∈ Sn and i ∈ [n], we have (π ↓ i) ↑π−1(i) i = π.Proof. This is clear from the definitions. Let us recall here a useful fact about iterated deletions of entries from a permutation.
Lemma 5 (Ginsburg [2, Lemma 1(vi)]) . Let π ∈ Sn and p, q ∈ [n] with p < q . Then
(π ↓ q) ↓ p = ( π ↓ p) ↓ (q − 1) .
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The existence of nontrivial monotone segments in a permutation is directly connected with the existence of cards of multiplicity greater than 1 in its deck.
Lemma 6 (Ginsburg [2, Lemma 1(iv)]) . Let π ∈ Sn and s, t ∈ [n] with s 6 t. Then
π − s = π − t if and only if π[s, t ] is a monotone segment in π.
Lemma 7. Let π ∈ Sn, I ⊆ [n], and D := deck I (π).
(i) The partial deck D comprises a single card τ with multiplicity |I| if and only if there is a monotone segment π[s, t ] of π such that I ⊆ π[s, t ].
(ii) Assume that the condition of statement (i) holds and |I| > 2. If π[s, t ] is an as-cending segment in π, then π[s, t − 1] = τ [s, t − 1] is an ascending segment in τ and
π = τ ↑s τ (s). If π[s, t ] is a descending segment in π, then π[s, t − 1] = τ [s, t − 1] is a descending segment in τ and π = τ ↑t τ (t − 1) .
(iii) Assume that the condition of statement (i) holds and |I| > dn/ 2e + 1 . Then τ
has a unique maximal monotone segment π[u, v ] of length at least dn/ 2e, and I ⊆
π[u, v + 1] .Proof. (i) This follows immediately from Lemma 6. (ii) This follows from a straightforward verification. (iii) By (i), π has a monotone segment π[s, t ] of length at least dn/ 2e + 1 such that
I ⊆ π[s, t ]; hence τ [s, t − 1] is a monotone segment in τ and has length at least dn/ 2e.If τ had two distinct maximal monotone segments of length at least dn/ 2e, then τ would have rank at least 2 dn/ 2e > n, a contradiction. The above Lemma 7 asserts that any card τ of multiplicity m > 2 in a partial deck of π arises by removing entries from a monotone segment of length > m + 1, and τ has consequently a monotone segment of length − 1 > m. The following lemma is a kind of converse statement and allows immediate reconstruction of π in the case when the partial deck contains a card with large multiplicity and a unique long monotone segment. For a segment σ = k1k2 . . . k q, denote by σ∗ the word obtained from σ by deleting its largest entry, and let σ− := ( k1 − 1)( k2 − 1) . . . (kq − 1).
Proposition 8. Let π ∈ Sn, and let D be a partial deck of π. Assume that D contains a card τ of multiplicity m > 3 such that τ has a unique maximal monotone segment
τ [u, v ] = σ = k1 . . . k q of length at least m − 1. If σ is ascending, then π = τ ↑u τ (u). If
σ is descending, then π = τ ↑u (τ (u) + 1) .Proof. Since τ has multiplicity m, it follows from Lemma 7(i) that π must contain a monotone segment λ of length at least m such that τ = π ↓ i for any i ∈ λ.Consequently, τ [u, v − 1] = λ∗ is a monotone segment in τ , and it has length at least
m − 1. Since τ has a unique maximal monotone segment of length at least m − 1, it holds that σ = λ∗. We have τ = π ↓ π(u). By Lemma 4, if σ (equivalently, λ) is ascending, then π = ( π ↓ π(u)) ↑u π(u) = τ ↑u k1 = τ ↑u τ (u). If σ (equivalently, λ) is descending, then π = ( π ↓ π(u)) ↑u π(u) = τ ↑u (k1 + 1) = τ ↑u (τ (u) + 1).
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We will now develop a result (Proposition 12) which, together with its dual, obtained by reversing permutations, allows us to reduce the reconstruction of a permutation π of rank n to the reconstruction of a permutation of rank n−q 6 n−2 whenever the cards of a partial deck of π contain the same maximal increasing or decreasing sequence of length q.
Definition 9. Let π ∈ Sn, and assume that π[u, v ] = k1k2 . . . k q =: σ and π[u′, v ′] =
k′
1
k′
2
. . . k ′
r
=: σ′ are maximal ascending segments in π. We say that i ∈ [n] is critical for (σ, σ ′) in π, if one of the following conditions holds: (1) u′ = v + 1, i = kq + 1, k′
1
= kq + 2, (2) u′ = v + 2, k′
1
= kq + 1, i = π(v + 1). In this case, we also say that i is ( upper ) critical for σ and ( lower ) critical for σ′.
Remark 10 . By definition, a maximal ascending segment π[u, v ] = k1k2 . . . k q may have at most two upper critical points, and there are exactly two upper critical points if and only if π(v + 1) = kq + 2 and π(v + 2) = kq + 1. In this case, kq + 2 is critical for ( π[u, v ], k q + 1) and kq +1 is critical for ( π[u, v ], k q +2); moreover, ( kq +2)( kq + 1) is a maximal descending segment in π, and hence π ↓ (kq + 1) = π ↓ (kq + 2). Similarly, π[u, v ] may have at most two lower critical points, and there are exactly two lower critical points if and only if
π(u − 1) = k1 − 2 and π(u − 2) = k1 − 1. In this case, k1 − 2 is critical for ( π[u, v ], k 1 − 1) and k1 −1 is critical for ( π[u, v ], k 1 −2); moreover, ( k1 −1)( k1 −2) is a maximal descending segment in π, and hence π ↓ (k1 − 1) = π ↓ (k1 − 2). Observe that if i is critical for ( μ, μ ′) in π, then in the card π ↓ i, the ascending segments μ and μ′ are merged and are part of a longer ascending segment. This will be made more precise in the following lemma.
Lemma 11. Let π ∈ Sn, and assume that π[u, v ] = k1k2 . . . k q =: σ is a maximal ascending segment in π.
(i) If q > 2, then for every i ∈ π[u, v ], (π ↓ i)[ u, v − 1] = σ∗ is a maximal ascending segment in π ↓ i.
(ii) If π−1(i) < u and i < k 1, then (π ↓ i)[ u − 1, v − 1] = σ−.
(iii) If π−1(i) < u and i > k q, then (π ↓ i)[ u − 1, v − 1] = σ.
(iv) If π−1(i) > v and i < k 1, then (π ↓ i)[ u, v ] = σ−.
(v) If π−1(i) > v and i > k q, then (π ↓ i)[ u, v ] = σ.
(vi) The ascending segments in π ↓ i described in (ii) –(v) are maximal precisely unless i
is critical for σ in π.
(vii) Consequently, σ∗ is an ascending segment in every card of π.
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(viii) If σ′ = k′
1
k′
2
. . . k ′
r
is another maximal ascending segment in π and i is critical for
(σ, σ ′), then π ↓ i has an ascending segment λ such that
• σσ ′− v λ if k′
1
= kq + 2 ;
• σσ ′ v λ if k′
1
= kq + 1 and i > k ′
r
;
• σ−σ′− v λ if k′
1
= kq + 1 and i < k 1.Proof. This follows from a straightforward verification.
Proposition 12. Assume n > 5, D is a partial deck of a permutation π ∈ Sn, and
|D| = H(n). Assume that the cards in D are not all equal, there is a card in D that has a maximal ascending segment κ = k1k2 . . . k q with 2 6 q 6 n − H(n), and in every card in
D, either κ or κ− is a maximal ascending segment. Then the following statements hold.
(i) There is a maximal ascending segment λ in π such that the cards in D are obtained by removing entries that do not lie in λ and are not critical for λ in π. Consequently, either λ or λ− is a maximal ascending segment in each card in D, and κ ∈ { λ, λ −}.
(ii) Let G := {gτ | τ ∈ D}, where
gτ :=
{
(k1, τ −1(k1)) , if κ v τ ,
(k1 − 1, τ −1(k1 − 1)) , if κ− v τ .Let A := {a | ∃ b (a, b ) ∈ G}, B := {b | ∃ a (a, b ) ∈ G}, a∗ := max A, b∗ := max B.Let
j :=
{
a∗ + 1 , if |A| = 1 and H(n) 6 a∗,
a∗, otherwise,
u :=
{
b∗ + 1 , if |B| = 1 and H(n) 6 b∗,
b∗, otherwise,
v := u + q − 1,
Then π[u, v ] is a maximal ascending segment in π with initial value j and length q.
(iii) Let D′ := 〈τ ′ | τ ∈ D〉, where
τ ′ :=
{
τ ↓ κ, if κ v τ ,
τ ↓ κ−, if κ− v τ .Then D′ is a partial (n − q − 1) -deck of π ↓ π[u, v ] ∈ Sn−q, where u and v are as in part (ii) .
