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How many subshells are in the n = 3 shell, How many orbitals are in the n = 3 shell?
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In the quantum mechanical model of the atom, the principal quantum number n n determines the energy level or shell of an electron. For n=3 n=3:
Number of Subshells: The number of subshells in a given shell is equal to n n. Therefore, for n=3 n=3, there are 3 subshells: s s, p p, and d d.
Number of Orbitals: Each subshell has a specific number of orbitals:
The s s subshell has 1 orbital.
The p p subshell has 3 orbitals.
The d d subshell has 5 orbitals.
To find the total number of orbitals in the n=3 n=3 shell, you add these together:
1\text{(from}s\text{)}+3\text{(from}p\text{)}+5\text{(1\text{(from}s\text{)}+3\text{(from}p\text{)}+5\text{(
Continue Reading
In the quantum mechanical model of the atom, the principal quantum number n n determines the energy level or shell of an electron. For n=3 n=3:
Number of Subshells: The number of subshells in a given shell is equal to n n. Therefore, for n=3 n=3, there are 3 subshells: s s, p p, and d d.
Number of Orbitals: Each subshell has a specific number of orbitals:
The s s subshell has 1 orbital.
The p p subshell has 3 orbitals.
The d d subshell has 5 orbitals.
To find the total number of orbitals in the n=3 n=3 shell, you add these together:
1(from s)+3(from p)+5(from d)=9 orbitals 1(from s)+3(from p)+5(from d)=9 orbitals
In summary:
Subshells in n=3 n=3: 3 (s, p, d)
Orbitals in n=3 n=3: 9
Upvote ·
Related questions
More answers below
How many subshells are in the n = 2 shell?
How many subshells and orbitals are in the principal shell with n=3?
What are the possible orbitals for n = 3?
How many subshells are associated with N equal to 3?
How many orbitals are there in the n = 4 shell?
Harsh Pundheer
B.sc(H) in Honours, Acharya Narendra Dev College, University of Delhi (Graduated 2013)
·6y
Shell is the energy level of an atom (n=1,2,3,etc.)
For n =3
An
s sub-shell only has one orbital.
p sub-shell has three orbitals.
d sub-shell has five orbitals.
So, the total number of orbitals in the third electron shell is:
1+3+5=9
Continue Reading
Shell is the energy level of an atom (n=1,2,3,etc.)
For n =3
An
s sub-shell only has one orbital.
p sub-shell has three orbitals.
d sub-shell has five orbitals.
So, the total number of orbitals in the third electron shell is:
1+3+5=9
Upvote ·
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Triveni Yeswin
Junior Lecturer (2013–present)
·7y
3 sub shells
n l sub shell
1 0 s
2 1 s,p
3 2 s,p,d
4 2 s,p,d,f
S contain 1 orbital
P contain 3 orbitals
d contains 5 orbitals
f contains 7 orbitals
n=3 means 3 subshell
9 orbitals (1+3+5=9)
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9 1
Braj Kishor
Always Learner! Stop Wishing Start Doing!
·8y
Thanks for asking question .
Configuration of any atom are written in this order 1s^2 ,2s^2,2p^6,3s^23p^6,4s^2,3d^10…………
For n=3 it mean that only we about that configuration 3s^2,3p^6,3d^10
Hence total number of subshell for n=3 is 9 ( 1for s & 3 for p & 5 for d ).
& Total number of orbit =3 (s,p,d orbit ).
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9 5
9 2
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More answers below
How many subshells are there in the n=4 principal shell?
What is the number of sub shells in the orbital n=5?
If n=1, then how many orbitals are there in that shell?
Which subshells are in the n=3 principal shell?
How many orbitals are possible for the subshell with n=5, l=3?
Md Yazdan Afzal
MBBS from College of Medicine and Sagore Dutta Hospital (Expected 2027) · Author has 93 answers and 408.1K answer views
·5y
Related
How many orbitals are there in the n = 4 shell?
To find the number of orbitals, there are 2 ways:
The 1st one (conceptual but long and boring)
This is the conventional way… First you have to find 'l' that is azimuthal quantum number, then from that you have to find ‘ml’ or 'm' that is magnetic quantum number.
n = 4 {given}
C a l c u l a t i o n C a l c u l a t i o n o f′l′o f′l′
Values of l (small letter L) = 0 to (n-1)= 0,1,2,3 {in this case}.
Magnitude of l = 4 {see it's equal to the value of n}
C a l c u l a t i o n C a l c u l a t i o n o f′m l′o f′m l′
Values of ml= -l to +l
When:
l=0, ml = 0
l=1, ml = -1,0,+1
l=2, ml = -2,-1,0,+1,+2 and so on…
Magnitude of ml = 2l +1
When:
l=0, magnitude of ml = 2×0 + 1 = 1
{you can actually
Continue Reading
To find the number of orbitals, there are 2 ways:
The 1st one (conceptual but long and boring)
This is the conventional way… First you have to find 'l' that is azimuthal quantum number, then from that you have to find ‘ml’ or 'm' that is magnetic quantum number.
n = 4 {given}
C a l c u l a t i o n C a l c u l a t i o n o f′l′o f′l′
Values of l (small letter L) = 0 to (n-1)= 0,1,2,3 {in this case}.
Magnitude of l = 4 {see it's equal to the value of n}
C a l c u l a t i o n C a l c u l a t i o n o f′m l′o f′m l′
Values of ml= -l to +l
When:
l=0, ml = 0
l=1, ml = -1,0,+1
l=2, ml = -2,-1,0,+1,+2 and so on…
Magnitude of ml = 2l +1
When:
l=0, magnitude of ml = 2×0 + 1 = 1
{you can actually count the number of values of ml for corresponding value of l to get the magnitude of ‘ml’ in each of the cases; you will get the same answer}
l=1, magn(ml) = 2×1 + 1 = 3
l=2, magn(ml) = 5
l=3, magn(ml) = 7
Total = 1+3+5+7 = 16
Hence the number of orbitals in n=4 shell is 16.
Huh, finally very tired….We don't have this much time for doing all this “conceptual” nonsense (pun intended)
So, now what??? Now comes the saviour, that is the 2nd Method. You can actually get the answer within 2 seconds, yes you read that right….
2nd method:
No. of orbitals = n² (yeah, that's it)
= 4² = 16 (Here you go, within 2 seconds, in my mind).
Hope this helps,
If it does, Please Upvote.
(I think this answer deserves ⬆️)
Thanks and regards!!!
Upvote ·
99 55
9 1
Hcbiochem
Former Professor of Biology and Chemistry (1982–2019) · Author has 4.6K answers and 1.9M answer views
·2y
Originally Answered: How many subshells and orbitals are in the principal shell with n=3? ·
If n = 3, then l = 0, 1, 2. This tells you that there are 3 subshells in the n=3 shell.
For the l = 0 subshell, there is only one value of ml (0), so this subshell contains only a single orbital.
For the l=1 subshell, there are 3 values of ml (-1, 0, 1), so this subshell contains 3 orbitals.
For the l=2 subshell, there are 5 values of ml (-2, -1, 0, 1, 2) so this subshell contains 5 orbitals.
So the total number of orbitals in the n=3 shell is 9.
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Krithvik V
Lives in India (1988–present)
·2y
Originally Answered: How many subshells and orbitals are in the principal shell with n=3? ·
The principal shell with n=3 has 3 subshells:
The subshell with ℓ=0 is called the s subshell, and can hold a maximum of 2 electrons.
The subshell with ℓ=1 is called the p subshell, and can hold a maximum of 6 electrons.
The subshell with ℓ=2 is called the d subshell, and can hold a maximum of 10 electrons.
So, the total number of orbitals in the principal shell with n=3 is 2+6+10=18.
Upvote ·
Norris Johnson
Studied Chemistry Math&Physics (Graduated 1965) · Author has 272 answers and 44.6K answer views
·2y
Originally Answered: How many subshells and orbitals are in the principal shell with n=3? ·
The 3s, 3p, and 3 d orbitals. Holding 2, 5, and 10electrons at a maximum. However, the 3d orbital does not have electrons until the 4s and 4p orbitals fill.
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Arno Vigen
Author, 3D Particle-Level Chemical Engineering · Author has 3.4K answers and 3.2M answer views
·4y
Related
What is a list of all the possible subshells and number of conceivable orbitals associated with the principle quantum “quantum” number n, if n = 3?
1
I think of quantum numbers as quantitized hemipsherical coordinates:
1st - r#, radial, shell, layer
2nd - theta#, inclination / longitude - starting with -0- at the poles (subshell-s) for a ring of electron a) at the same distance, and b) at the same angle relative to the nucleus and its nucleostatic axis
3rd - phi#, latitude - for electrons evenly spaced with a meridian chosen, A meridian of -0- is arbitrarily chosen, the count +1 and -1 from there. But, remember a) the other hemisphere and b) a 2nd layer at the same count, have that organized offset by 1/2 phase (180-degrees), so 2,8,8,18,18,3
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1
I think of quantum numbers as quantitized hemipsherical coordinates:
1st - r#, radial, shell, layer
2nd - theta#, inclination / longitude - starting with -0- at the poles (subshell-s) for a ring of electron a) at the same distance, and b) at the same angle relative to the nucleus and its nucleostatic axis
3rd - phi#, latitude - for electrons evenly spaced with a meridian chosen, A meridian of -0- is arbitrarily chosen, the count +1 and -1 from there. But, remember a) the other hemisphere and b) a 2nd layer at the same count, have that organized offset by 1/2 phase (180-degrees), so 2,8,8,18,18,32,32 and so on.
4th - z# - hemisphere which has 1/2 the energy, so not the above integers, but -1/2 and +1/2 in equations.
2
That means that Shells are 2 hemipsheres in one dimension x squared, from PIradius-squared, a circle, the tightest packing in the other two dimensions.
Shell-1: 2 x 1-squared = 2 x 1 = 2 electrons
Shell-2 & -3: 2 x 2-squared = 2 x 4 = 8 electrons
Shell-4 & -5: 2 x 3-squared = 2 x 9 = 18 electrons
Shell-6 & -7: 2 x 4-squared = 2 x 16 = 32 electrons
3
However, that means the counts in the subshells are based upon (x+1)-squared, which adds (2x+1) which is always ODD - - in each hemisphere.
That leads to the amazing, fun math of Odd-Count subshells rings building pole to equator.
Shell-1: 2 x (1) = 2 x 1 = 2 electrons
Shell-2 & -3: 2 x (1+3) = 2 x 4 = 8 electrons
Shell-4 & -5: 2 x (1+3+5) = 2 x 9 = 18 electrons
Shell-6 & -7: 2 x (1+3+5+7) = 2 x 16 = 32 electrons
Strange, amazing, yet simple to get the #1 graphic.
4
So, Layer 3 (N=3) has 18 electrons in three outer subshells and 28 total.
Subshell-1s
(1,0,0,-1/2) 1s1 at -0- degrees
(1,0,0,+1/2) 1s2 at -0- degrees
Subshell-2s
(2,0,0,-1/2) 2s1 at -0- degrees
(2,0,0,+1/2) 2s2 at -0- degrees
Subshell-2s
(2,1,0,-1/2) 2p1 at 70.5 degrees inclination and -0- chosen meridian
(2,1,0,+1/2) 2p2 at 70.5 degrees inclination and 180 2nd hemisphere meridian
(2,1,-1,-1/2) 2p1 at 70.5 degrees inclination and -120 (0–120) chosen meridian
(2,1,-1,+1/2) 2p2 at 70.5 degrees inclination and +60 (180–120) chosen meridian
(2,1,+1,-1/2) 2p1 at 70.5 degrees inclination and +120 (0+120) chosen meridian
(2,1,+1,+1/2) 2p2 at 70.5 degrees inclination and 300 (180+120) chosen meridian
Subshell-3s
(2,0,0,-1/2) 2s1 at -0- degrees
(2,0,0,+1/2) 2s2 at -0- degrees
Subshell-3s
(2,1,0,-1/2) 2p1 at 70.5 degrees inclination and 180 degrees
(2,1,0,+1/2) 2p2 at 70.5 degrees inclination and -0- 2nd hemisphere meridian
(2,1,-1,-1/2) 2p1 at 70.5 degrees inclination and +60 (180–120) chosen meridian
(2,1,-1,+1/2) 2p2 at 70.5 degrees inclination and -120 (0–120) chosen meridian
(2,1,+1,-1/2) 2p1 at 70.5 degrees inclination and 300 (180+120) chosen meridian
(2,1,+1,+1/2) 2p2 at 70.5 degrees inclination and +120 (0+120) chosen meridian
Note that Layer 3 is offset 1/2 phase!
Upvote ·
9 1
Kile Baker
Ph.D. in Physics&Astrophysics, Stanford University (Graduated 1978) · Author has 284 answers and 168.6K answer views
·5y
Related
For an electron in the n=3 energy level, what are all the possible subshell and orbital combinations in this level? Since each orbital can hold 2 electrons, how many electrons can exist in the n=3 energy level?
Well, let’s just count them up. We have two quantum numbers to consider, L, the total angular momentum and m l m l the z-component of the angular momentum.
L can run from 0 to n-1, so for n=3, L=0,1,or 2.
The m quantum number runs from -L to +L
So we have:
L = 0, m=0
L=1, m=1,0,-1
L=2, m=2,1,0,-1,-2
Count them up and you will find there are 9 possible quantum number. For each of the possibilities we can have two electrons, spin-up and spin-down. So this means we can have a total of 18 electrons in the n=3 energy level.
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Asad Hussain
Author has 84 answers and 196.2K answer views
·4y
Related
How many orbitals are there in n=3 shells?
Number of orbitals=?
But first you have to know about number of subshells in 3rd shell .
This is about 3rd shell so there should be exactly 3 subshells ..
Subshells s,p,d..
Now in each subshell orbitals are present.
S=has one orbtal ..which is (s ) as itself.
P=has 3 orbitals which are px, py,pz
d=has 5 orbital which are= dx,dy,dz,dyz, dz2. ..
Sumition of all..1+3+5=9 orbitals…
Upvote ·
9 3
Michael Flynn
BSc Hons Newcastle University · Author has 901 answers and 2.7M answer views
·7y
Related
How many orbitals are possible for the subshell with n=5, l=3?
If l = 3 then the magnetic quantum number m takes the values -l…..0…..+l .
This gives m = -3 , -2 , -1 , 0 , 1 , 2, 3
As you can see this means there are 7 orbitals, which in this case will be the 5f which can hold up to 14 electrons.
They look like this:
(Diagram courtesy of Wikipedia).
Continue Reading
If l = 3 then the magnetic quantum number m takes the values -l…..0…..+l .
This gives m = -3 , -2 , -1 , 0 , 1 , 2, 3
As you can see this means there are 7 orbitals, which in this case will be the 5f which can hold up to 14 electrons.
They look like this:
(Diagram courtesy of Wikipedia).
Upvote ·
9 6
Ron Morel
B.S. in Chemistry, United States Air Force Academy (Graduated 1979) · Author has 13.9K answers and 2.7M answer views
·2y
Related
For an electron in the n=3 energy level, what are all the possible subshell and orbital combinations in this level? Since each orbital can hold 2 electrons, how many electrons can exist in the n=3 energy level?
The s orbital can hold 2 electrons. The p orbital has 3 sub-orbitals (px, py, pz). Each sub-orbital can hold 2 electrons. At the n=4 level, the 3d orbital is found. It has 5 sub-orbitals. So 10 electrons. From the way your question is worded, I don't think the 3d orbital is considered but I was not sure.
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Using sealed sources, you can demonstrate most of the properties of alpha, beta and gamma radiation. The experiments in this collection allow students to see their ranges, penetrating powers and, in the case of beta radiation, how it is deflected in a magnetic field. They can link these properties to the nature of each type of radiation, and start to form a picture of why these types of radiation behave in the way they do.
The first experiment introduces the idea that radioactive atoms give out three distinct types of radiation, known as alpha, beta and gamma. The next four experiments allow you to investigate their properties in more detail. Each of these experiments is built around one type of radiation (and all its properties). However, you could equally choose to reshuffle these experiments and focus on each property in turn – i.e. look at the range in air for all three radiations and then move on to penetrating power.
Up next
Measuring the half-life of protactinium
Measuring the half-life of a radioactive isotope brings some of the wonder of radioactive decay into the school laboratory. Students can witness one element turning into another and hear (or see) the decrease in the radiation it gives out as it transmutes.
This demonstration uses a protactinium generator to show the exponential decay of protactinium-234, a grand-daughter of uranium. It has a half-life of just over a minute, which gives students the chance to measure and analyze the decay in a single lesson.
Apparatus and Materials
tray
Holder for Geiger-Müller tube
Geiger-Müller tube, thin window
Scaler
Stopclock
Retort stand, boss, and clamp
Ratemeter (OPTIONAL)
Protactinium generator
Health & Safety and Technical Notes
See the following guidance note:
Managing radioactive materials in schools
To limit the risk of radioactive liquids being spilt, there should be special instructions in the local rules for handling (and preparing) this source.
Read our standard health & safety guidance
Preparation of the protactinium generator
It is now possible to purchase the chemicals already made up in a sealed bottle. One supplier is TAAB Laboratories Equipment Ltd, 3 Minerva House, Calleva Park, Aldermaston, RG7 8NA. Tel: 0118 9817775. However, you can make your own if you prefer.
These quantities make a total volume of 20 cm3. You can scale them up if you have a larger bottle. (A '30 ml' bottle has a capacity of about 35 ml, so there is still room to shake the solution when the total volume is 30 ml.)
Dissolve 1 g of uranyl nitrate in 3 cm3 of water. Wash it into a small separating funnel or beaker with 7 cm3 of concentrated hydrochloric acid.
To this solution, add 10 cm3 of iso-butyl methyl ketone or amyl acetate.
Shake the mixture together for about five minutes. Then run the liquid into the polypropylene bottle and firmly screw down the cap. It can help to shield the lower half of the bottle with some lead.
Place the bottle in a tray lined with absorbent paper.
Once you have made the protactinium generator, you can store it with other radioactive materials, taking care to follow your school code of practice and local rules: see the Managing radioactive materials in schools guidance note:
Managing radioactive materials in schools
A polypropylene bottle is preferable to polythene because it is somewhat more resistant to attack by the acid and ketone. Nevertheless, polythene bottles can be used, provided no attempt is made to store the liquid in them for more than a few weeks.
The organic layer which separates out contains the protactinium-234. This decays with a half-life of about 70 seconds.
An alternative to protactinium: A new, effective and extremely low hazard system for measuring half-life is available from Cooknell Electronics Ltd, Weymouth, DT4 9TJ. This uses fabric gas mantles designed for camping lights. Each mantle contains a small quantity of radioactive thorium. More details are available on the Cooknell Electronics website:
Cooknell Electronics
Procedure
Support the Geiger-Muller tube holder in a clamp, so that the tube is facing downwards towards the neck of the bottle.
Allow the bottle to stand for at least ten minutes. Take the background count by running the counter for at least 30 seconds. This is done with the bottle in position, because some of the count will come from the lower layer. You can do this before the experiment or some time after it has finished.
Alternatively, the GM tube can be clamped horizontally with the window close to the upper layer.
Shake the bottle vigorously for about 15 seconds to thoroughly mix the layers.
Place the bottle in the tray.
As soon as the two layers have separated, start the count and start the stop-clock.
Record the time from the beginning of the experiment - i.e. the time of day for the sample.
Record the count every 10 seconds. Or record it for 10 seconds every 30 seconds.
Run the experiment for about five minutes, ample time to reveal the meaning of the term half-life and to illustrate the decay process.
Provided you leave a few minutes between each attempt, you can repeat the experiment. In 5 minutes the activity of the protactinium in the aqueous layer grows to 15/16 of its equilibrium value.
It is possible to record the growth to equilibrium. Do this by moving the GM tube so that the aqueous layer at the bottom of the bottle is immediately above the end window of the GM tube.
Teaching Notes
The chemistry of the experiment:
The first stages of the uranium-238 series are involved in this experiment.
The aqueous solution (at the bottom of the bottle) contains the uranium-238, its daughter thorium-234 and the short-lived granddaughter protactinium-234.
Uranium and protactinium both form anionic chloride complexes but thorium does not. At high hydrogen ion concentrations, these complexes will dissolve in the organic layer (which is floating on top of the aqueous solution).
When you shake the bottle, about 95% of the short-lived granddaughter (protactinium) and some of the uranium will be dissolved in the organic layer. The thorium stays in the aqueous layer.
Since radioactivity is a property of the innermost nucleus of the atom it is not affected by chemical combination.
The granddaughter (in the organic layer) decays without any more being produced by its parent (thorium) all of which is still in the aqueous layer. It emits beta particles which travel through the plastic wall of the bottle. Isolating the protactinium in the top (organic) layer allows it to decay without any top-up from its parent (thorium).
The radiation from the thorium and uranium should not interfere with the results, for two reasons:
The counter does not detect the alpha particles from the uranium or the low energy beta particles from the thorium. It only records the high energy (2 MeV) beta particles from the granddaughter (protactinium).
The uranium-238 decays with an extremely long half-life. It yields a meagre, almost constant, stream of low energy alpha particles. Its daughter, thorium-234, decays with a half-life of 24 days. During the length of this experiment the decay rate can be assumed to be constant. If these two isotopes contribute to the count at all, it will be accommodated in the background count. The stockpile of thorium is also constantly topped up in the aqueous layer as long as the protactinium is present with the thorium.
Table of count rate:
Get the students to make a table of count rate against time, and correct it for background count. The first 10-second reading should be allocated to a time of zero.
Plot a graph:
Get the students to plot a graph of count rate against time. They should draw a smooth curve through the points.
First point out the general pattern - that the count rate decreases with time. Then look for an exponential trend - that the best fit curve always takes the same amount of time to halve.
Get students to measure the half-life from the curve.
Point out the random nature of the points: although the decay follows a pattern, there is an element of randomness and it is not perfectly predictable.
How Science Works extension
This experiment provides an opportunity to assess the accuracy of the measured half-life value and how the random nature of decay affects the answer.
The accepted value for the half-life of protactinium is about 70 seconds.
Explore different ways in which a half-life value can be obtained from this apparatus:
Amend the procedure described above so that, instead of a scaler (counter), a ratemeter is used. One student just records the time it takes for the count-rate to halve. This will provide a very approximate value.
Repeat the experiment with several members of the class timing how long it takes for the count-rate to halve. There is likely to be considerable spread in results across the group and the mean result may differ from the accepted value for half-life. In each case, ask students to identify errors and uncertainties in their measurement(s) and to suggest ways in which these could be reduced.
For example, ask: "how does the random nature of the decay affect the measured count-rate when the count is low, or high, compared the background count?"
Either you or your students may suggest a graphical method as an improvement. The procedure described in the main experiment above could then be carried out, and then the accuracy of the half life value assessed and evaluated.
Radioactive materials raise significant safety issues, providing an opportunity to discuss the value and use of secondary data sources.
This experiment was safety-tested in February 2007
Up next
Simple model of exponential decay
Exponential Decay of Activity
Quantum and Nuclear
Simple model of exponential decay
Practical Activity
for 14-16
Class practical
In this activity, students model radioactive decay using coins and dice. By relating the results from the model to the experimental results in...
Measuring the half-life of protactinium
...students can see that the model helps to explain the way in which a radioactive substance decays. The model provides an insight into what might be happening within radioactive atoms.
This activity is a good analogy of radioactive decay as it is based on probability. The decaying trend will be noticeable and so too will the random nature.
Apparatus and Materials
Pennies or other coins, plentiful supply
Dice, plentiful supply (OPTIONAL)
Health & Safety and Technical Notes
Read our standard health & safety guidance
The more coins each student has, the better the analogy of radioactive decay. You could use as few as one per student to keep it simple. Any more than four is quite difficult to manage.
Small coins will turn around more in their cupped hands.
A canvas bag containing 500 plastic cubes (each side 10 mm), each with one face identified, is available in the UK from Lascells, order code 60-010.
Lascells
Procedure
Explain the procedure (as follows) to the class.
Each student has a number of coins. This could be between one and four. They hold them in their cupped hands.
On your instruction "shake", the students shake their coins for at least 5 seconds (they should ensure that the coins are moving around inside their cupped hands). On the instruction "stop", they stop shaking and open their hands with one hand flat and facing upwards so that they can see their coins.
If any coins come down heads, they take them out of their palm and place them on the desk.
On your instruction "show", they put up a number of fingers corresponding to the number of coins they took out of their palm.
Record this number on the board.
They keep the remaining coins in their hands and repeat from step 3. If you can arrange it that you take a reading once every minute, then you can record the readings against time. It will then give results very similar to protactinium.
Analyze the result by plotting a graph.
Teaching Notes
You might want to appoint a counter and a scribe to count the coins and record the results.
Take care with how you ask students to signal the numbers. They may be tempted to add their own (rude) gestures.
Draw out the similarities with the protactinium experiment. The trend is the same and there is also some randomness. The close match between the results from this model and the results from
Measuring the half-life of protactinium
show that radioactive atoms have a chance of decaying in any fixed time.
Use the activity to explain the downward trend of the decay curve. Only coins that are left can decay. As there are fewer of them each time, fewer will decay.
The activity raises the interesting question about how long a radioactive source will last and what happens to the last atom.
An alternative to shaking the coins in students' palms is to flick them. But this takes longer.
You could repeat the experiment with small dice to give a longer half-life. Combining results (as outlined here) makes for a smoother curve.
This experiment was safety-tested in May 2007
Up next
Developing a model of the atom: radioactive atoms
Exponential Decay of Activity
Quantum and Nuclear
Developing a model of the atom: radioactive atoms
Teaching Guidance
for 14-16
Initially, students may regard atoms as the fundamental chemical particles. True, electrons can be chipped off an atom, and possibly all an atom’s electrons stripped off to leave a bare nucleus; yet according to the simple story, the nucleus is still fixed and determines the element by its charge, Ze.
Therefore, to change one element into another, the alchemist’s dream of lead into gold, would require a change of nuclear charge. At first sight this seems impossible because the nucleus is buried deep in the atom bound together by tremendous forces. But it does happen in radioactive elements.
Soon after the discovery of radioactivity in 1896 by Becquerel, Marie Curie and her husband Pierre discovered a new element which they named radium. They extracted dangerously large samples of radium from vast quantities of rock and experimented on its radioactive behaviour.
You could say:
Radioactive atoms do not just stay there as atoms of ordinary copper do; they are completely different: they are unstable, they suddenly break up, flinging out a particle such as an alpha particle, becoming an atom of a different element.
A radium atom remains a radium atom, with the chemical behaviour of a heavy metal, until it suddenly hurls out this alpha particle. (The alpha particle has such a huge energy that it must come from the nucleus.) The remainder of the radium atom is no longer a heavy metal, but a quite different element. This ‘daughter’ of radium is an atom of a heavy inert gas, the end of the helium, neon, argon, krypton, xenon series. It is called radon. The atomic masses have been measured directly, radium-226, radon-222 (a difference of 4 suggesting that the lost alpha particle is a helium nucleus). Separate measurements confirm this.
When you have a mixture of a parent element and a daughter element which have different chemical properties, then they can be separated by ordinary chemical methods.
Radon gas is itself unstable and radioactive. Each of its atoms suddenly, at an unpredictable moment, hurls out an alpha particle. The remainder is a new atom, very unstable, which is called polonium, the ‘daughter’ of radon and the ‘granddaughter’ of radium. The series continues through several more radioactive elements and stops at a stable form of lead. The series does not begin with radium: it begins with uranium several stages earlier. Radioactive uranium (Z=92) has turned into lead (Z=82).
Making unstable atoms
A century ago, radioactivity was a peculiarity of a few mostly heavy, elements: the last few at the end of the Periodic Table. Nowadays scientists can bombard samples of lighter elements with high speed, high energy protons or neutrons, provided directly or indirectly by an accelerator. They can make unstable isotopes of every element in the periodic table. This has opened up the field of nuclear chemistry. Radioactive isotopes behave chemically like their stable isotopes and can be mixed with them. Their progress as radioactive tags can be traced, like luggage labels, following the progress of a ‘labelled’ isotope through the human body or an industrial process.
Half-lives
All the unstable members of these strange families have a constant, reliable characteristic: the atoms show no signs of ageing, or growing weaker, however long they last. Each radioactive element has a constant chance of breaking up in each succeeding second. This is described by a useful length of time, the ‘half-life’ of the radioactive element. For each individual atom the betting is 50:50 for and against its breaking up at any time during one half-life from now. The break-up seems to be controlled by pure chance. That chance does not change and make the break-up more likely for atoms that need to survive longer.
For radium the half-life is 1650 years. Start with 1000 mg of radium now and 1650 years later you will have only 500 mg left. After a further 1650 years only 250 mg will be left and so on. For radium’s daughter, radon -222, the half-life is 3.8 days. In less than four days half the radon gas will have disappeared. You will find helium gas there instead, with the solid products.
The instability appears to be something inherent in the nuclear structure. Nowadays, taking a wave view of the behaviour of nuclear particles, you can picture a stationary wave pattern defining the life of an alpha particle inside the nucleus. But the wave is not completely confined, it leaks through the potential barrier round the nucleus and runs on as a faint wave outside. The wave is interpreted as describing probabilities of locations. It is not a mechanical wave carrying energy and momentum.
While the alpha particle is expected to be found inside the nucleus, there is a chance of finding it one day outside, despite what would seem an insurmountable potential wall. That chance of the alpha particle being outside, being emitted, is definite and constant, a part of the defining wave property, as long as the nucleus lasts. It suggests that high energy alpha particles go with a short half-life of the parent nucleus.
Up next
Exponential decay of a radioactive substance
Exponential Decay of Activity
Quantum and Nuclear
Exponential decay of a radioactive substance
Teaching Guidance
for 14-16
One of the most important characteristics of radioactivity is that it decays exponentially. This has two basic mathematical implications at this level.
The rate falls by a constant ratio in a given time interval. The time it takes to fall by a half is always the same. It also falls to a tenth in equally regular, but longer, time intervals.
The rate of decay is proportional to the amount that is left. This can be seen in the experiment to model radioactive decay. The number of coins that decay in any ‘shake’ is proportional to the number that is left.
Radioactive decay experiment
From these features, you can argue, respectively, the following points.
The chance of an atom disintegrating is constant in time. Radioactive decay is a series of many chance events, all with an unalterable chance.
The rate of disintegrations is proportional to the total number of unchanged radioactive atoms at that moment. Both the rate and the stockpile itself die away exponentially with the same characteristic half-life.
Up next
Some useful equations for half-lives
Exponential Decay of Activity
Quantum and Nuclear
Some useful equations for half-lives
Teaching Guidance
for 14-16
The rate of decay of a radioactive source is proportional to the number of radioactive atoms (N) which are present.
dNdt = -λN
is the decay constant, which is the chance that an atom will decay in unit time. It is constant for a given isotope.
The solution of this equation is an exponential one where N0 is the initial number of atoms present.
N = N0e-λt (Equation 1)
Constant ratio
This equation shows one of the properties of an exponential curve: the constant ratio property.
The ratio of the value, N1, at a time t1 to the value, N2, at a time t2 is given by:
N1N2 = e-λt1e-λt2
N1N2 = e-λ(t1-t2)
In a fixed time interval, t2 – t1 is a constant. Therefore the ratio
N1N2 = a constant
So, in a fixed time interval, the value will drop by a constant ratio, wherever that time interval is measured.
Straight line log graph
Another test for exponential decay is to plot a log graph, which should be a straight line.
Since
N = N0e-λt
Taking natural logs of both sides:
lnN=lnN0-λt
Therefore a graph of N against t will be a straight line with a slope of QuantitySymbol{-λ}.
Half-life and decay constant
The half-life is related to the decay constant. A higher probability of decaying (bigger λ) will lead to a shorter half-life.
This can be shown mathematically.
After one half life, the number, N of particles drops to half of N0 (the starting value). So:
N = N02 when t=T½
By substituting this expression in equation (see above),
N02 = N0e-λT½
Taking natural logs of both sides gives:
ln½=-λT½
ln2 = +λT½
T½ = 0.693λ
Up next
Managing radioactive materials in schools
Ionising Radiation
Quantum and Nuclear
Managing radioactive materials in schools
Teaching Guidance
for 14-16
Countries have national laws to control how radioactive materials are acquired, used and disposed of. These laws follow internationally agreed principles of radiological protection.
The following principles apply to schools:
There should be a person designated to be responsible for the security, safety and proper use of radioactive sources.
Sealed radioactive sources should be of a safe design and type suitable for school science.
Sealed sources should be used whenever possible in preference to unsealed sources. Unsealed sources can only be justified when the scientific demonstrations would not be practicable using sealed sources.
Records of all radioactive sources should be properly kept, showing what they are, when they were bought, when and by whom they have been used, and eventually, how they were disposed of.
Radioactive sources should be used only when there is an educational benefit.
Radioactive sources should be handled in ways that minimize both staff and student exposures.
Sealed sources should be carefully checked periodically to make sure they remain in a safe condition.
The school should have a suitable radioactivity detector in good working order.
UK regulation & guidance
Generally, school employers will insist you obtain their permission before acquiring new radioactive sources.
You must follow your employer’s safety guidance relating to the use the radioactive sources. Most school employers will require you to use either SSERC or CLEAPSS safety guidance, as follows:
In Scotland, safety guidance for use of radioactive sources in schools is issued by the Scottish Schools Equipment Research Centre (SSERC) and is available to members through their website.
In the rest of the UK and British Isles Crown Dependencies, guidance is available from CLEAPSS, the School Science Service. Their guidance document, L93, is freely available from their website, even to non-members.
In the UK...
In classes where children are under the age of 16, the use of radioactive material shall be restricted to demonstrations by qualified science teachers, (which includes newly qualified teachers). However, closer inspection of devices containing low-activity sources such as diffusion cloud chambers is permitted provided the sources are fully enclosed within the devices and not removed during the inspection.
Young persons aged 16 and over may use radioactive sources under supervision. Although the use of radioactive material is regulated, it should not be used as an excuse to avoid practical work. As the ASE points out, "Using the small sources designed for school science gives a good opportunity to show the properties of radioactive emissions directly, and to discuss the radiation risks. Just as importantly, it is an opportunity to review pupils' perception of risks, as they are likely to have constructed their own understanding from a variety of sources, including science fiction films and internet sites. If the work is restricted just to simulations, it may reinforce exaggerated perceptions of risk from low-level radiation.”
Summary of legislation (UK)
Updated October 2008
The following summarizes the somewhat complicated legislative framework in which schools are expected to work with radioactive sources in the UK. However, teachers do not need to obtain and study this legislation; this has been done by CLEAPSS and SSERC, and it is incorporated into their guidance in plain English.
In the European Union, member states have implemented the 1996 EU Basic Safety Standards Directive (as amended) that in turn reflects the 1990 International Commission on Radiological Protection recommendations. In the UK, this has been done through the Radioactive Substances Act 1993 (RSA93), which controls the security, acquisition and disposal of radioactive material, and the Ionising Radiations Regulations 1999 (IRR99) which controls the use of radioactive material by employers. Transport of radioactive material is controlled by The Carriage of Dangerous Goods and Use of Transportable Pressure Equipment Regulations 2007.
There are exemptions from parts of the RSA93 and schools can make use of The Radioactive Substances (Schools etc.) Exemption Order 1963, The Radioactive Substances (Prepared Uranium and Thorium Compounds) Exemption Order 1962, and others. These exemption orders are conditional and to make use of them and avoid costly registration with the Environment Agency (or SEPA in Scotland, or the Environment and Heritage Service in Northern Ireland) you must adhere to the conditions. Note that currently, these exemption orders are being reviewed.
The way in which these laws are implemented in England, Wales, Northern Ireland and Scotland varies. The Department for Children, Schools and Families (DCSF) has withdrawn its guidance AM 1/92, and the associated regulations requiring this have been repealed. Consequently, purchase of radioactive sources by maintained schools in England is no longer regulated by the DCSF. The DCSF commissioned CLEAPSS to prepare and issue ‘Managing Ionising Radiations and Radioactive Substances in Schools, etc L93’ (September 2008) and has commended it to schools in England. Similar regulations relating to other educational institutions in the UK have not changed; English institutions for further education remain regulated through the Department for Innovation, Universities and Skills. Similarly, schools in Wales should follow the guidance from the Welsh Assembly Government Department for Children, Education, Lifelong Learning and Skills. Schools in Scotland should follow the guidance from the Scottish Government Education Directorate and associated guidance issued by SSERC. Schools in Northern Ireland should follow the guidance from the Department of Education Northern Ireland (DENI). The Crown Dependencies Jersey, Guernsey and Isle of Man are not part of the UK and schools and colleges should follow the guidance from their own internal government departments responsible for education.
In the UK, if an employer carries out a practice with sources of ionising radiations, including work with radionuclides that exceed specified activities (which is 100 kBq for Co-60, and 10 kBq for Sr-90, Ra-226, Th-232, Am-241 and Pu-239), the practice must be regulated according to the IRR99 and the employer must consult with a Radiation Protection Adviser (RPA). Since 2005, the RPA must hold a certificate of competence recognized by the Health and Safety Executive. Education employers are unlikely to have staff with this qualification, so the RPA will usually be an external consultant. Education employers need to notify the HSE 28 days before first starting work with radioactive sources. This is centralized at the HSE’s East Grinstead office.