(iv) Let u, v, and t be as in (ii) and D′ be as in (iii) . If θ := π ↓ π[u, v ] is reconstructible from D′, then π is reconstructible from D as θ ↑u π[u, v ]
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Proof. (i) Let γ be a card in D that has a maximal ascending segment κ satisfying the conditions of the proposition. The card γ must arise in one of the following ways: 1. There is a maximal ascending segment λ := k1k2 . . . k q(kq + 1) in π such that γ =
π ↓ i for some (any) i ∈ λ.2. There are maximal ascending segments μ and μ′ and a critical point i for ( μ, μ ′) in
π such that γ = π ↓ i and μμ ′− , μμ ′, or μ−μ′− is a subsegment of κ, according to Lemma 11(viii). 3. There is a maximal ascending segment λ in π such that κ ∈ { λ, λ −} and γ = π ↓ i
for some i ∈ [n] \ λ that is not critical for λ in π.We claim that cases 1 and 2 are impossible. Case 1 is not possible because of our assumption that the cards in D are not all equal. Namely, by Lemma 11, κ = λ∗ is an ascending segment in all cards, but it is not a maximal one in the cards distinct from γ, contradicting our hypothesis. Suppose now, to the contrary, that case 2 occurs. Consider first the case when the length of μ is at least 2. Since μ has at most 4 critical points in π and H(n) > 5, there is a card δ ∈ D such that δ = π ↓ j for some j that is not critical for μ in π. By Lemma 11,
μ, μ−, or μ∗ is a maximal ascending segment in δ. Since each one of μ, μ− and μ∗ has a nonempty intersection with both κ and κ− and is also shorter than them, neither κ
nor κ− is an ascending segment in δ, contradicting our assumptions. The case when the length of μ′ is at least 2 is treated similarly. Consider now the case when both μ and μ′ have length 1, say μ = a and μ′ = b (and hence b ∈ { a + 1 , a + 2 }); then a(a + 1) or ( a − 1) a is a subsegment of κ. Observe first that if D contained a card of the form π ↓ p for some p that is neither a nor a critical point of μ, then μ = a or μ− = a − 1 would be a maximal ascending segment in π ↓ p
by Lemma 11, and therefore neither κ nor κ− would be a maximal ascending segment in
π ↓ p, which would contradict the hypotheses of the proposition. Since μ has at most 4 critical points and H(n) > 5, we are left with the situation where H(n) = 5, μ has exactly 4 critical points, and π ↓ a ∈ D. Then necessarily ( a − 1)( a − 2) a(a + 2)( a + 1) v π; hence (a − 1)( a − 2)( a + 1) a v π ↓ a, so neither κ nor κ− is a maximal ascending segment in
π ↓ a. We have reached a contradiction also in this case. We conclude that case 3 is the only one possible. Furthermore, we see easily with the help of Lemma 11, by considering the length of the maximal ascending segment containing
λ∗, that all cards in D must be of the form π ↓ i for some i that neither belongs to λ nor is critical for λ. Therefore either λ or λ− is a maximal ascending segment in each card in
D, and κ ∈ { λ, λ −}.(ii) Let λ = 12 . . . ` q be the maximal ascending segment in π provided by part (i), and assume that π[u, v ] = λ. It is straightforward to verify that
• λ− v π − i and ( π − i)−1(1 − 1) = u − 1 if and only if i < u and π(i) < 1;
• λ− v π − i and ( π − i)−1(1 − 1) = u if and only if i < u and π(i) > q;
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• λ v π − i and ( π − i)−1(1) = u − 1 if and only if i > v and π(i) < 1;
• λ v π − i and ( π − i)−1(1) = u if and only if i > v and π(i) > q.Therefore, we have A ⊆ { 1, 1 − 1} and B ⊆ { u, u − 1}. It is clear that if |A| = 2, then
A = {k1, k 1 − 1} = {1, 1 − 1}, so j = a∗ = k1 = 1. Assume now that |A| = 1. Then all cards in D are of the form π − i with π(i) < 1, or they are all of the form π − i with
π(i) > q. The former case holds if and only if H(n) < 1, and the latter case holds if and only if H(n) < n − q (both inequalities cannot hold simultaneously because otherwise we would have n + 4 6 2H(n) < n − (q − 1) < n , a contradiction). In the former case, we have A = {1 − 1} and H(n) 6 1 − 1 = a∗, so j = a∗ + 1 =1. In the latter case, we have A = {1} and H(n)1 − 1, so j = a∗ = `1. In a similar way we can show that
u = π−1(1). Finally, since the length of κ equals that of λ, we have for v := u + q − 1that π[u, v ] = λ is a maximal ascending segment in π that has initial walue1 = j.(iii) Since each τ ∈ D is of the form π ↓ j for some j ∈ [n] \ [1, q], it follows from Lemma 5 that
• if j < k 1, then ( π ↓ j) ↓ λ− = ( π ↓ λ) ↓ j,
• if j > k q, then ( π ↓ j) ↓ λ = ( π ↓ λ) ↓ (j − q). Consequently, D′ is an ( n − q − 1)-deck of π ↓ λ ∈ Sn−q.(iv) Since H(n) > H(n − q), θ := π ↓ λ is reconstructible from D′. Then we obtain π
as θ ↑u (1 . . . q) = ( π ↓ λ) ↑u λ.
5 Reconstructing a permutation from a partial deck
In this section, we show that for n > 5, every permutation of rank n is uniquely deter-mined by any of its partial ( n − 1)-decks of cardinality H(n) := dn/ 2e + 2. The proof is constructive and can be turned into a reconstruction algorithm. The idea is to aim at determining π−1(i) and π ↓ i, for some i ∈ [n], from the given partial deck of π, because these would enable us to recover π by Lemma 4. If this is not directly possible, we are nevertheless able to determine π−1(1) and to reconstruct π ↓ 1 by a recursive application of the algorithm. For a permutation π ∈ Sn, the pair ( π−1(1) , π −1(2)) is referred to as the type of π. As described in the following lemma, the cards of π have only a few possible types, and only three of them may occur as types of multiple cards. Furthermore, the relative order of 1 and 2 in π and their adjacency can be determined from a few cards.
Lemma 13. Let n > 5, π ∈ Sn, p := π−1(1) and r := π−1(2) . For each t ∈ [n], let
pt := ( π − t)−1(1) and rt := ( π − t)−1(2) .
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(i) If p < r , then
(pt, r t) =
(p − 1, r − 1) if t < p ,
(r − 1, x ) if t = p,
(p, r − 1) if p < t < r ,
(p, y ) if t = r,
(p, r ) if r < t ,for some x, y ∈ [n] that satisfy x = y = π−1(3) 6 p − 1 or x = y = π−1(3) − 1 > r
or x = y − 1 with y = π−1(3) ∈ [p + 1 , r − 1] . If r < p , then
(pt, r t) =
(p − 1, r − 1) if t < r ,
(p − 1, x ) if t = r,
(p − 1, r ) if r < t < p ,
(r, y ) if t = p,
(p, r ) if p < t ,for some x, y ∈ [n] that satisfy x = y = π−1(3) 6 r − 1 or x = y = π−1(3) − 1 > p
or x = y − 1 with y = π−1(3) ∈ [r + 1 , p − 1] .
(ii) We have p < r if and only if pt < r t for all t ∈ [n] \ { p, r }. Similarly, r < p if and only if rt < p t for all t ∈ [n] \ { p, r }.
(iii) We have p = r − 1 if and only if pt = rt − 1 for all but at most two indices t ∈ [n].Moreover, p 6 = r − 1 if and only if pt 6 = rt − 1 for all but at most two indices t ∈ [n].Proof. (i) This follows from a straightforward verification. (ii) This follows immediately from (i). (iii) Necessity is clear by (i), because if p = r − 1, then pt = rt − 1 for all t ∈ [n] \ { p, r }
(note that the case p < t < r does not occur here). For sufficiency and for the last assertion we prove that p 6 = r−1 implies that pt 6 = rt −1for all but at most two indices t ∈ [n]. Since n > 5, this means that the inequality holds for at least three indices t ∈ [n]. So assume that p 6 = r − 1. We consider different cases. If r < p , then rt < p t and hence pt 6 = rt − 1 for all t ∈ [n] \ { p, r } by (ii). If p < r − 2, then pt < r t − 1 for all
t ∈ [n] \ { p, r } by (i). Consider finally the case p = r − 2. Then pt < r t − 1 for all
t ∈ [n] \ { p, p + 1 , r }, and pp+1 = rp+1 − 1. If pr = rr − 1, then π(p + 1) = 3, and it follows that ( pp, r p) = ( r − 1, p ) = ( p + 1 , p ), so pp > r p. If pp = rp − 1, then π(r + 1) = 3, and it follows that ( pr, r r) = ( p, r ) = ( p, p + 2), so pr < r r − 1. We conclude that in each case there are at most two indices t ∈ [n] such that pt = rt − 1. Note that we have also proved that p 6 = r − 1 if and only if pt 6 = rt − 1 for all but at most two indices t ∈ [n].
Remark 14 . By Lemma 13(ii), we may decide whether π−1(1) < π −1(2) or π−1(1) > π −1(2) by comparing the relative order of 1 and 2 in the cards of π; they are in the same order
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as in π with at most two exceptions. In fact, this condition can be checked by applying the majority rule to any five cards. Similarly, by Lemma 13(iii), we may decide whether 1 and 2 are adjacent in π, i.e., π−1(1) = π−1(2) − 1, by checking whether 1 and 2 are adjacent in all cards of π with at most two exception. Again, it is sufficient to check only five cards. Note that π−1(1) is known as soon as we know ( πr)−1(1). Since the pattern involvement relation is preserved and reflected under reversing permutations, we may consider πr in place of π and reverse all given cards when π−1(1) > π −1(2). Therefore, from now on we assume that π−1(1) < π −1(2). Given a permutation τ with τ −1(1) < τ −1(2), we denote by τ the length of the longest ascending segment of τ with initial value 1. Note thatτ = 1 if and only if 1 and 2 are not adjacent. Let us first consider the case where 1 and 2 are adjacent in π.
Lemma 15. Let n > 5, π ∈ Sn and I ⊆ [n] with |I| = H(n). For each i ∈ [n], let
pi := ( π ↓ i)−1(1) , ri := ( π ↓ i)−1(2) and i :=π↓i. Let L := 〈i | i ∈ I〉. Suppose thatπ > 1 and let π[p, t ] be the longest ascending segment with initial value 1 in π (i.e.,
p = π−1(1) and t = p + `π − 1) .
(i) For i ∈ [n], the following conditions are equivalent.
(a) π ↓ i = π ↓ 1.
(b) i ∈ π[p, t ].
(c) i =π − 1.
(ii) For all i ∈ [n] \ π[p, t ], we have i >π, and i ∈ { π(t + 1) , π (t) + 1 } whenever
i > π. Moreover, if i, j ∈ [n] are such that i 6 = j and i, j > π, theni = `j ,
π(t + 1) = π(t) + 2 , π(t + 2) = π(t) + 1 and π ↓ i = π ↓ j.