Note: For higher risk work with radioactive material, the IRR99 requires designated areas, called controlled areas and supervised areas, to be set up if special procedures are needed to restrict significant exposure – special means more than normal laboratory good practice. It should never be necessary for a school to designate an area as controlled, and only in special circumstances would it be necessary to designate an area as supervised. The normal use of school science radioactive sources, including the use of school science half-life sources, does not need a supervised or controlled area.
Disposal of sources in the UK
Sources that become waste because they are no longer in a safe condition, or are no longer working satisfactorily, or are of a type unsuitable for school science, should be disposed of. In England and Wales, the Environment Agency has produced a guidance document through CLEAPSS that explains the available disposal routes. Similarly, SSERC has produced guidance for schools in Scotland. Schools in Northern Ireland should refer to DENI.
Health and safety statement
See the health and safety notes in each experiment. This is general guidance.
Health and safety in school and college science affects all concerned: teachers and technicians, their employers, students, their parents or guardians, and authors and publishers. These guidelines refer to procedures in the UK. If you are working in another country you may need to make alternative provision.
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189703 | https://appliedradiology.com/Articles/bilateral-branchial-cleft-fistulae | Bilateral Branchial Cleft Fistulae | Applied Radiology
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Case StudyVascular Interventional
Bilateral Branchial Cleft Fistulae
Published Date:January 24, 2024
By PJ Preethi Philomina, MBBS; Vishwanath Vijay Joshi, MBBS; Jyotirmay Shyamsundar Hegde, MBBS
Images
Case Summary
A young adult presented with complaint of intermittent discharge from small openings located over both lateral aspects of the lower neck since childhood. On examination, two small openings were noted, one on each side of the lateral aspect of lower third of the neck along the anterior border of sternocleidomastoid, along with mildly inflamed margins and mucopurulent discharge. No opening was found in the oropharyngeal cavity.
Imaging Findings
Intravenous contrast-enhanced CT with additional 10 ml of diluted contrast introduced through both openings revealed bilateral linear fistulous tracts in the neck with the external opening in the skin at the lower one-third of the sternocleidomastoid muscle, at the level of C7-T1 (Figure 1). The tracts coursed cranially along the anterior borders of the sternocleidomastoid muscles, anterior to the carotid spaces with medial bends at the C3 level and passed between the internal and external carotid arteries (Figure 2,3). The internal openings were within the tonsillar fossae of the oropharynx with contrast seen pooling in the oropharyngeal lumen (Figure 4). There was no abscess formation. The patient underwent surgery of complete excision of the tracts (Figure 5).
Diagnosis
Bilateral second branchial cleft fistulae. The differential diagnoses include thyroglossal duct fistulae and acquired fistulae secondary to infection.
Discussion
Mesodermal condensations present in the side wall of the embryological pharynx give rise to branchial arches and their pouches. Second branchial arch and pouch anomalies are common anomalies of the branchial apparatus.1 The anomalies of branchial apparatus were first described by Von Ascheron.2
During embryological development, the branchial arch grows caudally, enveloping the third, fourth, and sixth arches and forming the cervical sinus by fusing with skin caudal to these arches. Normally the edges of cervical sinus fuse and the ectoderm within it regresses. Abnormal persistence of this ectoderm gives rise to a cyst. The fistula results from the breakdown of endoderm internally; most commonly this occurs in the second pouch.1,3
While anomalies of the second branchial cleft account for 90% of all developmental abnormalities of the branchial apparatus, complete second arch fistulae are rare and comprise only 2%.4 Branchial cysts and simple sinus openings ending blindly after a variable distance are more common than branchial fistulae.5
Complete branchial fistulae with an internal opening are very rare. They are usually present at birth with the tiny external opening often going initially unnoticed and commonly presenting in childhood or the second decade.
Intermittent or continuous mucoid discharge and recurrent infection, particularly following an upper respiratory tract infection are common presenting symptoms. Abscess with cellulitis may also be a complication. Owing to variability in the course of the tracts and the need for complete excision, preoperative radiologic visualization of the tract is a prerequisite to successful treatment.
Radiographic fistulograms show a smoothly marginated tract with variable width along the usual anatomic course. Neck CT with administration of contrast through cutaneous openings has been shown to be superior for fistula demonstration, over noncontrast or intravenous contrast-only CT and conventional fistulogram.6,7 However, contrast injected into the fistula may not flow freely when the tract is blocked by secretions or granulation tissue.
The advantages of contrast-enhanced CT with a sinogram or fistulogram over conventional imaging include the ability to reconstruct images in multiple planes, including curved reformations, to offer better delineation of the tract in relation to other neck structures; and accurate depiction of the tract course to help guide the surgeon.
A fistulogram can help confirm the clinical diagnosis, to estimate the length and course of the tract, to look for an associated cyst.
The typical course of a second branchial cleft fistula begins at the external opening in the middle or lower third of the neck along the anterior border of the sternocleidomastoid muscle, runs deep to the platysma across the carotid sheath, taking a medial bend, and passes between the internal and external carotid arteries before finally opening into the tonsillar fossa.9
Surgery is the treatment of choice, as these entities will not regress spontaneously and there is increased incidence of recurrent infections due to the external communication. Imaging assists in surgical planning to facilitate complete excision of the tract to prevent recurrence. Sclerosing agents carry the risk of necrosis and perforation.
Recurrence, secondary infection, hematoma, and injury to traversing nerves or the internal jugular vein and are examples of potential surgical complications.10
Conclusion
Lateral cutaneous openings with discharge should raise suspicion for branchial fistulae and radiologic investigation. A CT fistulogram may be valuable to guide surgery.
References
Ford GR, Balakrishnan A, Evans IN, et al. Branchial cleft and pouch anomalies.J Laryngol Otol. 1992;106:137-143.
De PR, Mikhail T. A combined approach excision of branchial fistula.J Laryngol Otol. 1995;109:999-1000.
Shinde K. Complete second branchial fistula: a study of four cases.Int J Head Neck Surg. 2013; 4 (3):129-132.
Ismail Y, Ozcan C, Nuri O, Fatih B, Beyhan D. Complete fistula of the second branchial cleft: case report of catheter aided total excision.Int J Ped Otorinolaryngol. 2004;68:1109-1113.
Kamal NR, Simi R, Dheeraj P, Joginder SG, Samar Pal Singh Y. Second branchial cleft fistula. Is fistulogram necessary for total excision.Int J Ped Otorinolaryngol. 2006;70:1027-1030.
Burton MG. Second branchial cleft cyst and fistula.Am J Radiol. 1980 May;134:1067-1069.
Ryu CW, Lee JH, Lee HK, Lee DH, Choi CG, Kim SJ. Clinical usefulness of multidetector CT fistulography of branchial cleft fistula.Clin Imaging. 2006; 30(5):339-342. doi: 10.1016/j.clinimag.2006.05.001. PMID: 16919556.
Sun Z, Fu K, Zhang Z, Zhao Y, Ma X. Multidetector computerized tomographic fistulography in the evaluation of congenital branchial cleft fistulae and sinuses.Oral Surg Oral Med Oral Pathol Oral Radiol. 2012;113(5):688-694. doi: 10.1016/j.oooo.2011.08.015. Epub 2012 Apr 12. PMID: 22668628.
Talaat M. Pull-through branchial fistulectomy: Technique for the otolaryngologist.Ann Otol Rhino Laryngol. 1992;101:501-502.
Francisco C, Agaton B, Cosmay GE. Diagnosis and treatment of branchial cleft cysts and fistulae. A retrospective study of 183 patients.Int J Oral Maxillofac Surg. 1996;25: 449-452.
Citation
PJ PP, J VV, J SH.Bilateral Branchial Cleft Fistulae. Appl Radiol.2024;(1):48g-48i.
January 24, 2024
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Divisibility rules/Rule for 11 proof
A number is divisible by 11 if the alternating sum of the digits is divisible by 11.
Proof
An understanding of basic modular arithmetic is necessary for this proof.
Let where the are base-ten numbers. Then
Note that . Thus
This is the alternating sum of the digits of , which is what we wanted.
Here is another way that doesn't require knowledge of modular arithmetic. Suppose we have a 3-digit number that is expressed in the form:
we then can transpose this into:
and that equals:
which equals
Since the first addend, will always be divisible by 11, we just need to make sure that is divisible by 11.
You can use this for any number. Here it is again, with an even-numbered digit number:
So you just need to check for divisibility with 11.
See also
Back to divisibility rules
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189706 | https://www.britannica.com/science/parallel-postulate | parallel postulate
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parallel postulate, One of the five postulates, or axioms, of Euclid underpinning Euclidean geometry. It states that through any given point not on a line there passes exactly one line parallel to that line in the same plane. Unlike Euclid’s other four postulates, it never seemed entirely self-evident, as attested by efforts to prove it through the centuries. The uniqueness of Euclidean geometry, and the absolute identification of mathematics with reality, was broken in the 19th century when Nikolay Lobachevsky and János Bolyai (1802–60) independently discovered that altering the parallel postulate resulted in perfectly consistent non-Euclidean geometries. |
189707 | https://dlnr.hawaii.gov/dar/files/2014/04/TaapeInteractions.pdf | INTERACTIONS OF NONINDIGENOUS BLUELINE SNAPPER (Taape) WITH NATIVE FISHERY SPECIES by James D. Parrish Greta S. Aeby Eric J. Conklin Gayla L. Ivey Brett D. Schumacher Hawaii Cooperative Fishery Research Unit University of Hawaii 2538 The Mall Honolulu, Hawaii 96822 31 August 2000 Final Report submitted to State of Hawaii Department of Land and Natural Resources Division of Aquatic Resources TABLE OF CONTENTS Abstract 1 I. Introduction 2 II. Materials and Methods – General 5 III. Results and Discussion 7 III.1 General 7 III.2 Handline Fishing and Catch Patterns 8 III.3. Frequency of Catch of Species and Co-occurrence in Catch 11 III.4. Distance Caught Above Bottom 15 III.5. Hook Size 17 III.6. Catches of Taape by NMFS Experimental Fishing 18 III.7. Underwater Observations 20 III.8. Diet Interactions 26 IV. Overall Assessment of Interactions 33 Acknowledgments 38 Literature Cited 39 Tables 42 Figures 56 Appendix A – Maps of scientific fishing locations 70 Appendix B – Description of submersible vehicles and equipment 75 ABSTRACT The blueline snapper or taape (Lutjanus kasmira) was intentionally introduced to Hawaii from French Polynesia some four decades ago. It has adapted quickly and successfully to the coastal waters of Hawaii, spread rapidly throughout the archipelago and over a large range of depths, and developed dense populations in some coastal waters of the high islands. Concern has been expressed that it may be producing negative effects on populations of native deep-water snappers or otherwise disrupting the existing valuable commercial/recreational handline fisheries for these species. The purpose of this project was to assess the potential for such negative effects. Fish specimens and quantitative data on characteristics of the fishing process for taape and the native snappers were collected by employing methods essentially similar to standard commercial handline fishing. Catch and effort data from existing NMFS files were extracted. Fish density and habitat use were examined by dives with a manned submersible and remotely operated vehicle, and data were extracted from records of previous dives in the areas of interest. Gut contents of specimens of all target species were examined for evidence of predation or diet overlap among the snappers. Our catch data indicated that taape were caught almost exclusively at significantly shallower depths than all the native snappers except opakapaka, and that taape fed considerably lower in the water column than opakapaka or kalekale (very close to the bottom) and primarily on benthos, while opakapaka and kalekale fed primarily on planktonic prey. The diel feeding schedule of taape was similar only to that of opakapaka. Taape were caught on the same line hauls and stations only with opakapaka. On almost all line hauls where taape were caught, with or without opakapaka, the catch did not saturate the available hooks on the line, and there was no evidence that an overabundance of taape was excluding native snappers by competition for hooks. There was considerable overlap in species caught on a common hook size, but some experimentation with a range of sizes suggested that careful selection of hooks could introduce some selectivity in catch. Underwater observations tended to confirm depth occupied, provided some information on habitats used, and revealed intimate association of taape with other benthic fishes, suggesting that they do not routinely display aggressive behavior or agonism. Diet studies indicated that taape clearly separated trophically from opakapaka and kalekale (feeding above the bottom) and from ehu, onaga and gindai (feeding at the bottom). The diet of taape contained some planktonic crustaceans and some fish, but was dominated by benthic invertebrates. Taape shared some specific fish prey with some native species, and there was tentative evidence of cannibalism and of consumption of three native snapper species (one specimen each) by taape. Fish prey specimens were too infrequent to quantify these trophic interactions. The overall impression is that the introduced taape shows little if any aggression toward native snappers, generally does not share the same depth and feeding habitat with most native species, overlaps little in diet, and is not a frequent predator or prey of the natives. This evidence does not imply strong negative effects of taape on adults of native fishery species in these habitats. It does not address the potential for interactions of taape with young stages of the native snappers or with native species in shallow-water coastal habitats. 1 INTERACTIONS OF NONINDIGENOUS BLUELINE SNAPPER (Taape) WITH NATIVE FISHERY SPECIES I. INTRODUCTION Many nonindigenous finfish species have been introduced, intentionally and accidentally, into fresh and brackish waters in much of the world. In many cases, such introductions have produced unexpected and/or undesirable effects on the native ecosystems, and significant attention has been focused on understanding and minimizing such effects through such management mechanisms as the ICES/EIFAC Code of Practice (Welcomme 1988). Introductions of fishes to marine waters have been less numerous, particularly in the tropics, and much less scientific effort has been applied to understanding the effects of such transfers on the receiving systems. The potential for further marine introductions is high, and there is a clear need to learn from experience about the present effects in order to assess the potential for future ecological hazards. The blueline snapper or taape (Lutjanus kasmira) is a species of marine lutjanid intentionally introduced to Hawaii from French Polynesia some four decades ago (Oda and Parrish 1982; Randall 1987). It has adapted quickly and successfully to the coastal waters of Hawaii, spread rapidly throughout the archipelago and over a large range of depths (to at least a few hundred meters), and developed dense populations in some coastal waters of the high islands. The populations have been exploited for commercial and recreational purposes for decades, but the taape is not a popular fish, and the catches and market value remain low. One incentive for initial introduction of the taape was that the lack of native, shallow-water, demersal snappers in Hawaii seemed to suggest a "vacant niche" that might be filled with minimal disturbance to the community. However, since the rapid and dramatic increase in numbers of taape, concern has been expressed (Tabata 1981; Maciolek 1984; Randall 1987) that it may be producing negative effects on populations of native food fishes or otherwise disrupting the existing fisheries for native species. Concern has been raised both for shallow, inshore waters with their native "reef fisheries" and for the handline fisheries of deeper waters. Little hard scientific evidence has been available to address these questions. The Hawaii Cooperative Fishery Research Unit (HCFRU) conducted a preliminary study on the diet and habitat ecology of the taape and a native soldierfish (menpachi) in waters to about 30 m deep (Oda and Parrish 1982). However, no focused study has been made of the taape in deeper waters farther offshore that would assess its potential interactions with the several species of lutjanid snappers of the subfamily Etelinae (Anderson 1987) that support the valuable Hawaiian deep-water bottomfishery. This group, here referred to as the etelines, includes: Aphareus furca (wahanui), Aphareus rutilans (lehi), Aprion virescens (uku), Etelis carbunculus (ehu), E. coruscans (onaga), Pristipomoides filamentosus (opakapaka), P. sieboldii (kalekale), and P. zonatus (gindai). Some information is available on basic life history characteristics (e.g., growth parameters, reproductive data, size data, and mortality rates) for some of the etelines (Ralston 1981; Everson 1984; Kikkawa 1984; Uchiyama et al. 1984; Uchiyama and Tagami 1984) and for taape (Rangarajan 1971, 1972; Suzuki and 2 Hioki 1979; Mizenko 1984; Morales-Nin and Ralston 1990). However, little is known about the deep-water habitat or how it is used by either the etelines or taape. A single initial study of eteline diets and some general habitat characteristics has been done in Hawaii (Haight et al. 1993), and the meager diet results for elsewhere in the Pacific have been collected recently (Parrish 1987; Seki and Callahan 1988). Diets of taape have been studied in deep-water habitats only in Western Samoa (Mizenko 1984), and there are no studies of the ecology of deep-water taape and the Hawaiian eteline species occurring together. Therefore, with the current knowledge of the species involved, it has not been clear whether the taape produces significant negative effects on the etelines and should be treated as a pest in that fishery. The Hawaiian bottomfishery has substantial economic and recreational value and affects many fishers as well as consumers. The retail market revenue from the 1995 catch of the entire archipelago is estimated at $3,600,000, about 4.6% of the total for all fisheries in the state. Over 750 licensed commercial bottom fishers are involved; the number of recreational fishers is unknown, but is believed to be much greater. The Hawaii Division of Aquatic Resources (HDAR) has recently implemented a new management regime to prevent further overfishing of some of the deep-water Hawaiian eteline stocks in State waters. Since one factor that may affect these populations is the growth of taape stocks and further spread into the habitats of the etelines, it seems important to assess the potential interactions with taape at this time. Major potential interactions between taape and etelines that could affect the fishery include: (1) predation by taape on etelines or vice versa, (2) competition for food between taape and etelines, and (3) co-occurrence and sharing of habitat. Predation by taape tends to reduce populations of etelines. Predation on taape improves nutrition of etelines and tends to increase their populations, while tending to control taape populations. Where there is much overlap between diets of taape and etelines, and limited food supply, competition for food may reduce nutrition and tend to restrain the populations of both groups. Wherever taape and etelines co-occur in a habitat, there is potential for competition for resources of the habitat (e.g. space, shelter, sites used for reproduction or other behavior), either by direct exploitation or behavioral interference with the other species. Such competition may restrain the populations of either or both species through reduced survival of present or potential spawners or reduced reproduction. All the above mechanisms (predation, food competition, habitat competition) may result in exclusion or displacement of taape or etelines from particular patches of habitat where the other group dominates, so that reduction in population occurs locally in ways that may or may not be representative of the population effects overall (e.g. archipelago-wide). The behavioral response to baited hooks by taape and etelines also affects their relative catch by the fishery and perceptions of their relative abundance (because line fishing is the most common way of sampling their populations, particularly in deep water). Experience fishing where taape occur suggests that they take bait very aggressively at times and may be superior competitors for hooks. This behavior may or may not reflect their competition with etelines for natural prey; and the relative catch of taape and etelines at a given location may not reflect their relative abundance there. 3 With funding from HDAR, the HCFRU performed a study between 1 Sep 97 and 31 Aug 00 to examine and evaluate factors in the ecology of taape and the eteline snappers in the fishery and characteristics of the fishing process that would help assess the nature and importance of these interactions between taape and these native species. This study had the following objectives: Assess the magnitude of predation by taape on important native eteline snappers (and vice versa) by quantitative analysis of diet. Assess the potential for competition for food resources between taape and important native eteline snappers by quantitative comparison of diets. Assess the potential for competition for important habitat by sampling the use of habitat and co-occurrence of taape and important native eteline snappers. These objectives were accomplished using the following basic approaches (see Materials and Methods sections for details): (1) Carefully controlled experimental fishing protocols provided fish specimens and detailed information on the fishing process, along with data on catch and effort (including time of day, depth of water, distance caught above bottom, exact geographic position, and co-occurrence with other species). (2) Results of experimental handline fishing by staff of the National Marine Fisheries Service (NMFS) and a fishing vessel chartered by the agency were extracted from their data base for comparison with our field results. (3) Visual and photographic observations by project staff and collaborators from a manned submersible vessel by day and from a remotely operated submersible vehicle by day and night provided sightings and counts of the target species and habitat information in typical snapper habitats of the fishery. (4) Records of similar relevant data from previous dives by other investigators using the manned submersible were examined and combined with results in item (2). (5) Gut contents of specimens collected in item (1) were examined for evidence of predation by any of the species on any of the others and for comparison of diets to assess the potential for competition. Results from all the diverse sources were synthesized to assess the nature and importance of interactions between the target species. 4 II. MATERIALS AND METHODS – General Fish Collection and Handling and Data on the Fishing Process All fish specimens and data on fishing were collected by handlining from small/medium-size vessels on the surface. Fishing was distributed over a wide range of the Main Hawaiian Islands, with considerable effort (especially in shallow depths) concentrated on windward (northeast) and leeward (southwest) sides of Oahu and Molokai (see Appendix A for maps). Since the project objectives were not primarily site specific, locations are reported only occasionally in Results where they may be of particular interest. Handlining methods were designed to be generally similar to those in use in the local commercial fishery, with some modifications required to obtain the detailed data needed and explore the effects of certain variables such as hook size (see report Section III.5). One of the most frequently used vessels and its captain, Mr. Gary Dill, are participants in the regular commercial fishery. Other experienced bottom fishermen participated on some other trips. A small number of different riggings of hooks on the line and combinations of lines were employed (see report Section III.4). All hooks were baited, almost always with cut squid. On some drops, palu (chum) was deployed when the lines were down, to attract fish from a greater distance. All lines were raised with haulers powered by either battery or hydraulics. Lines were lowered over the side of the vessel and the haulers allowed to free-spool until the lead touched bottom. The lines were then pulled up a short distance (see Section III.4) for fishing. Distance to the bottom was checked periodically while fishing the line by lowering to feel bottom with the lead and repositioning the height. In fishing done on trips in Dr. Chris Kelley’s project, two or three lines were fished together on a drop. Each line was fished for ~5 min, or less if a fish took the hook. The rate of hauling the lines was not closely controlled. In fishing done by Dill, either one or two lines were fished together on a drop. The duration of the drop varied widely, and the drop might be done by drifting or at anchor. In this mode, lines were hauled at a rate of about ½ m/sec – a speed that previous experience (Haight et al. 1993) had indicated tended to produce less barotrauma and eversion of the gut, with less attendant loss of gut contents. The following data about the fishing process were recorded thoroughly under Protocol 1 (see Section III.2): At both the beginning and end of each drop (or when each line was hauled): boat location by GPS; time; depth of bottom. At the beginning of each drop: general geographic location; habitat/bottom information (if available); bait used; hook size and type at each position on the lines. 5 At the end of each line haul: time spent hauling to the surface; fate of each hook (present/absent, bait/no bait/bait bitten/fish caught); for each target fish caught, the species and the hook on which caught (non-target species identified and thrown overboard). The following data were recorded for each target fish specimen caught: State of regurgitation (the extent to which the gut was moved forward or everted at the mouth due to barotrauma), recorded as a simple numerical code (condition 1-5). Unique identification code assigned to each specimen when removed from hook, and fish chilled at once. At end of fishing (several hours – two days later), chilled fish processed to include: Length measured to nearest 1 mm and weight by spring scale (typically to about 1%). Complete digestive tract and gills removed by ventral incision, preserved with any loose gut contents Found in mouth or elsewhere. Gonads collected (Protocol 1) or sex and gonad stage recorded (Protocol 2). Collected fish parts and diet items stored at freezing temperatures until analysis. Much of the same information was recorded on Protocol 2 collections, but because of time and logistic constraints of the major mission, the following items were often recorded incompletely or omitted: Regurgitation code; time spent hauling to the surface (rate not closely controlled); fate of hooks; time, depth and position, when fishing was fast-paced. Methods specific to individual parts of this study are reported separately under results for those individual sections. 6 III. RESULTS & DISCUSSION III.1 GENERAL Species Collected Besides the main target eteline species that were sampled in sizable numbers and discussed in the rest of this report, several species of interest were taken and observed in very small total numbers in the complete study. Four specimens of Aprion virescens (uku) were collected. Uku are usually caught shallower than most of this fishing was done, and collecting substantial numbers would have required targeting areas and depths that would probably have been unproductive of the other important etelines (Haight et al. 1993). Catch of uku is typically seasonal and often highly variable from year to year. Aphareus rutilans (lehi) (three specimens collected) provide a small component of the local commercial fishery and are normally caught incidentally while targeting other species. The congeneric Aphareus furca (wahanui) (two specimens caught) also have some commercial value, but often occur shallower than the main target species and contribute little to the total catch of etelines. Six specimens of Pristipomoides auricilla (yellowtail kalekale) were taken in only one drop at Kaula Rock, Niihau. Two Erythrocles scintillans (golden kalekale) were also taken in the entire study. These species have commercial value, but are relatively rare in the local fishery. Directed fishing efforts to obtain samples of useful size of these lesser species would have caused an unacceptable reduction in catch of the main species of interest. These species and other, non-eteline species caught or seen in small numbers (e.g. the deep-water commercial grouper Epinephelus quernus [hapuupuu]) are referred to in this report only incidentally to information on taape and the main target species. For almost all these incidental species, no data were taken other than notation of their occurrence, and specimens were not always retained. Organization of Results The following sections (III.2 – III.8) report results in several major areas of this investigation. Where appropriate, individual sections contain description of methods and/or discussion specific to their particular results. Section IV draws upon all previous sections for an integrated assessment of interactions by the species of interest. 7 III.2 HANDLINE FISHING & CATCH PATTERNS Material and Methods Data on fish catch and effort, details of the fishing process, and specimens for analysis came from two experimental fishing protocols used more or less concurrently (see Section II, Methods). In one of these, data were taken either (1) by our project staff accompanying fishing trips of a deep-water snapper project conducted by Dr. C. Kelley for DLNR, or (2) by an experienced fisherman (Mr. G. Dill) trained and equipped to take the data per our standard protocol on trips that were appropriate to our data needs. By either means, we were able to insure full and consistent data details of the fishing process and specimen collecting, handling and data recording. Such collections are here termed “Protocol 1” collections. Other data and specimens came incidentally from other fishing trips in Dr. Kelley’s project, in which our staff did not participate. That fishing was focused on catching onaga and ehu, and often occurred much deeper than most of the fishing under Protocol 1 (Fig. 1) - below depths where taape would be expected. It was done almost entirely in daylight hours. It was based on much different objectives, e.g., collection of live specimens for experiments and fresh tissues for biochemical analysis. Specimens were available to our project only on an opportunistic basis. These “Protocol 2” trips often provided useful specimens, and some of the basic catch data could usually be incorporated into some of our analyses. For other analyses where data were required on details such as specifics of fishing rigs, time of drops/hauls, exact fishing duration of each drop of each line, and fate of each hook in each drop, only the Protocol 1 data were suitable. The Protocol 2 trips were valuable for expanding the sample size for a number of analyses and for extending the geographic range covered. Reference will be made throughout this report to Protocol 1 and 2 data (also see Section II, Methods). Results Diel Patterns of Catch Fishing under Protocol 1 was done around the clock, with an attempt to give some coverage to all parts of the day (Fig. 2). The patterns of CPUE for the various target species were diverse (Fig. 3 and 4). Taape showed the highest recorded CPUE values (Fig. 4), but CPUE varied widely, with high values between ~1600 and 0200 and very low values between 0400 and 1600. Opakapaka also reached its highest values of CPUE between ~1600 and 0200, with lows during much of the daytime, but also showed a minor peak between ~0600 and 1000. Other eteline species were more erratic (and sample sizes were smaller), but all showed lowest values near midnight (Fig. 3). Depth of Catch The best overall estimate of the depth distribution of these snappers probably comes from combining the data from Protocol 1 and 2. The median depths of capture of all the target species, based on such pooled data, are shown in Fig. 5. Effort was focused shallower in Protocol 1 fishing and deeper in Protocol 2 fishing; this is indicated in the overall vertical distribution of effort shown in Fig. 1. Effort was 8 light shallower than 50 m, and trivial deeper than 350 m, but reasonably strong over the range 50-300 m. Because many of the results in this report are based only on data from Dill’s collections using Protocol 1, it is useful to consider the (rather different) depth distribution of that fishing effort (Fig. 1). A large majority of all fishing in the range 50-150 m (and thus a great deal of the effort in conventional taape depths) was done with this protocol. About half the total fishing at 150-200 m was done this way, and much less of the effort at greater depths. In order to examine CPUE by depth, catches by the two protocols were pooled in much the same way as effort in each depth range, and CPUE’s were calculated from these pooled catches and pooled efforts (Fig. 6). CPUE of taape was much the highest of all species at depths <100 m and declined to zero below 150 m. Opakapaka produced fairly high CPUE, peaking at about 100-150 m and declining rather steeply. Kalekale peaked broadly at 150-200 m with somewhat lower CPUE. Ehu built steadily to the highest peak of all etelines at 200-250 m and then declined rather sharply. Catches of onaga were very low shallower than 200 m and peaked at ~250-300 m. Similar calculations of CPUE using only Dill’s Protocol 1 collections indicated similar trends with depth (Fig. 7 and 8) except that CPUE values for taape in the shallowest depths were even higher, and all CPUE values for eteline species were somewhat lower. Because taape were essentially absent from catches and observations deeper than ~150 m, it seemed useful to compare CPUE of all the target species above and below 150 m. There was some difference in the pattern based on Protocol 1 data only (Fig 9) and the pattern based on all data pooled (Fig. 10). With both data sets, few opakapaka occurred deeper than 150 m, and the other four eteline species were much more abundant deeper than 150 m. Diel changes in patterns of depth of capture are also of interest. Only the data from Dill’s catches covered the hours of the day sufficiently for this breakout, and only taape and opakapaka at shallow depths produced sample sizes marginally large enough to be meaningful for examining 4-hr increments. For taape, the samples permitted division into three depth ranges (Fig. 11); for opakapaka, there was no catch <50 m, and only catches <150 m were used in this analysis for comparison with taape (Fig. 12). For both species, the general diel pattern was broadly similar in these different depth ranges. The very small total effort (particularly at 0-50 m) and the low total catches in the 0000-0400 period made the estimate of CPUE for these hours suspect. CPUE of taape was higher at shallower depths during the night and early morning hours, when nearly all catch occurred. The CPUE for 0000-0400 was actually the same as for 0000-0200, since no catch occurred during 0200-0400 (effort also approached zero). Examining the data for Fig. 11 and 12 using 25-m depth intervals showed little difference in pattern from the 50-m intervals shown here. Discussion Indications of Abundance Since no absolute measure of abundance or density of taape or eteline snappers is available, catch per unit effort (CPUE) has been used here as an index of abundance (more properly, density) to permit comparisons (e.g. Fig. 3, 4, 6, 7, 8, 11). The usual units here are number of individuals caught per hook- 9 hour. The assumption is made that each hook has the same fishing power for each individual of all species in a given situation when the fish are in feeding condition (e.g. active and receptive to prey). Thus CPUE may be considered proportional to the density of fish exposed to the baited hooks at the time. CPUE values for taape in Fig. 3-11 were in the range of ~0-6.5 individuals/hook-hr in this study, and were typically <1.0 averaged over larger ranges of time, depth, etc. For opakapaka, CPUE values were usually less in this study, in the range of 0-3.5 indiv./hook-hr, and typically <0.3 in similar situations. Overall values for these two species are best considered by major depth ranges, particularly since taape occurred only shallower than 150 m in this study. For taape (<150 m), the overall CPUE for the entire study was ~0.345 indiv./hook-hr (Fig 9 and 10). For opakapaka, in this depth range CPUE estimates were ~0.12 and ~0.22 indiv./hook-hr (Fig 9 and 10), and in the deep range the estimate was ~0.035 (Fig 10). In depths where taape occurred, the ratio of these values, CPUE taape/CPUEopakapaka , for the two species is estimated at 2.88-1.57. Little information is available from the literature or other sources to compare with these results. Haight et al. (1993) made a study on Penguin Bank involving experimental fishing somewhat similar to ours in this study for eteline snappers, but taape were not collected and were recorded only incidentally. Based on original data from that study, 21 taape and 232 opakapaka were caught in 2164 hook-hr of overall handlining, with 856 hook-hr at depths of 0-150 m, providing CPUE’s over all depths of 0.0097 indiv./hook-hr and 0.107 indiv./hook-hr for taape and opakapaka respectively: a ratio of 0.090. For the depth range 0-150 m (where nearly all taape seem to occur), 19 taape and 151 opakapaka were caught, and the CPUE values were 0.024 indiv./hook-hr for taape, 0.176 for opakapaka, and 0.136 for the ratio. Based on NMFS experimental fishing on 22 cruises (see Section III.6), catches in the total study included 31 taape and 609 opakapaka caught in 7514 hook-hr of handlining, or 21 taape and 416 opakapaka with 2130 hook-hr at 0-150 m. These numbers would result in CPUE’s overall of 0.0041 indiv./hook-hr for taape and 0.0810 for opakapaka (ratio 0.0506); for 0-150 m, the values would be 0.0099 for taape and 0.1953 for opakapaka (ratio 0.0506). The CPUE values for opakapaka from these three sources show some similarity. Estimates based on Haight et al. (1993) and on the 22 NMFS cruises are reasonably close for opakapaka shallow (~0.18 indiv./hook-hr) and overall (~0.09). Our estimates range above and below these shallow values, and our overall values seem significantly lower; they may have some value as another rough estimate of a point index of density of opakapaka in local waters. For taape, there is great diversity among the CPUE’s from the various sources. The estimate from NMFS is less than half that of Haight et al. (1993) and less than ours by a factor of >30. Effort by Haight et al. (1993) was intended to sample habitats on the “fingers” of Penguin Bank and nearby randomly, without effort to catch or avoid taape. Although the 22 NMFS cruises no doubt had diverse missions and methods, the large number of cruises probably provides some randomization and may leave the overall estimate fairly representative. In our study, since many taape specimens were needed, some fishing areas were selected for the prospects of good catches of taape, and it seems less likely that our CPUE values provide unbiased estimates of abundance of taape. (Such estimates were not an objective of the 10 study.) Comparison of all the above numbers seems to support the overall impression that distribution of taape abundance is very patchy (also see Discussion, Section III.7) Diel Patterns and Depth of Catch Good information is available on diel and depth distribution of catch for the five major eteline species plus uku, based on the experimental fishing on Penguin Bank by Haight et al. (1993). The trends of CPUE with depth are very similar in the two studies, although values for opakapaka and kalekale were higher, relative to the other etelines in the earlier study. The sample sizes of taape in the earlier study were inadequate for making quantitative diel or depth comparisons. The diel CPUE patterns for the etelines in our study show some general similarity to results from the previous Penguin Bank study. Results of the two studies differ considerably in detail for some species. It seems likely that the differences reflect sample size and random variability. One of the major indications from our results is the strong tendency of taape to concentrate feeding at night (although our sampling was much less thorough in the middle hours of daylight [Fig. 2]). This research is the first on diel activity of taape in Hawaii. The results are consistent with unpublished taape data from the earlier Penguin Bank project in which catch times of 19 taape specimens taken on handlines were recorded. The full range of times was 0955–2345, with 12 specimens collected during evening twilight to dark (1900-1942 in winter), six specimens late at night (2108-2345), and one in full daylight. This pattern is consistent with limited evidence on the feeding schedule of taape in shallow water (Oda and Parrish 1982). III.3 FREQUENCY OF CATCH OF SPECIES AND CO-OCCURRENCE IN CATCH Materials and Methods Records were kept of every individual line hauled and of the catch (number of individuals of each species) on each line, as well as the length of time the line was fished (a measure of effort). These data permitted determining the frequency of catch of each species in relation to the total number of line hauls or various subsets of line hauls, as well as the frequency of co-occurrence of a species on the lines with other species. Using the single line haul as the unit of sampling gave the most unequivocal measure and evidence of co-occurrence of fishes at a particular place and time. However, the size of the sample was severely limited (typically 4-8 hooks), and from an ecological and habitat perspective, co-occurrence of fishes over a somewhat larger space and time frame is certainly of interest. For this kind of sampling, such a more extended sample can best be taken by pooling lines that were closely adjacent in time and space. The most obvious and probably most reliable pooling would be of lines within the same “drop”, i.e. lines deployed and hauled from the same boat at the same time and place. Our data contained some multi-line drops (usually only two or three lines), but many of the drops used only one line. Occasionally two or more drops were made in quick succession at very nearly the same location (e.g. within a few tens of meters based on GPS readings), especially when the boat was anchored. In such cases, when the measured depths were very nearly the same and all other information suggested that fishing was in the same habitat and easily within the immediate range of the same group of fish, the catches from these 11 drops were pooled. These cases of pooled line hauls are referred to as fishing stations to distinguish them from individual line hauls (including individual hauls within the station). Results Of a total of 2823 line hauls in the complete study, 192 (<7%) caught taape. Taape were caught on 17.2% of all lines that caught fish, and on 24.4% of all lines that caught snappers (Table 1). Taape were never caught with the three deepest-living common native snappers: ehu, onaga and gindai. They were caught alone on 153 line hauls, on one line haul with kalekale, and 21 line hauls (10.9%) with opakapaka, as well as 17 line hauls with miscellaneous non-eteline species (~20 total individuals, four caught along with taape and opakapaka) (Table 2). The eteline snappers in the fishery were also caught on lines alone or in a variety of combinations with other species (Table 3). Ehu were caught on line hauls with all the other main commercial bottomfish species (including the grouper hapuupuu) except the shallow-living uku. Onaga were caught with three other etelines, kalekale with four others and the hapuupuu, and opakapaka with three others (one haul each). Gindai were caught with two other etelines and hapuupuu. Some combinations occurred on a good many line hauls (12-14). The shallow uku did not occur on hauls with other etelines. Taape and all the etelines except kalekale also occurred with a variety of non-snapper species, either caught with other snappers or alone. The “Miscellaneous only” column in Table 3 includes only cases where the only co-occurring fishes were non-snappers. For each species (row) in Table 3, percentages are of the figure in the “Total” column. Thus, they indicate what percent of all the hauls for the species are shared with each other species or group, or are not shared. Except for the shallow uku (with only four specimens), the range of shared hauls over the other six snapper species was about 19-26%. If interactions with the non-target species (in the “Miscellaneous only” column) are pooled with the “Alone” column, so that only co-occurrences of target species are involved, the range of shared hauls across the six species is ~12-25%, and taape shows the lowest co-occurrence. All together, ~27.8% of all hauls caught snapper, and ~ 70.4% of all hauls that caught fish caught snapper. Co-occurrence of target species was examined in a similar way using the “station” (consisting of more than one line haul at a site) as the sampling unit (see “Methods” above). The results appear in Table 4. Because of the difficulty in assigning many line hauls to stations with Dill’s data, this analysis included only the data from all Kelley’s collections. The patterns are generally rather similar. Overall, more interactions occurred among target species as the station took a larger sample at a site. For a few species pairs, the number of interactions decreased. A striking case was taape, which continued to occur only with kalekale (once) and with opakapaka. The 21 line hauls with only taape and opakapaka grouped into only nine stations. Two new species pair combinations occurred with low frequency: gindai with onaga and opakapaka. The net increase in co-occurrence among species resulted in somewhat lower percentages of occurrences of all species “Alone” at a station (Table 4), and in a wider range of values. Thus, the range of shared stations over the six snapper species was ~26-46%. For taape it was ~44%. If interactions with the non-target (“Miscellaneous only”) species are lumped with the “Alone” column so 12 that only co-occurrences of target species are involved, the range of co-occurrence across the six target species is ~20-44%, and taape is ~40%. For the above analysis by “stations”, a subset (Kelley’s collections) of the full data set was used that represented a relatively small amount of sampling at depths where taape are usually found, but a relatively large portion of the sampling at greater depths where most of the eteline species are taken. That subset of data was therefore divided into a block of all samples shallower than 150 m and a block deeper than 150 m. The shallow block included all the stations with taape present and most of the stations with opakapaka present. It seemed useful to examine the degree of co-occurrence of these two species within a depth habitat that they share, with the results “undiluted” by interactions with species that are relatively rarely caught in that habitat. Table 5 shows these results only. (The sample sizes for other etelines in this data block Kelley’s data <150 m deep+ were probably too small to be representative.) The entries for taape are identical with Table 4. For opakapaka, the total number of interactions is less; it occurs only with kalekale (once), at nine stations with taape (~16% of its total occurrences), and at 12 stations with miscellaneous non-target species (about 21.1% of occurrences). Interactions represent ~39% of occurrences at stations for opakapaka and about 44% for taape. If interactions with non-target species are pooled with the “Alone” category (as above), interactions occur in only ~17.5% of stations for opakapaka and ~40% for taape. To round out these comparisons of co-occurrence, the original data set with species grouped by line haul (as in Table 3) was examined again, but only the samples shallower than 150 m. The results appear in Table 6. For taape, these results were identical to the first analysis (see Table 3) because all taape collected occurred at depths <150 m. For opakapaka, the results were similar to the first analysis, including the frequency of interactions between taape and opakapaka and overall interactions of both species separately. Again, the sample sizes for the other etelines at these shallow depths were probably inadequate for interpretation. A potential negative impact of taape on the fishery for eteline snappers would occur if taape were sufficiently abundant and aggressive toward baited hooks in areas where etelines were being targeted that the taape saturated the hooks and reduced the catch of the desired etelines. Our data were examined for any evidence of such hook saturation. Because the data indicated virtually no co-occurrence of taape in catches of any species other than opakapaka, this analysis was limited to occurrence of taape and opakapaka. Of 153 taape and 148 opakapaka that were caught on line hauls without any other species, the breakdown of numbers caught and occurrence of remaining intact baited hooks appears in Table 7. These results indicate that in about 90% of all cases when a single fish of either species was caught on our lines, at least one intact baited hook remained (and on the average, 3.5 such hooks remained). Similarly, when 2 fish of the same species were caught together, at least one intact baited hook remained in 73% of the cases for taape and 57% for opakapaka (2.7 hooks and 2.1 hooks respectively). For larger multiple catches, hooks still remain, but the sample size may be too small for the numbers to be meaningful. Since for both species, only 9-12% of all hauls occurred with species other than taape 13 (Table 3), the results in Table 7 are also fairly representative of all hauls other than those with taape and opakapaka co-occurring. Since catches involving taape and opakapaka together are likely to be of greatest interest with regard to hook saturation, all such line hauls were identified and examined. As Table 3 indicates, 21 line hauls contained taape and opakapaka together; 14 of these hauls caught multiple taape and three hauls caught multiple opakapaka (Table 8). The three leftmost columns in each row of Table 8 together define a particular multi-species combination that occurred. The rightmost column indicates how many total line hauls produced this combination. The remaining columns indicate how many of these hauls ended with 1, 2, 3, 4, or 0 intact baited hooks remaining. Since the original number of hooks was not the same for every line haul, numbers in parentheses indicate the original number of baited hooks. In a few cases, competition by other species (all soldierfish/menpachi in these drops) reduced the number of intact hooks, and each of these effects is shown. The data set is small and there are no consistent overall trends. However, it is clear that in our collections, most hauls with both taape and opakapaka were not heavily overloaded with taape, and intact hooks usually remained except with the largest catches. Discussion Most of the 64 incidents of co-occurrence of target species on a line haul involved ehu, the species collected in largest numbers. Ehu were caught most often with gindai, kalekale and onaga in decreasing order of frequency; opakapaka and taape were much more abundant in the catch but co-occurred with ehu once and never, respectively. This may be largely a result of the relatively shallower distribution of opakapaka and taape. The co-occurrence of ehu with the deep-water benthic gindai and onaga seems reasonable; the high co-occurrence with the less bottom-oriented and more planktivorous kalekale seems less intuitive. The two shallowest snappers, opakapaka and taape, co-occurred most often with each other; for both species, about 11% of all catches were co-occurrences. These percentages were somewhat lower than the percentages of gindai and kalekale that co-occurred with ehu. It is not clear whether grouping species occurrence by line haul or by station is more enlightening. Relatively high percentages of the onaga, kalekale and gindai on line hauls co-occurred with ehu, and these percentages increased when co-occurrence was assessed on a station basis. Gindai also co-occurred rather strongly with kalekale when compared by line haul and especially by station. Grouping by station brought out its co-occurrence with onaga and opakapaka. Kalekale and onaga co-occurred by both ways of grouping, more strongly by station. Co-occurrence of miscellaneous non-target species was not especially high or very different by the two ways of grouping except for opakapaka (where grouping increased its percentage considerably) and taape (where the number and percentage of co-occurrences both decreased substantially). Both taape and opakapaka, and possibly a mix of non-target species, were probably the groups most strongly affected in the change from the full data set for line haul analysis to the partial data set (Kelley only) for station analysis. Some of these frequencies of co-occurrence seem rather high, e.g. ~11-19% by line haul and ~32-36% by station. The single shallow-water relationship – taape and opakapaka – was among the higher 14 frequencies by both groupings, and these species show strong ecological differences (e.g. diet, position in the water column, overall range of depth). Both species showed relatively low co-occurrence with any other species. In fact, given the full depth ranges of opakapaka and kalekale and their ecological similarities (e.g. diet, position in the water column), their relatively low co-occurrence is a little surprising. The overall co-occurrence of opakapaka with all species is among the lowest of all target species. Overall frequencies for taape cover a rather wide range (11.5-44%) depending on the way of grouping and on whether miscellaneous non-target species are pooled. The other strongest co-occurrences are found among deep-water pairs, e.g. ehu and gindai (~19-34% by the two ways of grouping), ehu and onaga (~10-32% by the two methods). These pairs seem to show strong similarities in ecological characteristics such as those mentioned above. It is not entirely clear what factors led to these patterns of co-occurrence. They may be affected by subtleties of habitat that we did not detect. The implications of the results relative to hook saturation will be discussed in the overall assessment in Section IV. III.4 DISTANCE CAUGHT ABOVE BOTTOM Material and Methods The depth at which fish encountered the bait was of interest for several reasons. For each drop made under Protocol 1, records were kept of the distance that the lead was above the bottom when fish struck, and of which hook(s) in the vertical array on that line caught fish. The distance above the lead was known for each hook. So the distance above the bottom at which each fish caught had struck the bait could be closely estimated. Since a greater variety of distances above bottom were fished in Dill’s fishing, those results are treated separately from results of fishing by Kelley under Protocol 1. In Dill’s fishing mode, there were typically 5-6 hooks in the vertical array spaced 1-2 m apart. The lead was fished at or just above the bottom while drifting and fished at the bottom or various known distances above the bottom while fishing at anchor. Kelley’s fishing mode was used primarily in considerably greater depths of water than Dill’s. The vertical array always consisted of one hook at each of four levels 2 m apart (the lowest hook just above the substrate), plus two hooks placed within the same general depth range for a few drops. This arrangement was less suitable for detecting fish higher above the bottom. Results Dill’s fishing Dill’s rigging and its temporal deployment resulted in the vertical distribution of fishing effort shown in Fig. 13. For taape, the sample size was large enough to estimate vertical location in 1-m increments (Fig. 14). Clearly, CPUE was much greater within about 5 m above bottom and declined sharply for hooks higher on the line or when the entire string of hooks was higher above the bottom. CPUE was very low more than 9 m above the bottom. The greatest distance above bottom where catch was reported was 12-14 m. 15 For other snapper species, sample sizes were generally adequate to estimate vertical location in 2-m increments. For ehu, onaga and gindai, catches appeared concentrated very close to the bottom (Fig. 15), with CPUE generally greatly reduced above ~4 m, although occasional catches occurred as high as 16 m in the water column. The distribution was considerably different for opakapaka and kalekale (Fig. 16). Catches of both species were rather widely distributed, relatively small near the bottom, and greatest at ~6-10 m up. For opakapaka, the peak was somewhat more pronounced. Kelley’s fishing Using this vertical array of hooks and fishing always very near the bottom resulted in the vertical distribution of fishing effort shown in Fig. 17. The vertical distribution of CPUE for six snapper species based on this fishing mode is shown in Fig. 18 and 19. At these water depths, the sample size of taape was small (44 total individuals), and for all species, distance above bottom is shown only in 2-m increments. Despite the small sample size, the pattern for taape was similar to that in Dill’s (shallower) fishing, with CPUE declining sharply more than 6 m above bottom (Fig. 18). CPUE for onaga and gindai also seemed to decline more or less continuously with increasing height above the bottom. (The apparent rise at 8-10 m may be an artifact of small absolute amount of catch [2-3 specimens] and effort.) For ehu, (Fig. 19), the steep decline in CPUE between 0-2 m and 6-8 m is consistent with results of Dill’s fishing. The subsequent increase in CPUE between 8 and 12 m is unexplained; although effort was too small to make accurate estimates, the raw catch numbers were high, and the trend must be real for this data set. When data from Dill’s and Kelley’s fishing are pooled, this increase in CPUE higher above the bottom is insignificant. For opakapaka and kalekale (Fig. 18), the trends were generally similar to those with Dill’s data. CPUE for opakapaka (Fig. 19) was low near the bottom, and increased from the 6-8 m level to the maximum height where data were usable (i.e. at least to 10-12 m). Kalekale (Fig. 11) showed an initial decrease in CPUE to ~2-4 m, then increased beginning at ~6-8 m to the maximum height where data were usable (i.e. at least 8-10 m). The height of maximum CPUE could not be determined for either species, but must be at least as high as that indicated by Dill’s data. 16 III.5 HOOK SIZE Materials and Methods One question of interest was the relative effectiveness of various hook sizes and types for taape and the various eteline snappers. A limited investigation of this question was made in fishing by Dill with Protocol 1. The hooks tested are shown in Table 9. All are standard commercial hooks. Several are commonly used in the commercial handline bottom fishery in Hawaii, and these were used most frequently in this study. However, other hooks less commonly used commercially were also tried opportunistically and less often. Results Figure 20 indicates the amount of fishing effort with each size class of hook (Table 9). All fishing with size classes 2, 3 and 6 and almost all fishing with size class 1 occurred at depths <150 m (depths where taape were caught) (Fig. 21). Hook size classes 5, 7, 8 and 9 were fished over a wide range of depths (Fig. 22), covering the habitat range of taape and of the deepest etelines. Although the effort with some size classes was too small to provide reliable results (e.g. classes 2, 4 and 9), it is clear that some sizes of hook were more productive for some species than others (Fig. 23). Size class 3 (mtc 6/0) was highly effective for taape only. Class 5 was also highly effective for taape and for opakapaka and was the hook used most often (Fig. 20). Classes 6, 7 and 8 were more productive for the other eteline snappers, and fairly good for opakapaka, but CPUE of taape reduced rapidly with hooks progressively larger than size 5. The apparent high CPUE of kalekale with the Class 6 hook, shown in Fig. 23, is somewhat suspect; the total effort involved was very small, applied at one location over a period of ~3 hr. 17 III.6 CATCHES OF TAAPE BY NMFS EXPERIMENTAL FISHING Materials and Methods The Honolulu Laboratory of the National Marine Fisheries Service has done considerable experimental handline fishing over a period of several years. This fishing may have had a variety of objectives related to commercial bottomfish, but the taape was probably never targeted, and the data may not have been examined from the perspective of the present study. The Honolulu Laboratory provided data from two different sources for our examination: (1) fishing by the laboratory staff, primarily from the NOAA vessel assigned to the laboratory, and (2) fishing by a commercial fisherman chartered by the laboratory. The laboratory’s standard catch data bank provided fishing data for 22 research cruises by NMFS staff in the Main Hawaiian Islands, carried out between June 1983 and September 1993. Collectively these cruises applied a total fishing effort of 7514 hook-hours, typically in the form of four lines fished simultaneously with four hooks per line. Data available for each commercial species and some incidental species included date, start and end time and start and end depth and location for each drop or drift, the specific fishing rig used, fishing effort (convertible to hook-hr), and for each fish caught, the species and size, depth and time caught. From these data, we calculated catch in number of individuals, corresponding effort, and CPUE for taape and each of the major eteline species. Detailed data were extracted on each taape caught and each eteline specimen caught on the same drop with a taape. The charter fishing was performed using the vessel Kaimi in October 1982 in the waters around Niihau and Penguin Bank. All fishing was done drifting using three lines, each with four Mustad #5 hooks. The data taken were much the same as collected in fishing by NMFS staff (above). From the Kaimi data, we calculated catch in number of individuals, corresponding effort, and CPUE for each of the major eteline species. Results from these two NMFS sources provided a supplement to our own fishing data and that from Kelley’s project and covered an earlier time period. Results NMFS Laboratory Fishing Of the 22 cruises from which data were used, 20 provided data on catches of snapper. Five of these included catches of taape. A total of 31 taape individuals was reported as part of a total catch of 1931 fish (see Table 10), i.e. ~1.6% of the total. The catch of taape at depths <150 m was 21 individuals from a total catch of 725 fish (~2.4%). However, the data for taape are not entirely credible. The six specimens reported caught at 150-200 m, the two specimens at 200-250 m, and the specimen reported at 300-350 m are possible but highly unusual compared to data from other sources. Catches of opakapaka at depths <150 m were high, and catches of gindai, ehu and kalekale were substantial (Table 10, Fig. 24). Effort in these depths was not high except in the 100-150 m range (Fig. 25). The result was a higher dominance of CPUE by opakapaka in this shallow range, with the other three etelines lower and similar (Fig. 26). CPUE of taape was relatively higher compared to all these etelines than in comparisons of catch alone, but except in the 0-50 m depth range, taape CPUE was low (e.g. <6% of opakapaka CPUE). By a depth of 150 18 m, CPUE of taape was near zero and remained so at greater depths. Effort peaked by 200 m as CPUE of opakapaka declined and CPUE of gindai, kalekale and ehu increased, then declined until CPUE of all species was near zero by depths of 350 m. The cruise data contain almost no information on habitat of capture. As with other handline data, aggregation of individuals and association between species can only be inferred by proximity of catches, so the data were examined for groupings of catches. On one evening drift of about 20 min, two taape and one opakapaka were caught on the same line at 73 m depth at 1955, and 5 min later 2 more taape were caught at a depth of 26 m on the same line (usually inferred from data on time of catch). On a particularly long drift (>2 hr), 8 taape were reported caught at depths between 150 and 203 m. Apparently all were on separate lines, with catches spaced apart from 1 min to 1½ h. In this drift, one opakapaka was caught shortly after a taape catch (but separately) at 207 m depth, and two opakapaka were caught on the same line with one of the eight taape just at the end of the drift at 2115 at a depth of 150 m. Several of these reported depths are surprisingly great. In two other cases, two taape were caught very close together on the same drift, once apparently on the same line (at ~1500 at depth 84 m), and once two catches within a minute (at 2130 at depth 95 m). In one other case, three taape were reported on the same line (at 2035 at depth 44 m). Another case of closely spaced catches of one taape and one opakapaka (within 3 min) occurred at 2345 at depth 157 m. Two taape were reported taken at 2100 on the same line at 57 m depth, and nine minutes later, an opakapaka was taken at the same depth. Five taape were taken in individual drifts with no other taape or etelines reported. Charter Fishing The total fishing effort of the Kaimi charter was ~109 hook-hours, distributed as shown in Table 11. Effort was focused in the depth range of ~100-150 m, a range in which taape are often encountered. The commercial species caught were ehu, gindai, opakapaka, kalekale, and the grouper hapuupuu. Total CPUE was much higher between 100 and 200 m (Fig. 27). Surprisingly, hapuupuu showed the highest values of CPUE, but only around 100 m depth. No taape were caught in this effort of 109 hook-hours applied strongly at depths where catch rates of taape in our recent project were sizable. 19 III.7 UNDERWATER OBSERVATIONS Materials and Methods Direct observations of taape and their habitat, with incidental observations of eteline snappers and other fish species present, were made on two cruises during the periods 16 Aug-7 Sep 98 and 15-27 Sep 99 using submersible vehicles of the Hawaii Undersea Research Laboratory (HURL), Pisces V and RCV-150, and the surface support vessel Kaimikai-o-Kanaloa. Pisces V is a manned submersible carrying a pilot and 2 observers, each observing the surroundings through an individual view port using external floodlights as needed at depth and recording data using still flash camera, digital video, and conventional video with simultaneous recording of observers’ voices. Time, depth and position of the submersible (based on GPS positions for the tending vessel and sonar fixes of the submersible’s position relative to the vessel) were available continuously or at frequent intervals. (See Appendix B for description, specifications and capabilities of Pisces V.) All diving with Pisces V was done in collaboration with other projects studying deep-water snappers. These projects were concerned with specific locations related to areas closed to fishing and corresponding control areas, or they required observations at greater depths than taape usually occur. However, taape were a target species of opportunity in all Pisces V dives, and all sightings were recorded with camera images and commentary. All Pisces V dives were made during daylight hours (~0800-1700). The RCV-150 is a remotely operated vehicle (ROV), connected to the support vessel by an umbilical cable with fiber optics for data signal transmission. Its attitude in the water and small-scale movements are adjusted by thrusters controlled from the tending vessel. It carries floodlights for operation at depth and provides signals of depth and video images, with simultaneous voice recording by an observer in the control room. Position can be constantly estimated from GPS position of the tending vessel, providing a record of the vehicle’s track with position, time and depth. (See Appendix B for description, specifications and capabilities of the RCV-150.) Diving with the RCV-150 was almost entirely devoted to the taape project, and except where noted in Results, was all done during hours of darkness. The general localities observed were mostly near those where Pisces V submersible diving was done, and were somewhat constrained by distance feasible to travel from the locations of daytime Pisces V dives. However, RCV-150 dives were concentrated in shallower waters to include the depths where taape occur most commonly. The entire period when the vehicle was submerged was recorded with videotape and commentary, including all sightings of taape and eteline snappers, with data on their depth, habitat and behavior. The RCV-150 and Pisces V have almost no capability for measuring objects in the water, and estimates of size and distance were based mostly on comparison with other objects in the field of view and (for fish) on general familiarity with the species morphology. Reports of observations of taape were also retrieved from submersible dives made previous to this project. The data archives at HURL were scanned to identify those dives most likely to provide accurate 20 sightings of taape. Dives were screened on the basis of location and depth, mission objectives, and observers’ backgrounds, familiarity with taape and eteline snappers, and level of professional interest in these fish that would have led to reporting such sightings. A relatively small fraction of all previous dives met these criteria; all these dives employed the 2-person submersible Makalii, predecessor of Pisces V. Makalii provided generally similar capabilities for observation and identification of fish in the depths of interest. Project staff examined the full transcript of the audiotapes and the logs of still photos from the most promising dives (made in 1981-1987) found in the archives. The data of interest for taape are reported in Results. Results Project Dives (1998-1999) On the 1998 cruise, the submersible Pisces V made 19 dives for a total of ~123 hours of underwater observation. Taape were observed on only one dive at ~0900 31 Aug 98 at a depth of ~110 m. A school of 100 or more ~15 cm long were seen on the southern face near the east end of the “Third Finger” of Penguin Bank (off the southwest tip of Molokai), a little below the top of the feature. The habitat consisted of a rocky sloping wall with a dusting of sand. Several balistids were sighted in the area at the same time. On this 1998 cruise, a total of 22 dives (37.7 hr) of underwater observation were made with the RCV-150: 5 dives (4.8 hr) at depths <160 m and 21 dives (32.9 hr) at depths >160m. No taape was ever seen or caught using any equipment at depths >160m in the entire study. In 1998, the only sightings of taape from the ROV were also on “Third Finger” of Penguin Bank, on a track run diagonally across the top of the feature near the middle at ~2015 on 29 Aug 98. The substrate at a depth of 96 m was mostly flat, hard sand with small depressions (<1 m across) scattered rather closely and containing knobby rock/rubble encrusted with algae and sessile invertebrates, with perhaps 15 cm of relief. One taape was motionless under a small ledge, and two were seen immediately afterwards resting on bottom against a rock face. Several “reef fish” species were seen relatively motionless close to the taape, including two Heniochus diphreutes, one Chaetodon fremblii, and two Myripristis chryseres – the latter sheltering within centimeters of the two taape against the same rock. In all the 1998 ROV diving, no eteline snapper was seen at depths <160 m. Deeper than 160 m, three total individuals of ehu (Etelis carbunculus) were seen on two dives at depths of 199 and 300 m, and one onaga was seen at 200 m (see Table 12). The only sightings of taape from Pisces V on the 1999 cruise were in the same general area on “Third Finger” of Penguin Bank near the east end of the southern face, near the shoulder of the steep drop from the top of the feature at ~110 m depth. The first sighting, at ~0900 15 Sep 99, was of one taape ~18 cm long, near the shoulder, with a relatively flat sandy top and a slope with about half the surface made of exposed rock, with ~15-30 cm relief. Other species seen in the same area at the time were a school of ~25 Naso sp. 30-45 cm long, Chaetodon modestus, C. miliaris, Desmoholacanthus arcuatus, and possibly the snapper Aprion virescens (which usually occurs at shallower depths). The other Pisces V sighting in 1999 was made close to this location five days later at ~1500 at a depth of ~115 m. Two taape 21 were seen swimming with Symphysanodon typus about ½ m above the bottom in a habitat that was ~95% rock with relief to ~1/2 m, and many ledges and caves. Other species seen in the same immediate area at the time were the deep-living snapper Aphareus furca (and possibly also opakapaka), the jacks Caranx melampygus and Seriola dumerili, Naso sp., Chromis struhsakeri, Holanthias fuscipinnis, Pseudanthias fucinus, and Myripristis chryseres. In total, 3 taape individuals were sighted in only two of the 13 Pisces V dives in 1999 and only in the 110-115 m depth range. The available observation time for these dives was 72 hr; only 3 hr were spent at depths shallower than 160 m. Over the same observation periods, many more sightings were made of native eteline snappers (Table 13). All except 23 of the estimated number of fish observed were seen at depths >160 m, probably reflecting a preference of most of these species for deeper habitat, as well as the much greater observation time >160 m. Opakapaka and kalekale are generally considered to have the shallowest distributions of these five eteline species, and only these two species were reported at depths <160 m. The low number of total sightings and individuals of opakapaka was somewhat unexpected and may reflect a behavioral avoidance of the submersible as well as the higher position above the bottom that this species is believed to occupy commonly (see Section III.4). On this 1999 cruise, a total of 31 dives (38 hr) of underwater observation were made with the ROV: 25 dives (21.1 hr) at depths < 160 m and 24 dives (16.9 hr) at depths >160 m. Many more sightings of all species occurred during these dives in 1999 than in the 1998 dives (see Table 14). Of these, all taape and opakapaka and over half the kalekale were seen shallower than 160 m; all except two ehu and all the onaga were seen at greater depths (~165-325 m). Taape were seen from the ROV in 1999 in six general areas: east tip of “Third Finger” (Penguin Bank), west end of “Third Finger”, between “Second Finger” and “Third Finger” on the edge of the Bank, top of “Second Finger”, southernmost tip of Penguin Bank, and an area northeast of Mokapu Point on the windward coast of Oahu. Considerable ROV surveying was also done in an area on Oahu east of the Mokulua Islands (mostly in somewhat deeper water), but no taape was seen. A track was run 17 Sep 99 ~0200 northeasterly across a small pinnacle situated just east of the tip of “Third Finger” (Penguin Bank). Near the base, at 153 m depth, two taape ~15 cm and 22 cm long rested motionless ~30 cm apart on a flat sand bottom, with occasional rocks, but no substantial shelter or bottom relief. No other species was seen. On 18 Sep 99 at ~2000, a track was run northeasterly from the northwest corner where “Third Finger” joins Penguin Bank, roughly at the top of the feature along the edge of the Bank. Two taape ~25 cm or smaller were seen at ~100m depth at the edge of a steep drop from the edge of the feature in a rocky habitat with relief of ~1/3-1 m. They were under a ledge, very close to (possibly touching) a goatfish and two unidentified fishes. Three taape (lengths ~15, 22, 22 cm) were seen later at a depth of 71 m in an area of mostly sandy bottom with rocky outcrops and ~5-10 cm relief (no obvious ledges or caves). On 15 Sep 99, between ~1900 and 2300, two tracks were run end-to-end in a generally easterly direction along a portion of the rim near the top of the southward facing slope of Penguin Bank roughly halfway between the bases of “Third Finger” and “Second Finger”. A total of eight taape were seen in six 22 separate sightings; one individual was estimated at about 30 cm long and the others at <25 cm long. All were seen between 91 and 94 m. depth in broadly similar habitat with mostly hard substrate and a dusting of sand cover. Much of the substrate was rocky, often with knobby rock encrusted with benthic invertebrates and algae and several centimeters of relief. Occasionally boulders up to 2-3 m in size were present. Three taape (three sightings) were within several centimeters of the bottom and very close to small ledges that could provide some shelter. Two taape (two sightings) were on the bottom very close to small caves or holes in rocks. Two taape (two sightings) were swimming within several centimeters of the bottom in similar habitat, but not near any features that appeared to offer shelter. At two sightings, a pair of taape were 2-3 m apart. At one sighting, 2 Myripristis chryseres and one Sargocentron or Neoniphon species sheltered under the same ledge, and at another sighting, an Acanthurus dussumieri swam near a taape at the same ledge. At another, a M. chryseres was under a small ledge just below a taape. At another, Neoniphon aurolineolatus was close to a potential shelter hole in the ~1-m high feature that a taape swam close over; a Dardanus brachiops was also nearby. At another sighting, an Arothron hispidus swam in the water column ~2 m from a taape at the opening of a cave, and two Myripristis chryseres sheltered 6-7 m away in another hole. On 19 Sep 99 between about 0100 and 0200, a track was run starting on the top of “Second Finger” near the west end and running northeast diagonally across the top of the feature and part way down over the north facing slope. The substrate and benthic habitat were much like that described above for the 15 Sep 99 dive. Five taape were observed in 3 sightings within a period of about 7 min on or near the top of the feature at depths of 107-113 m. Sizes were probably in the middle of the range of others reported here. One taape was sighted along with three Myripristis chryseres in a cave with an opening of about 30 cm in an area of knobby, encrusted rocky bottom with up to ~30 cm relief. Three other taape were seen in similar habitat, swimming together several centimeters above the bottom, next to a ledge about 30 cm high. A single taape was reported “low down in algae” on the bottom. Ten minutes after the last taape sighting, an ehu was sighted at a depth of 160 m. Two dives were made at the southernmost tip of Penguin Bank on 17-18 Sep 99. A dive from ~2200 17 Sep through ~0100 18 Sep explored the southeastern corner of this tip on tracks that ran roughly northeasterly, sometimes above or near the top of the steep slope of the bank and sometimes on or far down that slope. Two onaga were seen far down the slope (210 m), but all the taape were seen at the top of the slope on a relatively flat, sand/rubble substrate that in most places provided little relief or cover. Seven taape were observed in five separate sightings over about 1½ hr. One taape was resting on the bottom in an area with no relief or cover; another was resting on bottom in a rubble patch with a few centimeters of relief but no real shelter. One taape swam a few centimeters above the bottom close to one of several scattered carbonate ledges that occasionally emerged half a meter or more from the sand bottom. Three taape were seen together a few centimeters above the substrate in one of several scattered rocky patches, each patch ~2 m in radius, that provided ~15 cm of relief; this patch also contained (in close proximity) Desmoholacanthus arcuatus and Myripristis sp. The other dive on this southernmost tip of Penguin Bank was at the southwestern corner, where tracks were run in a northerly direction up the slope and for some distance across the relatively flat top of the bank. Taape were seen only on top of the bank at depths of 45-50 m on substrate that was basically rock 23 and hard sand, much of it with knobby cobbles encrusted with sessile invertebrates and algae, providing several centimeters of relief. Five taape were observed in 3 separate sightings within about 6 min on a dive between ~2000 and 2200 17 Sep 99. Two sightings were of individual taape resting openly in this knobby bottom, not sheltering. In one of these cases, an acanthurid and four Myripristis species were sheltering under a small ledge near the taape. At the other sighting, three taape swam just above the bottom, meters apart, with no recognizable shelter in sight. At Mokapu Point on the windward coast of Oahu, taape were seen on three of the nine ROV dives made. A dive was made at dusk (~1730-1830) 24 Sep 99 in depths a little over 100 m over a hard sand slope. At one location at a depth of 113 m, emergent rock created a deeply undercut ledge with a vertical opening of ~2 m. A large group of taape, estimated to number ~250 fish, schooled near the ledge. Large schools of the small deep-water snapper, Symphysanodon typus were also present nearby. Some 13 min later on the same track, a single kalekale was seen at a depth of 122 m. In a daylight dive the same afternoon, a track was run between ~1600 and 1700 at slightly shallower depths nearby, over the same kind of hard sand with little relief. At 105 m and 103 m depths, rather large groups of opakapaka were sighted about 3 min apart. Later in the track, a series of large boulders (some 3 m or more in height) was encountered at depths of 99-101 m. The boulders provided high relief, with caves, cracks, crevices and ledges for shelter. Within a period of 4 min, three apparently separate groups of taape were sighted with numbers estimated at 27, 15 and 100. They schooled at the edge of the group of boulders and among them. Throughout much of this boulder area, large clouds of Luzonichthys sp. were also present. A school of 30 Naso sp. was present ~5 m from where the first taape school was encountered, as well as a school of carangids. Chaetodon miliaris was also common throughout the immediate vicinity. Heniochus sp. and some wrasses were also present. On 23 Sep 99 at about 0230-0400, a night dive was made nearby, beginning in what seemed a generally similar habitat (i.e. hard, relatively flat sand with little relief). During the track, the topography became more rough, uneven and rocky, and taape and etelines were sighted: three taape at a depth of 107 m, one opakapaka each at depths of 90 and 121 m, one kalekale at 107 m, and one ehu each at 200 and 228 m. The taape and kalekale were seen at the same depth and within 1 min; one of the opakapaka was seen 2 min later and 14 m deeper. The above account includes all sightings of taape in both years with both underwater vehicles. Previous Dives (from HURL archives) All the most promising records found were from dives on Penguin Bank by the submersible Makalii. Useful data were examined from a total of 51 dives (Table 15); 29 dives were classified as deep (significant time at depths >160 m), and 23 dives were classified as shallow (significant time at depth <160 m). No sighting of taape was reported >160 m. Twenty-one of the 29 deep dives (72%) reported sightings of etelines. Of the shallow dives, 19 of 23 (83%) reported etelines and seven (30%) reported taape. In only one dive, taape were explicitly reported once in the same sighting with an eteline species (uku). From the 7 dives with taape observed, a total of 19 individual reports of taape occurred. Each may have represented a separate sighting, but the information on location and time suggests (but does not 24 confirm) that some may have occurred at essentially the same place at very nearly the same time, and may have been multiple counts of some of the same individuals. Records from this group of 23 shallow dives contained 80 individual reports of sightings of snappers (including taape and etelines). The fraction of sightings of taape (19/80 = 24%) is similar to the fraction of all shallow dives (7/23 = 30%) that provided sightings of taape. Of the dives on which taape were reported, four dives (including 13 sightings) were made on the extensive, relatively flat, sandy top of Penguin Bank at a few sites where artificial reefs had been placed within the previous few years. A few of the sightings could have occurred on the open sand between these reefs, but the available records and discussions with scientists involved in these dives strongly suggest that all or nearly all these taape were in fact seen at the artificial reefs. All these dives were made at depths of 58-61 m and in full daytime between 1025 and 1224. As suggested above, the relevant substrate and habitat are probably the reef materials and structure. For example, at one sighting, the taape (reported as ~6 individuals) were described as hiding in holes in the pipes of the reef structure. That was the only sighting for that dive and the only sighting where even an approximate count was available. Another dive in this series reported four sightings; based on time and depth, at least two were very likely the same reef and group of fish. Apparently at least two other artificial reefs were visited and taape reported in that dive. Another dive provided taape sightings from at least two (probably three) reefs. At one sighting they were described as “several good size”. The last dive apparently made sightings on at least four reefs, without record of numbers or sizes of taape or habitat. The remaining three dives (including 6 sightings) retrieved from the archives were made at two or more general locations away from the artificial reefs. These dives were apparently independent of the artificial reef program and of each other. These sightings were also made in late morning (1102-1200). One dive on the “First Finger” of Penguin Bank at a depth of 76 m sighted a “big school” of taape <10 in. long over rubble and sand bottom. “Lots of fish” of other species were reported present, including Chromis sp., Pseudanthias sp. and other anthiids. Opakapaka and/or uku may also have been seen elsewhere on this dive. Another dive on Penguin Bank, probably near the west end, at 107 m depth reported a “few” taape over hard bottom. The third dive also occurred near the west end of Penguin Bank at depths of 107-110 m and produced four sightings, each with “many” taape. The first sighting also reported the co-occurrence with 2 uku individuals, without further details. Many opakapaka and uku were also reported elsewhere on this dive (not with taape present). Depth and time data suggest that at least the last three of the reported sightings (and very possibly all) from this dive may have been of the same school of taape. If the total number of taape sightings is adjusted for the probable duplications (multiple reportings of the same individuals) suggested above, the actual total is probably about 15 or 16 sightings. This amounts to an average of a little over two sightings of taape on each of the seven dives where they were sighted, i.e. two sightings over the course of a dive of several hours at depths of 58-107 m. Based on all dives within the “shallow” range, this amounts to roughly 0.65-0.70 sightings per dive of several hours. 25 Discussion The greater number of sightings/effort with the ROV in 1999 than in 1998 was due partly to improvement in technology and increased operating experience with the vehicle (first put into service in 1998). Results may also have been affected by differences in areas surveyed. Both cruises involved major effort on the “fingers” of Penguin Bank (Molokai) and at stations east of the Mokulua Islands on the windward coast of Oahu. However, the 1999 cruise included dives northeast of Mokapu Point on windward Oahu. These included five daylight dives with 2 hr 10 min total at depths <160 m and 2 hr 50 min total at depths >160 m. A few of the dives in this area encountered some rather large schools of taape (estimated at 20-250 individuals) that accounted for most of the individuals counted in 1999 surveys. The tracks in the Mokapu area ran through some terrain with high relief that seemed to be prime habitat, and in fact was obviously used as shelter by many fish. However, other areas surveyed (e.g. the Penguin Bank “fingers”) also included high relief substrate that appeared to be very attractive fish habitat, and a school of ~100 taape was sighted in this area also. Large schools may be infrequent enough that this limited sampling cannot give a useful estimate of their frequency. What is known about the distribution of taape suggests that they are extremely patchy. For example, in a large series of shallow-water visual censuses in Hanalei Bay, Kauai, taape was the second most abundant species by number of individuals and biomass, but occurred in only 22% of all censuses (Friedlander et al. 1997). III.8 DIET INTERACTIONS Materials and Methods For fish specimens collected under Protocol 1 and many of those collected under Protocol 2, the complete alimentary tract plus any associated food material found was preserved under refrigeration beginning very shortly after collection (see general “Materials and Methods” section). For most of this time, this material was fully frozen until analysis in the laboratory. Guts were thawed and dissected to remove the contents. Procedures for analysis of the diets were generally similar to those described by Hyslop (1980), Harrison et al. (1983), Parrish et al. (1985), and Haight et al. 1993). Material from the gills, stomach, pyloric caeca and the entire intestinal tract was rinsed in 8% saline solution and examined under a dissecting microscope. Potentially recognizable items were removed, cleaned, described, counted and measured as appropriate (using an ocular micrometer where needed), and stored in labeled vials. Fish otoliths were stored dry, cephalopod beaks were stored in 70% ethanol, and the remainder of the items were stored in 10% formalin solution. Prey items were visually identified under magnification to the lowest feasible taxon, using an extensive taxonomic literature and reference specimens of potential prey items or their parts. In particular, fish otoliths, scales and other fish parts were compared to reference specimens of local deep-water species and related fishes obtained by our own fishing and by loan from colleagues, museums and other collections. Whole prey organisms and parts were counted and used to estimate the total number of prey individuals represented by the contents of each gut. 26 Diet composition was expressed in terms of number of prey items (individuals) and frequency of occurrence of predator guts containing the prey category. Also, an index of prey importance (IPI) for each prey category in the diet was calculated as: IPI = F x N where: F = percentage of all predator individuals that contained that prey category, and N = percentage of total prey individuals that were of that category . These variables (F, N, and IPI) were compared for taape and the eteline species as a means of assessing the similarities and differences of their diets. These diet comparisons were also quantified by doing pairwise analyses using the following measures of overlap (Krebs 1999): (a) Pianka’s measure, Ojk : (1) (b) Percentage overlap, Pjk: (2) (c) Simplified Morisita-Horn index, CH: (3) (d) Horn’s index, Ro : (4) where: pij = proportion that prey category i is of the total prey eaten by species j. 27 pik = proportion that prey category i is of the total prey eaten by species k. n = total number of prey categories . These overlap analyses (1) through (4) were implemented using the software package Ecological Methodology, Version 5 (for IBM PC running Windows) by Exeter Software, 47 Route 25A, Suite 2, Setauket, NY 11733. Results For the purposes of this study, trophic analysis is focused on the diet of taape and the diets of the eteline snappers in the fishery. These diets are characterized and compared in an attempt to estimate the nature and general magnitude of ecological effects of any of these species on any other. For comparisons and analyses of fish diets, prey items could be pooled at a wide variety of levels into systematic and/or ecological groups. Pooling was done at several levels that seemed potentially interesting, and those that seemed to produce best insight into questions of diet interactions between taape and the eteline species are presented and discussed here. Table 16 provides the most complete and detailed form of diet data presented. It contains much information on specific prey types for each of the above target predators and is organized to facilitate comparisons on the basis of frequency of occurrence in predator diet, numbers of prey individuals, or IPI. For each predator species, the value of %N at each level in Table 16 is the actual number of prey individuals of that level (category) as a percentage of all prey individuals (i.e. N in the IPI equation in the Methods section). The value of %F at any level is simply the number of individuals of the predator that ate prey of that category as a percentage of the number of predator individuals examined (i.e. F in Methods). Sample sizes of all species except onaga appear to be adequate for broad comparisons of the overall diet, and their results in Table 16 are generally credible (see Discussion section). The small sample size of onaga appears to have been inadequate and produced results that are believed to be non- representative of the population; our onaga data were replaced for subsequent analysis (e.g. Table 17 and 18) with the data from Haight et al. (1993). For more rigorous and quantitative comparisons of the diets of these species, attention focuses on specific levels of aggregation (pooling) of the prey. In Table 17, prey are aggregated into 12 groups at very high systematic/ecological levels (not necessarily closely related to traditional “trophic” levels). These are the levels indicated in bold print in Table 16. The general pattern for the six species is conspicuous in Table 17, particularly with %IPI. Diets of opakapaka and kalekale were heavily dominated by planktonic animals, especially crustaceans, but also molluscs and urochordates. Some fish were eaten by most individuals, but they were quantitatively much less important. In contrast, ehu, gindai and onaga (the latter based on data from Haight et al 1993) were heavily dependent on fish as prey, in terms of large numbers of piscivorous individuals and heavy consumption (%IPI). Each of the three species had a second, much less important food source in benthic crustaceans, in terms of %F and %IPI, and a third minor but significant source was cephalopods. Planktonic urochordates were of some importance to onaga, and pelagic crustaceans were significant for gindai. Taape were more or less intermediate 28 between these planktivorous and piscivorous groups, overlapping most of these major categories to some extent, but they were clearly most dependent on benthic invertebrates, primarily crustaceans and mollusks. Many taape ate fish and a good many ate planktonic crustaceans, but the contributions of both categories to the total diet were much less. Quantitatively, pairwise comparisons between diet compositions of predator species by four common overlap indices (see Methods section) produced the index values of Table 18 for %IPI. (Computations using %F and %N produced generally similar patterns.) The same trends appear with all four indices in Table 18. Absolute values are not especially meaningful or directly comparable across the four indices. It is clear that taape has very low and similar overlaps with opakapaka and kalekale, and has moderate and similar overlaps with ehu, onaga and gindai. Opakapaka and kalekale show very high overlap with each other and low overlaps with the other four species (highest with gindai). Ehu, onaga and gindai show high overlaps with each other. These overall results are consistent with the results for specific groups discussed in the previous and following paragraphs. The data permit comparisons for specific diet overlap at some lower systematic levels (Table 16). Large benthic crustaceans are clearly the dominant category for taape. The most important benthic crustacean category for taape was crabs, which are a much smaller component of the benthic crustacean diet and the total diet of all the etelines. The same is true for the anomuran and brachyuran divisions within the crab category. Benthic mollusks were very significant in the diet of taape and trivial in the diets of all the etelines. This was true of both gastropods and bivalves. Taape also ate planktonic molluscs (mostly pteropods), including several of the same species as opakapaka and kalekale, but much less abundantly (more nearly as the other three etelines did). All the snappers ate pelagic urochordates, including larvaceans and especially thaliaceans. It seems that these groups may be fairly important prey for opakapaka , kalekale, and even onaga, but they were relatively minor groups for taape. All the snappers ate some planktonic crustaceans, but the quantities were trivial for ehu and onaga (when an adequate sample is used as in Haight et al. 1993). For taape, this group was somewhat more important, but a clearly minor component by comparison. Gindai in our study ate many more pelagic crustaceans than the Etelis species and taape, and consumption by the opakapaka and kalekale was much greater. The taape diet seemed to contain several of the common planktonic crustacean subgroups, and no major qualitative separations are obvious between the fish predators’ diets at these levels. Cephalopods were a minor component of the taape diet. They were trivial in the diets of opakapaka and kalekale, somewhat more important for onaga, and provided several percent of the diet for gindai and ehu. Based on this comparison, taape could hardly compete strongly for the available cephalopod resources. Of the six squid species identified, taape ate three species (probably four individuals); one of these species was also eaten by ehu and opakapaka. These limited data at the species level do not suggest much commonality in cephalopod diets of these predators. Further species identifications may be feasible and may provide a clearer pattern. 