(iii) I has a nonempty intersection with both π[1 , p − 1] and π[p, n ] if and only if there exist i, j ∈ I such that pi 6 = pj . In this case p = max {pi, p j } and p−1 = min {pi, p j }.Moreover, for any i ∈ I with pi = p − 1, it holds that i =π.
(iv) I ⊆ π[1 , p − 1] if and only if there is an a ∈ [n − 1] with a > H(n) such that pi = a
for all i ∈ I. In this case, p = a + 1 and there is no i ∈ I such that π ↓ i = π ↓ 1.
(v) I ⊆ π[p, n ] if and only if there is an a ∈ [n − 1] with a 6 n − H(n) + 1 such that
pi = a for all i ∈ I. In this case p = a.Proof. Recall that |I| = H(n) and note that π > 1 implies that π−1(2) = p + 1. (i) The equivalence of (a) and (b) follows from Lemma 6, and the equivalence of (b) and (c) is clear. (ii) By Lemma 11, we havei > π whenever i ∈ [n] \ π[p, t ]. Since π(p) = 1, the strict inequalityi > ` π holds only if i is an upper critical point for π[p, t ], which implies by Definition 9 that i ∈ { π(t + 1) , π (t) + 1 }. If π[p, t ] has two upper critical points, then, by
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Remark 10, π(t + 1) = π(t) + 2, π(t + 2) = π(t) + 1, and π ↓ π(t + 1) = π ↓ π(t + 2), from which the claim follows immediately. (iii) Recall that pi = ( π ↓ i)−1(1) and ri = ( π ↓ i)−1(2). Since π−1(2) = p + 1, by applying Lemma 13(i), we see that pi ∈ { p−1, p } for any i ∈ I. Moreover, for any i, j ∈ I,it holds that p − 1 = pi < p j = p if and only if π−1(i) < p 6 π−1(j), which is equivalent to i ∈ π[1 , p − 1] and j ∈ π[p, n ]. Finally, for any i ∈ π[1 , p − 1], we have i =π by part (ii). (iv) and (v) By Lemma 13(i), the following holds:
• If I ⊆ π[1 , p − 1], then p − 1 > H(n) and pi = p − 1 for all i ∈ I. This shows the necessity of the condition in (v) with a = p − 1.
• If I ⊆ π[p, n ], then n − p + 1 > H(n) (and so n − H(n) + 1 > p) and pi = p for all
i ∈ I. This shows the necessity of the condition in (iv) with a = p.Conversely, assume that there is an a ∈ [n − 1] such that pi = a for all i ∈ I. By part (iii), we have I ⊆ π[1 , p − 1] (and so a > H(n)) or I ⊆ π[p, n ] (and so a 6 n − H(n) + 1). Note that the conditions a > H(n) and a 6 n − H(n) + 1 cannot hold simultaneously, because if H(n) 6 a 6 n − H(n) + 1, then n > 2H(n) − 1 = 2( dn/ 2e + 2) − 1 > n + 3, which is absurd. Therefore a > H(n) implies I ⊆ π[1 , p − 1] and a 6 n − H(n) + 1 implies
I ⊆ π[p, n ]. It follows immediately from (i) that when a > H(n) there is no i ∈ I such that
π ↓ i = π ↓ 1. Let us now consider the case when 1 and 2 are not adjacent in π.
Lemma 16. Let n > 5, π ∈ Sn, I ⊆ [n] with |I| = H(n). Let p := π−1(1) and r := π−1(2) ,and assume that p < r −1. For each k ∈ [n], let pk := ( π ↓ k)−1(1) and rk := ( π ↓ k)−1(2) .
(I) For all k ∈ [n], it holds that pk = r − 1 if and only if k = 1 . In other words, π ↓ 1
is the unique card of π with 1 at position r − 1.
(II) If α, β, γ, δ ∈ [n] are distinct indices such that
(pα, r α) = ( pβ , r β ) 6 = ( pγ , r γ ) = ( pδ, r δ),
then p = max {pα, p γ } and r − 1 = min {rα, r γ }.
(III) Assume that there exists a unique pair (a, b ) ∈ [n − 1] ×[n − 1] such that there exist distinct i, j ∈ I with (pi, r i) = ( pj , r j ) = ( a, b ). Let E := {(pk, r k) | k ∈ I}{ (a, b )}.
(1) Assume |E| > 3.
• If {a, a + 1 } ⊆ { x | ∃ y (x, y ) ∈ E} and {(a, a + 1) , (a − 1, a + 1) } ∩ E 6 = ∅,then p = a, r = b.
the electronic journal of combinatorics 28(3) (2021), #P3.41 15
• Otherwise
p = max {x 6 a + 1 | ∃ y (x, y ) ∈ E}
and
r =
b if (a − 1, b − 1) ∈ E or
a + 1 6 = b − 1 ∈ { x | ∃ y (x, y ) ∈ E},
b + 1 otherwise.
(2) Assume |E| = m 6 2. Then there exist x, y ∈ [n − 1] such that the following statements hold.
(i) ( a, b ) = ( p, r ) if and only if
(A) H(n) − m 6 n − b + 1 and E ⊆ { (a, b − 1) , (a − 1, b − 1) , (a, x ),
(b − 1, y )}.
(ii) ( a, b ) = ( p, r − 1) if and only if
(B) H(n) − m 6 b − a + 1 and E ⊆ { (a, b + 1) , (a − 1, b ), (a, x ), (b, y )}.
(iii) ( a, b ) = ( p − 1, r − 1) if and only if
(C) H(n) − m 6 a and E ⊆ { (a + 1 , b + 1) , (a + 1 , b ), (a + 1 , x ), (b, y )}.Proof. (I) This follows immediately from Lemma 13(i). (II) Since, by Lemma 13(i), there may exist only one index i ∈ [n] such that ( pi, r i) = (r − 1, x ) for some x ∈ [n], and only one index j ∈ [n] such that ( pj , r j ) = ( p, y ) for some
y ∈ [n] \ { r − 1, r }, we necessarily have
{(pα, r α), (pγ , r γ )} ⊆ { (p, r ), (p, r − 1) , (p − 1, r − 1) }.
It follows that p = max {pα, p γ } and r − 1 = min {rα, r γ }.(III) We necessarily have ( a, b ) ∈ { (p, r ), (p, r − 1) , (p − 1, r − 1) }. Table 3 shows the possible types of cards in terms of a and b in each case. Note that, by Lemma 13(i), the values of x and y are fixed and only depend on the positions of entries with values 1, 2, and 3 in π.(p, r ) (p, r − 1) (p − 1, r − 1) (p, x ) (r − 1, y )Case 1 (a, b ) (a, b − 1) (a − 1, b − 1) (a, x ) (b − 1, y )Case 2 (a, b + 1) (a, b ) (a − 1, b ) (a, x ) (b, y )Case 3 (a + 1 , b + 1) (a + 1 , b ) (a, b ) (a + 1 , x ) (b, y )Table 3: The possible types of cards in the different cases of the proof of Lemma 16(III). By our hypothesis we have p < r − 1 and then b − a > 2 clearly holds in Cases 1 and 3. In Case 2 we have ( a, b ) = ( p, r − 1) and since there are at least two cards of this type, we have r − p − 1 > 2, so b − a = r − p − 1 > 2 holds also in this case. Thus b − a > 2always holds.
the electronic journal of combinatorics 28(3) (2021), #P3.41 16
(1) Assume |E| > 3. By reading off from Table 3, we can first note that the conditions (a, b ) = ( p, r ) and b − 1 = a + 1 ∈ { x | ∃ y (x, y ) ∈ E} hold simultaneously if and only if the conditions {a, a + 1 } ⊆ { x | ∃ y (x, y ) ∈ E} and {(a, a + 1) , (a − 1, a + 1) } ∩ E 6 = ∅
also hold. Now suppose {a, a + 1 } {x | ∃ y (x, y ) ∈ E} or {(a, a + 1) , (a − 1, a + 1) } ∩ E = ∅.In Case 1 either {x | ∃ y (x, y ) ∈ E} = {a − 1, a } or b − 1 > a + 1 and in both cases
p = a = max {x 6 a + 1 | ∃ y (x, y ) ∈ E} and r = b. In Cases 2 and 3, we can immediately conclude from Table 3 that p = max {x 6 a + 1 | ∃ y (x, y ) ∈ E} and since neither (a − 1, b − 1) ∈ E nor b − 1 ∈ { x | ∃ y (x, y ) ∈ E} with b − 1 6 = a + 1, we also have r = b + 1. (2) Assume |E| = m 6 2. We prove first the necessity of conditions (A), (B), and (C). If ( a, b ) = ( p, r ), then the cards of type ( a, b ) are among the cards π − i with r < i
and, in the case with π(r + 1) = 3, also π ↓ 2 = π − r. Therefore, the inequality
H(n) − m 6 n − r + 1 = n − b + 1 holds. Furthermore, by reading off from Table 3, we see that E ⊆ { (a, b − 1) , (a − 1, b − 1) , (a, x ), (b − 1, y )} for some x, y ∈ [n − 1]. If ( a, b ) = ( p, r − 1), then the cards of type ( a, b ) are among the cards π − i with
p < i < r and, in the case when π(r − 1) = 3, also π ↓ 2 = π − r. Therefore, the inequality
H(n) − m 6 r − p = b − a + 1 holds. Furthermore, by reading off from Table 3, we see that E ⊆ { (a, b + 1) , (a − 1, b ), (a, x ), (b, y )} for some x, y ∈ [n − 1]. If ( a, b ) = ( p − 1, r − 1), then the cards of type ( a, b ) are among the cards π − i with
i < p ; therefore the inequality H(n) − m 6 p − 1 = a holds. Moreover, by reading off from Table 3, we see that E ⊆ { (a + 1 , b + 1) , (a + 1 , b ), (a + 1 , x ), (b, y )} for some x, y ∈ [n − 1]. In order to prove the sufficiency of conditions (A), (B), and (C), it suffices to show that they are mutually exclusive. In the case when m = 2, this is clear, because for any x, y ∈
[n − 1], the intersection of any two of the sets {(a, b − 1) , (a − 1, b − 1) , (a, x ), (b − 1, y )},
{(a, b + 1) , (a − 1, b ), (a, x ), (b, y )} and {(a + 1 , b + 1) , (a + 1 , b ), (a + 1 , x ), (b, y )} has at most one element. Now assume that m 6 1, and suppose, to the contrary, that some two of the inequalities H(n) − m 6 n − b + 1, H(n) − m 6 b − a + 1, and H(n) − m 6 a hold simultaneously. Since 1 6 a < b 6 n − 1 and b − a > 2, it follows that
n + 2 6 2( H(n) − m) 6 max {n − a + 2 , n − (b − a) + 1 , b + 1 } 6 n + 1 ,
a contradiction. If none of the H(n) cards equals π ↓ 1, then the next result assures that the ( n − 2)-patterns obtained by deleting the entry 1 from each card constitutes a partial deck of
π ↓ 1, and we can make use of this partial deck to reconstruct π ↓ 1. Consequently, we can reconstruct π.