29 The second most important prey category for taape was fish. Although fish were clearly more important to ehu, onaga, and gindai by all available measures, it seems useful to compare diets of the snappers at the lowest systematic levels feasible. Table 19 shows all fish species that were at least tentatively identified in the diets. A total of 93 fish prey individuals are included (of 446 prey individuals with possible potential for identification). Of immediate qualitative interest is occurrence of any of the snappers in the gut of any other. Only in guts of taape were tentative identifications of snapper prey made, based on scales found in the guts that appear similar to scale reference material for snappers (4 prey species, 6 predator individuals, 6 prey individuals, each prey identified based on a single scale). Three of the prey individuals (scales) appeared most like taape, the other three most like wahanui, lehi and opakapaka (one each). Six of the 30 identified individuals in this sample of taape guts (and six of the 93 identified individuals in this sample of all snapper guts) may have been snappers eaten by taape. Based on the tentative prey identifications to date, diet overlap on fish at these low systematic levels appears between: Taape and ehu - 2 prey species, once each Taape and gindai - 2 prey species/genera, once each Taape and opakapaka - 1 prey species, once Ehu and gindai - 1 prey species, several times Opakapaka and kalekale - 1 prey family, once In this sample of 93 prey individuals, of 31 taxa identified, 7 are shared between some pair of snapper predators and 24 are not. Discussion Although the sample sizes for diets are not large for some species, they are probably at least minimally adequate for high-level comparisons for all species except onaga. Certainly the sample would be expected to give representative results for taape, the species of greatest interest and the one for which least data existed in these habitats previously. For opakapaka (arguably the next most critical species in this study), the sample is clearly adequate, and it seems to be so for ehu and kalekale. For these last three species, the diet compositions are credible and compare favorably with results from reasonable sample sizes in the trophic study on Penguin Bank by Haight et al. (1993). The sample size for gindai is less impressive, but the results are in good agreement with the only six specimens reported in Hawaii previously (Haight et al. 1993) and with a more extensive sample in the Mariana Islands (Seki and Callahan 1988). The sample of 9 specimens of onaga in the present study is clearly inadequate, particularly when no good measure of food bulk consumed is available, and the results (as presented in Table 16) are inconsistent with the results of Haight et al. (1993, based on a much larger sample) and with our total knowledge of ecology of onaga (Parrish 1987). Because these results are believed to be non-representative, data for onaga from Haight et al. (1993) are used in subsequent reporting (Table 17 30 and 18). For the other strictly deep-water species (etelines), the diet results in this study are generally consistent with the existing (limited) published information. For taape, these results are concordant with available information on its diet in shallow waters (Rangarajan 1972b, Oda and Parrish 1982, Parrish 1987, J.D. Parrish, unpublished data). For evaluating the importance of predation or the co-occurrence of particular prey taxa (relative to the potential for food competition) at low systematic levels, much larger sample sizes are typically required. This is partly the result of the relatively low yield of identifiable prey organisms (or parts) and difficulty of identification at low systematic levels. For some important questions, e.g. quantitative co-occurrence of prey species or comparison of predation among two or more predators, sample sizes of the predator species should be approximately balanced where feasible as well as large. Neither the size nor the balance required was feasible with the resources of this study, and its objectives were limited to detecting co-occurrence qualitatively and obtaining a general idea of the magnitude of predation on key species. Comparison of numbers in Table 19 among the snappers is not meaningful unless considered in relation to the very different sample sizes of the snapper species involved. Clearly if 180 specimens of taape are examined, occurrence of a specimen of prey species x (or any arbitrary number of species x) is more likely than if 32 specimens of kalekale are examined, even if the actual rate of consumption of prey species x is the same by both predator species. Comparing variables such as the number of individual predators with a particular prey and the number of individuals of a particular prey eaten are probably better compared on a “per predator sampled” basis (e.g. dividing the variable values by the sample sizes in parentheses in Table 19. It is less clear how variables such as the number of prey taxa can be normalized since (like the “species area curve”) this relationship is unknown but probably nonlinear. When these variables in Table 19 are compared on a “per predator sampled” basis, the order of the resulting numbers for these predator species is much the same as the order of importance of fish as a whole in their diets. Kalekale is the exception (with normalized numbers placing it in the order above taape and similar to onaga). Notably, kalekale has a rather low sample size (32). Clearly none of the normalized variable values for taape are out of line (high) relative to the other predator species These considerations apply directly to the issue of co-occurrence. With the raw numbers in Table 19, it appears that taape share considerably more prey taxa with other snappers than do the other etelines. The number of prey taxa adjusted to a “per predator sampled” basis for the five snappers with usable data are: ehu = 5/42 = 0.119, taape = 19/180 = 0.106, opakapaka = 0.075, kalekale = 0.187, gindai = 0.130. On this basis, taape would take the second lowest number of prey taxa and almost certainly would share fewer prey taxa with other snappers. (As noted above, this particular adjustment is probably unrealistic because the relationship is probably nonlinear, but the exercise demonstrates that the co-occurrence of taape is exaggerated by its large sample size.) The key point is that these data give little reason to believe that taape is likely to share more prey species (with other snappers) than any other snapper, and that (large) samples of comparable size of all the snapper species would show this directly. Obtaining a large sample size of taape was a priority because (1) almost nothing was known about its diet in this environment and (2) any information about shared prey taxa (and potential for competition) or about prey taxa of particular interest (such as eteline snappers) was viewed as important. Now that 31 some level of prey species/genus overlap has been shown by this (feasible) unbalanced sampling program, it is important to bear the limitations of this result in mind and not make interpretations beyond what the data support. These considerations are also highly relevant to the finding of some scales in a few taape that resemble snapper scales and may represent as many as four species of interest. If taken at face value, this finding is a first and is certainly qualitatively interesting. If it can be fully confirmed, it will be meaningful (although not particularly surprising) to know that there is strong physical evidence that taape do sometimes eat other snappers. These few scales are a very recent find, coming after examining many scales from many fish. The sample size of scales identified as snappers is very small, the basis for comparison not optimum, and further confirmation (either way) seems likely to be feasible eventually. There may also be opportunity to examine other specimens. An important characteristic of this particular study is that relatively little fish prey material has been found in really good condition, and a large fraction of identifications have been based on otoliths and scales. These parts can be very useful, but especially with fish species as poorly known as most of these prey, better prey specimens would lead to higher confidence in identifications. Interpretation of the taape scales in the taape predator is particularly troublesome (although it would not be surprising to find that the taape is cannibalistic). There are a number of ways in which a diet sample can become contaminated by scales from the predator. Although our staff was aware of the potential problems and took precautions, a finding of scales of the predator’s own species tends to undermine confidence in identifications. In particular, where regurgitation is a problem (as in this study) and every effort is made to recover food items that may be found outside the fish, the risk of contamination with the predator’s scales is probably higher. Examination of 180 taape specimens has produced six scales in six fish of four species that may be useful in establishing the qualitative fact of these predatory interactions for taape. In accordance with the discussion above, since less than a third of this sample size has been examined for any of the other five snapper species, the fact that no snapper material has been identified in their guts probably gives little indication about whether any of these species consume snappers. The particular interactions that these scales imply are interesting for several reasons. Certainly our results suggest that sizable populations of opakapaka occupy more or less the same habitat as taape routinely as adults. Adult opakapaka are too large to be prey for taape. Provided that appropriate sizes of (younger) opakapaka are present where adult taape are present, such predation seems feasible. There may be no evidence of such sharing of habitat (see Section IV, final segment). Wahanui seem to frequent depths appropriate for interacting with taape as adults, although they are not known to be especially abundant in the habitats we studied. Probably little or nothing is known about their juvenile habitats (at sizes when they would seem to be vulnerable). Adult lehi are taken mostly at considerable depths and are not commonly caught or observed from submersibles. Opportunities for interaction between adult taape and adult lehi would seem minimal because of the difference in depth of habitat and probable low abundance of lehi. Again, knowledge of juvenile lehi habitat must be very scarce. Although the habitat of juvenile taape in these areas has probably never been studied, experience with taape elsewhere suggests that juveniles can often be found in the same general areas as adults. There are no known barriers against cannibalism. 32 IV. OVERALL ASSESSMENT OF INTERACTIONS All relevant factors in the preceding results were considered together to reach an overall evaluation of the interactions between taape and the native eteline species in terms of what interactions occur and the probable general magnitude of their effects. If these effects can be rigorously demonstrated quantitatively, it would almost certainly require a very large-scale and intensive experimental program that is far beyond the scope of the present study. The conclusions reached here are based on trends and extrapolation from limited data, but the accumulation of evidence from a variety of sources is convincing. Abundance and Depth Range of Taape Indications of the abundance and depth distribution of taape over a rather wide range of locations in the Main Hawaiian Islands came from (1) underwater observations (several hundred hours) in our project and previous submersible missions in the archives of HURL, and (2) handline fishing (>9000 hook-hr) in our project (and Kelley’s collaborating project) and previous cruises by NMFS vessels and a NMFS charter (see Sections III.2, III.3, III.6, III.7). These observations and collections involved sufficient effort distributed over a wide enough range of area, depth, season and time of day to provide a reasonable picture of the depth range occupied routinely by adults of the target species. In all the handlining performed or reviewed, only 11 taape were reported taken at depths greater than ~150 m. (None were reported deeper in fishing or submersible observations in this project). Although taape have occasionally been reported anecdotally at greater depths, it seems clear that in these waters, the adult taape population occurs almost entirely at depths <150 m. Our methods did not permit any direct estimates of abundance or density of any fish species. As is commonly assumed in fisheries biology, catch or sightings per unit effort may provide a reasonable index of density for some purposes, particularly for comparison between species in a location/habitat. Fig. 9 and 10 show these indices from handlining for each target species, and Fig. 5 shows the median depth of catch for each. It is clear that the abundance of all the major etelines in the fishery except opakapaka (and uku) is concentrated at depths greater than where taape are common. Of the others, only kalekale and ehu were taken in any quantity shallower than 150 m (Fig. 6, 9 and 10), and they were caught in much greater numbers at greater depths. On the other hand, a very large fraction of the opakapaka occur in depths <150 m. The relative CPUE values in these figures suggest that in depths <150 m, taape may be about 1½ to 3 times as abundant as opakapaka – the only eteline with which it overlaps significantly in depth. Underwater observations from submersible vehicles (e.g. Table 12, 13 and 14) are not easily reconciled with these catch results nor among themselves; i.e. the results from Pisces V in these depths (only ~3 hr of observations) in 1999 are not very comparable with the results from the ROV in 1998, and both are considerably different from the results of the ROV in 1999. The main conclusion from Table 12 and 13 would be that taape are seldom seen from these vehicles. The ROV results in 1999 (Table 14) may be the best controlled for comparisons, and they also suggest that taape are at least several times as abundant as opakapaka at depths <150 m. (For example, the number of sightings of taape from the ROV in 1999 was about three times the number of opakapaka sightings.) However, all results confirm that the spatial distribution of taape is patchy, varying between widely dispersed individuals and occasional widely scattered groups with numbers from dozens to hundreds. In 33 summary, adult taape occur almost entirely at depths <150 m, where they co-occur primarily with adult opakapaka, and seem to be considerably more abundant than opakapaka in this core depth range of both species. Activity/Feeding/Catch with Depth and Time Catch data (Section III.2, Fig. 3 and 4) indicate that the diel pattern of CPUE is considerably different among the target species, implying that their diel cycles of feeding – and probably activity generally – are different. Although the results for the etelines are somewhat different from those based on larger samples in earlier work (Haight et al. 1993), it appears that taape and opakapaka both tend to be highly active in the late hours of the night and much less so during much of the daytime. In our data, the other etelines tend to have a generally opposite trend, with low activity late at night and higher CPUE between early morning and late evening. This low overlap in activity period between these etelines and taape could tend to reduce interactions between these species, e.g. competition (especially interference competition) for food. However, there appears to be rather strong temporal overlap with the only eteline species that also shows high overlap in general depth range, i.e. opakapaka (Fig. 4). Interestingly (although sample sizes are somewhat small in some time and depth groups), at the extreme hours in the middle of the night when the activity of both species seems to increase greatly, the increase for opakapaka is primarily in the depth range of 100-150 m (Fig. 12), and for taape it is primarily in the depth range of 0-50 m (Fig. 11). This may suggest that these two species seek different habitats at key times of intense foraging. In any case, it implies that the most intense feeding by these two species does not occur at the same time and depth. Visual underwater observations provided little evidence regarding the diel cycle of activity and foraging (Section III.7). Very few sightings of taape were made from the submersible (all in daylight), and almost all sightings from the ROV were made at night. The daytime sightings of taape showed cases of schooling in sizable numbers and cases of one or two individuals moving together. The several nighttime observations and one dive at dusk showed taape alone or in groups of 2-3, sometimes swimming in the open near the bottom, often resting motionless on the bottom or under ledges or other cover. Individual observations were very brief; no behavior could be characterized as either torpor (sleep) or active foraging. Our results about distance above the bottom at which the various species feed (Section III.4, Fig. 13 – Fig. 19 ) may be relevant to the degree of interaction between the target species, particularly when they are feeding. These results clearly showed higher CPUE (thus probably higher feeding activity) for taape, gindai and the Etelis species in the 3-5 meters immediately above the substrate; in contrast, they indicated an increase in CPUE for the other two Pristipomoides species, especially opakapaka, beginning several meters above the substrate and extending to at least 10 m above it. This strongly suggests a vertical partitioning of the feeding habitat, and probably of the available assortment of prey types. (See following section on “Trophic Interactions”). The two species that show strongest overlap in depth of habitat – taape and opakapaka – show a large displacement in their distance of feeding above the bottom. 34 Co-occurrence of Taape with other Species Co-occurrence relates to issues of interaction among species in a number of ways. One issue is whether one species is sufficiently aggressive, dangerous or disagreeable to neighboring species that they avoid it, i.e. give it a wide berth and are not found together with it in close proximity or in a resting situation. This kind of avoidance might be characteristic of behavior toward a generalized piscivorous predator. Such behavior toward taape might suggest that it has some sort of negative effect on other fishes. Our only source of evidence regarding this possibility is underwater observations (Section III.7, text description of observations). Although such observations were not very numerous, taape were usually seen either schooling rather calmly (often relatively close to a variety of fish species), or as individuals moving slowly or resting on the bottom, sometimes quite close to other demersal fishes. These observations of co-occurrence with other species gave no evidence of interactions or effects (positive or negative) between taape and other species, but they can be interpreted as limited behavioral evidence against agonistic relations. We made essentially no observations of taape and etelines remaining close together. The closest spacings were one nighttime observation of three taape and then of one kalekale, all at depths of ~107 m, within 1 minute, followed by one opakapaka 2 min later and 14 m deeper. All other sightings of taape were considerably more widely spaced from eteline sightings. The fairly frequent observations on some dives and frequent collection of some of the eteline species without taape present suggest that taape are neither extremely abundant nor obligate users of the same habitat or set of resources. If taape co-occurred almost always with an eteline species or were very abundant and highly competitive for bait on handlines, a likely result would be catch of taape on a very large fraction of all fishing drops that caught the eteline species, and perhaps many more taape than etelines on these drops. Such co-occurrence was not observed in this study. The results of analyzing all lines hauled for co-occurrence (Section III.3), showed no case where a taape was caught on the same line as ehu, onaga, gindai, or uku (Table 1, 3 and 4). (This must be at least partly because taape seem to be almost absent from the main depth ranges of these species.) A taape was caught only once on a line that caught kalekale (1.1% of all lines with kalekale), and taape were caught on 21 lines (10.9%) that caught opakapaka. Twelve cases (line hauls) were found where two or more taape were caught on a line with one opakapaka, as well as two cases where two or more taape were caught with two opakapaka. There were seven cases of one taape with one or two opakapaka (Table 2). Usually four or more hooks were fished on a drop, and there were few if any cases where all hooks on a line were filled with opakapaka and taape. On most line hauls, even with multiple taape and/or opakapaka caught, intact baited hooks remained on the line (Table 7 and 8). On the basis of co-occurrence on “stations”, taape co-occurred with opakapaka in 36% of the stations where it occurred and in ~12% of the stations where opakapaka occurred. This incidence of co-occurrence of taape on lines and at stations with the etelines and multiple occurrences on a line does not indicate a great abundance of taape where the etelines were fished or extreme bait competition. Considering co-occurrence both at the level of the line haul and of the “station”, with and without including co-occurrence with non-target species, the overall frequencies for taape (with all other species combined) were in the range of 11.5%-44%. Compared with frequencies of the main eteline target species, taape was within the overall range, from near the low end to near the high end (Section III.3). Therefore, based on data from handline catches, particularly 35 when grouped by line haul, taape may be no more “interactive” with other species on lines than are the main eteline target species. The degree of co-occurrence of different species on handlines may be influenced by the vulnerability of the various species to the hook, its presentation, and the bait. The number and location of hooks was largely standardized in this study, and cut squid was used as bait almost uniformly. Palu was used occasionally and not confined to any particular depth range, habitat or situation. Experience fishing for all the target species indicates that squid is an effective bait (e.g. Haight et al. 1993), and all the species seem to respond positively to use of palu. It does not seem likely that the type of bait caused a major bias for or against any of the target species. Results of Section III.5 indicate that hook size (and possibly design) can create bias in the catch, and that careful selection of hooks can reduce the incidental catch of taape. However, the above results on co-occurrence of the main target species on lines are based heavily on the use of hooks that were productive for both etelines and taape, so the occurrence of taape in these catches is not less than would be expected in commercial fishing. Trophic Interactions The most obvious kind of interaction to be considered is predation by any of the target species on any other. Our findings about the diet composition of taape (Table 16, 17, and 19) indicate that the species is significantly piscivorous (although apparently less strongly than onaga, ehu and gindai), but that benthic invertebrates are the dominant prey, and pelagic invertebrates are of some importance. Therefore, predation by taape on etelines small enough to be consumed is possible, although a wide range of alternative fish and invertebrate prey is certainly consumed. As mentioned above, the strong tendency toward separation of habitat depth between several of the species reduces the opportunity for realized diet overlap and potential food competition. Opakapaka was the only eteline that seemed to overlap strongly in depth with taape in this study. Differences in height of feeding above the bottom may also reduce the realized diet overlap even for species that occur in the same general depth habitat (e.g. taape and opakapaka). Results of diet studies in this project (Section III.8) indicated some qualitative differences and large quantitative differences between the diets of some of the target species. Diets of opakapaka and kalekale were dominated by plankton and were rather similar, but they were clearly different enough from the more demersal species – ehu, gindai and taape - to make serious diet competition unlikely. They were also markedly different from the heavily piscivorous diet of onaga in a previous study with an adequate sample size (Haight 1993). The diet of taape, heavily dominated by benthic invertebrates and to a lesser extent by fish, was radically different from that of the opakapaka with which it overlaps heavily in depth distribution, and also from that of the kalekale with which it overlaps slightly in depth distribution. These diet differences are clearly demonstrated quantitatively by the small overlap indices of taape with opakapaka and kalekale (Table 18). Diet overlap indices with taape are moderate and similar for gindai (the largest), ehu, and onaga (using onaga data from the adequate sample of Haight et al. ). Pairwise examination of diets for shared use of particular prey items produced diverse results. Overall, for invertebrate prey groups at intermediate systematic levels, taape did not appear to overlap strongly 36 with other snappers for groups that were important in its diet. Identification of invertebrates at the lowest systematic levels generally did not permit comparison. Some squid could be identified to low systematic levels, but cephalopods were a minor diet item for taape. Relatively few fish (~21%) could be identified to low systematic levels (as usual). Of those, diet overlap among the six snappers was found for taape with ehu, gindai and opakapaka (one or two prey species or genera, once each), for opakapaka with kalekale (one prey family, once), and for ehu with gindai (one prey species, several times). The relative incidence, numbers, etc. among the snapper species have little meaning because of widely different predator sample sizes. For a total sample of prey composed of 30 identified low level taxa, the degree of overlap of the snappers as a whole seemed moderate. The finding, in the guts of six taape, of six fish scales that look like snapper scales (possibly taape, wahanui, lehi and opakapaka) may turn out to provide the first solid evidence of predation by taape on snappers. The most salient points regarding this result are: (1) the evidence is strongly suggestive but wants confirmation; (2) if confirmed, it is the first known evidence of these trophic interactions, but not inconsistent with what is known generally about feeding of the species; (3) the identified sample size of this prey is very small, drawn from a rather large sample of predators; (4) it provides almost no quantitative clue to the frequency of such predation; (5) obtaining useful estimates of such frequency or magnitude or other details on this interaction will probably require extraordinary efforts; (6) there is still no information to put this finding in perspective with other predation (if any) between snappers. Missing Evidence on Early Life Stages As expected, this study produced little or no evidence about interactions of younger life stages of any of the target species with any life stage of other target species. It is possible that young stages of some of the target species are represented in the gut contents examined, but if so, they could not be identified as such. No young juveniles were caught nor recognized in underwater observations. The locations and habitats used by young stages of the etelines are very poorly known. For opakapaka, there is information from elsewhere (Parrish et al. 1997) that the nursery grounds for juveniles well beyond the postlarval pelagic stage are rather far removed from the known habitat and traditional fishing grounds of the adults and provide much different environments (e.g. shallower, open, sandy, relatively featureless bottoms). Unpublished submersible observations by Kelley of HURL and R. Moffitt of NMFS have produced some preliminary information about the habitat of juvenile ehu, which may be more similar to that of adults. The habitat of adult taape studied here does not appear to include either of these kinds of juvenile nurseries. No larvae of taape or etelines were recognized in the present study. Although eggs that may have been fish eggs were found in the guts of a good many taape and some opakapaka, kalekale and a gindai, identification of the eggs does not seem feasible. The possibility that interactions involving young stages occur between the species cannot be eliminated by results of this study. If studies that are developing now on culture of some of the etelines begin to produce good reference material for identification of young stages, and knowledge of the habitats that they use accumulates, an assessment of such interactions may become feasible. Summary Assessment The overall impression is that the introduced taape shows little if any aggression toward native snappers, generally does not share the same depth and feeding habitat with most native species, 37 overlaps little in diet, and is not a frequent predator or prey of the natives. This evidence does not imply strong negative effects of taape on adults of native fishery species in these habitats. It does not address the potential for interactions of taape with young stages of the native snappers or with native species in shallow-water coastal habitats. 38 ACKNOWLEDGMENTS This work was funded by the Hawaii Department of Land and Natural Resources (DLNR) as a research project to the Hawaii Cooperative Fishery Research Unit (HCFRU). Major administrative and logistic support was provided by HCFRU, which is supported by the U. S. Geological Survey, University of Hawaii and DLNR. Dr. Chris Kelley collaborated throughout the life of this project as it ran concurrently with his own deep-water snapper project funded by DLNR. His willingness to share personnel, equipment, data and sea time on vessels permitted this project to accomplish much more than would otherwise have been possible. Gary Dill was indispensable to the project, providing fish specimens and high quality fishing data with his vessel for much of the project’s needs, as well as valuable practical advice and “know-how”. Equipment, facilities and technical personnel support for the deep diving was provided by Hawaii Undersea Research Laboratory (HURL). The highly dedicated and professional performance of the vehicle operators and support staff, as well as the ship’s crew, are greatly appreciated. Support in the laboratory, processing data and specimens, was provided by Ryan Fu, Lucy Kida and Kualani Richardson. Many people contributed to identification of animals, alive, fresh, preserved, or retrieved from fish guts. This assistance included loan of literature, specimens, and other reference materials, and personal examination of some of our specimens. Some important contributors included C. E. Birkeland, D. Catania, C. D. Kelley, J. M. Leis, R. B. Moffitt, B. C. Mundy, M. Parry, R. L. Pyle, M. P. Seki, A. Suzimoto, and R. E. Young. Data from their archives of research cruises were generously provided by the National Marine Fisheries Service Honolulu Laboratory and HURL. 39 LITERATURE CITED Anderson, W. D., Jr. 1987. Systematics of the fishes of the family Lutjanidae (Perciformes: Percoidei), the snappers. pp. 1-31 In: J.J. Polovina and S. Ralston (Eds.) Tropical Snappers and Groupers: Biology and Fisheries Management. Westview Press, Inc., Boulder, CO. 659 pp. DeMartini, E. E., F. A. Parrish, and D. M. Ellis. 1996. Barotrauma-associated regurgitation of food: implications for diet studies of Hawaiian pink snapper, Pristipomoides filamentosus (family Lutjanidae). Fishery Bulletin 94:250-256. Everson, A. R. 1984. Spawning and gonadal maturation of the ehu, Etelis carbunculus, in the Northwestern Hawaiian Islands. pp.128-148 In: R. W. Grigg and K.Y. Tanoue (Eds). Proceedings of the 2nd Symposium on Resource Investigations in the Northwestern Hawaiian Islands Vol. 2. University of Hawaii Sea Grant Misc. Report UNIHI-SEAGRANT-MR-84-01. Honolulu. Friedlander, A. M., R. C. DeFelice, J. D. Parrish, and J. L. Frederick. 1997. Habitat resources and recreational fish populations at Hanalei Bay, Kauai. Final project report to Hawaii Department of Land and Natural Resources by Hawaii Cooperative Fishery Research Unit. 320 pp. Haight, W. R., J. D. Parrish, and T. A. Hayes. 1993. Feeding ecology of deepwater lutjanid snappers at Penguin Bank, Hawaii. Transactions American Fisheries Society 122:328-347. Harrison, C. S., T. S. Hida, and M. P. Seki. 1983. Hawaiian seabird feeding ecology. Wildlife Monographs 85:71. Hyslop, E. J. 1980. Stomach contents analysis: a review of methods and their application. Journal of Fish Biology. 17:411-429. Kikkawa, B. S. 1984. Maturation, spawning, and fecundity of opakapaka, Pristipomoides filamentosus, in the Northwestern Hawaiian Islands. pp.149-160 In: R.W. Grigg and K.Y. Tanoue (Eds.) Proceedings of the 2nd Symposium on Resource Investigations in the Northwestern Hawaiian Islands Vol. 2. University of Hawaii Sea Grant Misc. Report UNIHI-SEAGRANT-MR-84-01. Honolulu. Krebs, C. J. 1999. Ecological methodology. Second Edn., Addison Wesley Longman, Menlo Park, CA. 620 pp. Maciolek, J. A. 1984. Exotic fishes in Hawaii and other islands of Oceania. pp. 131-161 In: W. R. Courtenay, Jr. and J. R. Stauffer, Jr. (Eds.) Distribution, Biology and Management of Exotic Fishes. Johns Hopkins Univ. Press, Baltimore. 446 pp. Mizenko, D. 1984. The biology of Western Samoan reef-slope snapper (Pisces: Lutjanidae) populations of Lutjanus kasmira, Lutjanus rufolineatus, and Pristipomoides multidens. M.S. Thesis. University of Rhode Island, Kingston, RI. 66 pp. 40 Morales-Nin, B. and S. Ralston. 1990. Age and growth of Lutjanus kasmira (Forskal) in Hawaiian waters. Journal of Fish Biology 36:191-203. Oda, D. K. and J. D. Parrish. 1982. Ecology of commercial snappers and groupers introduced to Hawaiian reefs. Proceedings of the 4th International Coral Reef Symposium. 1:59-67. Parrish, F. A., E. E. DeMartini, and D. M. Ellis. 1997. Nursery habitat in relation to production of juvenile pink snapper, Pristipomoides filamentosus, in the Hawaiian archipelago. Fishery Bulletin 95:137-148. Parrish, J. D. 1987. The trophic biology of snappers and groupers. pp. 405-463 In: J.J. Polovina and S. Ralston (Eds.) Tropical Snappers and Groupers: Biology and Fisheries Management. Westview Press, Inc., Boulder, CO. 659 pp. Parrish, J. D., M. W. Callahan, and J. E. Norris. 1985. Fish trophic relationships that structure reef communities. Proceedings 5th International Coral Reef Congress 4:73-78. Ralston, S. V. D. 1981. A study of the Hawaiian deep sea hand line fishery with special reference to the population dynamics of opakapaka, Pristipomoides filamentosus (Pisces: Lutjanidae). PhD Thesis, University of Washington, Seattle. Randall, J. E. 1987. Introductions of marine fishes to the Hawaii Islands. Bulletin of Marine Science 41:490-502. Rangarajan, K. 1971. Maturity and spawning of the snapper, Lutjanus kasmira (Forskal), from the Andaman Sea. Indian Journal of Fisheries 18:114-125. Rangarajan, K. 1972. Food and feeding habits of the snapper, Lutjanus kasmira (Forskal), from the Andaman Sea. Indian Journal of Fisheries 17:43-52. Seki, M. P. and M. W. Callahan. 1988. The feeding habits of two deep slope snappers, Pristipomoides zonatus and P. auricilla, at Pathfinder Reef, Mariana Archipelago. Fishery Bulletin 86(4):807-811. Suzuki, K. and S. Hioki. 1979. Spawning behavior, eggs, and larvae of the lutjanid fish, Lutjanus kasmira, in an aquarium. Japanese Journal of Ichthyology 26:161-165. Tabata, R. S. 1981. Taape: What needs to be done? Transcripts of a workshop. University of Hawaii Sea Grant College Program Working Paper 46. Uchiyama, J. H., S. M. Kuba and D. T. Tagami. 1984. Length-weight and standard length-fork length relationships of deep sea handline fishes of the Northwestern Hawaiian Islands. pp. 209-225 In: R. W. Grigg and K. Y. Tanoue (Eds.) Proceedings of the 2nd Symposium on Resource Investigations in the Northwestern Hawaiian Islands Vol. 2. University of Hawaii Sea Grant Misc. Report UNIHI-SEAGRANT-MR-84-01. Honolulu. 41 Uchiyama, J. H. and D. T. Tagami. 1984. Life history, distribution and abundance of bottomfishes in the Northwestern Hawaiian Islands. pp. 229-247 In: R. W. Grigg and K. Y. Tanoue (Eds.) Proceedings of the 2nd Symposium on Resource Investigations in the Northwestern Hawaiian Islands Vol. 1. University of Hawaii Sea Grant Misc. Report UNIHI-SEAGRANT-MR-84-01. Honolulu. Welcomme, R. L. (comp.). 1988. International introductions of inland aquatic species. FAO Fisheries Technical Paper 294. 318 pp. 42 43 44 45 46 Table 9. Hooks compared in fishing by Dill under Protocol 1. Size class designations are arbitrary and indicate increasing hook size with size class numbers. Manufacturer's designation Size Class Limmirick #5 1 Limmirick #3 2 mtc 6/0 3 mtc 7/0 4 mtc 8/0 5 maruto #18 5 maruto #22 6 mtc 10/0 7 mtc 11/0 8 mtc 12/0 9 Table 10. Total catches and catches shallower than 150 m for taape and five species of eteline snappers taken in experimental handline collections by the NMFS Honolulu Laboratory in waters of the Main Hawaiian Islands between June 1983 and Sept 1993. Taape Opakapaka Kalekale Gindai Ehu Onaga Catch at depths <150 m 21 416 76 111 99 0 Catch between 150m-550m 10 193 212 315 442 34 Total Catch (0-550 m) 31 609 288 426 541 34 47 Table 11. Fishing effort by depth for experimental handline fishing by the NMFS charter vessel, Kaimi, in waters of the Main Hawaiian Islands in October of 1982. Depth strata (m) Effort (hook-hours) 0-50 8.8 50-100 4.4 100-150 58.2 150-200 36.2 200-250 1.4 Total 109 Table 12. Summary of snapper sightings from remotely operated vehicle dives conducted in 1998. "No. dives" = total no. of dives on which a given species was seen; "No. sightings" = total no. of separate incidents when individuals of a species were seen; "Total no. of ind." = total no. of individuals of a given species seen, across all sightings. Taape Opakapaka Kalekale Gindai Ehu Onaga No. dives 1 0 0 0 2 1 No. sightings 1 0 0 0 3 1 Total no. of ind. 1 0 0 0 3 1 Depths (m) 96 na na na 199-300 200 48 Table 13. Snappers observed from the Pisces V research submersible in 1999. Numbers in the table are derived by assigning each fish sighting to a standard abundance group and summing within species over all sightings. Sums are broken down into depth categories, with the amount of submersible observation time within depth categories noted in parentheses. Taape Opakapaka Kalekale Gindai Ehu Onaga < 160m (3 hours) 3 12 11 0 0 0 > 160m (69 hours) 0 9 292 81 634 277 Table 14. Summary of snapper sightings from remotely operated vehicle dives conducted in 1999. "No. dives" = total no. of dives on which a given species was seen; "No. sightings" = total no. of separate incidents when individuals of a species were seen; "Total no. of ind." = total no. of individuals of a given species seen, across all sightings. Taape Opakapaka Kalekale Gindai Ehu Onaga ROV observations at <160m depth No. dives 10 6 4 0 2 0 No. sightings 24 8 6 0 2 0 Total no. of ind. ~ 425 ~ 26 7 0 2 0 Depths (m) 45-153 86-152 113-159 na 150-160 na ROV Observations at >160m depth No. dives 0 0 5 0 8 6 No. sightings 0 0 5 0 17 10 Total no. of ind. 0 0 ~ 14 0 25 29 Depths (m) na na 162-170 na 165-325 165-300 49 Table 15. Breakdown of previous (1981-1987) dives by HURL submersible Makalii examined for sightings of taape and eteline snappers. Total no. dives 51 No. dives with snapper sightings 34 Depths <160 m >160 m Dives with snapper sightings 20 21 Dives with eteline sightings 14 21 Dives with taape sightings 7 0 Dives with eteline and taape sightings 2 0 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 APPENDIX A Locations of handline fishing in the Main Hawaiian Islands that produced data for this project. Each black dot indicates the location of a fishing drop (occasionally more than one drop at about the same time and location). All fishing under both protocols by Dill (our project) and Kelley’s project are included. Fish specimens were available from all the main islands (few from Maui). The depth contour shown is 100 fathoms (~183 m). 71 72 73 74 75 APPENDIX B Description of deep diving vehicles and support equipment/facilities of Hawaii Undersea Research Laboratory (HURL) used in this project. (Extracted from HURL information packet for principal investigators). 1.3 EQUIPMENT AND FACILITIES Center Facilities The HURL operations center is located on the Makai Research Pier at Makapuu Point on the east coast of the island of Oahu. The center is approximately 15 miles from the city of Honolulu. A submersible shed is located on the pier, which also houses the operations office, diving locker, machine and electronic shops. Ship operations are managed by the UH Marine Center at Snug Harbor, on Sand Island Access Road in Honolulu. The ROV facility also operates at this location. The HURL administrative offices, data processing center, and labs are located on the University of Hawai’i at Manoa campus in Honolulu. Pisces V Submersible Pisces V is a 3-person, 1-atmosphere submersible that has a depth capability of 2,000m (6,560 ft.). Pisces V usually operates with one pilot and two observers, although there are two pilots and one observer for Loihi dives and certain other dives. The pressure hull is 2.13m (7 ft.) in diameter. The submersible is equipped with two hydraulic manipulators and a sonar ranging device. Dive duration is 6 to 8 hours with emergency life support for 72 hours. Dives occur during daylight hours. Equipment carried by or used in conjunction with the submersible includes: - sample storage baskets - color 8mm and digital video cameras, monitors, and recorders - flood and video camera lights - 35mm still camera and strobes - sediment grab samplers - temperature probes - rotating Niskin water samplers (18) - directional antenna for site relocation - titanium water samplers - dictaphone tape recorder - CTD recorder for salinity, temperature density and depth - Mini-Ranger navigation system - short-baseline submersible tracking system - suction sampler with rotating specimen containers 76 Additional equipment must be supplied by the scientist or may be fabricated by HURL, provided the request is made at least 3 months prior to use. Weight in air must not exceed 100 lbs. There are also size constraints, determined on a case-by-case basis by the Submersible Operations Director. Contact the Science Director for further information. RCV-150 Remotely Operated Vehicle The RCV-150 system consists of the vehicle and launching garage, a transportable winch/A-frame unit, and associated power and control consoles. The vehicle’s compact hydrodynamic design and neutrally buoyant tether cable permit close-up inspections and a high degree of maneuverability at speeds up to 3 knots. The vehicle can operate to depths of 800 m (2,625 ft.) and in currents up to 1.5 knots. Color video and a single manipulator are standard equipment on the RCV-150. A standard VHS record, including vehicle data (depth, heading, etc.), is made continuously during each dive, verbally annotated by the investigator in real time. A two-hour S-VHS cassette is available for higher quality real-time recording of highlights at the investigator’s discretion. Other equipment and sensors may be adapted for use on the ROV, as well as sample baskets installed on the tethered garage. Small instrument packages can be carried to the ocean floor aboard the vehicle garage. RCV-150 operates day and night, but not during submersible dives. R/V Ka’imikai-o-Kanaloa The R/V Ka’imikai-o-Kanaloa is the University of Hawai’i mother-ship for Pisces V and the RCV-150. The vessel is 222 feet in length. There are facilities for 18-20 scientists and technicians (including the submersible and ROV crews) and a ship crew of 14. The ship has A-frame launch capability, wet and dry labs, and photographic processing facilities. It is equipped with a CTD winch and rosette system for water column sampling. The ship can remain at sea for up to 50 days. In addition, it has a hull-mounted SeaBeam TM 210 (hybrid) multibeam bathymetric mapping system which consists of the original 16 hardware-former narrow beams capable of ensonifying a swath roughly 70% of water depth. SeaBeam TM 210 uses the modern SeaBeam TM 2100 projectors and receivers and Silicon Graphics-based (UNIX/IRIX) shipboard post-processing software packages including the standard “GMT-System” and “MB-System”. Large and small format color plotters are available for map generation at-sea. SeaBeam TM 210 does not presently have sidescan backscatter capability. |
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What is the Value of Cos 90?
Why Is Cos 90 Important in Trigonometry?
Trigonometry is one of the most important mathematical tools used for day-to-day measurements. Trigonometry is used to measure the angles and sides of a triangle which are not known. However this holds only for triangles which are right angled. It uses ratios and reciprocals to find the measurements of unknown values. The three main components of trigonometry are Sin, cosine and tangent, based on which all calculations and measurements are done. Cosine or cos is one of those important trigonometric ratios which is actually the ratio of the lengths of adjacent sides and hypotenuses. To know about these important trigonometric functions we start here with the value of Cos 90 degree, the laws relevant to it, the reciprocals and also the derivations relevant to it with the help of unit circle. The value of cos 90 degrees has been derived to be 0.
Cos 90 Value
To define the cosine function of an acute angle, we will consider the right angle triangle with an angle of interest and the sides of a triangle. The three sides of a triangle are defined as:
(Image will be Uploaded soon)
The opposite side of a right-angle triangle is the side opposite to the angle of interest.
The longest side of a right-angle triangle i.e. hypotenuse is the opposite side of a right angle in the triangle.
The remaining side of a triangle i.e. an adjacent side forms between both the angle of interest and the right angle.
The cosine function of an angle is defined as the ratio of the length of the adjacent side of a right-angle triangle and hypotenuse.
[Cos Ө = \frac{\text{Adjacent Side}}{\text{Hypotenuse side}} ]
Derivation of Cos 90 Degrees Using Unit Circle
Now, we will calculate the value of cos 90° using the unit circle with the radius 1 unit and the center of the circle placed at the origin of the coordinate axis ‘x’ and ‘y’. Let us take the point P (a,b) anywhere in the circle that forms an angle AOP= x radian. It implies that the length of the arc AP is equivalent to x. With this, we will define the following value,
(Image will be Uploaded soon)
Cos x= a and Sin x =b.
With the unit circle, we will now consider the right angle triangle OMP.
Through Pythagorean Theorem, we get
OM2 + MP2 = OP2
or
a2 + b2 = 1
Thus, each point on the unit circle is defined as
A2 +B2 = 1
or
Cos2 x + Sin2 x = 1
Note- One complete revolution subtends an angle of 2 π radian at the center of the circle, and from the unit circle. It is defined as:
[\angle AOB = \frac{\pi}{2}]
[\angle AOC = \pi] and [\angle AOD = 3\frac{\pi}{2}]
All the angles of a triangle are the integral multiples of π/2 and it is usually known as quadrant angle. The coordinates of point A, B, C and D are stated as (1, 0), (0, 1), (-1,0) and (0.-1) respectively. We will get the Cos 90 value through the quadrant angle.
Therefore, the value of Cos 90° is 0
Cos 90 = 0
It can be seen that the value of the sine and cosine function does not change if the x and y values are the integral multiples of π/2. If we will consider one complete revolution from the point P, it will come back again to the same point. For triangle ABC, with sides a, b, and c opposite to the ∠A, ∠B, and ∠C respectively, the cosine law will be defined.
With ∠C, the law of sine is stated as
[c^{2} = a^{2} + b^{2} - 2ab Cos (c) ]
With this, it will be easy to remember 0°,30°, 45° 60° and,90° as all these values are present in the first quadrant. Each sine and cosine function in the first quadrant takes the form [\sqrt{\frac{n}{2}}] or [\sqrt{\frac{n}{4}}].
We can easily find the value of the cosine function if we know the values of the sine function.
Sin 0° = [\sqrt{\frac{0}{4}}].
Sin 30° = [\sqrt{\frac{1}{4}}].
Sin 45° = [\sqrt{\frac{2}{4}}].
Sin 60° = [\sqrt{\frac{3}{4}}].
Sin 90° = [\sqrt{\frac{4}{4}}].
Now through the sine value, we can find the cosine value easily because
Cos 0° =Sin 90° =1
Cos 30° = Sin 60° = [\sqrt{\frac{3}{4}}] = [\frac{\sqrt{3}}{2}]
Cos 45° = Sin 45 ° = [\sqrt{\frac{2}{4}}] = [\frac{1}{\sqrt{2}}]
Cos 60° = Sin 30° = [\sqrt{\frac{1}{4}}] = [\frac{1}{2}]
Cos 90° = Sin 0° = 1
So accordingly, the value of Cos 90 degree or Cos 90 =0
Similarly, values of other degrees of trigonometry functions can be found out.
Solved Example
Evaluate the following
a. [ 4 (Sin^{2}30° + Cos^{2}60°) – 3(Cos^{2}45° - Sin^{2} 90°)]
Solution- Value of Sin 30° =[\frac{1}{2}]
Value of Cos 60° = [\frac{1}{2}]
Value of Cos 45° = [\frac{1}{\sqrt{2}}]
Value of Sin 90° = 1
= [4 ( (\frac{1}{2})^{2} + (\frac{1}{2})^{2} ) – 3 ((\frac{1}{\sqrt{2}})^{2} -1^{2}) ]
=[4 (\frac{1}{4} + \frac{1}{4}) - 3(\frac{1}{2} -1) ]
=[ (4 \times \frac{2}{4}) - (3 \times \frac{-1}{2}) ]
= [ (4 \times \frac{1}{2}) - ( \frac{-3}{2}) ]
=[ (\frac{4}{2}) + ( \frac{3}{2}) ]
=[ \frac{4+3}{2} ]
=[ \frac{7}{2} ]
b. If Cos Ө or Sin Ө =[\sqrt{2}]Cos Ө, show that Cos Ө –Sin Ө =[\sqrt{2}] Sin Ө.
Solution: Cos Ө or Sin Ө =[\sqrt{2}] Cos Ө
= Sin Ө =[\sqrt{2}] Cos Ө – Cos Ө
= Sin Ө = ([\sqrt{2}] -1) Cos Ө
(Multiply with both side ‘[\sqrt{2}] + 1 ‘)
= ([\sqrt{2}] + 1) Sin Ө = ([\sqrt{2}] + 1) ([\sqrt{2}] - 1) Cos Ө
= [\sqrt{2}] Sin Ө + Sin Ө = ([\sqrt{2}^{2}] - 1) Cos Ө = (2 - 1) Cos Ө = Cos Ө
= [\sqrt{2}] Sin Ө = Cos Ө –Sin Ө
Hence, Proved
Fun Facts
In trigonometry, the law of cosines is also known as cosine formula, cosine rule or al-Kashi’s theorem.
The position of the ship can be determined through trigonometry and Marine chronometer.
Hipparchus, who compiled the first trigonometry table, is also known as “Father of Trigonometry”.
Quiz Time
For any acute angle, cosine would be equal to
a. –Cos (180°- Ө)
b. Cos (180° - Ө)
c. –Cos (180° + Ө)
d. Cos (180° +Ө)
The Cosine Rule is also known as
a. Sine triangle
b. Cosine Triangle
c. Cosine Area
d. Cosine Formula
Trigonometry is based on
a. Squares
b. Rectangle
c. Triangle
d. Octagons
FAQs on What is the Value of Cos 90?
What is the exact value of cos 90 degrees?
The exact value of cos 90 degrees (cos 90°) is 0. This is a fundamental value in trigonometry, representing the cosine of a right angle, and it is a key point in the NCERT syllabus for Class 10 and 11 Maths.
Why is the value of cos 90 equal to 0?
The value of cos 90° is 0 because the cosine of an angle (θ) in a right-angled triangle is defined as the ratio of the adjacent side to the hypotenuse. As the angle θ approaches 90°, the length of the adjacent side effectively shrinks to zero. Therefore, the ratio becomes 0 / Hypotenuse, which results in 0.
How can you prove that cos 90 is 0 using the unit circle?
On a unit circle (a circle with a radius of 1), any point on the circumference has coordinates (cos θ, sin θ). For an angle of 90°, the point lies exactly on the positive y-axis. The coordinates of this specific point are (0, 1). Since the x-coordinate always represents the cosine value for the angle, this visually proves that cos 90° is 0.
What is the value of cos 90 in radians and as a fraction?
The value of cos 90 remains 0 regardless of the notation used:
What is the common confusion between the values of cos 90 and cos 0?
A very common mistake students make is swapping the values of cos 90° and cos 0°. It's critical to remember their distinct values:
How is cos 90 related to the sine function through co-function identities?
Cos 90° is directly related to the sine function using the co-function identity: cos θ = sin (90° - θ). By substituting θ = 90° into this formula, we get cos 90° = sin (90° - 90°), which simplifies to sin 0°. Since we know that sin 0° is 0, this provides another confirmation that cos 90° must also be 0.
What is the practical importance of the cos 90 value in Physics?
The value of cos 90° is extremely important in Physics, particularly for calculating Work Done. The formula for work is W = Fd cos θ, where θ is the angle between the applied force (F) and the displacement (d). If a force is applied perpendicular (at 90°) to the direction of movement, cos 90° = 0, making the total work done zero. A real-world example is carrying a heavy suitcase and walking horizontally; no work is done on the suitcase by the lifting force.
NCERT Study Material
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Even Black Square Chessboard Rooks [duplicate]
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I'm stuck on this problem: "There are $8$ rooks on the chessboard such that no two of them they don't hit each other. Prove that the black squares contain an even number of rooks."
What I've tried: I first realized that proving for black or white squares is the same thing, because if you have even white, you must have even black.
I noticed that when you pick a square for a rook, you can't place any other rook on the same row or column similar to sudoku with numbers. I imagined that when a rook is placed, the row and column gets cut and removed and combining the separated rows/columns to create a square 1 size smaller.
I put the first rook on the first column.
I started looking at smaller cases like a $2\times2$, in the first row no matter what square you pick, you will be left with a $1\times1$ with the same colour (because the adjacent squares have opposite colours and diagonal have the same). So $2\times2$ works. I tried extending it to $3\times3$, if you place a rook on any of the corners, it will turn into a $2\times2$ again, so I hypothesised $n\times n$, when $n$ is odd will result in odd number of black or white squares (depending which one has $5$ and the other $4$ in the $3\times3$) and even $n$ will result in even black and white.
Now when I put a rook in the outer edge of the $3\times3$, I get a $2\times2$ but with the rows of same colour, this gives the same result.
Now I tried a $4\times4$, putting a rook in the corners will turn into a $3\times3$, let's say WLOG black was on top left corner, now it will become a $3\times3$ with $5$ blacks, meaning there will be an odd number of black, and we will have $1$+odd which will be even.
Now I want to do the same process with a $4\times4$ and larger squares, but there are too many cases especially putting the first rook in squares that are not corners, because then you will get weird colour configurations.
Also I think my overall process is not the "proper" way of doing it and would love to know how to tackle problems like these.
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edited Sep 1, 2024 at 10:40
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As mention by HackR, this question is similar to parity of rooks
Maybe this explanation is more simple :
We have a number for each column (c=1 to 8); a number for each row (r=1 to 8). We can previously rotate the chessboard, so that the first cell (1,1) is white.
We define a value for each cell $(r,c)$ : $v=r+c$ ; So all white cells have a value which is an even-number, and all black cells have a value which is an odd-number.
If we put the rooks so that 2 of them cannot hit each other, we are sure that we have one rook in each column, and one in each row. So, if we compute sum of values of the 8 cells where we have a rook, we obtain $2(1+2+3+4+5+6+7+8)$.
This number is even.
When we compute sum of integers, the sum is even only when the number of odd-integers is an even-number. So to obtain a total which is even, we need to select an even-number of black cells (=cells with odd value).
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edited Sep 1, 2024 at 9:01
answered Sep 1, 2024 at 7:56
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Syphilis
This article has a correction. Please see:
What you need to know
Incidence rates of syphilis have increased substantially around the world, mostly affecting men who have sex with men and people infected with HIV
Have a high index of suspicion for syphilis in any sexually active patient with genital lesions or rashes
Primary syphilis classically presents as a single, painless, indurated genital ulcer (chancre), but this presentation is only 31% sensitive; lesions can be painful, multiple, and extra-genital
Diagnosis is usually based on serology, using a combination of treponemal and non-treponemal tests. Syphilis remains sensitive to benzathine penicillin G
Staging syphilis is important because it is the basis of management (treatment, expected treatment response, follow-up periods, and partner follow-up)
Patients with syphilis should be screened for HIV, gonorrhoea, and chlamydia
Caused by the bacteria Treponema pallidum,12 syphilis is transmitted through direct (usually sexual) contact with infected mucosal lesions. Other bodily fluids are also infectious when patients are bacteraemic. With infectivity up to 10-30% per sexual contact or 60% per relationship,3 syphilis rates have risen 300% since 2000 in many Western countries.4567 While most infections involve men who have sex with men, infections among people with opposite sex partners also occur.4567 In addition to increasing rates, syphilis can cause early complications such as irreversible loss of vision, so awareness of the infection is important for primary care clinicians.8910
What symptoms should alert me to this diagnosis?
While syphilis causes protean symptoms (box 1), the diagnosis should be considered in any sexually active patient with genital lesions or with rashes.161718
Symptoms of syphilis by stage of infection (see fig 1)
Primary
Symptoms appear 10-90 days (mean 21 days) after exposure
Main symptom is a <2 cm chancre:
Progresses from a macule to papule to ulcer over 7 days
Painless, solitary, indurated, clean base (98% specific, 31% sensitive)
On glans, corona, labia, fourchette, or perineum
A third are extragenital in men who have sex with men and in women
Localised painless adenopathy
Secondary
Symptoms appear 2 weeks to 6 months (mean 2-12 weeks) after exposure. Can be concurrent with, or up to 8 weeks after, chancre
Rash—Diffuse, symmetric, on trunk (often subtle or atypical)
Mucus lesions, condylomata lata
Patchy alopecia (4-11%)
Fever, headaches, generalised painless adenopathy
Neurologic symptoms—Cranial nerve palsies (II,VIII), eye redness or pain, meningitis, changes to mental status or memory
Latent
No symptoms
In early latent stage (<12 months or <24 months† after exposure) 25% of subjects relapse to secondary syphilis (90% of these in first year, 94% within 2 years)
In late latent stage (>12 or 24† months after exposure), no relapse and not infectious
Tertiary
1-46 years after exposure
Neurologic—paresis, tabes dorsalis
Cardiovascular—aortitis
Gummatous—necrotic granulomatous lesions
According to US,11 UK,12 and Canadian13 guidelines
†According to World Health Organization14 and European15 guidelines
Primary syphilis—Patients with primary syphilis (fig 1) have a chancre at the site of inoculation—classically a solitary, painless, indurated, non-exudative ulcer (fig 2).1719 While often on the glans, corona, labia, fourchette, or perineum, it may occur in the mouth (fig 3), rectum, or vagina.17 Chancres can be inconspicuous (fig 4) and resolve in 3-10 weeks, possibly explaining why 60% of patients do not recall this lesion. Chancres may be multiple, painful, or atypical due to coinfection with other bacteria or herpesvirus.20 Depending on inoculum size, chancres appear 10-90 days after exposure (mean 21 days).17 Localised painless adenopathy may occur.17
Stages of syphilis
Chancre (sore) on penis due to syphilis
Syphilis in the mouth
Inconspicuous syphilitic chancre on penis
Secondary syphilis—Secondary syphilis is a manifestation of bacterial dissemination and classically presents as a diffuse, symmetric, copper, maculopapular, possibly pruritic rash of any morphology except vesicular (fig 5).172021 A rash on the palms or soles is common (11-70%, fig 6). Mucus lesions (fig 7), patchy alopecia, fever, headaches, and generalised painless adenopathy may also occur.1718192021 Early neurosyphilis develops in 25-60% of people (box 1).9171819202122 Secondary symptoms appear 2-24 weeks after infection, concurrently with or up to eight weeks after chancres, and disappear spontaneously after several weeks with or without marking.1719
Rash associated with secondary syphilis
Secondary syphilis on palms of hands
Condylomata lata in secondary syphilis
Latent syphilis—Syphilis then becomes latent, although symptoms of secondary syphilis recur in 25% of people, mostly (90%) within one year of acquiring the infection.17 Latent syphilis has early and late stages.17 Early latent disease includes the period of potential symptom relapse, classified by the WHO14 and European15 guidelines as <2 years since inoculation and as <1 year by US,11 UK,12 and Canadian13 guidelines. As symptom relapse indicates bacterial replication, early latent disease can be infectious. Late latent syphilis occurs >1-2 years after acquisition and is non-infectious (see fig 1).
Tertiary syphilis—Without treatment, 14-40% of people with syphilis progress to tertiary disease—irreversible damage to any organ—within 1-46 years. The damage is primarily neurologic, cardiovascular, or gummatous (necrotic granulomatous lesions pathognomonic of tertiary syphilis).217
Which diagnostic test should be done (table 1)?
Diagnostic tests for syphilis232425
Treponema pallidum may be visualised from lesions using dark field microscopy, direct fluorescent antibody testing, or polymerase chain reaction.1112131415 Because these tests are not widely available, diagnosis predominantly relies on serology.172627 While serologic tests and laboratory algorithms vary, testing usually begins with a screening treponemal test, such as an enzyme or chemiluminescence immunoassay (EIA or CLIA) to detect treponemal antibodies. A positive screening test should be followed by a confirmatory treponemal test, typically the T pallidum particle agglutination (TPPA). If both tests are positive, infection with syphilis is confirmed. Thereafter, the rapid plasma reagin (RPR) test (a quantitative non-treponemal test) should be used to measure disease activity and to track response to treatment (although 15-41% of patients remain reactive even after successful treatment).28
Test timing
Screening treponemal tests (EIA or CLIA) usually become reactive first, often within two weeks of the chancre. However, patients with negative results who have syphilis-like symptoms or who report a high risk contact should be re-tested after a further two to four weeks.13
The RPR test may remain non-reactive for up to four weeks after the chancre, so it is often negative in primary syphilis, but it is 98-100% sensitive in secondary syphilis. However, because the RPR is a test of non-specific tissue damage, it may be positive for reasons other than syphilis.15
In the absence of treatment, a negative non-treponemal test three months after potential exposure effectively rules out a new syphilis infection.
Note that treponemal tests cannot distinguish active from treated infections and generally remain positive for life (see table 2).
Interpretation of results from diagnostic tests for syphilis
Staging of syphilis
Staging of syphilis cannot be done based on laboratory results alone, and requires history and examination. Primary and secondary syphilis are symptomatic; early and late latent syphilis are generally asymptomatic. Careful examination to identify any symptoms not noticed by the patient is important and should include thorough anogenital and dermatologic inspection.21
The staging criteria for early latent syphilis are given in box 2. Asymptomatic patients with positive serology who do not fulfil the criteria of early latent syphilis could be staged as latent syphilis or as having syphilis of unknown duration.
Staging criteria for early latent syphilis1112131415
Patients with early latent syphilis are asymptomatic, with one of the following:
New positive serology with a documented negative test within previous 12 or 24† months
≥4-fold increase in the RPR titre relative to a previous test done within 12 or 24† months
Unequivocal symptoms of primary or secondary syphilis within the previous 12 or 24† months
Only one possible exposure, which occurred within previous 12 or 24† months
According to US,11 UK,12 and Canadian13 guidelines
†According to World Health Organization14 and European15 guidelines
What should I do with inconclusive results?
Generally, inconclusive results arise in early infection or from waning antibody levels in late infection. The most common combinations are:
A positive RPR with negative treponemal screening (EIA/CLIA) and confirmatory tests (TPPA) suggests the RPR result is a false positive
A positive screen (EIA/CLIA) with negative confirmatory test (TPPA) and negative RPR is likely a false positive but could indicate early infection
A positive screen (EIA/CLIA) with indeterminant confirmatory test (TPPA) and negative RPR could represent waning antibody levels after a previous, treated infection or a new infection.
When results are inconclusive, clinicians should inquire about previous syphilis infection and treatment, and, if early syphilis is possible, retest in two to four weeks.13 If results are unchanged, interpretation is based on history—consider the possibility of late untreated infection, treated infection, or non-venereal treponemal disease in adults from endemic countries in South and Central America, South-East Asia, and Africa.
What are the recommended treatment options? (box 3)
Recommended treatments for syphilis
Primary, secondary, and early latent disease
First line treatment
Benzathine penicillin G 2.4×106 units, single intramuscular dose†
Doxycycline 100 mg, taken orally twice daily for 14 days†
Alternate treatments
Ceftriaxone 1 g, intravenous or intramuscular once daily for 10 days
Procaine penicillin G 1.2×106 units, intramuscular once daily for 10 days†
Azithromycin 2 g, single oral dose†
Late latent disease
First line treatment
Benzathine penicillin G 2.4×106 units, intramuscular dose once weekly for 3 weeks†
Doxycycline 100 mg, taken orally twice daily for 28 days†
Alternate treatments
Ceftriaxone 1 g, intravenous or intramuscular once daily for 10 days
Procaine penicillin G 1.2×106 units, intramuscular once daily for 14-21 days†
According to US,11 UK,12 and Canadian13 guidelines
†According to WHO,14 UK,12 and European15 guidelines
For primary, secondary, and early latent syphilis, in the absence of neurologic findings, first-line treatment is benzathine penicillin G (BPG) 2.4×106 units as a single intramuscular injection. Late latent syphilis is treated with the same dose of BPG weekly for three weeks with no more than 14 days between doses (no more than 7 days for pregnant women).1112131415 Additional BPG doses do not improve treatment outcomes for patients with early syphilis,282930 although some guidelines suggest pregnant women with early syphilis can receive up to two doses.13 BPG has not been evaluated in controlled trials, but remains the first-line treatment because it is long acting (so covering the long dividing time of T pallidum) and because penicillin-resistant syphilis has not been documented in 60 years of the drug’s use.283031
Non-pregnant patients who are allergic to penicillin can be given doxycycline 100 mg by mouth twice daily for two weeks for primary, secondary, and early latent syphilis, or for four weeks for late latent syphilis,1112131415 although doxycycline, compared with BPG, may yield more treatment failures (defined according to the CDC as a fourfold or higher increase in RPR titre).283031 There is no alternative treatment to BPG for pregnant women.
Counsel patients about the possibility of Jarish-Herxheimer reactions, which start two to four hours after treatment and usually resolve within 24 hours. Symptoms include fever and systemic symptoms (such as chills, rigors, myalgias, arthralgias) with worsening rash or chancre.17 While UK guidelines12 state prednisolone can be used for symptom management, other guidelines11131415 recommend only over-the-counter antipyretics.
What about follow-up?
Because syphilis has no test-of-cure, conversion to a non-reactive RPR is the best evidence of successful treatment.28 Patients should be tested at the start of treatment and monitored at six and 12 months. No clinical data guide interpretation of RPR titres after treatment, and guidelines are based on expert opinion. See table 3 for recommendations from European, UK, US, Canadian, and WHO guidelines.1112131415
Recommendations for assessment of treatment of syphilis
When should I consider evaluation of cerebrospinal fluid?
Indications for lumbar puncture and evaluation of cerebrospinal fluid (CSF) include neurologic symptoms or tertiary disease (table 3).1112131415 CSF evaluation can also be considered for the 10-20% of patients with earlier disease who do not achieve a fourfold decline in RPR titres by 6-12 months after treatment.1112131415
Because BPG poorly penetrates CSF,11 neurosyphilis should be treated with aqueous penicillin G, 4×106 units intravenously every 4 hours for 10-14 days. If neurosyphilis is ruled out, optimal management is unclear.11 Clinicians may monitor the RPR titre until it is low or non-reactive, or repeat the treatment for early or late latent syphilis.11 Factors limiting post-treatment RPR declines in the absence of neurosyphilis include prior infection, longer duration of infection, older age, HIV co-infection, and low pre-treatment titres.11121415
Referral to secondary care may be necessary for patients requiring CSF evaluation and should be considered for those with uncertain diagnoses or poor response to treatment.
Are there specific considerations for patients with HIV?
Syphilis and HIV infection often co-exist.3233 Patients with syphilis should be screened for HIV and, if negative, offered pre-exposure prophylaxis.323334 They should also be screened for gonorrhoea and chlamydia. HIV-positive patients have additional indications for CSF evaluation (see table 3). Otherwise, diagnosis and treatment are unchanged.31353637
How should I manage contact tracing?
Contacts (people who have had sex with a person diagnosed with infectious (early) syphilis) within 90 days should receive treatment with one dose of BPG even if their serology results are is negative; asymptomatic contacts who had sex with an infected person more than 90 days ago could defer treatment until their serology results are available, but only if follow-up is assured. Discussions about contact tracing should be non-stigmatising and sensitive to patients’ concerns about confidentiality. Explain that contact tracing has important benefits for the individual concerned and their contacts. It helps to limit ongoing transmission of a serious infection and prevent re-infection. Patients need help and support to notify contacts confidentially.
How this article was created
This article was created based on a review of international guidelines, expert opinion (local public health unit, STI clinic, and infectious disease department), and through a review of Medline and CINAHL, using the search term “syphilis.” We also undertook a manual review of the reference lists of identified articles.
Education into practice
Do you consider syphilis as a differential diagnosis of genital lesions and rashes among sexually active patients?
How would you approach a conversation about contact tracing with a young man, recently diagnosed?
How patients were involved in this article
We reviewed the contents of this material with Max Ottawa, a local “community-based organisation that focuses on maximising the health and wellness of gay, bisexual, Two-spirit, queer, and other guys who are into guys, both cis and trans.”
Acknowledgments
PO’B holds a research chair in public health and HIV prevention from the Ontario HIV Treatment Network.
Footnotes
Competing interests: We have read and understood the BMJ policy on declaration of interests and have no relevant interests to declare.
Provenance and peer review: Commissioned; externally peer reviewed.
This is an Open Access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited and the use is non-commercial. See:
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Information |
189711 | https://www.youtube.com/watch?v=BaQXFstxCMo | Linear Equations in Standard Form
Khan Academy
9090000 subscribers
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Description
411884 views
Posted: 6 Apr 2010
Linear Equations in Standard Form
More free lessons at:
123 comments
Transcript:
Let's do some examples dealing
with equations of lines in standard form. So, so far we've had two
other forms. We've had slope-intercept, which
is of the form, y is equal to mx plus b. That's actually this
right here. This is in slope-intercept
form. We've seen point-slope form
in the last video. That's of the form, y minus some
y-value on the line being equal to the slope times x
minus some x-value on the line, when you have
that y-value. So the point x1, y1
is on the line. This right here is an example
of point-slope form. And now we're going to talk
about the standard form. And the standard form-- let me
write it here-- standard form is essentially putting all of
the x and y terms onto the left-hand side of
the equation. So you get ax plus
by is equal to c. I want to really emphasize that
all of these are just different ways of writing
the same equation. If you're given this, you can
out algebraically manipulate it to get to that or to that. If you're given that, you
can get to that or that. These are all different ways
of writing the exact same relationship, the
exact same line. So let's do a couple of
examples of this. So here we have a
line right here. We have an equation written
in slope-intercept form. The slope is 3, the y-intercept
is negative 8. Let's put it into
standard form. So we just have to get
the 3x onto the other side of the equation. And the best way I can think of
doing that-- let me rewrite the equation, y is equal to 3x
minus 8-- let's some subtract 3x from both sides
of the equation. So if you subtract 3x from both
sides-- so you subtract 3x, subtract 3x-- what do the
left- and right-hand sides of the equation become? The left-hand side becomes
negative 3x plus y being equal to-- the 3x and the negative 3x
cancel out-- being equal to negative 8. We're done. That's standard form
right there. Standard form, I guess people
like it because it has both the coefficients on the
left-hand side. But it's kind of useless in
trying to figure out slope and y-intercept. I don't know what the slope and
y-intercept is when I look at it in standard form. My favorite is slope-intercept
form, because it tells you exactly the slope and
an intercept. Point-slope, easy to get to,
and you can look at it and figure out the slope. But y-intercept, you have to
do a little bit of work to figure it out. But at least you can just go
immediately from the slope and a point to it. But anyway, let's go from this
equation, which is written in point-slope form, and get
it to the standard form. So we want to get it to the
standard form, to the same type of standard form. So a good thing to do, let's
just distribute things out. y minus 7 is equal to negative 5
times x, negative 5x, plus negative 5, times negative
12, which is positive 60. Now, we want all of the variable
terms on the left, all of the constant terms
on the right. So let's add 7 to both sides
of this equation. So plus 7 to both sides
of this equation. What does it become? Well, the minus 7 disappears,
because negative 7 plus 7. So you're just left with
a y being equal to negative 5x plus 67. Now, if we want this x term on
the left-hand side, we could add 5x to both sides. So let's add 5x to both sides
of this equation. And we will get y plus 5x
is equal to-- these cancel out-- 67. Now, this is pretty much
standard form. If you really want to be a
stickler for it, you can rearrange these two. So it'd be 5x plus
y is equal to 67. And you are done. Let's do one more of these. So this is in neither
point-slope nor in slope-intercept form. It's just in some type
of intermediary mixed form right there. This looks like some type of
point-slope, but this looks like something different. So it's really not
point-slope. Let's see if we can
algebraically manipulate it to the standard form. So we get 3y plus 5. Let's distribute out this 4. So it's equal to 4x minus 36. Let's do exactly what
we did in the last. I'm using different notation on
purpose, to expose you to different things. So instead of doing it this way,
I'm going to subtract 5 from both sides, but I'm going
to do it on the same line. So I'm going to subtract
5 from both sides. And so the left-hand side of
this equation becomes 3y, because these two guys cancel
out, and that is equal to 4x. And then what is minus
36 minus 5? That's minus 41. And now we want the x terms
of the left-hand side. So let's subtract 4x from both
sides of this equation. So negative 4x plus,
and then minus 4x. What does our equation become? Well, the left-hand side just
stays negative 4x plus 3y. And the right-hand, the reason
why we subtracted 4x is so it cancels out with that. You just have a negative 41. And we're done. We are in standard form. Now, let's go the other way. Let's start with some equations
in standard form and figure out their slope
and y-intercept. And the best way I know to
figure out the slope and y-intercept is to put it into
slope-intercept form. So we want to put these
equations right here into the form, y is equal to mx plus b. So we're essentially
solving for y. Let's do that. So the best thing to do here--
so let me rewrite it. 5x minus 2y is equal to 15. Let's subtract 5x
from both sides. So minus 5x plus, you
have a minus 5x. These cancel out. And so you're left with
negative 2y is equal to 15 minus 5x. And now, let's divide everything
by negative 2. If you divide everything by
negative 2, what do we get? The left-hand side
just becomes a y. y is equal to-- 15 divided by
negative 2 is negative 7.5. And then negative 5 divided by
negative 2-- you can imagine I'm distributing the negative
1/2 if you will. I'm dividing both of these
by negative 2. So negative 5 divided by
negative 2 is positive 2.5x. And if you really wanted to put
it in the slope-intercept form, you could say that y is
equal to-- you could just rearrange these--
2.5x minus 7.5. You want the slope. It's right here. That is our slope. You want the y-intercept. Actually, let me be careful. It is right there. It is negative 7.5. That is the y-intercept. And now this would be a form
that's actually pretty straightforward to
graph it in. Let's do this one. So once again, we just
need to solve for y. So let's subtract 3x
from both sides. So you get 6y is equal
to 25 minus 3x. And then you can divide
both sides by 6. So you're left with y is equal
to 25 over 6 minus 3 over 6, or minus 1/2x. If you really want it in this
from, you just rearrange this. y is equal to negative
1/2x plus 25 over 6. Where is the slope? Here is the slope. Negative 1/2, that
is the slope. Where is the y-intercept? That's the y-intercept. That is our b. The point 0, 25 over
6 is on the line. Let's do one more of these. So we get 9x minus
9y is equal to 4. Just for fun, let's just start
off by dividing both sides of the equation by 9. You don't have to
do it that way. But this is kind of a fun way
to do it, because the coefficients here will
immediately become 1. So if you divide both sides of
the equation by 9, if you divide everything by 9, it
becomes-- actually, well, let's divide everything. Let's divide everything by
negative 9, even better. I'm just doing this for fun. So this first term will
become negative x. The second term, you have a
negative 9 divided by a negative 9, it will
be a plus y. And then this last term
will just become a negative 4 over 9. Actually, let me write
this out here. Negative 4 over 9. I'm giving some space there. Now, we want the x on the
right-hand side, so let's add x to both sides of
this equation. These cancel out. And then the equation becomes
y is equal to x minus 4/9. Where is the slope? The slope is the coefficient
on the x term. The slope is equal to 1. Where is the y-intercept? The y-intercept is
right there. It is negative 4/9. |
189712 | https://pubs.acs.org/doi/10.1021/acsami.7b03005 | Enhanced Kinetics of Electrochemical Hydrogen Uptake and Release by Palladium Powders Modified by Electrochemical Atomic Layer Deposition | ACS Applied Materials & Interfaces
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Download Hi-Res ImageDownload to MS-PowerPointCite This:ACS Appl. Mater. Interfaces 2017, 9, 21, 18338-18345
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Research Article April 28, 2017
Enhanced Kinetics of Electrochemical Hydrogen Uptake and Release by Palladium Powders Modified by Electrochemical Atomic Layer Deposition
Click to copy article link Article link copied!