Lemma 17. Let π ∈ Sn. If 1 < i 1 < i 2 < · · · < i 6 n, then 〈(π ↓ ij ) ↓ 1 | j ∈ []〉 is a partial deck of π ↓ 1.Proof. The multiset 〈(π ↓ 1) ↓ (ij − 1) | j ∈ [`]〉 is clearly a partial deck of π ↓ 1, and (π ↓ ij ) ↓ 1 = ( π ↓ 1) ↓ (ij − 1) holds by Lemma 5.
Theorem 18. For every permutation π of rank n, with n > 5, the value π−1(1) is uniquely determined by any partial (n − 1) -deck D of π with H(n) cards not all equal. Moreover,
the electronic journal of combinatorics 28(3) (2021), #P3.41 17
from the partial deck D alone, it is also possible to decide whether any of the cards in D
is equal to π ↓ 1, and, if positive, to identify such a card. Proof. Let D := 〈τ1, . . . , τ H(n)〉 be a partial n−1-deck D of π with H(n) cards and assume that D has at least two distinct cards. Let p := π−1(1) and r := π−1(2). (Recall that we are given only the partial deck D, so p and r are unknown quantities.) By Lemma 13(ii) and (iii), we may determine whether p < r or p > r and whether p = r − 1 by comparing the positions of 1 and 2 in the cards in D; in fact, it is enough to check only five cards (see Remark 14). Henceforth we may assume that p < r . Otherwise we may consider
Dr := 〈τ r1 , . . . , τ r
H(n)
〉 instead of D and reconstruct πr, which can then be reversed to obtain π. We now consider different cases and subcases. Case 1: Assume p = r − 1. By Lemma 15(i), π ↓ i = π ↓ 1 if and only if π↓i =π − 1, so our aim is to determine π−1(1) and π, if possible. Then it will be straightforward to check whether or not D contains a card τk withτk = `π − 1; such a card equals π ↓ 1. Let P := {τ −1
k
(1) | k ∈ [H(n)] }. Then ∅ 6 = P ⊆ { p − 1, p }. We consider two subcases according to the cardinality of P .Case 1.1: Assume |P | = 2. Then P = {p − 1, p }, and we have τk =π for any k such that τ −1
k
(1) = p − 1, by Lemma 15(iii). Then it is easy to detect whether or not there is a card τj such that τj =π − 1, i.e., τj = π ↓ 1. Case 1.2: Assume |P | = 1, say P = {u}. Observe that, by Lemma 15, statements (iv) and (v), either u > H(n) or n − (u − 1) > H(n), and these cases are mutually exclusive. Case 1.2.1: Assume u > H(n). Then p = u+1 and π ↓ 1 is not in D by Lemma 15(iv). Case 1.2.2: Assume n − (u − 1) > H(n). Then p = u by Lemma 15(iv) and every card in D must be of the form π − t with t > p. Let π[p, v ] be the maximal monotone segment of length π in π, let L := 〈τj : j ∈ [H(n)] 〉, and let k ∈ [H(n)] be such that
τk = min Supp( L). Observe that there is a number a > 1 such that ∅ 6 = Supp( L) ⊆ {π − 1, π, π + a}
with χL(π + a) 6 2 by Lemma 15(i) and (ii). If χL(π) = 0 holds, then χL(`π − 1) =
H(n) − χL(π + a) > H(n) − 2; since χL(π − 1) 6 π, this impliesτk = π − 1 > H(n) − 3. If χL(π − 1) = 0 holds, then H(n) = χL(π) + χL(π + a) 6 n − v = n − (p − 1 + π)and soτk = `π 6 n − H(n) − p + 1. Finally notice that we cannot have simultaneously
τk > H(n) − 3 andτk 6 n − H(n) − p + 1 since n − H(n) − p + 1 > H(n) − 3 implies
n − p + 1 > 2H(n) − 3 > n + 1. Hence one of the following cases occurs. Case 1.2.2.1: Assume |Supp( L)| = 3. Then τk =π − 1 and we have τk = π ↓ 1 ∈ D.Case 1.2.2.2: Assume |Supp( L)| 6 2 and τk > H(n) − 3. Thenτk = `π − 1. Thus
τk = π ↓ 1 ∈ D.Case 1.2.2.3: Assume |Supp( L)| 6 2 and τk 6 n − H(n) − p + 1. Thenτk = `π. Thus
π ↓ 1 /∈ D.Case 1.2.2.4: Assume |Supp( L)| 6 2 and neither of the inequalities `τk > H(n) − 3 and
τk 6 n−H(n)−p+1 holds. By the above observations, we must then have χL(π −1) > 0and χL(π) > 0; hence L = {π − 1, π} andτk = `π − 1. Thus τk = π ↓ 1 ∈ D.Case 2: Assume p < r − 1. By Lemma 16(I), π − t = π ↓ 1 if and only if ( π − t)−1(1) =
the electronic journal of combinatorics 28(3) (2021), #P3.41 18
r − 1, so our aim is to determine π−1(1) = p and π−1(2) = r, if possible. Then it will be straightforward to check whether or not D contains a card τk with τ −1
k
(1) = r − 1; such a card equals π ↓ 1. Let T := 〈(τ −1
k
(1) , τ −1
k
(2)) | k ∈ H(n)〉 be the multiset of the types of the cards in D.By Lemma 13, the only possible elements of T are ( p, r ), ( p, r − 1), ( p − 1, r − 1), ( r − 1, x ), and ( p, y ), for some x, y ∈ [n], and the only elements that may have multiplicity greater than 1 are ( p, r ), ( p, r − 1), and ( p − 1, r − 1). Case 2.1: Assume T contains all five possible types. Since p − 1 < p < r − 1, we can immediately determine both p and r − 1, and clearly D contains a card τk with
τ −1
k
(1) = r − 1 and so π ↓ 1 = τk.Case 2.2: Assume T contains at most four different types. Since H(n) > 5, this means that some type has multiplicity at least 2 in T .Case 2.2.1: Assume ( a, b ) and ( c, d ) are distinct types with multiplicity at least 2 in
T . Then, by Lemma 16(II), we have p = max {a, c } and r − 1 = min {b, d }.Case 2.2.2: Assume there is only one type ( a, b ) of multiplicity at least 2 in T . Now the values of p and r − 1 are obtained by applying Lemma 16(III).
Theorem 19. For n > 5, every permutation of rank n is reconstructible from H(n) cards. Proof. If all cards in D are equal, say τk = τ for all k ∈ [H(n)], then by Lemma 7(iii)
τ contains a unique maximal monotone segment π[u, v ] of length at least dn/ 2e, and by Lemma 7(ii) we can immediately conclude that π = τ ↑u τ (u) if π[u, v ] is ascending and
π = τ ↑v+1 τ (v) if π[u, v ] is descending. From now on we may assume that D has at least two distinct cards. We prove the claim by induction on n. For n = 5, we have H(n) = 5, and the claim holds by the results of Smith [6, Theorem 2.3] and Raykova [5, Lemma 3.3]. Assume now that every permutation of rank m (m > 5) is reconstructible from H(m) cards. Let π ∈ Sm+1 , and let D = 〈τ1, . . . , τ H(m+1) 〉 be a partial m-deck of π with H(m + 1) cards. By Theorem 18 we can find π−1(1) and we can determine if π ↓ 1 /∈ D.Case 1: Assume that π ↓ 1 /∈ D. In this case, D′ := 〈τk ↓ 1 | k ∈ [H(m + 1)] 〉 is a partial deck of π ↓ 1 by Lemma 17. Since |D′| = H(m + 1) > H(m), our inductive hypothesis asserts that we can reconstruct π ↓ 1 from D′, and then we recover π as
π = ( π ↓ 1) ↑π−1(1) 1 by Lemma 4. Case 2: Assume that π ↓ 1 ∈ D. In this case we may identify π ↓ 1 among the cards in D, by Theorem 18. Then we recover π as π = ( π ↓ 1) ↑π−1(1) 1 by Lemma 4. The proofs of Theorems 18 and 19, along with Lemmas 15 and 16, give readily rise to a reconstruction algorithm for an unknown permutation π if any of its partial decks of size H(n) is given. We may also make use of the shortcuts provided by Propositions 8 and 12. The following examples illustrate the reconstruction method.
Example 20.