David M. Benson†
Chu F. Tsang†
Joshua D. Sugar‡
Kaushik Jagannathan†
David B. Robinson‡
Farid El Gabaly‡
Patrick J. Cappillino‡§
John L. Stickney†
View Author Information View Author Information
†Department of Chemistry, University of Georgia, Athens, Georgia 30602, United States
‡Sandia National Laboratories, Livermore, California 94550, United States
§Department of Chemistry and Biochemistry, University of Massachusetts Dartmouth, North Dartmouth, Massachusetts 02747, United States
E-mail: stickney@uga.edu
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ACS Applied Materials & Interfaces
Cite this: ACS Appl. Mater. Interfaces 2017, 9, 21, 18338–18345
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Published April 28, 2017
Publication History
Received
1 March 2017
Accepted
28 April 2017
Published
online 16 May 2017
Published
in issue 31 May 2017
research-article
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Abstract
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Electrochemical atomic layer deposition (E-ALD) is a method for the formation of nanofilms of materials, one atomic layer at a time. It uses the galvanic exchange of a less noble metal, deposited using underpotential deposition (UPD), to produce an atomic layer of a more noble element by reduction of its ions. This process is referred to as surface limited redox replacement and can be repeated in a cycle to grow thicker deposits. It was previously performed on nanoparticles and planar substrates. In the present report, E-ALD is applied for coating a submicron-sized powder substrate, making use of a new flow cell design. E-ALD is used to coat a Pd powder substrate with different thicknesses of Rh by exchanging it for Cu UPD. Cyclic voltammetry and X-ray photoelectron spectroscopy indicate an increasing Rh coverage with increasing numbers of deposition cycles performed, in a manner consistent with the atomic layer deposition (ALD) mechanism. Cyclic voltammetry also indicated increased kinetics of H sorption and desorption in and out of the Pd powder with Rh present, relative to unmodified Pd.
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Atomic layer deposition
Deposition
Electrochemical cells
Granular materials
Palladium
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E-ALD
SLRR
palladium
rhodium
powder
electrodeposition
UPD
flow cell
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Marta Moreno-Gonzalez, Aoxue Huang, Phil A. Schauer, Ke Hu, Brian Lam, James W. Grayson, Adam Bottomley, David J. Dvorak, David K. Fork, Curtis P. Berlinguette. Sulfuric Acid Electrolyte Impacts Palladium Chemistry at Reductive Potentials. Chemistry of Materials2020, 32 (21) , 9098-9106.
Sita Gurung, David B. Robinson, Patrick J. Cappillino. Palladium-Coated Platinum Powders with Tunable, Nanostructured Surfaces for Applications in Catalysis. ACS Applied Nano Materials2020, 3 (1) , 530-537.
Aliya Sarsenbayeva, Selenay Sadak, Ipek Kucuk, Leila Kudreyeva, Abu Moldir Bakytzhanovna, Bengi Uslu. Molybdenum-Based Electrochemical Sensors for Breast Cancer Biomarker Detection: Advances and Challenges. Critical Reviews in Analytical Chemistry2025, 14, 1-21.
A. C. Mkhohlakali, X. Fuku, R. M. Modibedi, L. E. Khotseng, M. K. Mathe. Electroformation of Pd‐modified Thin Film Electrocatalysts Using E‐ALD Technique. Electroanalysis2021, 33 (7) , 1746-1760.
László Péter. Ultrathin Layers. 2021, 79-131.
Ivan Korolev, Pelin Altınkaya, Petteri Halli, Pyry-Mikko Hannula, Kirsi Yliniemi, Mari Lundström. Electrochemical recovery of minor concentrations of gold from cyanide-free cupric chloride leaching solutions. Journal of Cleaner Production2018, 186, 840-850.
Long Chao, Nian Liu, Xiujuan Xiong, Fang He, Ting Huang, Qingji Xie, Shouzhuo Yao. Preparation of an ultrathin Pt electrocatalyst via a galvanic replacement reaction of electrodeposited CuCl for the oxidation of methanol in an alkaline medium. Chemical Communications2018, 54 (30) , 3743-3746.
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ACS Applied Materials & Interfaces
Cite this: ACS Appl. Mater. Interfaces 2017, 9, 21, 18338–18345
Click to copy citation Citation copied!
Published April 28, 2017
Publication History
Received
1 March 2017
Accepted
28 April 2017
Published
online 16 May 2017
Published
in issue 31 May 2017
Copyright © 2017 American Chemical Society
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Article Views
634
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Citations
7
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Recommended Articles
### Hydrogen Adsorption, Absorption, and Desorption at Palladium Nanofilms formed on Au(111) by Electrochemical Atomic Layer Deposition (E-ALD): Studies using Voltammetry and In Situ Scanning Tunneling Microscopy July 22, 2013 The Journal of Physical Chemistry C Leah B. Sheridan , Youn-Geun Kim , Brian R. Perdue , Kaushik Jagannathan , John L. Stickney ,and David B. Robinson
### Atomic-Layer Electroless Deposition: A Scalable Approach to Surface-Modified Metal Powders April 16, 2014 Langmuir Patrick J. Cappillino , Joshua D. Sugar , Farid El Gabaly , Trevor Y. Cai , Zhi Liu , John L. Stickney ,and David B. Robinson
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189713 | https://chemistrytalk.org/standard-reduction-potentials-made-easy/ | Skip to content
Standard Reduction Potentials made easy
Standard reduction potentials are very useful in chemistry. They are also known as standard cell potentials, or standard electrode potentials. They are measured in volts, and they tell you how likely an element or ion is to be reduced by gaining electrons. We explain the concepts clearly, and give you a list of standard reduction potentials.
For example, fluorine, at the top of the table shown at the bottom of this article, has an extremely high affinity for electrons and will literally rip them off almost any other molecule it comes in contact with. The fluorine molecule will gain 2 electrons, one for each atom, with the result being two fluoride ions.
This table is often consulted when trying to determine a powerful enough reducing or oxidizing agent for a redox reaction, or determine which metals will displace others. They also represent the voltage of an electrochemical half-reaction.
All of the reactions listed in the table below are half-reactions, which means they need another half-reaction that transfers electrons in the opposite direction, to make a complete reaction.
What is a standard reduction potential?
Let’s define what a standard reduction potential is. Spontaneous redox (oxidation-reduction reactions) can produce electrical energy. A voltaic (galvanic) cell is an electrochemical cell that can capture this energy, using two half-cells, an anode, a cathode, and a salt bridge – similar to the photo at the top of this page.
The SRP is measured for a half-reaction by using hydrogen half-cell, known as a SHE (standard hydrogen electrode). The hydrogen half-cell is an arbitrary electrode that serves as a reference, and the cell potential for the hydrogen half-cell is set at zero.
What standard potentials tell you, and don’t tell you
When you calculate the standard potential for a reaction, a positive result tells you that the reaction should be spontaneous. It does not mean it will be, and it doesn’t tell you how fast the reaction will occur.
For example, based on calculations, aluminum should displace hydrogen from water. However, under normal conditions, it does not due to a passive aluminum oxide micro-coating that forms. But when you mix aluminum with gallium, you allow water to bypass the oxide coating, and you can see the reaction occur.
You can also use the standard potential of a reaction, to calculate the equilibrium constant for the reaction. You can calculate the Gibbs free energy change of the reaction. And you can use the Nernst equation to calculate the potential, free energy change, and equilibrium constant under non-standard conditions, such as the reactants and the products being a different concentration.
Reduction Potential Tips
Here are some useful tips when reading this table:
All standard reduction potentials are relative to hydrogen, which is assigned a value of zero. The potentials for all other reactions are measured by using what is known as a standard hydrogen electrode.
All reactions are listed as reduction reactions, where an element or ion gains electrons and takes on a more negative charge.
Note that elemental fluorine, at the top of the table, is the most powerful oxidizer, and that nothing in the table can oxidize fluoride to a positive ion.
Note that elemental lithium, at the bottom of the list of standard reduction potentials, is the most powerful reducer. You can even see lithium reduce potassium ions to potassium metal. When lithium acts a reducing agent, you reverse the equation and the charge listed in the table, giving the reaction a potential of +3.05 volts.
The more positive the potential, the more likely the half-reaction is to occur
These potentials are all measured in standard conditions (25 Celsius, 1 atm pressure, 1 molar concentration solution).
Reactions with H+ ions in the equation take place in acidic solutions. Reactions with OH- ions take place in a basic solution.
To get the potential of the reverse reaction, known as an “oxidation potential”, simply reverse the sign of the potential. For example, standard oxidation potential for the half reaction of fluoride ions to elemental fluorine has a potential of -2.87 volts (which means it is very difficult to make this reaction occur).
This table can be used to predict if a metal will replace another metal in solution. For example, to see if zinc metal will replace copper ions in solution, producing elemental copper, check if the potential of Zn->Zn+2 (+0.76) added to the potential of Cu+2->Cu (+0.34) is greater than zero (it is!)
Standard Reduction Potentials for common half-reactions
Common oxidizing and reducing agents
If you scan down the list, you will notice many common strong oxidizing agents in the lab near the top of the list, such as hydrogen peroxide, the peroxydisulfate ion, the permanganate ion, and the hypochlorite ion. The higher up in the list, the more powerful of an oxidizing agent the element/ion on the left side of the half-reaction is.
You’ll also see the nitrate ion, NO3– in the list. The nitrate ion is a medium-powered oxidizing agent, which is why nitric acid can react with many metals like copper that do not react with a non-oxidizing acid like hydrochloric acid.
Further down the list, you will see common reducing agents such as zinc metal, the tin (II) ion, and both the sulfite and thiosulfate ion which have similar reducing capability. The further down the list of standard reduction potentials, the more powerful the reducing agent on the right side of the half-reaction is.
Redox reactions are some of the most exciting reactions in chemistry. We hope you consult this table often!
What is the strongest oxidizing agent?
Fluorine, ozone, hydrogen peroxide, and the permanganate ion are considered some of the strongest oxidizing agents. That said, elemental fluorine is almost never used as an oxidizing agent due to how dangerous it is.
List of Standard Reduction Potentials
Standard Reduction Potential Table
| | |
--- |
| F2(g) + 2e-1 ———> 2F-1(aq) | +2.87 |
| O3(g) + 2H+1(aq) + 2e-1 ———> O2(g) + H2O(l) | +2.08 |
| S2O82-(aq) + 2e-1 ———> 2 SO42-(aq) | +2.05 |
| Au1+(aq) + e-1 ———> Au(s) | +1.83 |
| Co3+(aq) + e-1 ———> Co2+(aq) | +1.82 |
| H2O2(aq) + 2 H+1(aq) + 2e-1 ———> 2 H2O(l) | +1.77 |
| MnO4-1(aq) + 4 H+1(aq) + 3e-1 ———> MnO2(s) + 2 H2O(l) | +1.695 |
| PbO2(s) + SO42-(aq) + 4 H+1(aq) + 2e-1 ———> PbSO4(s) + 2 H2O(l) | +1.69 |
| 2 HOCl(aq) + 2 H+1(aq) + 2e-1 ———> Cl2(g) + 2 H2O(l) | +1.63 |
| Mn3+(aq) + e-1 ———> Mn2+(aq) | +1.51 |
| MnO4-1(aq) + 8 H+1(aq) + 5e-1 ———> Mn2+(aq) + 4 H2O(l) | +1.49 |
| ClO3-1(aq) + 12 H+1(aq) + 10e-1 ———>Cl2(g) + 6 H2O(l) | +1.49 |
| PbO2(s) + 4 H+1(aq) + 2e-1 ———> Pb2+(aq) + 2 H2O(l) | +1.46 |
| BrO3-1(aq) + 6 H+1(aq) + 6e-1 ———> Br-1(aq) + 3 H2O(l) | +1.44 |
| Ce4+(aq) + e-1 ———> Ce3+(aq) | +1.44 |
| Au3+(aq) + 3e-1 ———> Au(s) | +1.42 |
| Cl2(g) + 2e-1 ———> 2 Cl-1(aq) | +1.36 |
| Cr2O72-(aq) + 14 H+1(aq) + 6e-1 ———> 2 Cr3+(aq) + 7 H2O(l) | +1.33 |
| O3(g) + H2O(l) + 2e-1 ———> O2(g) + 2 OH-1(aq) | +1.24 |
| MnO2(s) + 4 H+1(aq) + 2e-1 ———> Mn2+(aq) + 2 H2O(l) | +1.23 |
| O2(g) + 4 H+1(aq) + 4e-1 ———> 2 H2O(l) | +1.23 |
| Pt2+(aq) + 2e-1 ———> Pt(s) | +1.20 |
| IO3-1(aq) + 5H+1(aq) + 4e-1 ———HIO(aq) + 2 H2O(l) | +1.13 |
| Br2(aq) + 2e-1 ———> 2 Br-1(aq) | +1.07 |
| NO3-1(aq) + 4 H+1(aq) + 3e-1 ———> NO(g) + 2 H2O(l) | +0.96 |
| NO3-1(aq) + 3 H+1(aq) + 2e-1 ———> HNO2(g) + H2O(l) | +0.94 |
| 2 Hg2+(aq) + 2e-1 ———> Hg22+(aq) | +0.91 |
| HO2-1(aq) + H2O(l) + 2e-1 ———> 3 OH-1(aq) | +0.87 |
| 2 NO3-1(aq) + 4 H+1(aq) + 2e-1 ———> 2 NO2(g) + 2H2O(l) | +0.80 |
| Ag+1(aq) + e-1 ———> Ag(s) | +0.80 |
| Fe3+(aq) + e-1 ———> Fe2+(aq) | +0.77 |
| O2(g) + 2H+1(aq) + 2e-1 ———> H2O2(aq) | +0.69 |
| I2(s) + 2e-1 ———> 2 I-1(aq) | +0.54 |
| NiO2(s) + 2 H2O(l) + 2e-1 ———> Ni(OH)2 + 2 OH-1(aq) | +0.49 |
| SO2(aq) + 4 H+1(aq) + 4e-1 ———> S(s) + 2 H2O(l) | +0.45 |
| O2(g) + 2 H2O(l) + 4e-1 ———> 4 OH-1(aq) | +0.401 |
| Cu2+(aq) + 2e-1 ———> Cu(s) | +0.34 |
| Hg2Cl2(s) + 2e-1 ———> 2 Hg(l) + 2 Cl-1(aq) | +0.27 |
| PbO2(s) + H2O(l) + 2e-1 ———> PbO(s) + 2 OH-1(aq) | +0.25 |
| AgCl(s) + e-1 ———> Ag(s) + Cl-1(aq) | +0.2223 |
| SO42-(aq) + 4H+1(aq) + 2e-1 ———> H2SO3(aq) + H2O(l) | +0.172 |
| S4O62-(aq) + 2e-1 ———> 2 S2O32-(aq) | +0.169 |
| Cu2+(aq) + e-1 ———> Cu+1(aq) | +0.16 |
| Sn4+(aq) + 2e-1 ———> Sn2+(aq) | +0.15 |
| S(s) + 2H+1(aq) + 2e-1 ———> H2S(g) | +0.14 |
| AgBr(s) + e-1 ———> Ag(s) + Br-1(aq) | +0.07 |
| 2 H+1(aq) + 2e-1 ———> H2(g) | 0.00 |
| Pb2+(aq) + 2e-1 ———> Pb(s) | -0.13 |
| Sn2+(aq) + 2e-1 ———> Sn(s) | -0.14 |
| AgI(s) + e-1 ———> Ag(s) + I-1(aq) | -0.15 |
| Ni2+(aq) + 2e-1 ———> Ni(s) | -0.25 |
| Co2+(aq) +2e-1 ———> Co(s) | -0.28 |
| In3+(aq) + 3e-1 ———> In(s) | -0.34 |
| Tl+1(aq) + e-1 ———> Tl(s) | -0.34 |
| PbSO4(s) + 2e-1 ———> Pb(s) + SO42-(aq) | -0.36 |
| Cd2+(aq) + 2e-1 ———> Cd(s) | -0.40 |
| Fe2+(aq) + 2e-1 ———> Fe(s) | -0.44 |
| Ga3+(aq) + 3e-1 ———> Ga(s) | -0.56 |
| PbO(s) + H2O(l) + 2e-1 ———> Pb(s) + 2 OH-1(aq) | -0.58 |
| Cr3-(aq) + 3e-1 ———> Cr(s) | -0.74 |
| Zn2+(aq) + 2e-1 ———> Zn(s) | -0.76 |
| 2 H2O(l) + 2e-1 ———> H2(g) + 2 OH-1(aq) | -0.83 |
| Fe(OH)2(s) + 2e-1 ———> Fe(s) + 2 OH-1(aq) | -0.88 |
| Cr2+(aq) + 2e-1 ———> Cr(s) | -0.91 |
| N2(g) + 4 H2O(l) + 4e-1 ———> N2O4(aq) +4 OH-1(aq) | -1.16 |
| V2+(aq) + 2e-1 ———> V(s) | -1.18 |
| ZnO2-1(aq) + 2 H2O(l) + 2e-1 ———> Zn(s) + 4OH-1(aq) | -1.216 |
| Ti2+(aq) + 2e-1 ———> Ti(s) | -1.63 |
| Al3+(aq) + 3e-1 ———> Al(s) | -1.66 |
| U3+(aq) + 3e-1 ———> U(s) | -1.79 |
| Sc3+(aq) + 3e-1 ———> Sc(s) | -2.02 |
| Er3+(aq) + 3e-1 ———> Er(s) | –2.33 |
| Ce3+(aq) + 3e-1 ———> Ce(s) | –2.34 |
| Pr3+(aq) + 3e-1 ———> Pr(s) | -2.35 |
| La3+(aq) + 3e-1 ———> La(s) | -2.36 |
| Y3+(aq) + 3e-1 ———> Y(s) | -2.37 |
| Mg2+(aq) + 2e-1 ———> Mg(s) | -2.37 |
| Na+1(aq) + e-1 ———> Na(s) | -2.71 |
| Ca2+(aq) + 2e-1 ———> Ca(s) | -2.76 |
| Sr2+(aq) + 2e-1 ———> Sr(s) | -2.89 |
| Ba2+(aq) + 2e-1 ———> Ba(s) | -2.90 |
| Cs+1(aq) +e-1 ———> Cs(s) | -2.92 |
| K+1(aq) + e-1 ———> K(s) | -2.92 |
| Rb+1(aq) + e-1 ———> Rb(s) | -2.93 |
| Li+1(aq) + e-1 ———> Li(s) | -3.05 |
List of standard reduction potentials
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189714 | https://www.math.purdue.edu/~zhan4306/spring_2022_la/sec_4_1.pdf | Math 26500 - Zecheng Zhang, Spring 2022 Section 4.1 Vector Spaces and Subspaces Definition (Vector space): Examples: Facts: For each u in V and scalar c, 0u = 0 c0 = 0 −u = (−1)u 1 addit µ Jiu V is unique w ut w e 0 -↳ -new) is scalar , muitip " °" also unique .
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= an = 0, PCH = 0 , zero polynomial , (zero in the Ipn) addition : qltl = bot bit-1 - .
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Math 26500 - Zecheng Zhang, Spring 2022 Example 1: Let V be the first quadrant in the xy-plane; that is, tlet V = ⇢ x y # : x ≥ 0,y ≥0 $ (1) If u and v are in V, is u+v in V?
(2) Find a specific vector u in V and a specific scalar c such that cu is not in V.
Definition (Subspace): Examples: 1. The set consisting of only the zero vector in a vector space V is a subspace of V , called the zero subspace and written as {0}.
2. Let P be the set of all polynomials with real coefficients, with operations in P defined as for functions. Then P is a subspace of the space of all real-valued functions defined on R.
3. The vector space R2 is not a subspace of R3 because R2 is not even a subset of R3.
4. A plane in R3 not through the origin is not a subspace of R3.
2 i' + i' = ( "+ " ñ=(¥1,420,410 yay,) b/c ✗ith 70 , Y ,-14270 F- (¥1 , ✗no , %:O ⇒ ñtv ' c-J, J is closed under addition .
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1121=11×11, ✗Ya c-IR) ↳in:{ III.✗4138414 Math 26500 - Zecheng Zhang, Spring 2022 Example 2: Determine if the given set is a subspace of Pn for an appropriate value of n.
(1) All polynomials of the form p(t) = at2, where a is in R (2) All polynomials of the form p(t) = a+t2, where a is in R Theorem: If v1,v2,···,vp are in a vector space V, then Span{v1,···,vp} is a subspace of V. We call Span{v1,···,vp} the subspace spanned (or generated) by v1,···,vp.
Example 3: Let W be a set of all vectors of the form 2 6 6 4 2s+4t 2s 2s−3t 5t 3 7 7 5. Show that W is a subspace of R4.
3 1h32) = { } ①If we set a -- 0 , pct) = at ' = 8 ⇒É is in this set .
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a, HR , q= azt, GEIR , Piaf C- V, ptf-la.tw# c- ✓✓ ③ p= at ' , CHR , cp = act ' ⇒cp c-V ⇒ closed .
TIR v:| } No, zero vector of Ipu is not in ✓⇒ V is not a subspace .
( 2s-10 ÷¥)= 1%1+1 " -¥+1 as-15T = si:/ + +1¥) -un u EIR " ✓c-1124 = span fñ , F) -1=4 this a subspace of 1124 .
Math 26500 - Zecheng Zhang, Spring 2022 Example 4: Let v1 = 2 4 1 0 −1 3 5, v2 = 2 4 2 1 3 3 5, v3 = 2 4 4 2 6 3 5, and w = 2 4 3 1 2 3 5.
(1) Is w in {v1,v2,v3}? How many vectors are in {v1,v2,v3}?
(2) How many vectors are in Span{v1,v2,v3}?
(3) Is w in the subspace spanned by {v1,v2,v3}?
4 |
189715 | https://openstax.org/books/physics/pages/12-4-applications-of-thermodynamics-heat-engines-heat-pumps-and-refrigerators | Skip to Content
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Physics
12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
Physics
12.4
Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
Search for key terms or text.
Section Learning Objectives
By the end of this section, you will be able to do the following:
Explain how heat engines, heat pumps, and refrigerators work in terms of the laws of thermodynamics
Describe thermal efficiency
Solve problems involving thermal efficiency
Teacher Support
Teacher Support
The learning objectives in this section will help your students master the following standards:
(6) Science concepts. The student knows that changes occur within a physical system and applies the laws of conservation of energy and momentum. The student is expected to:
(G) analyze and explain everyday examples that illustrate the laws of thermodynamics, including the law of conservation of energy and the law of entropy.
Section Key Terms
| cyclical process | heat engine | heat pump |
| thermal efficiency | | |
Teacher Support
Teacher Support
[BL][OL][AL]
Return again to the discussion of efficiency that was begun at the start of the module. Review the ideal gas law, laws of thermodynamics, and entropy.
[OL] Ask students whether they can explain the limits on efficiency in terms of what they have now learned.
Heat Engines, Heat Pumps, and Refrigerators
In this section, we’ll explore how heat engines, heat pumps, and refrigerators operate in terms of the laws of thermodynamics.
One of the most important things we can do with heat is to use it to do work for us. A heat engine does exactly this—it makes use of the properties of thermodynamics to transform heat into work. Gasoline and diesel engines, jet engines, and steam turbines that generate electricity are all examples of heat engines.
Figure 12.13 illustrates one of the ways in which heat transfers energy to do work. Fuel combustion releases chemical energy that heat transfers throughout the gas in a cylinder. This increases the gas temperature, which in turn increases the pressure of the gas and, therefore, the force it exerts on a movable piston. The gas does work on the outside world, as this force moves the piston through some distance. Thus, heat transfer of energy to the gas in the cylinder results in work being done.
Figure
12.13
(a) Heat transfer to the gas in a cylinder increases the internal energy of the gas, creating higher pressure and temperature. (b) The force exerted on the movable cylinder does work as the gas expands. Gas pressure and temperature decrease during expansion, indicating that the gas’s internal energy has decreased as it does work. (c) Heat transfer of energy to the environment further reduces pressure in the gas, so that the piston can more easily return to its starting position.
To repeat this process, the piston needs to be returned to its starting point. Heat now transfers energy from the gas to the surroundings, so that the gas’s pressure decreases, and a force is exerted by the surroundings to push the piston back through some distance.
A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. All heat engines use cyclical processes.
Heat engines do work by using part of the energy transferred by heat from some source. As shown in Figure 12.14, heat transfers energy, Qh, from the high-temperature object (or hot reservoir), whereas heat transfers unused energy, Qc, into the low-temperature object (or cold reservoir), and the work done by the engine is W. In physics, a reservoir is defined as an infinitely large mass that can take in or put out an unlimited amount of heat, depending upon the needs of the system. The temperature of the hot reservoir is Th, and the temperature of the cold reservoir is Tc .
Figure
12.14
(a) Heat transfers energy spontaneously from a hot object to a cold one, as is consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the energy transferred by heat to do work. The hot and cold objects are called the hot and cold reservoirs. Qh is the heat out of the hot reservoir, W is the work output, and Qc is the unused heat into the cold reservoir.
As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system’s internal energy, U, is the same at the beginning and end of every cycle—that is, ΔU=0
. The first law of thermodynamics states that ΔU=Q−W, where Q is the net heat transfer during the cycle, and W is the net work done by the system. The net heat transfer is the energy transferred in by heat from the hot reservoir minus the amount that is transferred out to the cold reservoir ( Q=Qh−Qc ). Because there is no change in internal energy for a complete cycle ( ΔU=0
), we have
0=Q−W,
12.19
so that
W=Q.
12.20
Therefore, the net work done by the system equals the net heat into the system, or
W=Qh−Qc
12.21
for a cyclical process.
Because the hot reservoir is heated externally, which is an energy-intensive process, it is important that the work be done as efficiently as possible. In fact, we want W to equal Qh, and for there to be no heat to the environment (that is, Qc=0 ). Unfortunately, this is impossible. According to the second law of thermodynamics, heat engines cannot have perfect conversion of heat into work. Recall that entropy is a measure of the disorder of a system, which is also how much energy is unavailable to do work. The second law of thermodynamics requires that the total entropy of a system either increases or remains constant in any process. Therefore, there is a minimum amount of Qh that cannot be used for work. The amount of heat rejected to the cold reservoir, Qc, depends upon the efficiency of the heat engine. The smaller the increase in entropy, ΔS
, the smaller the value of Qc, and the more heat energy is available to do work.
Heat pumps, air conditioners, and refrigerators utilize heat transfer of energy from low to high temperatures, which is the opposite of what heat engines do. Heat transfers energy Qc from a cold reservoir and delivers energy Qh into a hot one. This requires work input, W, which produces a transfer of energy by heat. Therefore, the total heat transfer to the hot reservoir is
Qh=Qc+W.
12.22
The purpose of a heat pump is to transfer energy by heat to a warm environment, such as a home in the winter. The great advantage of using a heat pump to keep your home warm rather than just burning fuel in a fireplace or furnace is that a heat pump supplies Qh=Qc+W . Heat Qc comes from the outside air, even at a temperature below freezing, to the indoor space. You only pay for W, and you get an additional heat transfer of Qc from the outside at no cost. In many cases, at least twice as much energy is transferred to the heated space as is used to run the heat pump. When you burn fuel to keep warm, you pay for all of it. The disadvantage to a heat pump is that the work input (required by the second law of thermodynamics) is sometimes more expensive than simply burning fuel, especially if the work is provided by electrical energy.
The basic components of a heat pump are shown in Figure 12.15. A working fluid, such as a refrigerant, is used. In the outdoor coils (the evaporator), heat Qc enters the working fluid from the cold outdoor air, turning it into a gas.
Figure
12.15
A simple heat pump has four basic components: (1) an evaporator, (2) a compressor, (3) a condenser, and (4) an expansion valve. In the heating mode, heat transfers Qc to the working fluid in the evaporator (1) from the colder, outdoor air, turning it into a gas. The electrically driven compressor (2) increases the temperature and pressure of the gas and forces it into the condenser coils (3) inside the heated space. Because the temperature of the gas is higher than the temperature in the room, heat transfers energy from the gas to the room as the gas condenses into a liquid. The working fluid is then cooled as it flows back through an expansion valve (4) to the outdoor evaporator coils.
The electrically driven compressor (work input W) raises the temperature and pressure of the gas and forces it into the condenser coils that are inside the heated space. Because the temperature of the gas is higher than the temperature inside the room, heat transfers energy to the room, and the gas condenses into a liquid. The liquid then flows back through an expansion (pressure-reducing) valve. The liquid, having been cooled through expansion, returns to the outdoor evaporator coils to resume the cycle.
The quality of a heat pump is judged by how much energy is transferred by heat into the warm space ( Qh ) compared with how much input work (W) is required.
Teacher Support
Teacher Support
Misconception Alert
Remember that refrigerators and air conditioners do not create cold. They merely transfer heat from the inside to the outside.
Revisit the ideal gas law, laws of thermodynamics, and entropy. Use these to understand the workings of air conditioners and refrigerators. This will also give you the opportunity to assess your understanding of these concepts. Both refrigerators and air conditioners use chemicals that can easily change phase from liquid to gas and back. The chemical is present in a closed circuit of tubing. Initially, it is in a gaseous state. The compressor works to squeeze the gas particles of the chemical closer together, creating high pressure. Following the ideal gas law, as pressure increases, so does temperature. This hot, dense gas spreads out in the small pipes or fins of the condenser, which is located on the outside part of the air conditioner (and backside of a refrigerator). The fins come in contact with outside air, which is cooler than the compressed chemical, and hence, as entropy indicates, heat transfers energy from the hot condenser to the relatively cooler air. The result is that the gas cools and condenses into a liquid. This liquid is then allowed to go to an evaporator through a tiny, narrow hole. On the other side of the hole, the gas spreads out (entropy increases), and its pressure drops. Consequently, obeying the ideal gas law, its temperature decreases as well. A fan blows air over this now-cool evaporator and into the room or refrigerator (Figure 12.16).
Figure
12.16
Heat pumps, air conditioners, and refrigerators are heat engines operated backward. Almost every home contains a refrigerator. Most people don’t realize that they are also sharing their homes with a heat pump.
Air conditioners and refrigerators are designed to cool substances by transferring energy by heat Qc out of a cool environment to a warmer one, where heat Qh is given up. In the case of a refrigerator, heat is moved out of the inside of the fridge into the surrounding room. For an air conditioner, heat is transferred outdoors from inside a home. Heat pumps are also often used in a reverse setting to cool rooms in the summer.
As with heat pumps, work input is required for heat transfer of energy from cold to hot. The quality of air conditioners and refrigerators is judged by how much energy is removed by heat Qc from a cold environment, compared with how much work, W, is required. So, what is considered the energy benefit in a heat pump, is considered waste heat in a refrigerator.
Thermal Efficiency
In the conversion of energy into work, we are always faced with the problem of getting less out than we put in. The problem is that, in all processes, there is some heat Qc that transfers energy to the environment—and usually a very significant amount at that. A way to quantify how efficiently a machine runs is through a quantity called thermal efficiency.
We define thermal efficiency, Eff, to be the ratio of useful energy output to the energy input (or, in other words, the ratio of what we get to what we spend). The efficiency of a heat engine is the output of net work, W, divided by heat-transferred energy, Qh, into the engine; that is
Eff=WQh.
An efficiency of 1, or 100 percent, would be possible only if there were no heat to the environment ( Qc=0
).
Tips For Success
All values of heat ( Qh and Qc ) are positive; there is no such thing as negative heat. The direction of heat is indicated by a plus or minus sign. For example, Qc is out of the system, so it is preceded by a minus sign in the equation for net heat.
Q=Qh−Qc
12.23
Solving Thermal Efficiency Problems
Worked Example
Daily Work Done by a Coal-Fired Power Station and Its Efficiency
A coal-fired power station is a huge heat engine. It uses heat to transfer energy from burning coal to do work to turn turbines, which are used then to generate electricity. In a single day, a large coal power station transfers 2.50×1014J by heat from burning coal and transfers 1.48×1014J by heat into the environment. (a) What is the work done by the power station? (b) What is the efficiency of the power station?
Strategy
We can use W=Qh−Qc
to find the work output, W, assuming a cyclical process is used in the power station. In this process, water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators and then condensed back to water to start the cycle again.
Solution
Work output is given by
W=Qh−Qc.
12.24
Substituting the given values,
W=2.50×1014J−1.48×1014J=1.02×1014J.
12.25
Strategy
The efficiency can be calculated with Eff=WQh, because Qh is given, and work, W, was calculated in the first part of this example.
Solution
Efficiency is given by
Eff=WQh.
12.26
The work, W, is found to be 1.02×1014J, and Qh is given ( 2.50×1014J ), so the efficiency is
Eff=1.02×1014J2.50×1014J=0.408, or 40.8%.
12.27
Discussion
The efficiency found is close to the usual value of 42 percent for coal-burning power stations. It means that fully 59.2 percent of the energy is transferred by heat to the environment, which usually results in warming lakes, rivers, or the ocean near the power station and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants—including plants fired by nuclear fuel, oil, and natural gas—the energy transferred by heat to the environment could be, and sometimes is, used for heating homes or for industrial processes.
Practice Problems
17
.
A heat engine is given 120J by heat and releases 20J by heat to the environment. What is the amount of work done by the system?
−100J
−60J
60J
100J
A heat engine takes in 6.0 kJ from heat and produces waste heat of 4.8 kJ. What is its efficiency?
25 percent
2.50 percent
2.00 percent
20 percent
Check Your Understanding
Teacher Support
Teacher Support
Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.