Let π ∈ S10 and take D to be the following partial deck of π:
〈(9 7 8 6 1 2 3 5 4) 3, (9 8 7 1 2 3 6 5 4) 2, (9 7 8 1 2 3 6 5 4) 1, (8 9 7 1 2 3 6 5 4) 1〉.
the electronic journal of combinatorics 28(3) (2021), #P3.41 19
Hence D = deck I (π) for some I ⊆ with |I| = 7 = H(10). The sequence σ = 1 2 3 is a maximal ascending segment in every card in D. We apply Proposition 12. From part (ii) we obtain the following quantities: G = {(1 , 5) , (1 , 4) }, A = {1},
B = {4, 5}, a∗ = 1, b∗ = 5, t = a∗ = 1, u = b∗ = 5, v = 5 + 3 − 1 = 7. Consequently,
π[5 , 7] = 1 2 3 is a maximal ascending segment in π.Part (iii) yields the partial deck D′ = 〈(6 4 5 3 2 1) 4, (6 5 4 3 2 1) 2, (5 6 4 3 2 1) 1〉 of
π ↓ π[5 , 7]. We apply Lemmas 6 and 7(ii) to the card τ = 6 4 5 3 2 1 of multiplicity 4, and we obtain π ↓ π[5 , 7] = τ ↑7 τ (6) = 6 4 5 3 2 1 ↑7 1 = 7 5 6 4 3 2 1. Hence
π = ( π ↓ π[5 , 7]) ↑5 π[5 , 7] = 7 5 6 4 3 2 1 ↑5 1 2 3 = 10 8 9 7 1 2 3 6 5 4. 2. Let π ∈ S11 and let D be the partial deck of π comprising
τ1 = 10 2 4 6 8 3 1 7 5 9 , τ2 = 10 1 3 6 8 4 2 7 5 9 ,τ3 = 10 1 3 5 6 9 4 2 7 9 , τ4 = 10 1 2 4 6 8 3 7 5 9 ,τ5 = 10 1 3 5 7 8 4 2 6 9 , τ6 = 10 1 3 4 6 8 2 7 5 9 ,τ7 = 1 3 5 7 9 4 2 8 6 10 , τ8 = 10 1 3 5 7 9 4 2 8 6 .
First we compare the positions of 1 and 2 in the cards. With the help of Lemma 13(ii) and (iii) we conclude that p := π−1(1) < π −1(2) =: r and p < r − 1. The multi-set of pairs ( τ −1(1) , τ −1(2)) with τ ∈ D is 〈(1 , 7) 1, (2 , 3) 1, (2 , 7) 2, (2 , 8) 3, (7 , 2) 1〉.This multiset contains two distinct elements of multiplicity greater than 1, namely (2 , 3) and (2 , 7), so we can apply Lemma 16(II) and we obtain ( π−1(1) , π −1(2)) = (max {2, 2}, min {3, 7}+1) = (2 , 8). Since τ −11 (1) = 7 = π−1(2) −1 we have τ1 = π ↓ 1by Lemma 16(I), and then π = τ1 ↑2 1 = 11 1 3 5 7 9 4 2 8 6 10. 3. Let π ∈ S8, and let D = 〈(5 7 4 1 3 2 6) 3, (6 7 5 1 4 3 2) 1, (5 7 4 3 2 1 6) 1, (5 7 1 4 3 2 6) 1〉.Applying Lemma 13, we conclude that p := π−1(1) < π −1(2) =: r and p < r − 1. The multiset of pairs ( τ −1(1) , τ −1(2)) with τ ∈ D is 〈(4 , 6) 3, (4 , 7) 1, (6 , 5) 1, (3 , 6) 1〉.This multiset has a unique element of multiplicity greater than 1, namely (4 , 6), and three elements of multiplicity 1, so we apply Lemma 16(III)(1) and we obtain
π−1(1) = p = 4 and π−1(2) = r = 7. Since τ = 5 7 4 3 2 1 6 ∈ D satisfies τ −1(1) = 6 = r − 1, we have π = τ ↑4 1 by Lemma 16(I). Hence π = 6 8 5 1 4 3 2 7. 4. Let D be the following partial deck of π ∈ S9:
〈(1 7 2 8 3 4 5 6) 1, (7 1 2 8 3 4 5 6) 1, (8 1 7 2 3 4 5 6) 1, (7 1 6 2 8 3 4 5) 4〉.
By applying Lemma 13 we conclude that p := π−1(1) < π −1(2) =: r and p < r − 1. The multiset of pairs ( τ −1(1) , τ −1(2)) with τ ∈ D is 〈(1 , 3) 1, (2 , 3) 1, (2 , 4) 5〉. This multiset has a unique element of multiplicity greater than 1, namely ( a, b ) := (2 , 4), and m := 2 elements of multiplicity 1, so with the help of Lemma 16(III)(2) we conclude that ( p, r ) = (2 , 4) since n − b = 9 − 4 = 5 > 7 − 2 = H(n) − m and there is no ( x, y ) ∈ D with x = b = 4. No card in D has 1 at position r − 1 = 3 and so
π ↓ 1 /∈ D by Lemma 16(I).
the electronic journal of combinatorics 28(3) (2021), #P3.41 20
Now take
D′ := 〈(π ↓ i) ↓ 1 | i ∈ I〉 = 〈(6 1 7 2 3 4 5) 2, (7 6 1 2 3 4 5) 1, (6 5 1 7 2 3 4) 4〉;
D′ is a partial deck of π ↓ 1 ∈ S8 of cardinality H(8) + 1. Theorem 19 guarantees that θ = π ↓ 1 is reconstructible from the multiset
Dθ = 〈(6 1 7 2 3 4 5) 1, (7 6 1 2 3 4 5) 1, (6 5 1 7 2 3 4) 4〉 ⊆ D′.
We apply Proposition 8 to the card τ = 6 5 1 7 2 3 4 that has multiplicity 4 and contains a unique maximal monotone segment of length 3, namely τ [5 , 7] = 2 3 4 =:
σ. Since σ is ascending, we get θ = τ ↑5 τ (5) = τ ↑5 2 = 7 6 1 8 2 3 4 5. Finally we obtain π = θ ↑p 1 = θ ↑2 1 = 8 1 7 2 9 3 4 5 6.
6 Concluding remarks and open problems
We have taken the first steps of answering Ginsburg’s Problem 2 by showing that H1(n) =
dn/ 2e + 2 for all n > 5. It remains an open problem to determine values of Hk(n) when the parameter k is greater than 1. Another curious problem, which may be a necessary step for approaching Problem 2, is to generalize Ginsburg’s Lemma 6 and determine necessary and sufficient conditions for the equality of two ( n − k)-cards of an n-permutation when k > 2. In this case, the situation looks much more complicated. It is possible that two cards coincide even if entries are not removed from the same monotone segments, as illustrated by the following example: there are no nontrivial monotone segments in the permutation π = 52413, yet we have π ↓ { 1, 3} = π ↓ { 1, 4} = π ↓ { 2, 4} = π ↓ { 2, 5} = 312.
References
M. B´ ona. Combinatorics of Permutations. Discrete Math. Appl. (Boca Raton). Chap-man & Hall/CRC, Boca Raton, 2004. J. Ginsburg. Determining a permutation from its set of reductions. Ars Combin. ,82:55–67, 2007. P. J. Kelly. On Isometric Transformations. Ph.D. thesis, University of Wisconsin, 1942. S. Kitaev. Patterns in Permutations and Words. Monogr. Theoret. Comput. Sci. EATCS Ser. Springer, Heidelberg, 2011. M. Raykova. Permutation reconstruction from minors. Electron. J. Combin. , 13:#R66, 2006. R. Smith. Permutation reconstruction. Electron. J. Combin. , 13:#N11, 2006. S. M. Ulam. A Collection of Mathematical Problems. Interscience Tracts in Pure and Applied Mathematics, no. 8. Interscience Publishers, New York, 1960. |
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A cervical cerclage is a treatment that involves temporarily sewing the cervix closed with stitches. This may help the cervix hold a pregnancy in the uterus. A cerclage is done in the second trimester of pregnancy to prevent preterm birth.
Sometimes the cervix isn’t strong enough to stay closed as the pregnancy grows. This weakness is called cervical insufficiency (formerly called an incompetent cervix). Cervical weakness may cause a quick delivery and preterm birth. This can happen with only mild contractions or without any pain or other signs of preterm labor. The goal of a cerclage is to prevent cervical weakness from causing preterm birth.
Your obstetrician–gynecologist (ob-gyn) may diagnose cervical weakness and recommend placing a cerclage if you have had
pregnancy loss due to painless cervical dilation in the second trimester
multiple second trimester pregnancy losses or preterm births
a previous cerclage due to painless cervical dilation in the second trimester
Your ob-gyn also may recommend placing a cerclage based on
painless cervical dilation in the second trimester
an ultrasound exam that shows a short cervix, if you also have other risk factors for preterm birth
There are two types of cerclage:
Transvaginal cerclage. Ob-gyns most often do a cerclage by reaching through the vagina to place stitches in the cervix.
Transabdominal cerclage. This is when a cut is made in the abdomen (belly) to reach the cervix and place stitches. This can be done with open surgery (laparotomy) or with a very small cut and a camera (laparoscopy). Sometimes transabdominal cerclage is done if you
have had surgeries on your cervix in the past
had a previous transvaginal cerclage that didn’t prevent the loss of a pregnancy
Cerclage is usually done in a hospital. It typically does not require an overnight stay.
A transvaginal cerclage is typically removed around 37 weeks of pregnancy. This might be done at an office visit or in the hospital.
A transabdominal cerclage is typically left in place until delivery through cesarean birth. A transabdominal cerclage can also be left in place between pregnancies, to prevent future preterm birth or pregnancy loss.
After a cerclage, you may have
some bleeding or spotting for up to 3 days
more clear discharge from the vagina than before
a few days of light cramping
If you have spotting or discharge, wear a sanitary pad and change as often as needed. Do not put anything in your vagina. Avoid heavy activity for a few days and check with your ob-gyn before having sex.
Overall, there is a low risk of complications from a cerclage. Possible complications include
preterm prelabor rupture of membranes (PPROM)
infection of fetal membranes and the uterus
tears in the cervix
the stitch moving from the correct place
bleeding during or after the procedure
Cerclage: [sur-KLAHZH]: A procedure that closes the cervix with stitches to prevent or delay preterm birth.
Cervical Dilation:Widening the opening of the cervix.
Cervical Insufficiency: The inability of the cervix to hold a pregnancy in the uterus in the second trimester.
Cervix: The lower, narrow end of the uterus at the top of the vagina.
Cesarean Birth: Birth of a fetus from the uterus through an incision (cut) made in the abdomen.
Complications: Diseases or conditions that happen as a result of another disease or condition. An example is pneumonia that develops with the flu. An example of a pregnancy complication is preterm labor.
Laparoscopy [lap-uh-RAH-skuh-pee]: A surgical procedure using a thin, lighted telescope called a laparoscope. The laparoscope is inserted through a small incision (cut) in the abdomen and used to view the pelvic organs. Other long, thin instruments can be used with it to perform surgery.
Laparotomy [lap-uh-RAH-tuh-mee]: A surgical procedure that uses a large incision (cut) in the abdomen.
Obstetrician–Gynecologist (Ob-Gyn): A doctor with medical and surgical training and education in the female reproductive system.
Preterm: Less than 37 weeks of pregnancy.
Preterm Prelabor Rupture of Membranes (PPROM): Rupture of the amniotic membranes that happens before labor begins and before 37 weeks of pregnancy.
Trimester: A time period of 3 months. There are three trimesters in pregnancy: the first trimester, second trimester, and third trimester.
Ultrasound Exam: A test that uses sound waves to examine inner parts of the body. During pregnancy, ultrasound can be used to check the fetus. Also called ultrasonography or sonography.
Uterus: A muscular organ in the female pelvis. During pregnancy, this organ holds and nourishes the fetus. Also called the womb.
Vagina: A tube-like structure surrounded by muscles. The vagina leads from the uterus to the outside of the body.
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FAQ526
Published: April 2024
Last reviewed: November 2023
Topics:
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188494 | https://www.geeksforgeeks.org/maths/angle-of-elevation/ | Angle of Elevation
Angle of Elevation is defined as the upward angle measured from a horizontal line or plane to a point situated above it. It is an important geometrical concept used in trigonometry to find the height and distance of the object. This elevation angle is commonly used to describe the viewing perspective of an object and plays a significant role in activities such as navigation and surveying. The angle of elevation refers to the upward angle formed between an observer's line of sight and a horizontal plane.
The concept of angle of elevation holds fundamental significance in trigonometry and has diverse practical applications across different fields. In this article, we will discuss the 'Angle of Elevation' in detail. We will also look at solved examples and practice problems on the concept of angle of elevation.
Table of Content
What is Angle of Elevation?
Angle of Elevation refers to the angle formed between a horizontal line and the line of sight when an observer looks upward. The angle of elevation is an important geometrical concept indicating the vertical inclination of an observer's line of sight to an object or point positioned above a horizontal reference plane. It is commonly used in trigonometry and surveying to measure the height or distance of an object.
In the world of trigonometry, the angle of elevation is typically measured in degrees and is closely associated with the tangent function. When an individual looks upward at an object, the angle formed between the horizontal line of sight and the observer's line of sight is referred to as the angle of elevation. This measurement is valuable for determining distances, heights and the overall geometric characteristics of the observed object.
Read More Trigonometry and Height and Distance.
Angle of Elevation Definition
Angle of Elevation defines the inclination between a horizontal reference line and the observer's line of sight when gazing upwards
Terms Related to Angle of Elevation
Below listed are the major terms related to Angle of Elevation:
Angle of Elevation Example
In practical applications, surveyors use the angle of elevation to calculate the height of structures like buildings or towers, utilizing trigonometric principles for accurate assessments. Similarly, astronomers and geographers apply this concept to gain insights into the positions and dimensions of celestial objects and geographical features.
Angle of Elevation Formula
The formula for the angle of elevation (θ) is given by:
sin θ = Perpendicular/Hypotenuse
cos θ = Base/Hypotenuse
tan θ = Perpendicular / Base
How to Find Angle of Elevation?
To find the angle of elevation, measure the height and horizontal distance then use the formula mentioned earlier.
Step 1: Identify the Triangle
Determine the right-angled triangle formed by the horizontal line of sight and the line of sight to the object.
Step 2: Identify Sides
Label the sides of the triangle: the side opposite the angle of elevation, the adjacent side, and the hypotenuse.
Step 3: Apply Trigonometry
Use either the trigonometric ratios such as sine, cosine, tangent function depending on the known sides to find the angle of elevation.
Calculate Angle of Elevation
Depending on the known sides, use either the sine, cosine or tangent function to find the angle of elevation. When examining angles of elevation, trigonometric functions come into play. These functions help us understand the relationship between the angle of elevation and the sides of a right-angled triangle.
To determine the angle of elevation measure the vertical height and the horizontal distance, then use the formula:
Using Tangent (tan θ) = Perpendicular/Base
Using Sine (sin θ) = Perpendicular/Hypotenuse
Using Cosine (cos θ) = Base/Hypotenuse
Example: A boy standing 20 m away from a pole sees the top of the pole of height 20√3 m. Find the angle of elevation.
Solution:
Let the Pole be AB of height 20√3 m and the boy is standing at point C 20 away from the pole. If we join all the three points we will get a right triangle ABC.
In triangle ABC let the angle of elevation i.e. ∠ACB be θ
tan θ = Perpendicular/Base
⇒ tan θ = AB/AC = 20√3/20 = √3
Now since, tan θ = √3
Hence, θ = tan-1(√3) = 60°
Hence, angle of elevation is 60°.
Angle of Elevation and Depression
Angle of Elevation and Angle of Depression both these geometric concept are used in various fields including architecture, navigation and physics.
While the angle of elevation is formed when looking upward, the angle of depression is formed when looking downward from a horizontal line. When observing an object above eye level such as a mountain or a building, the angle of elevation helps determine the angle at which the observer must look to see the top of the object. Similarly, when observing an object below eye level such as a submarine or a water-plants, the angle of depression helps determine the angle at which the observer must look to see the bottom of the object.
Below is the tabular difference between Angle of Elevation and Angle of Depression
| Angle of Elevation | Angle of Depression |
--- |
| The angle formed when looking upward from a horizontal line or plane. | The angle formed when looking downward from a horizontal line or plane. |
| The angle at which the observer must look to see the top of the object. | The angle of depression helps determine the angle at which the observer must look to see the bottom of the object. |
| Example: Observing an object above eye level such as a mountain or a building | Example: Observing an object below eye level such as a submarine or a water-plants |
Angle of Elevation
Angle of Depression
The angle formed when looking upward from a horizontal line or plane.
The angle formed when looking downward from a horizontal line or plane.
The angle at which the observer must look to see the top of the object.
The angle of depression helps determine the angle at which the observer must look to see the bottom of the object.
Example: Observing an object above eye level such as a mountain or a building
Example: Observing an object below eye level such as a submarine or a water-plants
Also, Check
Solved Examples on Angle of elevation
Example 1. A person standing 100 meters away from a tower observes the top of the tower at an angle of elevation of 45 degrees. Determine the height of the tower.
Solution:
Distance of tower from person = 100 meters
⇒ tan(45°) = height of the tower/ distance of tower from person
⇒ height of the tower = 100 × tan(45°)
⇒ Height of the tower = 100 meter
Example 2. An archer aims an arrow at a 30-degree angle of elevation. If the arrow travels a horizontal distance of 200 meters, find the arrow's vertical displacement.
Solution:
Horizontal Distance from archer = 200 meters
⇒ tan(30°) = vertical distance of archer/ horizontal distance of archer
⇒ Vertical Displacement= 200× tan (30°)
⇒ Vertical Displacement= 200/√3
Example 3. From a point on the ground, the angle of elevation to the top of a cliff is 60 degrees. If the cliff is 80 meters distance, calculate the height of the cliff.
Solution:
Distance of cliff from person = 80 meters
⇒ tan(60°) = height of the cliff/ distance of cliff from person
⇒ height of the cliff = 80 × tan(60°)
⇒ Height of the cliff = 80√3 meter
Example 4. A person on a boat sees the top of a lighthouse at a 30 degree angle of elevation. If the lighthouse is 30 meters tall, calculate the distance between the boat and the lighthouse.
Solution:
Height of Lighthouse = 30 meters
⇒ tan(30°) = Height of Lighthouse/ distance between the boat and the lighthouse
⇒ distance between the boat and the lighthouse = Height of Lighthouse/ tan(30°)
⇒ distance between the boat and the lighthouse = 30√3 meter
Example 5. From the top of a building, the angle of elevation to the top of a taller building is 15 degrees. If the distance between the two buildings is 50 meters, find by what is extra height of the taller building.
Solution:
Distance between two buildings = 50 meters
⇒ tan(45°) = Extra height of the taller building/ Distance between two buildings
⇒ Extra height of the taller building = tan(45°) × 50
⇒ Extra height of the taller building = 50 meters.
Practice Problems on Angle of Elevation
Q1: A person at the beach looks up at a kite with an angle of elevation of 60 degrees. If the kite is 100 meters above the ground, calculate the horizontal distance from the person to the kite.
Q2: An observer on a mountain peak measures the angle of elevation to a lower peak as 45 degrees. If the horizontal distance between the peaks is 500 meters, find the difference in their heights.
Q3: A sniper aims at a target with an angle of elevation of 30 degrees. If the bullet travels a horizontal distance of 500 meters, determine the vertical displacement of the bullet.
Q4: An airplane is flying at an altitude of 10,000 meters. If the angle of elevation from the ground is 20 degrees, determine the horizontal distance from the airplane to the observer.
Q5: A ladder leans against a wall forming a 60-degree angle of elevation. If the base of the ladder is 15 meters away from the wall, find the length of the ladder.
Q6: What is the relationship between the angle of elevation of a cloud, the height of the cloud, and the distance from the observer?