19
.
What is a heat engine?
A heat engine converts mechanical energy into thermal energy.
A heat engine converts thermal energy into mechanical energy.
A heat engine converts thermal energy into electrical energy.
A heat engine converts electrical energy into thermal energy.
Give an example of a heat engine.
A generator
A battery
A water pump
A car engine
21
.
What is thermal efficiency?
Thermal efficiency is the ratio of work input to the energy input.
Thermal efficiency is the ratio of work output to the energy input.
Thermal efficiency is the ratio of work input to the energy output.
Thermal efficiency is the ratio of work output to the energy output.
22
.
What is the mathematical expression for thermal efficiency?
Eff=QhQh−Qc
Eff=QhQc
Eff=QcQh
Eff=Qh−QcQh
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189716 | https://rosettacode.org/wiki/Magic_squares_of_odd_order | Published Time: 2025-07-21T13:20:21Z
Magic squares of odd order - Rosetta Code
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Beginning
1 11l
2 360 Assembly
3 Ada
4 ALGOL 68
5 ALGOL W
6 APL
7 AppleScript
8 Arturo
9 AutoHotkey
10 AWK
11 BASICToggle BASIC subsection
11.1 Applesoft BASIC
11.2 BASIC256
11.3 Chipmunk Basic
11.4 FreeBASIC
11.5 GW-BASIC
11.6 IS-BASIC
11.7 Liberty BASIC
11.8 MSX Basic
11.9 PureBasic
11.10 QB64
11.11 uBasic/4tH
11.12 VBA
11.13 Visual Basic
11.14 Visual Basic .NET
11.15 Yabasic
12 Batch File
13 bc
14 BCPL
15 Befunge
16 BQN
17 C
18 C++
19 CLU
20 Common Lisp
21 Cowgol
22 DToggle D subsection
22.1 Alternative Version
23 Delphi
24 Draco
25 EasyLang
26 EchoLisp
27 EDSAC order code
28 Elixir
29 ERRE
30 Factor
31 Fortran
32 Frink
33 Go
34 HaskellToggle Haskell subsection
34.1 Translating imperative code
34.2 Transpose . cycled
34.3 Siamese method
35 Icon and Unicon
36 J
37 Java
38 JavaScriptToggle JavaScript subsection
38.1 ES5
38.2 ES6
38.2.1 Cycled . transposed . cycled
38.2.2 Traditional 'Siamese' method
39 jq
40 Julia
41 Kotlin
42 Lua
43 Mathematica /Wolfram Language
44 Maxima
45 Nim
46 Oforth
47 PARI/GP
48 PascalToggle Pascal subsection
48.1 improved
49 PascalABC.NET
50 Perl
51 Phix
52 PicatToggle Picat subsection
52.1 Testing a larger instance
53 PicoLisp
54 PL/I
55 Pluto
56 PythonToggle Python subsection
56.1 Procedural
56.2 Composition of pure functions
57 R
58 Racket
59 Raku
60 REXX
61 Ring
62 RPL
63 Ruby
64 Rust
65 Scala
66 Seed7
67 Sidef
68 Stata
69 Swift
70 TclToggle Tcl subsection
70.1 TI-83 BASIC
71 VBScript
72 V (Vlang)
73 VTL-2
74 Wren
75 XPL0
76 zkl
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From Rosetta Code
Magic squares of odd order
You are encouraged to solve this task according to the task description, using any language you may know.
A magic square is an NxN square matrix whose numbers (usually integers) consist of consecutive numbers arranged so that the sum of each row and column, and both long (main) diagonals are equal to the same sum (which is called the magic number or magic constant).
The numbers are usually (but not always) the first N2 positive integers.
A magic square whose rows and columns add up to a magic number but whose main diagonals do not, is known as a semimagic square.
816
357
492
Task
For any odd N, generate a magic square with the integers 1 ──► N, and show the results here.
Optionally, show the magic number.
You should demonstrate the generator by showing at least a magic square for N = 5.
Related tasks
Magic squares of singly even order
Magic squares of doubly even order
See also
MathWorld™ entry: Magic_square
Odd Magic Squares (1728.org)
11l
Translation of: Python
F magic(n)
L(row) 1..n
print(((1..n).map(col -> @n ((@row + col - 1 + @n I/ 2) % @n)
+ ((@row + 2 col - 2) % @n) + 1)).map(cell -> String(cell).rjust(String(@n ^ 2).len)).join(‘ ’))
print("\nAll sum to magic number #.".format((n n + 1) n I/ 2))
L(n) (5, 3, 7)
print("\nOrder #.\n=======".format(n))
magic(n)
Output:
Order 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
All sum to magic number 65
Order 3
8 1 6
3 5 7
4 9 2
All sum to magic number 15
Order 7
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
All sum to magic number 175
360 Assembly
Translation of: C
Magic squares of odd order - 20/10/2015
MAGICS CSECT
USING MAGICS,R15 set base register
LA R6,1 i=1
LOOPI C R6,N do i=1 to n
BH ELOOPI
LR R8,R6 i
SLA R8,1 i2
LA R9,PG pgi=@pg
LA R7,1 j=1
LOOPJ C R7,N do j=1 to n
BH ELOOPJ
LR R5,R8 i2
SR R5,R7 -j
A R5,N +n
BCTR R5,0 -1
XR R4,R4 clear high reg
D R4,N /n
LR R5,R4 //n
M R4,N n
LR R2,R5 (i2-j+n-1)//nn
LR R5,R8 i2
AR R5,R7 -j
S R5,=F'2' -2
XR R4,R4 clear high reg
D R4,N /n
AR R2,R4 +(i2+j-2)//n
LA R2,1(R2) +1
XDECO R2,PG+80 (i2-j+n-1)//nn+(i2+j-2)//n+1
MVC 0(5,R9),PG+87 put in buffer
LA R9,5(R9) pgi=pgi+5
LA R7,1(R7) j=j+1
B LOOPJ
ELOOPJ XPRNT PG,80
LA R6,1(R6) i=i+1
B LOOPI
ELOOPI XR R15,R15 set return code
BR R14 return to caller
N DC F'9' <== input
PG DC CL92' ' buffer
YREGS
END MAGICS
Output:
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
Ada
with Ada.Text_IO, Ada.Command_Line;
procedure Magic_Square is
N: constant Positive := Positive'Value(Ada.Command_Line.Argument(1));
subtype Constants is Natural range 1 .. NN;
package CIO is new Ada.Text_IO.Integer_IO(Constants);
Undef: constant Natural := 0;
subtype Index is Natural range 0 .. N-1;
function Inc(I: Index) return Index is (if I = N-1 then 0 else I+1);
function Dec(I: Index) return Index is (if I = 0 then N-1 else I-1);
A: array(Index, Index) of Natural := (others => (others => Undef));
-- initially undefined; at the end holding the magic square
X: Index := 0; Y: Index := N/2; -- start position for the algorithm
begin
for I in Constants loop -- write 1, 2, ..., NN into the magic array
A(X, Y) := I; -- write I into the magic array
if A(Dec(X), Inc(Y)) = Undef then
X := Dec(X); Y := Inc(Y); -- go right-up
else
X := Inc(X); -- go down
end if;
end loop;
for Row in Index loop -- output the magic array
for Collumn in Index loop
CIO.Put(A(Row, Collumn),
Width => (if NN < 10 then 2 elsif NN < 100 then 3 else 4));
end loop;
Ada.Text_IO.New_Line;
end loop;
end Magic_Square;
Output:
./magic_square 3
8 1 6
3 5 7
4 9 2
./magic_square 11
68 81 94 107 120 1 14 27 40 53 66
80 93 106 119 11 13 26 39 52 65 67
92 105 118 10 12 25 38 51 64 77 79
104 117 9 22 24 37 50 63 76 78 91
116 8 21 23 36 49 62 75 88 90 103
7 20 33 35 48 61 74 87 89 102 115
19 32 34 47 60 73 86 99 101 114 6
31 44 46 59 72 85 98 100 113 5 18
43 45 58 71 84 97 110 112 4 17 30
55 57 70 83 96 109 111 3 16 29 42
56 69 82 95 108 121 2 15 28 41 54
ALGOL 68
construct a magic square of odd order
PROC magic square = ( INT order ) [,]INT:
IF NOT ODD order OR order < 1
THEN
# can't make a magic square of the specified order #
LOC [ 1 : 0, 1 : 0 ]INT
ELSE
# order is OK - construct the square using de la Loubère's #
# algorithm as in the wikipedia page #
[ 1 : order, 1 : order ]INT square;
FOR i TO order DO FOR j TO order DO square[ i, j ] := 0 OD OD;
# as square [ 1, 1 ] if the top-left, moving "up" reduces the row #
# operator to advance "up" the square #
OP PREV = ( INT pos )INT: IF pos = 1 THEN order ELSE pos - 1 FI;
# operator to advance "across right" or "down" the square #
OP NEXT = ( INT pos )INT: ( pos MOD order ) + 1;
# fill in the square, starting from the middle of the top row #
INT col := ( order + 1 ) OVER 2;
INT row := 1;
FOR i TO order order DO
square[ row, col ] := i;
IF square[ PREV row, NEXT col ] /= 0
THEN
# the up/right position is already taken, move down #
row := NEXT row
ELSE
# can move up and right #
row := PREV row;
col := NEXT col
FI
OD;
square
FI # magic square # ;
prints the magic square
PROC print square = ( [,]INT square )VOID:
BEGIN
INT order = 1 UPB square;
# calculate print width: negative so a leading "+" is not printed #
INT width := -1;
INT mag := order order;
WHILE mag >= 10 DO mag OVERAB 10; width MINUSAB 1 OD;
# calculate the "magic sum" #
INT sum := 0;
FOR i TO order DO sum +:= square[ 1, i ] OD;
# print the square #
print( ( "maqic square of order ", whole( order, 0 ), ": sum: ", whole( sum, 0 ), newline ) );
FOR i TO order DO
FOR j TO order DO write( ( " ", whole( square[ i, j ], width ) ) ) OD;
write( ( newline ) )
OD
END # print square # ;
test the magic square generation
FOR order BY 2 TO 7 DO print square( magic square( order ) ) OD
Output:
maqic square of order 1: sum: 1
1
maqic square of order 3: sum: 15
8 1 6
3 5 7
4 9 2
maqic square of order 5: sum: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
maqic square of order 7: sum: 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
ALGOL W
begin
% construct a magic square of odd order - as a procedure can't return an %
% array, the caller must supply one that is big enough %
logical procedure magicSquare( integer array square ( , )
; integer value order
) ;
if not odd( order ) or order < 1 then begin
% can't make a magic square of the specified order %
false
end
else begin
% order is OK - construct the square using de la Loubère's %
% algorithm as in the wikipedia page %
% ensure a row/col position is on the square %
integer procedure inSquare( integer value pos ) ;
if pos < 1 then order else if pos > order then 1 else pos;
% move "up" a row in the square %
integer procedure up ( integer value row ) ; inSquare( row - 1 );
% move "accross right" in the square %
integer procedure right( integer value col ) ; inSquare( col + 1 );
integer row, col;
% initialise square %
for i := 1 until order do for j := 1 until order do square( i, j ) := 0;
% initial position is the middle of the top row %
col := ( order + 1 ) div 2;
row := 1;
% construct square %
for i := 1 until ( order order ) do begin
square( row, col ) := i;
if square( up( row ), right( col ) ) not = 0 then begin
% the up/right position is already taken, move down %
row := row + 1;
end
else begin
% can move up/right %
row := up( row );
col := right( col );
end
end for_i;
% sucessful result %
true
end magicSquare ;
% prints the magic square %
procedure printSquare( integer array square ( , )
; integer value order
) ;
begin
integer sum, w;
% set integer width to accomodate the largest number in the square %
w := ( order order ) div 10;
i_w := s_w := 1;
while w > 0 do begin i_w := i_w + 1; w := w div 10 end;
for i := 1 until order do sum := sum + square( 1, i );
write( "maqic square of order ", order, ": sum: ", sum );
for i := 1 until order do begin
write( square( i, 1 ) );
for j := 2 until order do writeon( square( i, j ) )
end for_i
end printSquare ;
% test the magic square generation %
integer array sq ( 1 :: 11, 1 :: 11 );
for i := 1, 3, 5, 7 do begin
if magicSquare( sq, i ) then printSquare( sq, i )
else write( "can't generate square" );
end for_i
end.
Output:
maqic square of order 1 : sum: 1
1
maqic square of order 3 : sum: 15
8 1 6
3 5 7
4 9 2
maqic square of order 5 : sum: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
maqic square of order 7 : sum: 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
APL
Works with: Dyalog APL
Translation of: C
magic←{⍵{+/1,(1 ⍺⍺)×⍺(⍺⍺|1+⊢+2×⊣)⍵,⍺⍺-⍵+1}/¨⎕IO-⍨⍳⍵ ⍵}
Output:
magic¨ 1 3 5 7
1 2 9 4 2 23 19 15 6 2 45 39 33 27 21 8
7 5 3 14 10 1 22 18 18 12 6 49 36 30 24
6 1 8 21 17 13 9 5 34 28 15 9 3 46 40
8 4 25 16 12 43 37 31 25 19 13 7
20 11 7 3 24 10 4 47 41 35 22 16
26 20 14 1 44 38 32
42 29 23 17 11 5 48
AppleScript
Translation of: JavaScript
Composing functions ( cycleRows . transpose . cycleRows ), and lifting AppleScript handlers into first class script objects, to allow for first class functions and closures.
---------------- MAGIC SQUARE OF ODD ORDER ---------------
-- oddMagicSquare:: Int ->
on oddMagicSquare(n)
if 0 < (n mod 2) then
cycleRows(transpose(cycleRows(table(n))))
else
missing value
end if
end oddMagicSquare
--------------------------- TEST -------------------------
on run
-- Orders 3, 5, 11
-- wikiTableMagic:: Int -> String
script wikiTableMagic
on |λ|(n)
formattedTable(oddMagicSquare(n))
end |λ|
end script
intercalate(linefeed & linefeed, map(wikiTableMagic, {3, 5, 11}))
end run
-- table:: Int ->
on table(n)
set lstTop to enumFromTo(1, n)
script cols
on |λ|(row)
script rows
on |λ|(x)
(row n) + x
end |λ|
end script
map(rows, lstTop)
end |λ|
end script
map(cols, enumFromTo(0, n - 1))
end table
-- cycleRows:: ->
on cycleRows(lst)
script rotationRow
-- rotatedList:: [a] -> Int -> [a]
on rotatedList(lst, n)
if n = 0 then return lst
set lng to length of lst
set m to (n + lng) mod lng
items -m thru -1 of lst & items 1 thru (lng - m) of lst
end rotatedList
on |λ|(row, i)
rotatedList(row, (((length of row) + 1) div 2) - (i))
end |λ|
end script
map(rotationRow, lst)
end cycleRows
-------------------- GENERIC FUNCTIONS -------------------
-- intercalate:: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- map:: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn:: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- enumFromTo:: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
-- splitOn:: Text -> Text -> [Text]
on splitOn(strDelim, strMain)
set {dlm, my text item delimiters} to {my text item delimiters, strDelim}
set xs to text items of strMain
set my text item delimiters to dlm
return xs
end splitOn
-- transpose:: ->
on transpose(xss)
script column
on |λ|(_, iCol)
script row
on |λ|(xs)
item iCol of xs
end |λ|
end script
map(row, xss)
end |λ|
end script
map(column, item 1 of xss)
end transpose
----------------------- WIKI DISPLAY ---------------------
-- formattedTable:: -> String
on formattedTable(lstTable)
set n to length of lstTable
set w to 2.5 n
"magic(" & n & ")" & linefeed & linefeed & wikiTable(lstTable, ¬
false, "text-align:center;width:" & ¬
w & "em;height:" & w & "em;table-layout:fixed;")
end formattedTable
-- wikiTable:: [Text] -> Bool -> Text -> Text
on wikiTable(xs, blnHdr, strStyle)
script wikiRows
on |λ|(lstRow, iRow)
set strDelim to cond(blnHdr and (iRow = 0), "!", "|")
set strDbl to strDelim & strDelim
linefeed & "|-" & linefeed & strDelim & space & ¬
intercalate(space & strDbl & space, lstRow)
end |λ|
end script
linefeed & "{| class=\"wikitable\" " & ¬
cond(strStyle ≠ "", "style=\"" & strStyle & "\"", "") & ¬
intercalate("", ¬
map(wikiRows, xs)) & linefeed & "|}" & linefeed
end wikiTable
-- cond:: Bool -> a -> a -> a
on cond(bool, f, g)
if bool then
f
else
g
end if
end cond
Output:
magic(3)
8 3 4
1 5 9
6 7 2
magic(5)
17 23 4 10 11
24 5 6 12 18
1 7 13 19 25
8 14 20 21 2
15 16 22 3 9
magic(11)
68 80 92 104 116 7 19 31 43 55 56
81 93 105 117 8 20 32 44 45 57 69
94 106 118 9 21 33 34 46 58 70 82
107 119 10 22 23 35 47 59 71 83 95
120 11 12 24 36 48 60 72 84 96 108
1 13 25 37 49 61 73 85 97 109 121
14 26 38 50 62 74 86 98 110 111 2
27 39 51 63 75 87 99 100 112 3 15
40 52 64 76 88 89 101 113 4 16 28
53 65 77 78 90 102 114 5 17 29 41
66 67 79 91 103 115 6 18 30 42 54
Arturo
oddMagicSquare: function [n][
ensure -> and? odd? n
n >= 0
map 1..n 'i [
map 1..n 'j [
(n ((i + (j - 1) + n / 2) % n)) +
(((i - 2) + 2 j) % n) + 1
]
]
]
loop [3 5 7] 'n [
print ["Size:" n ", Magic sum:" n(1+nn)/2 "\n"]
loop oddMagicSquare n 'row [
loop row 'item [
prints pad to :string item 3
]
print ""
]
print ""
]
Output:
Size: 3 , Magic sum: 15
8 1 6
3 5 7
4 9 2
Size: 5 , Magic sum: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Size: 7 , Magic sum: 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
AutoHotkey
msgbox % OddMagicSquare(5)
msgbox % OddMagicSquare(7)
return
OddMagicSquare(oddN){
sq := oddN2
obj := {}
loop % oddN
obj[A_Index] := {} ; dis is row
mid := Round((oddN+1)/2)
sum := Round(sq(sq+1)/2/oddN)
obj[mid] := 1
cR := 1 , cC := mid
loop % sq-1
{
done := 0 , a := A_index+1
while !done {
nR := cR-1 , nC := cC+1
if !nR
nR := oddN
if (nC>oddN)
nC := 1
if obj[nR][nC] ;filled
cR += 1
else cR := nR , cC := nC
if !obj[cR][cC]
obj[cR][cC] := a , done := 1
}
}
str := "Magic Constant for " oddN "x" oddN " is " sum "`n"
for k,v in obj
{
for k2,v2 in v
str .= " " v2
str .= "`n"
}
return str
}
Output:
Magic Constant for 5x5 is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic Constant for 7x7 is 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
AWK
syntax: GAWK -f MAGIC_SQUARES_OF_ODD_ORDER.AWK
BEGIN {
build(5)
build(3,1) # verify sum
build(7)
exit(0)
}
function build(n,check, arr,i,width,x,y) {
if (n !~ /^[0-9]$/ || n < 3) {
printf("error: %s is invalid\n",n)
return
}
printf("\nmagic constant for %dx%d is %d\n",n,n,(nn+1)n/2)
x = 0
y = int(n/2)
for (i=1; i<=(nn); i++) {
arr[x,y] = i
if (arr[(x+n-1)%n,(y+n+1)%n]) {
x = (x+n+1) % n
}
else {
x = (x+n-1) % n
y = (y+n+1) % n
}
}
width = length(nn)
for (x=0; x<n; x++) {
for (y=0; y<n; y++) {
printf("%s ",width,arr[x,y])
}
printf("\n")
}
if (check) { verify(arr,n) }
}
function verify(arr,n, total,x,y) { # verify sum of each row, column and diagonal
print("\nverify")
horizontal
for (x=0; x<n; x++) {
total = 0
for (y=0; y<n; y++) {
printf("%d ",arr[x,y])
total += arr[x,y]
}
printf("\t: %d row %d\n",total,x+1)
}
vertical
for (y=0; y<n; y++) {
total = 0
for (x=0; x<n; x++) {
printf("%d ",arr[x,y])
total += arr[x,y]
}
printf("\t: %d column %d\n",total,y+1)
}
left diagonal
total = 0
for (x=y=0; x<n; x++ y++) {
printf("%d ",arr[x,y])
total += arr[x,y]
}
printf("\t: %d diagonal top left to bottom right\n",total)
right diagonal
x = n - 1
total = 0
for (y=0; y<n; y++ x--) {
printf("%d ",arr[x,y])
total += arr[x,y]
}
printf("\t: %d diagonal bottom left to top right\n",total)
}
Output:
magic constant for 5x5 is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
magic constant for 3x3 is 15
8 1 6
3 5 7
4 9 2
verify
8 1 6 : 15 row 1
3 5 7 : 15 row 2
4 9 2 : 15 row 3
8 3 4 : 15 column 1
1 5 9 : 15 column 2
6 7 2 : 15 column 3
8 5 2 : 15 diagonal top left to bottom right
4 5 6 : 15 diagonal bottom left to top right
magic constant for 7x7 is 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
BASIC
Applesoft BASIC
Even if the code works for any odd number, N=9 is the maximum for a 40 column wide screen. Line 130 is a user defined modulo function, and 140 helps calculate the addends for the number that will go in the current position.
100 :
110 REM MAGIC SQUARE OF ODD ORDER
120 :
130 DEF FN MOD(A) = A - INT (A / N) N
140 DEF FN NR(J) = FN MOD((J + 2 I + 1))
200 INPUT "ENTER N: ";N
210 IF N < 3 OR (N - INT (N / 2) 2) = 0 GOTO 200
220 FOR I = 0 TO (N - 1)
230 FOR J = 0 TO (N - 1): HTAB 4 (J + 1)
240 PRINT N FN NR(N - J - 1) + FN NR(J) + 1;
250 NEXT J: PRINT
260 NEXT I
270 PRINT "MAGIC CONSTANT: ";N (N N + 1) / 2
Output:
ENTER N: 5
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
MAGIC CONSTANT: 65
BASIC256
Translation of: Liberty BASIC
arraybase 1
global m
call magicSquare(5)
call magicSquare(17)
end
subroutine magicSquare(n)
redim m(n,n)
inc = 1
cont = 1
row = 1
col = (n+1) / 2
while cont <= nn
m[row,col] = cont
cont += 1
if inc < n then
inc += 1
row -= 1
col += 1
if row <> 0 then
if col > n then col = 1
else
row = n
end if
else
inc = 1
row += 1
end if
end while
call printSquare(n)
end subroutine
subroutine printSquare(n)
Arbitrary limit to fit width of A4 paper
if n < 23 then
print
print n; " x "; n; " Magic Square --- ";
print "Magic constant is "; int((nn+1)/2n)
for row = 1 to n
for col = 1 to n
print rjust(string(m[row,col]),4);
next col
print
next row
else
print "Magic Square will not fit on one sheet of paper."
end if
end subroutine
Output:
5 x 5 Magic Square --- Magic constant is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
17 x 17 Magic Square --- Magic constant is 2465
155 174 193 212 231 250 269 288 1 20 39 58 77 96 115 134 153
173 192 211 230 249 268 287 17 19 38 57 76 95 114 133 152 154
191 210 229 248 267 286 16 18 37 56 75 94 113 132 151 170 172
209 228 247 266 285 15 34 36 55 74 93 112 131 150 169 171 190
227 246 265 284 14 33 35 54 73 92 111 130 149 168 187 189 208
245 264 283 13 32 51 53 72 91 110 129 148 167 186 188 207 226
263 282 12 31 50 52 71 90 109 128 147 166 185 204 206 225 244
281 11 30 49 68 70 89 108 127 146 165 184 203 205 224 243 262
10 29 48 67 69 88 107 126 145 164 183 202 221 223 242 261 280
28 47 66 85 87 106 125 144 163 182 201 220 222 241 260 279 9
46 65 84 86 105 124 143 162 181 200 219 238 240 259 278 8 27
64 83 102 104 123 142 161 180 199 218 237 239 258 277 7 26 45
82 101 103 122 141 160 179 198 217 236 255 257 276 6 25 44 63
100 119 121 140 159 178 197 216 235 254 256 275 5 24 43 62 81
118 120 139 158 177 196 215 234 253 272 274 4 23 42 61 80 99
136 138 157 176 195 214 233 252 271 273 3 22 41 60 79 98 117
137 156 175 194 213 232 251 270 289 2 21 40 59 78 97 116 135
Chipmunk Basic
Works with: Chipmunk Basic version 3.6.4
Translation of: FreeBASIC
100 cls
110 sub magicsq(size,filename$ = "")
120 if (size and 1) = 0 or size < 3 then
130 print
140 print "error: size is not odd or size is smaller then 3"
160 exit sub
170 endif
180 ' filename$ <> "" then save magic square in a file
190 ' filename$ can contain directory name
200 ' if filename$ exist it will be overwriten, no error checking
210 dim sq(size,size)' array to hold square
220 ' start in the middle of the first row
230 nr = 1
240 x = size-int(size/2)
250 y = 1
260 max = sizesize
270 ' create format string for using
280 for c = 1 to len(str$(max))+1 : frmt$ = frmt$+"#" : next c
290 'main loop for creating magic square
300 do
310 if sq(x,y) = 0 then
320 sq(x,y) = nr
330 if nr mod size = 0 then
340 y = y+1
350 else
360 x = x+1
370 y = y-1
380 endif
390 nr = nr+1
400 endif
410 if x > size then
420 x = 1
430 do while sq(x,y) <> 0
440 x = x+1
450 loop
460 endif
470 if y < 1 then
480 y = size
490 do while sq(x,y) <> 0
500 y = y-1
510 loop
520 endif
530 loop until nr > max
540 ' printing square's bigger than 19 result in a wrapping of the line
550 print "Odd magic square size: ";size;"";size
560 print "The magic sum = ";int((max+1)/2)size
570 print
580 for y = 1 to size
590 for x = 1 to size
600 print using "####";val(sq(x,y));
610 next x
620 print
630 next y
640 print
650 ' output magic square to a file with the name provided
660 if filename$ <> "" then
670 nr = freefile
680 open filename$ for output as #1
690 print #1,"Odd magic square size: ";size;"";size
700 print #1,"The magic sum = ";int((max+1)/2)size
710 print #1,
720 for y = 1 to size
730 for x = 1 to size
740 print #1,using frmt$;sq(x,y);
750 next x
760 print #1,
770 next y
780 endif
790 close #1
800 end sub
810 input "Enter N: ",number
820 magicsq(number)
830 end
FreeBASIC
' version 23-06-2015
' compile with: fbc -s console
Sub magicsq(size As Integer, filename As String ="")
If (size And 1) = 0 Or size < 3 Then
Print : Beep ' alert
Print "error: size is not odd or size is smaller then 3"
Sleep 3000,1 'wait 3 seconds, ignore key press
Exit Sub
End If
' filename <> "" then save magic square in a file
' filename can contain directory name
' if filename exist it will be overwriten, no error checking
Dim As Integer sq(size,size) ' array to hold square
' start in the middle of the first row
Dim As Integer nr = 1, x = size - (size \ 2), y = 1
Dim As Integer max = size size
' create format string for using
Dim As String frmt = String(Len(Str(max)) +1, "#")
' main loop for creating magic square
Do
If sq(x, y) = 0 Then
sq(x, y) = nr
If nr Mod size = 0 Then
y += 1
Else
x += 1
y -= 1
End If
nr += 1
End If
If x > size Then
x = 1
Do While sq(x,y) <> 0
x += 1
Loop
End If
If y < 1 Then
y = size
Do While sq(x,y) <> 0
y -= 1
Loop
EndIf
Loop Until nr > max
' printing square's bigger than 19 result in a wrapping of the line
Print "Odd magic square size:"; size; " "; size
Print "The magic sum ="; ((max +1) \ 2) size
Print
For y = 1 To size
For x = 1 To size
Print Using frmt; sq(x,y);
Next
Print
Next
print
' output magic square to a file with the name provided
If filename <> "" Then
nr = FreeFile
Open filename For Output As #nr
Print #nr, "Odd magic square size:"; size; " "; size
Print #nr, "The magic sum ="; ((max +1) \ 2) size
Print #nr,
For y = 1 To size
For x = 1 To size
Print #nr, Using frmt; sq(x,y);
Next
Print #nr,
Next
End If
Close
End Sub
' ------=< MAIN >=------
magicsq(5)
magicsq(11)
' the next line will also print the square to a file called: magic_square_19.txt
magicsq(19, "magic_square_19.txt")
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
Odd magic square size: 5 5 Odd magic square size: 11 11
The magic sum = 65 The magic sum = 671
17 24 1 8 15 68 81 94 107 120 1 14 27 40 53 66
23 5 7 14 16 80 93 106 119 11 13 26 39 52 65 67
4 6 13 20 22 92 105 118 10 12 25 38 51 64 77 79
10 12 19 21 3 104 117 9 22 24 37 50 63 76 78 91
11 18 25 2 9 116 8 21 23 36 49 62 75 88 90 103
7 20 33 35 48 61 74 87 89 102 115
19 32 34 47 60 73 86 99 101 114 6
31 44 46 59 72 85 98 100 113 5 18
43 45 58 71 84 97 110 112 4 17 30
55 57 70 83 96 109 111 3 16 29 42
Only the first 2 square shown. 56 69 82 95 108 121 2 15 28 41 54
GW-BASIC
Works with: PC-BASIC version any
Works with: BASICA
Works with: Chipmunk Basic
Works with: QBasic
Works with: MSX BASIC
Translation of: IS-BASIC
100 REM Magic squares of odd order
110 INPUT "The square order: ", N
115 'INPUT "The square order:"; N ' for MSX Basic
120 IF (N AND 1) = 0 OR N < 3 THEN PRINT "error: size is not odd or size is smaller then 3" : GOTO 100
130 FOR I = 1 TO N
140 FOR J = 1 TO N
150 PRINT USING " ###"; ((I2-J+N-1) MOD N) N + ((I2+J-2) MOD N) + 1;
160 NEXT J
170 PRINT
180 NEXT I
190 PRINT "The magic number is: "; N (N^2+1) / 2
IS-BASIC
100 PROGRAM "MagicN.bas"
110 DO
120 INPUT PROMPT "The square order: ":N
130 LOOP UNTIL MOD(N,2)>0 AND INT(N)=N AND N>0
140 FOR I=1 TO N
150 FOR J=1 TO N
160 PRINT USING " ###":MOD((I2-J+N-1),N)N+MOD(I2+J-2,N)+1;
170 NEXT
180 PRINT
190 NEXT
200 PRINT "The magic number is:";N(N^2+1)/2
Liberty BASIC
Dim m(1,1)
Call magicSquare 5
Call magicSquare 17
End
Sub magicSquare n
ReDim m(n,n)
inc = 1
count = 1
row = 1
col=(n+1)/2
While count <= nn
m(row,col) = count
count = count + 1
If inc < n Then
inc = inc + 1
row = row - 1
col = col + 1
If row <> 0 Then
If col > n Then col = 1
Else
row = n
End If
Else
inc = 1
row = row + 1
End If
Wend
Call printSquare n
End Sub
Sub printSquare n
'Arbitrary limit to fit width of A4 paper
If n < 23 Then
Print n;" x ";n;" Magic Square --- ";
Print "Magic constant is ";Int((nn+1)/2n)
For row = 1 To n
For col = 1 To n
Print Using("####",m(row,col));
Next col
Print
Print
Next row
Else
Notice "Magic Square will not fit on one sheet of paper."
End If
End Sub
Output:
5 x 5 Magic Square --- Magic constant is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
17 x 17 Magic Square --- Magic constant is 2465
155 174 193 212 231 250 269 288 1 20 39 58 77 96 115 134 153
173 192 211 230 249 268 287 17 19 38 57 76 95 114 133 152 154
191 210 229 248 267 286 16 18 37 56 75 94 113 132 151 170 172
209 228 247 266 285 15 34 36 55 74 93 112 131 150 169 171 190
227 246 265 284 14 33 35 54 73 92 111 130 149 168 187 189 208
245 264 283 13 32 51 53 72 91 110 129 148 167 186 188 207 226
263 282 12 31 50 52 71 90 109 128 147 166 185 204 206 225 244
281 11 30 49 68 70 89 108 127 146 165 184 203 205 224 243 262
10 29 48 67 69 88 107 126 145 164 183 202 221 223 242 261 280
28 47 66 85 87 106 125 144 163 182 201 220 222 241 260 279 9
46 65 84 86 105 124 143 162 181 200 219 238 240 259 278 8 27
64 83 102 104 123 142 161 180 199 218 237 239 258 277 7 26 45
82 101 103 122 141 160 179 198 217 236 255 257 276 6 25 44 63
100 119 121 140 159 178 197 216 235 254 256 275 5 24 43 62 81
118 120 139 158 177 196 215 234 253 272 274 4 23 42 61 80 99
136 138 157 176 195 214 233 252 271 273 3 22 41 60 79 98 117
137 156 175 194 213 232 251 270 289 2 21 40 59 78 97 116 135
MSX Basic
Works with: MSX BASIC version any
Translation of: IS-BASIC
100 REM Magic squares of odd order
110 INPUT "The square order:"; N
120 IF (N AND 1) = 0 OR N < 3 THEN PRINT "error: size is not odd or size is smaller then 3" : GOTO 100
130 FOR I = 1 TO N
140 FOR J = 1 TO N
150 PRINT USING " ###"; ((I2-J+N-1) MOD N) N + ((I2+J-2) MOD N) + 1;
160 NEXT J
170 PRINT
180 NEXT I
190 PRINT "The magic number is:"; N (N^2+1) / 2
PureBasic
Translation of: Pascal
N=9
Define.i i,j
If OpenConsole("Magic squares")
PrintN("The square order is: "+Str(#N))
For i=1 To #N
For j=1 To #N
Print(RSet(Str((i2-j+#N-1) % #N#N + (i2+j-2) % #N+1),5))
Next
PrintN("")
Next
PrintN("The magic number is: "+Str(#N(#N#N+1)/2))
EndIf
Input()
Output:
The square order is: 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic number is: 369
QB64
_Title "Magic Squares of Odd Order"
'$Dynamic
DefLng A-Z
Dim Shared As Long m(1, 1)
Call magicSquare(5)
Call magicSquare(15)
Sleep
System
Sub magicSquare (n As Integer)
Dim As Integer inc, count, row, col
If (n < 3) Or (n And 1) <> 1 Then n = 3
ReDim m(n, n)
inc = 1
count = 1
row = 1
col = (n + 1) / 2
While count <= n n
m(row, col) = count
count = count + 1
If inc < n Then
inc = inc + 1
row = row - 1
col = col + 1
If row <> 0 Then
If col > n Then col = 1
Else
row = n
End If
Else
inc = 1
row = row + 1
End If
Wend
Call printSquare(n)
End Sub
Sub printSquare (n As Integer)
Dim As Integer row, col
'Arbitrary limit ensures a fit within console window
'Can be any size that fits within your computers memory limits
If n < 21 Then
Print "Order "; n; " Magic Square constant is "; Str$(Int((n n + 1) / 2 n))
For row = 1 To n
For col = 1 To n
Print Using "####"; m(row, col);
Next col
Print
' Print
Next row
End If
End Sub
Output:
Order 5 Magic Square constant is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Order 15 Magic Square constant is 1695
122 139 156 173 190 207 224 1 18 35 52 69 86 103 120
138 155 172 189 206 223 15 17 34 51 68 85 102 119 121
154 171 188 205 222 14 16 33 50 67 84 101 118 135 137
170 187 204 221 13 30 32 49 66 83 100 117 134 136 153
186 203 220 12 29 31 48 65 82 99 116 133 150 152 169
202 219 11 28 45 47 64 81 98 115 132 149 151 168 185
218 10 27 44 46 63 80 97 114 131 148 165 167 184 201
9 26 43 60 62 79 96 113 130 147 164 166 183 200 217
25 42 59 61 78 95 112 129 146 163 180 182 199 216 8
41 58 75 77 94 111 128 145 162 179 181 198 215 7 24
57 74 76 93 110 127 144 161 178 195 197 214 6 23 40
73 90 92 109 126 143 160 177 194 196 213 5 22 39 56
89 91 108 125 142 159 176 193 210 212 4 21 38 55 72
105 107 124 141 158 175 192 209 211 3 20 37 54 71 88
106 123 140 157 174 191 208 225 2 19 36 53 70 87 104
uBasic/4tH
Translation of: FreeBASIC
' ------=< MAIN >=------
Proc _magicsq(5)
Proc _magicsq(11)
End
_magicsq Param (1) Local (4)
' reset the array
For b@ = 0 to 255
@(b@) = 0
Next
If ((a@ % 2) = 0) + (a@ < 3) + (a@ > 15) Then
Print "error: size is not odd or size is smaller then 3 or bigger than 15"
Return
EndIf
' start in the middle of the first row
b@ = 1
c@ = a@ - (a@ / 2)
d@ = 1
e@ = a@ a@
' main loop for creating magic square
Do
If @(c@a@+d@) = 0 Then
@(c@a@+d@) = b@
If (b@ % a@) = 0 Then
d@ = d@ + 1
Else
c@ = c@ + 1
d@ = d@ - 1
EndIf
b@ = b@ + 1
EndIf
If c@ > a@ Then
c@ = 1
Do While @(c@a@+d@) # 0
c@ = c@ + 1
Loop
EndIf
If d@ < 1 Then
d@ = a@
Do While @(c@a@+d@) # 0
d@ = d@ - 1
Loop
EndIf
Until b@ > e@
Loop
Print "Odd magic square size: "; a@; " "; a@
Print "The magic sum = "; ((e@+1) / 2) a@
Print
For d@ = 1 To a@
For c@ = 1 To a@
Print Using "____"; @(c@a@+d@);
Next
Print
Next
Print
Return
Output:
Odd magic square size: 5 5
The magic sum = 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Odd magic square size: 11 11
The magic sum = 671
68 81 94 107 120 1 14 27 40 53 66
80 93 106 119 11 13 26 39 52 65 67
92 105 118 10 12 25 38 51 64 77 79
104 117 9 22 24 37 50 63 76 78 91
116 8 21 23 36 49 62 75 88 90 103
7 20 33 35 48 61 74 87 89 102 115
19 32 34 47 60 73 86 99 101 114 6
31 44 46 59 72 85 98 100 113 5 18
43 45 58 71 84 97 110 112 4 17 30
55 57 70 83 96 109 111 3 16 29 42
56 69 82 95 108 121 2 15 28 41 54
0 OK, 0:64
VBA
Translation of: C
Works with Excel VBA.