R
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188496 | http://vassarstats.net/tab2x2.html | For a 2x2 Contingency Table:
·Phi Coefficient of Association ·Chi-Square Test of Association ·Fisher Exact Probability Test For a table of frequency data cross-classified according to two categorical variables, X and Y, each of which has two levels or subcategories, this page will
calculate the Phi coefficient of association;T
perform a chi-square test of association, if the sample size is not too small; andT
perform the Fisher exact probability test, if the sample size is not too large. [Although the Fisher test is traditionally used with relatively small samples, the programming for this page will handle fairly large samples, up to about n=1000, depending on how the frequencies are arrayed within the four cells.]T For intermediate values of n, the chi-square and Fisher tests will both be performed.
To proceed, enter the values of X0Y1, X1Y1, etc., into the designated cells. When all four cell values have been entered, click the «Calculate» button. To perform a new analysis with a new set of data, click the «Reset» button.
The logic and computational details of the Chi-Square and Fisher tests are described in Chapter 8 and Subchapter 8a, respectively, of Concepts and Applications. A briefer account of the Fisher test will be found toward the bottom of this page.
Data EntryT
| | | | | | | | | | | | | |
--- --- --- --- --- ---
| | X | | | Expected Cell Frequencies per Null Hypothesis| | | | | --- --- | | | 0 | 1 | Totals | | Y | 1 | | 0 | | Totals | |
Calculate Reset
| | | | | | | |
--- --- ---
| | Chi-Square | | Chi-square is calculated only if all expected cell frequencies are equal to or greater than 5. The Yates value is corrected for continuity; the Pearson value is not. Both probability estimates are non-directional.| | | | --- | Phi | Yates | Pearson | |
Fisher Exact Probability Test:T
| | |
--- |
| P | one-tailed |
| two-tailed |
| | |
--- |
| Home | Click this link only if you did not arrive here via the VassarStats main page. |
©Richard Lowry 2001- All rights reserved. Fisher Exact Probability Test: Logic and Procedure Consider a 2x2 contingency table of the sort described above, with the cell frequencies represented by a, b, c, d, and the marginal totals represented by a+b, c+d, a+c, b+d, and n.
| | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- |
| | 0 | 1 | Totals| 1 | a | b | a+b| | | | | | | | | --- --- --- --- | | 0 | c | d | c+d| | | | | --- --- | | Totals | a+c | b+d | n | | | |
If there were no systematic association between the variables A and B within the population from which the cell frequencies are randomly drawn, the probability of any particular possible array of cell frequencies, a, b, c, d, given fixed values for the marginal totals a+b, c+d, etc., would be given by the hypergeometric rule
which for computational purposes reduces to Also, the degree of disproportion within any array of cell frequencies—in effect, the degree of ostensible association between variables A and B within the sample—can be measured by the absolute difference For any particular observed array of cell frequencies, the programming embedded in this page calculates the probability of that particular array plus the probabilities of all other possible arrays whose degree of disproportion is equal to or greater than that of the observed array. Thus, for the observed array
| | | | | | | | | |
--- --- --- ---
| 2 | 7 | 9| | | | | | | --- --- --- | | 8 | 2 | 10| | | | --- | 10 | 9 | 19 | | |
the one-tailed probability would be the sum of the separate probabilities for the arrays
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | probability| | 2 | 7 | | | | 8 | 2 | | 0.01754| | | | 1 | 8 | | | | 9 | 1 | | 0.00097| | | | 0 | 9 | | | | 10 | 0 | | 0.00001| | | | | sum = 0.01852 | (one-tailed probability) | | | | |
And the two-tailed probability would be that sum plus the sum of the separate probabilities for the arrays of equal or greater disproportion at the other extreme:
| | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- |
| | | | probability| | 8 | 1 | | | | 2 | 8 | | 0.00438| | | | 9 | 0 | | | | 1 | 9 | | 0.00011| | | | | sum = 0.00449 | | | |
two-tailed probability = 0.01852 + 0.00449 = 0.02301 Return to Top ©Richard Lowry 1998- All rights reserved. |
188497 | https://mathoverflow.net/questions/25067/given-n-k-element-subsets-of-n-is-there-a-small-subset-a-of-n-which-intersects | Skip to main content
Given n k-element subsets of n, is there a small subset A of n which intersects them all?
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I'm looking for an answer to the following question. (An answer to a slightly different question would be good as well, since it could be useful for the same purpose.)
Given a set C consisting of n
subsets of {1, 2, ..., n}, each of
size k, does there exist some small A
⊂ {1, 2, ..., n} such that
A intersects all (or all except a small number) of the sets in C?
Preferably, "small" will be ϵn where ϵ can be made arbitrarily small, as long as n and k are sufficiently large.
I'm hoping the answer is yes. Here is why some such A might exist: on average, each element of {1, 2, ..., n} intersects k sets in C, so one might hope to make do with A of size on the order of n/k.
This smells a bit like some version of Ramsey's theorem to me, or like the Erdős–Ko–Rado theorem, but it doesn't (as far as I can tell) follow directly from either.
co.combinatorics
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asked May 17, 2010 at 22:29
skeptical scientistskeptical scientist
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I believe, reading the abstract, that the paper "Transversal numbers of uniform hypergraphs", Graphs and Combinatorics 6, no. 1, 1990 by Noga Alon answers your question in the affirmative, for some definition of ``your question''. Namely, the worst case is that A has to have size about 2logk/k times n, and this multiplier tends to zero as k tends to infinity.
Here's a free copy of the paper.
I'm certainly no expert on these matters and my advice would be to look at this and related literature on transversals of hypergraphs. Your collection C of sets is the same thing as a k-uniform hypergraph, and the property that you want from A is equivalent to it being a transversal.
Reading Alon's paper a little more I see that what you want is the easier direction of his argument (which gives a tight dependence on k). The basic idea is to choose your transversal randomly by picking elements of {1,…,n} with an appropriate probability p. That way, with high probability, you'll hit most of the sets from your collection C, and then you just add in one extra element of A for each un-hit set from C.
Reading a little further still, I see that the upper bound is probabilistic as well: that is, to make a collection C which is ``bad'', the best plan is to choose sets in C at random from amongst all k-element subsets of {1,…,n}.
There's probably literature on your ``almost transveral'' question, but I'll leave someone else to find it. My guess is that random does best in both directions there too.
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edited May 17, 2010 at 23:51
answered May 17, 2010 at 23:19
Ben GreenBen Green
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Do you want to know the size of A without having to look at what is in C, just looking at n and k?
If you do, you can easily find a few bounds on the size of A.
Lets call the sets in C, C1,C2 etc. and if we define C1={1,2,...k} and then C2={k+1,k+2,...2k} etc obviously upto C⌊nk⌋ and define the rest of them to be whatever you like (as long as you use the last few elements). Then if A intersects all of the sets in C, then it must contain at least 1 element from the first k, one from the second k etc. so must have at least n/k elements.
If you actually have particular sets of C and would like to find the A as small as possible, then you could try some algorithm that tries to home in on the set. Something such as:
Relabel 1, 2, ... n such that 1 appears in more (or the same) number of sets in C as 2 does, and so on.
So set A1={1,2,...n}
So given Ai, take the highest element x of Ai such that for each set Ci with x in Ci then A∩Ci≥2.
Then set Ai+1=Ai−x.
I hope those two ideas help.
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answered May 19, 2010 at 16:20
Karl WaughKarl Waugh
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Section 7.3 : Trig Substitutions
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Use a trig substitution to evaluate ( \displaystyle \int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}}).
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Start Solution
The first thing we’ll need to do here is complete the square on the polynomial to get this into a form we can use a trig substitution on.
[\begin{align}40 - 6z - {z^2} & = - \left( {{z^2} + 6z - 40} \right) = - \left( {{z^2} + 6z + 9 - 9 - 40} \right) = - \left[ {{{\left( {z + 3} \right)}^2} - 49} \right]\ & = 49 - {\left( {z + 3} \right)^2}\end{align}]
The integral is now,
[\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} = \int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)}^{\frac{3}{2}}}}}\,dz}}]
Now we can proceed with the trig substitution.
Show Step 2
It looks like we’ll need to the following trig substitution.
[z + 3 = 7\sin \left( \theta \right)]
Next let’s eliminate the root.
[{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)^{\frac{3}{2}}} = {\left[ {\sqrt {49 - \left( {z + 3} \right)^{2}} } \right]^3} = {\left[ {\sqrt {49 - 49{{\sin }^2}\left( \theta \right)} } \right]^3} = {\left[ {7\sqrt {{{\cos }^2}\left( \theta \right)} } \right]^3} = 343{\left| {\cos \left( \theta \right)} \right|^3}]
Next, because we are doing an indefinite integral we will assume that the cosine is positive and so we can drop the absolute value bars to get,
[{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)^{\frac{3}{2}}} = 343{\cos ^3}\left( \theta \right)]
For a final substitution preparation step let’s also compute the differential so we don’t forget to use that in the substitution!
[\left( 1 \right)dz = 7\cos \left( \theta \right)\,d\theta \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}dz = 7\cos \left( \theta \right)\,d\theta ]
Recall that all we really need to do here is compute the differential for both the right and left sides of the substitution.
Show Step 3
Now let’s do the actual substitution.
[\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} = \int{{\frac{{16807{{\sin }^5}\left( \theta \right)}}{{343{{\cos }^3}\left( \theta \right)}}\left( {7\cos \left( \theta \right)} \right)\,d\theta }} = 343\int{{\frac{{{{\sin }^5}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}}\,d\theta }}]
Do not forget to substitute in the differential we computed in the previous step. This is probably the most common mistake with trig substitutions. Forgetting the differential can substantially change the problem, often making the integral very difficult to evaluate.
Show Step 4
We now need to evaluate the integral. Here is that work.
[\begin{align}\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} & = 343\int{{\frac{{{{\left[ {1 - {{\cos }^2}\left( \theta \right)} \right]}^2}}}{{{{\cos }^2}\left( \theta \right)}}\sin \left( \theta \right)\,d\theta }}\hspace{0.25in}\hspace{0.25in}u = \cos \left( \theta \right)\ & = - 343\int{{\frac{{{{\left[ {1 - {u^2}} \right]}^2}}}{{{u^2}}}\,du}} = - 343\int{{{u^{ - 2}} - 2 + {u^2}\,du}}\ & = - 343\left( { - {u^{ - 1}} - 2u + \frac{1}{3}{u^3}} \right) + c\ & = - 343\left( { - \frac{1}{{\cos \left( \theta \right)}} - 2\cos \left( \theta \right) + \frac{1}{3}{{\cos }^3}\left( \theta \right)} \right) + c\end{align}]
Don’t forget all the “standard” manipulations of the integrand that we often need to do in order to evaluate integrals involving trig functions. If you don’t recall them you’ll need to go back to the previous section and work some practice problems to get good at them.
Every trig substitution problem reduces down to an integral involving trig functions and the majority of them will need some manipulation of the integrand in order to evaluate.
Show Step 5
As the final step we just need to go back to (z)’s. To do this we’ll need a quick right triangle. Here is that work.
From the substitution we have,
[\sin \left( \theta \right) = \frac{{z + 3}}{7}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{hyp}}}}} \right)]
From the right triangle we get,
[\cos \left( \theta \right) = \frac{{\sqrt {49 - {{\left( {z + 3} \right)}^2}} }}{7}]
The integral is then,
[\int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2401}}{{\sqrt {49 - {{\left( {z + 3} \right)}^2}} }} + 98\sqrt {49 - {{\left( {z + 3} \right)}^2}} - \frac{{{{\left( {49 - {{\left( {z + 3} \right)}^2}} \right)}^{\frac{3}{2}}}}}{3} + c}}]
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Covet, Covert & Overt
Covet vs. Covert vs. Overt: Untangling the Differences ✨📖
Introduction
Have you ever felt the urge to covet something? 🤔 Or maybe you’ve heard of a covert operation? 🕵️♂️ And what about an overt gesture? 👀 If these words have ever made you pause and wonder what they truly mean, you’re not alone. “Covet,” “covert,” and “overt” may sound similar, but their meanings are quite different. Understanding the distinctions between these words can help you use them with confidence and precision. Let’s break them down together in a fun and straightforward way, so that next time you encounter these words, you’ll know exactly how to use them. Whether you’re reading a spy novel, discussing someone’s ambitions, or observing a public action, these words can help convey precise meanings. So, let’s dive in and explore these words in detail! 🌟
Definitions
Covet: To yearn to possess something, especially something that belongs to someone else. 💭 It usually implies a strong desire that may not be entirely appropriate.
Covert: Something that is hidden or concealed. 🔒 This word often implies secrecy, like covert operations or actions that are done out of sight or without others knowing.
Overt: Something that is done openly and is not hidden. 🔓 It is the complete opposite of covert, representing actions or gestures that are visible and meant to be noticed. 👁️
Pronunciation
Covet: KUV-it. 🔊 The emphasis is on the first syllable, making it sound direct and simple.
Covert: KOH-vurt. 🔊 The pronunciation hints at something subtle, as if the word itself is hiding a secret.
Overt: oh-VURT. 🔊 The stress on the second syllable makes it sound bold and open, just like its meaning.
Etymology
Covet: The word covet comes from the Old French “coveitier,” which means to desire. 💡 It’s often associated with a kind of longing that crosses boundaries. Historically, it has been used to describe an intense, sometimes morally questionable desire for something that belongs to someone else, like in the phrase “Thou shalt not covet.”
Covert: Covert comes from the Old French “covrir,” meaning to cover. 🛑 Think of something that is deliberately kept secret or out of sight. This word has been used in contexts involving stealth and secrecy for centuries, often in relation to espionage or secret operations.
Overt: The origin of overt can also be traced back to Old French “overt,” meaning open or uncovered. 🚪 It refers to actions that are done openly without any attempt to conceal. This word has always carried a sense of transparency, making it perfect for describing actions that are intended to be seen by others.
Usage Examples
Covet: He couldn’t help but covet his neighbor’s new car. 🚗✨ The car was sleek, fast, and everything he had ever dreamed of, and though he knew it was wrong, the feeling of envy was strong.
Covert: The agent was on a covert mission to gather intelligence. 🕶️ He moved through the shadows, ensuring that no one noticed his presence, as the success of the mission depended on complete secrecy.
Overt: She made an overt display of affection by hugging her friend in front of everyone. 🤗 It was clear that she wanted to show her support, and there was nothing hidden about her actions.
Synonyms & Antonyms
Covet: Synonyms: desire, yearn for, crave, envy; Antonyms: reject, spurn, disregard. 🔄
Covert: Synonyms: secret, hidden, undercover, clandestine; Antonyms: open, public, overt, obvious. 🔍
Overt: Synonyms: obvious, open, clear, blatant; Antonyms: covert, hidden, secret, concealed. 👀
Comparison and Contrast
It’s easy to mix up these words, but here’s a simple way to keep them straight. ⚖️ Covet is about wanting something, often something that’s not yours to take. It implies a deep desire that may even border on envy. Covert is all about secrecy—if it’s covert, it’s hidden, not meant to be seen. 🤐 Meanwhile, overt is the opposite of covert—it’s done openly, with nothing to hide, and meant to be noticed. 🌟
Imagine a scenario: Your friend makes an overt statement about wanting the same job you’re applying for—no secrets, just openness. 🗣️ On the other hand, someone might covet that job, quietly desiring it with a hint of envy. 😔 And if they make covert moves to get it, they’re keeping it all hush-hush, working behind the scenes to achieve their goal without anyone noticing. 🤫
Another example could involve a celebrity. 🌟 A celebrity might covet the privacy that they no longer have, longing for a life away from the cameras. 📸 They might take covert actions, like using disguises, to avoid paparazzi. On the other hand, their fans might make overt displays of admiration, openly showing their love by cheering or holding up signs. 🎉
Contextual Usage
“John would covet the promotion, but Jane made it overtly clear that she was aiming for it, while Tom made covert efforts to win over the boss.” 💼👥
“The organization carried out a covert investigation to gather evidence, while the rival group made overt claims about their innocence in the media.” 📢🕵️♀️
“She couldn’t help but covet her friend’s talent, while her own efforts to improve were covert, and her admiration was overt when she applauded her friend’s performance.” 👏🎭
Mnemonic Devices
Covet: Covet is like “crave it,” which sounds like craving or desiring. If you covet something, you crave it so much that you might cross a line. 🚧
Covert: Covert means covered—something covered is hidden. 🛏️ Imagine a blanket covering something secret; that’s covert.
Overt: Overt rhymes with open, which can help you remember that it’s something done out in the open. 🚪 Imagine a door that is wide open—nothing is hidden, everything is visible. 👁️
Related Confusing Word Pairs
If you found this word trio confusing, you might also want to check out pairs like “complement vs. compliment” 📝 or “affect vs. effect.” 🌊⚡ Each of these sets involves similar sounding words with very different meanings. Other word pairs that often confuse people include “imply vs. infer,” “discreet vs. discrete,” and “elicit vs. illicit.” Learning these distinctions can help you become a more precise communicator and avoid common pitfalls. 🧠📚
Conclusion
Congratulations! 🎉 You’ve just mastered three tricky words that often trip people up. Remember: covet is about desire, often with a hint of envy; covert means hidden or secret, usually involving stealth; and overt is openly done, with nothing to hide. With these distinctions in your toolkit, you’re ready to use these words accurately and impressively in conversation and writing. ✍️ The more you practice, the easier it becomes to remember these nuances, and soon you’ll be using these words with the confidence of a language expert. 🌟💬
Test Your Knowledge: Covet, Covert & Overt Quiz
1. He couldn’t help but ___ his neighbor’s new car. 🚗💭
Covert 😶Overt 🌟Covet 😍
Correct answer: Covet 😍. “Covet” means to desire something, especially something that belongs to someone else.
2. The agent was on a ___ mission. 🕶️🎯
Covet 😍Covert 🕵️♂️
Correct answer: Covert 🕵️♂️. “Covert” refers to something secret or hidden, as highlighted in the context.
3. Overt means something is hidden. ❌
True ❓False ✔️
Correct answer: False ✔️. “Overt” means open or not hidden, so the statement is false.
4. Which word is a synonym of ‘Covet’? 💭
Hide 🙈Desire 😍Open 🚪Cover 🛡️
Correct answer: Desire 😍. “Desire” is the correct synonym for “Covet,” referring to wanting something intensely.
5. The boss made an ___ show of support for the new policies. 📢
Covet 😍Covert 🕶️Overt 🎉
Correct answer: Overt 🎉. “Overt” means openly displayed, fitting the context of the sentence.
6. Which of the following words refers to something done openly? 🌟
Covet 😍Covert 🕶️Overt 🎉
Correct answer: Overt 🎉. “Overt” actions are done openly and clearly visible to others.
7. She ___ her friend’s talent but made a(n) ___ effort to support her. 💭👏
Covet 😍, Overt 🌟Covert 🕶️, Covet 😍Overt 🌟, Covet 😍
Correct answer: Covet 😍, Overt 🌟. She desires her friend’s talent (covet), but her efforts to support are open and obvious (overt).
8. Which word has its origins in Latin meaning ‘to cover’? 📜
Covet 😍Covert 🕶️
Correct answer: Covert 🕶️. “Covert” comes from the Latin word “coopertus,” meaning “to cover.”
9. Covet means to hide something. ❌
True ❓False ✔️
Correct answer: False ✔️. “Covet” means to desire something, not to hide it.
10. She made ___ efforts to be noticed by the manager, while others did it more ___. 💼🔍
Overt 🌟, Covert 🕶️Covert 🕶️, Overt 🌟
Correct answer: Overt 🌟, Covert 🕶️. She made obvious efforts (overt), while others did it secretly (covert).
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