Sub magicsquare()
'Magic squares of odd order
Const n = 9
Dim i As Integer, j As Integer, v As Integer
Debug.Print "The square order is: " & n
For i = 1 To n
For j = 1 To n
Cells(i, j) = ((i 2 - j + n - 1) Mod n) n + ((i 2 + j - 2) Mod n) + 1
Next j
Next i
Debug.Print "The magic number of"; n; "x"; n; "square is:"; n (n n + 1) \ 2
End Sub 'magicsquare
Visual Basic
Translation of: C
Works with: Visual Basic version VB6 Standard
Sub magicsquare()
'Magic squares of odd order
Const n = 9
Dim i As Integer, j As Integer, v As Integer
Debug.Print "The square order is: " & n
For i = 1 To n
For j = 1 To n
v = ((i 2 - j + n - 1) Mod n) n + ((i 2 + j - 2) Mod n) + 1
Debug.Print Right(Space(5) & v, 5);
Next j
Debug.Print
Next i
Debug.Print "The magic number is: " & n (n n + 1) \ 2
End Sub 'magicsquare
Output:
The square order is: 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic number is: 369
Visual Basic .NET
Works with: Visual Basic .NET version 2011
Sub magicsquare()
'Magic squares of odd order
Const n = 9
Dim i, j, v As Integer
Console.WriteLine("The square order is: " & n)
For i = 1 To n
For j = 1 To n
v = ((i 2 - j + n - 1) Mod n) n + ((i 2 + j - 2) Mod n) + 1
Console.Write(" " & Right(Space(5) & v, 5))
Next j
Console.WriteLine("")
Next i
Console.WriteLine("The magic number is: " & n (n n + 1) \ 2)
End Sub 'magicsquare
Output:
The square order is: 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic number is: 369
Yabasic
Translation of: Liberty BASIC
magicSquare(5)
magicSquare(17)
end
sub magicSquare(n)
redim m(n,n)
inc = 1
cont = 1
row = 1
col = (n+1) / 2
while cont <= nn
m(row,col) = cont
cont = cont + 1
if inc < n then
inc = inc + 1
row = row - 1
col = col + 1
if row <> 0 then
if col > n col = 1
else
row = n
end if
else
inc = 1
row = row + 1
end if
end while
printSquare(n)
end sub
sub printSquare(n)
//Arbitrary limit to fit width of A4 paper
if n < 23 then
print "\n", n, " x ", n, " Magic Square --- ";
print "Magic constant is ", int((nn+1)/2n)
for row = 1 to n
for col = 1 to n
print m(row,col) using("####");
next col
print
next row
else
print "Magic Square will not fit on one sheet of paper."
end if
end sub
Output:
5 x 5 Magic Square --- Magic constant is 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
17 x 17 Magic Square --- Magic constant is 2465
155 174 193 212 231 250 269 288 1 20 39 58 77 96 115 134 153
173 192 211 230 249 268 287 17 19 38 57 76 95 114 133 152 154
191 210 229 248 267 286 16 18 37 56 75 94 113 132 151 170 172
209 228 247 266 285 15 34 36 55 74 93 112 131 150 169 171 190
227 246 265 284 14 33 35 54 73 92 111 130 149 168 187 189 208
245 264 283 13 32 51 53 72 91 110 129 148 167 186 188 207 226
263 282 12 31 50 52 71 90 109 128 147 166 185 204 206 225 244
281 11 30 49 68 70 89 108 127 146 165 184 203 205 224 243 262
10 29 48 67 69 88 107 126 145 164 183 202 221 223 242 261 280
28 47 66 85 87 106 125 144 163 182 201 220 222 241 260 279 9
46 65 84 86 105 124 143 162 181 200 219 238 240 259 278 8 27
64 83 102 104 123 142 161 180 199 218 237 239 258 277 7 26 45
82 101 103 122 141 160 179 198 217 236 255 257 276 6 25 44 63
100 119 121 140 159 178 197 216 235 254 256 275 5 24 43 62 81
118 120 139 158 177 196 215 234 253 272 274 4 23 42 61 80 99
136 138 157 176 195 214 233 252 271 273 3 22 41 60 79 98 117
137 156 175 194 213 232 251 270 289 2 21 40 59 78 97 116 135
Batch File
@echo off
rem Magic squares of odd order
setlocal EnableDelayedExpansion
set n=9
echo The square order is: %n%
for /l %%i in (1,1,%n%) do (
set w=
for /l %%j in (1,1,%n%) do (
set /a v1=%%i2-%%j+n-1
set /a v1=v1%%nn
set /a v2=%%i2+%%j+n-2
set /a v2=v2%%n
set /a v=v1+v2+1
set v= !v!
set w=!w!!v:~-5!)
echo !w!)
set /a w=n(nn+1)/2
echo The magic number is: %w%
pause
Output:
The square order is: 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic number is: 369
Press any key to continue ...
bc
Works with: GNU bc
define magic_constant(n) {
return(((n n + 1) / 2) n)
}
define print_magic_square(n) {
auto i, x, col, row, len, old_scale
old_scale = scale
scale = 0
len = length(n n)
print "Magic constant for n=", n, ": ", magic_constant(n), "\n"
for (row = 1; row <= n; row++) {
for (col = 1; col <= n; col++) {
x = n ((row + col - 1 + (n / 2)) % n) + \
((row + 2 col - 2) % n) + 1
for (i = 0; i < len - length(x); i++) {
print " "
}
print x
if (col != n) print " "
}
print "\n"
}
scale = old_scale
}
temp = print_magic_square(5)
Output:
Magic constant for n=5: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
BCPL
Translation of: C
get "libhdr"
let cell(n, x, y) = f(n, n-x-1, y)n + f(n, x, y) + 1
and f(n, x, y) = (x + y2 + 1) rem n
let magic(n) be
$( writef("Magic square of order %N with constant %N:N", n, (nn+1)/2n)
for y = 0 to n-1
$( for x = 0 to n-1 do writed(cell(n, x, y), 4)
wrch('N')
$)
wrch('N')
$)
let start() be for n = 1 to 7 by 2 do magic(n)
Output:
Magic square of order 1 with constant 1:
1
Magic square of order 3 with constant 15:
2 9 4
7 5 3
6 1 8
Magic square of order 5 with constant 65:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
Magic square of order 7 with constant 175:
2 45 39 33 27 21 8
18 12 6 49 36 30 24
34 28 15 9 3 46 40
43 37 31 25 19 13 7
10 4 47 41 35 22 16
26 20 14 1 44 38 32
42 29 23 17 11 5 48
Befunge
Translation of: C
The size, n, is specified by the first value on the stack.
500p0>:::00g%00g-1-\00g/2+1+00g%00g\:00g%v
@<$<_^#!-:g00:,+9!%g00:+1.+1+%g00+1+2/g00\<
Output:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
BQN
Translation of: C
Magic ← {𝕏{+´1∾1‿𝕗×𝕨(𝕗|1+⊢+2×⊣)𝕩∾𝕗-𝕩+1}´¨↕2⥊𝕩}
Magic¨ ⟨1,3,5,7⟩
Output:
┌─
· ┌─ ┌─ ┌─ ┌─
╵ 1 ╵ 2 9 4 ╵ 2 23 19 15 6 ╵ 2 45 39 33 27 21 8
┘ 7 5 3 14 10 1 22 18 18 12 6 49 36 30 24
6 1 8 21 17 13 9 5 34 28 15 9 3 46 40
┘ 8 4 25 16 12 43 37 31 25 19 13 7
20 11 7 3 24 10 4 47 41 35 22 16
┘ 26 20 14 1 44 38 32
42 29 23 17 11 5 48
┘
┘
C
Generates an associative magic square. If the size is larger than 3, the square is also panmagic.
include
include
int f(int n, int x, int y)
{
return (x + y2 + 1)%n;
}
int main(int argc, char argv)
{
int i, j, n;
//Edit: Add argument checking
if(argc!=2) return 1;
//Edit: Input must be odd and not less than 3.
n = atoi(argv);
if (n < 3 || (n%2) == 0) return 2;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++)
printf("% 4d", f(n, n - j - 1, i)n + f(n, j, i) + 1);
putchar('\n');
}
printf("\n Magic Constant: %d.\n", (nn+1)/2n);
return 0;
}
Output:
$ ./magic 5
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
Magic Constant: 65.
C++
include
include
include
include
include
using namespace std;
class MagicSquare
{
public:
MagicSquare(int d) : sqr(dd,0), sz(d)
{
assert(d&1);
fillSqr();
}
void display()
{
cout << "Odd Magic Square: " << sz << " x " << sz << "\n";
cout << "It's Magic Sum is: " << magicNumber() << "\n\n";
ostringstream cvr;
cvr << sz sz;
int l = cvr.str().size();
for( int y = 0; y < sz; y++ )
{
int yy = y sz;
for( int x = 0; x < sz; x++ )
cout << setw( l + 2 ) << sqr[yy + x];
cout << "\n";
}
cout << "\n\n";
}
private:
void fillSqr()
{
int sx = sz / 2, sy = 0, c = 0;
while( c < sz sz )
{
if( !sqr[sx + sy sz] )
{
sqr[sx + sy sz]= c + 1;
inc( sx ); dec( sy );
c++;
}
else
{
dec( sx ); inc( sy ); inc( sy );
}
}
}
int magicNumber()
{ return sz ( ( sz sz ) + 1 ) / 2; }
void inc( int& a )
{ if( ++a == sz ) a = 0; }
void dec( int& a )
{ if( --a < 0 ) a = sz - 1; }
bool checkPos( int x, int y )
{ return( isInside( x ) && isInside( y ) && !sqr[sz y + x] ); }
bool isInside( int s )
{ return ( s < sz && s > -1 ); }
vector sqr;
int sz;
};
int main()
{
MagicSquare s(7);
s.display();
return 0;
}
Output:
Odd Magic Square: 7 x 7
It's Magic Sum is: 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
CLU
magic_square = cluster is create, unparse, magic_number
rep = array[array[int]]
create = proc (order: int) returns (cvt) signals (invalid)
if order<1 cor order//2 = 0 then signal invalid end
sq: rep := rep$fill_copy(1, order, array[int]$fill(1, order, 0))
x: int := (order+1)/2
y: int := 1
for i: int in int$from_to(1, order2) do
sq[y][x] := i
next_x: int := inc(sq,x)
next_y: int := dec(sq,y)
if sq[next_y][next_x]=0
then x, y := next_x, next_y
else y := inc(sq,y)
end
end
return(sq)
end create
inc = proc (sq: rep, co: int) returns (int)
order: int := rep$size(sq)
if co=order then return(1) else return(co+1) end
end inc
dec = proc (sq: rep, co: int) returns (int)
order: int := rep$size(sq)
if co=1 then return(order) else return(co-1) end
end dec
unparse = proc (sq: cvt) returns (string)
order: int := rep$size(sq)
col_size: int := string$size(int$unparse(order 2)) + 1
ss: stream := stream$create_output()
for y: int in int$from_to(1, order) do
for x: int in int$from_to(1, order) do
stream$putright(ss, int$unparse(sq[y][x]), col_size)
end
stream$putl(ss, "")
end
return(stream$get_contents(ss))
end unparse
magic_number = proc (sq: cvt) returns (int)
order: int := rep$size(sq)
n: int := 0
for x: int in int$from_to(1, order) do n := n + sq[x] end
return(n)
end magic_number
end magic_square
print_magic_square = proc (order: int)
po: stream := stream$primary_output()
ms: magic_square := magic_square$create(order)
stream$putl(po, "Magic square of order "
|| int$unparse(order)
|| " with magic number "
|| int$unparse(magic_square$magic_number(ms))
|| ": ")
stream$putl(po, magic_square$unparse(ms))
end print_magic_square
start_up = proc ()
for n: int in int$from_to_by(1, 7, 2) do
print_magic_square(n)
end
end start_up
Output:
Magic square of order 1 with magic number 1:
1
Magic square of order 3 with magic number 15:
8 1 6
3 5 7
4 9 2
Magic square of order 5 with magic number 65:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic square of order 7 with magic number 175:
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
Common Lisp
(defun magic-square (n)
(loop for i from 1 to n
collect
(loop for j from 1 to n
collect
(+ ( n (mod (+ i j (floor n 2) -1)
n))
(mod (+ i ( 2 j) -2)
n)
1))))
(defun magic-constant (n)
( n
(/ (1+ ( n n))
2)))
(defun output (n)
(format T "Magic constant for n=~a: ~a~%" n (magic-constant n))
(let ((size (length (write-to-string ( n n))))
(format-str (format NIL "~~{~~{~~~ad~~^ ~~}~~%~~}~~%" size)))
(format T format-str (magic-square n))))
Output:
(output 5)
Magic constant for n=5: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Cowgol
Translation of: C
include "cowgol.coh";
sub magic(n: uint16) is
sub f(x: uint16, y: uint16): (r: uint16) is
r := (x + y2 + 1) % n;
end sub;
sub cell(x: uint16, y: uint16): (c: uint16) is
c := f(n-x-1, y)n + f(x, y) + 1;
end sub;
var y: uint16 := 0;
while y < n loop
var x: uint16 := 0;
loop
print_i16(cell(x, y));
x := x + 1;
if x == n then
print_nl();
break;
else
print_char('\t');
end if;
end loop;
y := y + 1;
end loop;
print_nl();
end sub;
var n: uint16 := 1;
while n <= 7 loop
print("Magic square of order ");
print_i16(n);
print(" with constant ");
print_i16((nn+1)/2n);
print(":\n");
magic(n);
n := n + 2;
end loop;
Output:
Magic square of order 1 with constant 1:
1
Magic square of order 3 with constant 15:
2 9 4
7 5 3
6 1 8
Magic square of order 5 with constant 65:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
Magic square of order 7 with constant 175:
2 45 39 33 27 21 8
18 12 6 49 36 30 24
34 28 15 9 3 46 40
43 37 31 25 19 13 7
10 4 47 41 35 22 16
26 20 14 1 44 38 32
42 29 23 17 11 5 48
D
Translation of: Python
void main(in string[] args)
{
import std.stdio, std.conv, std.range, std.algorithm, std.exception;
immutable n = args.length == 2 ? args.to!uint : 5;
enforce(n > 0 && n % 2 == 1, "Only odd n > 1");
immutable len = text(n ^^ 2).length.text;
// writeln(len);
foreach (immutable r; 1 .. n + 1)
{
foreach (immutable c; 1 .. n + 1)
{
auto a = (n ((r + c - 1 + (n / 2)) % n)) + ((r + (2 c) - 2) % n) + 1;
// n(( I + J - 1 + ( n / 2 ) ) mod n ) + (( I + 2J - 2 ) mod n ) + 1
// writeln("n = ",n, " r = ",r," c = ",c, " a = ",a );
writef("%" ~ len ~ "d%s",a, " ");
}
writeln("");
}
;
writeln("\nMagic constant: ", ((n n + 1) n) / 2);
}}
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic constant: 65
Alternative Version
Translation of: C
import std.stdio, std.conv, std.string, std.range, std.algorithm;
uint[][] magicSquare(immutable uint n) pure nothrow @safe
in {
assert(n > 0 && n % 2 == 1);
} out(mat) {
// mat is square of the right size.
assert(mat.length == n);
assert(mat.all!(row => row.length == n));
immutable magic = mat.sum;
// The sum of all rows is the same magic number.
assert(mat.all!(row => row.sum == magic));
// The sum of all columns is the same magic number.
//assert(mat.transposed.all!(col => col.sum == magic));
assert(mat.dup.transposed.all!(col => col.sum == magic));
// The sum of the main diagonals is the same magic number.
assert(mat.enumerate.map!(ir => ir[ir]).sum == magic);
//assert(mat.enumerate.map!({i, r} => r[i]).sum == magic);
assert(mat.enumerate.map!(ir => ir[ir]).sum == magic);
} body {
enum M = (in uint x) pure nothrow @safe @nogc => (x + n - 1) % n;
auto m = new uint[];
uint i = 0;
uint j = n / 2;
foreach (immutable uint k; 1 .. n ^^ 2 + 1) {
m[i][j] = k;
if (m[M(i)][M(j)]) {
i = (i + 1) % n;
} else {
i = M(i);
j = M(j);
}
}
return m;
}
void showSquare(in uint[][] m)
in {
assert(m.all!(row => row.length == m.length));
} body {
immutable maxLen = text(m.length ^^ 2).length.text;
writefln("%(%(%" ~ maxLen ~ "d%)\n%)", m);
writeln("\nMagic constant: ", m.sum);
}
int main(in string[] args) {
if (args.length == 1) {
5.magicSquare.showSquare;
return 0;
} else if (args.length == 2) {
immutable n = args.to!uint;
if (n > 0 && n % 2 == 1) {
n.magicSquare.showSquare;
return 0;
}
}
stderr.writefln("Requires n odd and larger than 0.");
return 1;
}
Output:
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
Magic constant: 65
Delphi
See Pascal.
Draco
proc inc(word n, order) word: if n=order-1 then 0 else n+1 fi corp
proc dec(word n, order) word: if n=0 then order-1 else n-1 fi corp
proc odd_magic_square([,]word square) void:
word order, x, nx, y, ny, i;
order := dim(square,1);
for x from 0 upto order-1 do
for y from 0 upto order-1 do
square[x,y] := 0
od
od;
x := order/2;
y := 0;
for i from 1 upto orderorder do
square[x,y] := i;
nx := inc(x,order);
ny := dec(y,order);
if square[nx,ny] = 0 then
x := nx;
y := ny
else
y := inc(y,order)
fi
od
corp
proc digit_count(word n) word:
word count;
count := 0;
while n > 0 do
count := count + 1;
n := n / 10
od;
count
corp
proc print_magic_square([,]word square) void:
word order, max, col_size, magic, x, y;
order := dim(square,1);
max := orderorder;
col_size := digit_count(max) + 1;
magic := 0;
for x from 0 upto order-1 do magic := magic + square[x,0] od;
writeln("Magic square of order ",order," with magic number ",magic,":");
for y from 0 upto order-1 do
for x from 0 upto order-1 do write(square[x,y]:col_size) od;
writeln()
od;
writeln()
corp
proc main() void:
[1,1]word sq1;
[3,3]word sq3;
[5,5]word sq5;
[7,7]word sq7;
odd_magic_square(sq1);
odd_magic_square(sq3);
odd_magic_square(sq5);
odd_magic_square(sq7);
print_magic_square(sq1);
print_magic_square(sq3);
print_magic_square(sq5);
print_magic_square(sq7)
corp
Output:
Magic square of order 1 with magic number 1:
1
Magic square of order 3 with magic number 15:
8 1 6
3 5 7
4 9 2
Magic square of order 5 with magic number 65:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic square of order 7 with magic number 175:
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
EasyLang
Translation of: C
func f n x y .
return (x + y 2 + 1) mod n
.
numfmt 3 0
proc msqr n .
for i = 0 to n - 1
for j = 0 to n - 1
write f n (n - j - 1) i n + f n j i + 1
.
print ""
.
.
msqr 5
Output:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
EchoLisp
The make-ms procedure allows to construct different magic squares for a same n, by modifying the grid filling moves. (see MathWorld reference)
(lib 'matrix)
;; compute next i,j = f(move,i,j)
(define-syntax-rule (path imove jmove)
(begin (set! i (imove i n)) (set! j (jmove j n))))
;; We define the ordinary and break moves
;; (1 , -1), (0, 1) King's move
(define (inext i n) (modulo (1+ i) n))
(define (jnext j n) (modulo (1- j) n))
(define (ibreak i n) i)
(define (jbreak j n) (modulo (1+ j) n))
(define (make-ms n)
(define n2+1 (1+ ( n n)))
(define ms (make-array n n))
(define i (quotient n 2))
(define j 0)
(array-set! ms i j 1)
(for ((ns (in-range 2 n2+1)))
(if (zero? (array-ref ms (inext i n ) (jnext j n )))
(path inext jnext) ;; ordinary move if empty target
(path ibreak jbreak)) ;; else break move
(if (zero? (array-ref ms i j))
(array-set! ms i j ns)
(error ns "illegal path"))
)
(writeln 'order n 'magic-number (/ ( n n2+1) 2))
(array-print ms))
Output:
(make-ms 7)
order 7 magic-number 175
30 38 46 5 13 21 22
39 47 6 14 15 23 31
48 7 8 16 24 32 40
1 9 17 25 33 41 49
10 18 26 34 42 43 2
19 27 35 36 44 3 11
28 29 37 45 4 12 20
;; Changing the moves allow to generate other magic squares
;; (2 ,1) (1,-2) Knight's move !
(define (inext i n) (modulo (+ 2 i) n))
(define (jnext j n) (modulo (1+ j) n))
(define (ibreak i n) (modulo (1+ i) n))
(define (jbreak j n) (modulo (- j 2) n))
(make-ms 7)
order 7 magic-number 175
37 48 3 14 18 22 33
11 15 26 30 41 45 7
34 38 49 4 8 19 23
1 12 16 27 31 42 46
24 35 39 43 5 9 20
47 2 13 17 28 32 36
21 25 29 40 44 6 10
;; (2 ,1) (1,-1)
(define (inext i n) (modulo (+ 2 i) n))
(define (jnext j n) (modulo (1+ j) n))
(define (ibreak i n) (modulo (1+ i) n))
(define (jbreak j n) (modulo (1- j) n))
(make-ms 7)
order 7 magic-number 175
48 22 3 33 14 37 18
30 11 41 15 45 26 7
19 49 23 4 34 8 38
1 31 12 42 16 46 27
39 20 43 24 5 35 9
28 2 32 13 36 17 47
10 40 21 44 25 6 29
EDSAC order code
[Magic squares of odd order, for Rosetta Code.
EDSAC program, Initial Orders 2.]
[The entries in a magic square of side n can be written as nu + v + 1,
where u and v range independently over 0, ..., n - 1.
Let the cells be labelled by (x, y) coordinates, where
x = column (left = 0), y = row (bottom = 0).
If n is odd then magic squares can be constructed by setting
u = cx + dy + h (mod n)
v = ex + fy + k (mod n)
where c, d, e, f, h, k are suitable constants.
Define m = (n - 1)/2. The values of c, ..., k for various methods
of construction are as follows:
c, d, e, f, h, k
Bachet: m + 1, m, m + 1, m + 1, m, 0
De la Loubere: 1, 2m, 2, 2m, m, 0
Conway (lozenge): 1, 2m, 1, 1, m, m + 1
Rosetta Code C: 2m, 2m - 1, 1, 2m - 1, 2m - 1, 2m
------------------------------------------------------------------------------]
[Arrange the storage]
T45K P56F [H parameter: subroutine to print string]
T46K P100F [N parameter: subroutine to print number]
T47K P200F [M parameter: main routine + high-level subroutine]
[Main routine + non-library subroutine]
E25K TM GK
[Rows are printed in the order y = n - 1 (top) to y = 0 (bottom).
Row y = n is a fictitious row used during initialization.]
[Locations set up by main routine; some are changed by subroutine]
PF [m]
PF [n]
PF [n^2]
PF [n 2^11]
PF [c, changed to n(n - c) = dec to nu on inc(x)]
PF [d, changed to nd = dec to nu on dec(y)]
PF [e, changed to n - e = dec to v on inc(x)]
PF [f = dec to v when dec(y)]
PF [h, changed to nu for start of current row]
PF [k, changed to v for start of current row]
[Locations used only by subroutine]
PF [nu]
PF [v]
PF [x count]
PF [y count]
[Subroutine to print magic square, using parameters set up by main routine.]
A3F T77@ [plant return link as usual]
A80@ T1F [set to print leading zeros as spaces]
A1@ S6@ T6@ [replace e by n - e]
A1@ S4@ T4@ [replace c by n - c]
[Multiply certain values by n. To maintain the integer scaling,
products have to be shifted 16 left before storing.]
H3@ V8@ [acc := (n << 11)h]
L8F T8@ [shift 5 more left and store nh]
V5@ L8F T5@ [similarly nd]
V4@ L8F T4@ [similarly n(n - c)]
[Loop round rows y := n - 1 down to 0. At the moment y = n.]
S1@ T13@ [initialize negative count of rows (y values)]
[Start of a row. Here acc = 0]
S1@ T12@ [inititialize negative count of columns (x values)]
A8@ S5@ [decrement nu by nd]
E42@ [skip if nu >= 0]
A2@ [else inc nu by n^2]
U8@ [store updated nu for next time]
T10@ [also copy to initialize this row]
A9@ S7@ [decrement v by f]
E48@ [skip if v >= 0]
A1@ [else inc v by n]
U9@ [store updated v at for next time]
U11@ [also copy to initialize this row]
[Next column. Here acc = v]
A10@ A78@ TF [cell value v + nu + 1 to 0F for printing]
A53@ GN [call subroutine to print cell value]
A12@ A78@ [increment negative column count]
E70@ [jump if row is complete]
T12@ [else update count]
A10@ S4@ [dec nu by nn - c)]
E63@ [skip if nU >= 0]
A2@ [else inc nu by n^2]
T10@ [store updated nu for next time]
A11@ S6@ [dec v by n - e]
E68@ [skip if v >= 0]
A1@ [else inc v by n]
U11@ [store updated v, keep v in acc]
E50@ [loop back for next cell in row]
[Row finished]
O81@ O82@ [print CR LF]
A13@ A78@ [inc negative row count]
E77@ [exit if done all rows]
T13@ E36@ [else update count and loop back]
ZF [(planted) jump back to caller]
[Constants]
PD [17-bit 1]
K4096F [null]
!F [space]
@F [carriage return]
&F [line feed]
P10F [for testing number of phone pulses]
[Strings for printing. K2048F sets letters mode; K4096F is EDSAC null.]
K2048FMFAFGFIFCF!FSFQFUFAFRFEF!FOFFF!FOFRFDFEFRF!F#FWFFMF#FZFQF@F&FK4096F
K2048FDFIFAFLF!FMF!F#FKFPF!FFTFOF!FCFAFNFCFEFLF#FLF@F&FK4096F
K2048FBFAFCFHFEFTF#FCF@F&FK4096F
K2048FDFEF!FLFAF!FLFOFUFBFEFRFEF#FCF@F&FK4096F
K2048FCFOFNFWFAFYF#FCF@F&FK4096F
K2048FRFOFSFEFTFTFAF!FCFOFDFEF!FCF#FCF@F&FK4096F
[Enter with acc = 0]
A207@ GH A84@ [print heading]
A210@ GH A117@ [prompt user to dial m, where n = 2m + 1]
ZF [halt program; restarts when user dials]
[Here acc holds number of pulses in address field.
Number of pulses = 10 if user dialled '0', else = number that user dialled.]
S83@ E292@ [test for '0', jump to exit if so]
A83@ [restore acc after test]
L512F [shift m to top 5 bits for printing]
UF [temp to 0F]
OF O81@ O82@ [print digit m, plus CR LF]
R512F [restore m in address field, same as 2m right-justified]
A78@ U1@ [make and store n = 2m + 1, right justified]
RD T@ [make and store m right-justified]
A1@ L512F T3@ [make and store n << 11]
H3@ V1@ [acc := (n << 11)n]
L8F [shift 5 left for integer scaling]
T2@ [store n^2]
[Bachet's method]
A234@ GH A144@ [print name of method]
A@ U5@ U8@ [d, h := m]
A78@ U4@ U6@ T7@ [c, e, f := m + 1]
T9@ [k := 0]
A245@ G14@ [call s/r to print square]
[De la Loubere's (miscalled Siamese) method]
A247@ GH A156@ [print name of method]
A78@ U4@ [c := 1]
LD T6@ [e := 2]
A@ U8@ [h := m]
LD U5@ T7@ [d, f := 2m]
T9@ [k := 0 ]
A260@ G14@ [call s/r to print square]
[Conway's lozenge method turns out to be of this type]
A262@ GH A175@ [print name of method]
A78@ U4@ U6@ U7@ [c, e, f := 1]
A@ T9@ [k := m + 1]
A@ U8@ [h := m]
LD T5@ [d := 2m]
A275@ G14@ [call s/r to print square]
[C solution on Rosetta Code website]
A277@ GH A187@ [print name of method]
A78@ T6@ [e := 1]
A@ LD U4@ U9@ [c, k := 2m]
S78@ U5@ U7@ T8@ [d, f, h := 2m - 1]
A290@
G14@ [call s/r to print square]
O79@ [done; print null to flush teleprinter buffer]
ZF [halt the machine]
E25K TH
[Subroutine to print a string.
Input: A order for first character must follow subroutine call (G order).
String is terminated with EDSAC null, which is sent to the teleprinter.]
GKA18@U17@S19@T4@AFT6@AFUFOFE12@A20@G16@TFA6@A2FG5@TFZFU3FU1FK2048F
E25K TN
[Subroutine to print non-negative 17-bit integer.
Parameters: 0F = integer to be printed (not preserved)
1F = character for leading zero (preserved)
Workspace: 4F..7F, 38 locations]
GKA3FT34@A1FT7FS35@T6FT4#FAFT4FH36@V4FRDA4#FR1024FH37@E23@O7FA2F
T6FT5FV4#FYFL8FT4#FA5FL1024FUFA6FG16@OFTFT7FA6FG17@ZFP4FZ219DTF
[================ M parameter again ================]
E25K TM GK
E207Z [define entry point]
PF [acc = 0 on entry]
Output:
MAGIC SQUARE OF ORDER 2M+1
DIAL M (0 TO CANCEL)
2
BACHET:
3 16 9 22 15
20 8 21 14 2
7 25 13 1 19
24 12 5 18 6
11 4 17 10 23
DE LA LOUBERE:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
CONWAY:
18 24 5 6 12
22 3 9 15 16
1 7 13 19 25
10 11 17 23 4
14 20 21 2 8
ROSETTA CODE C:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
Elixir
Translation of: Ruby
defmodule RC do
def odd_magic_square(n) when rem(n,2)==1 do
for i <- 0..n-1 do
for j <- 0..n-1, do: n rem(i+j+1+div(n,2),n) + rem(i+2j+2n-5,n) + 1
end
end
def print_square(sq) do
width = List.flatten(sq) |> Enum.max |> to_char_list |> length
fmt = String.duplicate(" ~#{width}w", length(sq)) <> "~n"
Enum.each(sq, fn row -> :io.format fmt, row end)
end
end
Enum.each([3,5,11], fn n ->
IO.puts "\nSize #{n}, magic sum #{div(nn+1,2)n}"
RC.odd_magic_square(n) |> RC.print_square
end)
Output:
Size 3, magic sum 15
8 1 6
3 5 7
4 9 2
Size 5, magic sum 65
16 23 5 7 14
22 4 6 13 20
3 10 12 19 21
9 11 18 25 2
15 17 24 1 8
Size 11, magic sum 671
73 86 99 101 114 6 19 32 34 47 60
85 98 100 113 5 18 31 44 46 59 72
97 110 112 4 17 30 43 45 58 71 84
109 111 3 16 29 42 55 57 70 83 96
121 2 15 28 41 54 56 69 82 95 108
1 14 27 40 53 66 68 81 94 107 120
13 26 39 52 65 67 80 93 106 119 11
25 38 51 64 77 79 92 105 118 10 12
37 50 63 76 78 91 104 117 9 22 24
49 62 75 88 90 103 116 8 21 23 36
61 74 87 89 102 115 7 20 33 35 48
ERRE
PROGRAM MAGIC_SQUARE
!$INTEGER
PROCEDURE Magicsq(size,filename$)
LOCAL DIM sq[25,25] ! array to hold square
IF (size AND 1)=0 OR size<3 THEN
PRINT PRINT(CHR$(7)) ! beep
PRINT("error: size is not odd or size is smaller then 3")
PAUSE(3)
EXIT PROCEDURE
END IF
! filename$ <> "" then save magic square in a file
! filename$ can contain directory name
! if filename$ exist it will be overwriten, no error checking
! start in the middle of the first row
nr=1 x=size-(size DIV 2) y=1
max=sizesize
! create format string for using
frmt$=STRING$(LEN(STR$(max)),"#")
! main loop for creating magic square
REPEAT
IF sq[x,y]=0 THEN
sq[x,y]=nr
IF nr MOD size=0 THEN
y=y+1
ELSE
x=x+1
y=y-1
END IF
nr=nr+1
END IF
IF x>size THEN
x=1
WHILE sq[x,y]<>0 DO
x=x+1
END WHILE
END IF
IF y<1 THEN
y=size
WHILE sq[x,y]<>0 DO
y=y-1
END WHILE
END IF
UNTIL nr>max
! printing square's bigger than 19 result in a wrapping of the line
PRINT("Odd magic square size:";size;"";size)
PRINT("The magic sum =";((max+1) DIV 2)size)
PRINT
FOR y=1 TO size DO
FOR x=1 TO size DO
WRITE(frmt$;sq[x,y];)
END FOR
PRINT
END FOR
! output magic square to a file with the name provided
IF filename$<>"" THEN
OPEN("O",1,filename$)
PRINT(#1,"Odd magic square size:";size;" ";size)
PRINT(#1,"The magic sum =";((max+1) DIV 2)size)
PRINT(#1,)
FOR y=1 TO size DO
FOR x=1 TO size DO
WRITE(#1,frmt$;sq[x,y];)
END FOR
PRINT(#1,)
END FOR
END IF
CLOSE(1)
END PROCEDURE
BEGIN
PRINT(CHR$(12);) ! CLS
Magicsq(5,"")
Magicsq(11,"")
!----------------------------------------------------
! the next line will also print the square to a file
! called 'magic_square_19txt'
!----------------------------------------------------
Magicsq(19,"msq_19.txt")
END PROGRAM
Output:
Same as FreeBasic version
Odd magic square size: 5 5 Odd magic square size: 11 11
The magic sum = 65 The magic sum = 671
17 24 1 8 15 68 81 94 107 120 1 14 27 40 53 66
23 5 7 14 16 80 93 106 119 11 13 26 39 52 65 67
4 6 13 20 22 92 105 118 10 12 25 38 51 64 77 79
10 12 19 21 3 104 117 9 22 24 37 50 63 76 78 91
11 18 25 2 9 116 8 21 23 36 49 62 75 88 90 103
7 20 33 35 48 61 74 87 89 102 115
19 32 34 47 60 73 86 99 101 114 6
31 44 46 59 72 85 98 100 113 5 18
43 45 58 71 84 97 110 112 4 17 30
55 57 70 83 96 109 111 3 16 29 42
Only the first 2 square shown. 56 69 82 95 108 121 2 15 28 41 54
Factor
This solution uses the method from the paper linked in the J entry:
USING: formatting io kernel math math.matrices math.ranges
sequences sequences.extras ;
IN: rosetta-code.magic-squares-odd
: inc-matrix ( n -- matrix )
[ 0 ] dip dup [ 1 + dup ] make-matrix nip ;
: rotator ( n -- seq ) 2/ dup [ neg ] dip [a,b] ;
: odd-magic-square ( n -- matrix )
[ inc-matrix ] [ rotator [ rotate ] 2map flip ] dup tri ;
: show-square ( n -- )
dup "Order: %d\n" printf odd-magic-square dup
[ [ "%4d" printf ] each nl ] each first sum
"Magic number: %d\n\n" printf ;
3 5 11 [ show-square ] tri@
Output:
Order: 3
8 1 6
3 5 7
4 9 2
Magic number: 15
Order: 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic number: 65
Order: 11
68 81 94 107 120 1 14 27 40 53 66
80 93 106 119 11 13 26 39 52 65 67
92 105 118 10 12 25 38 51 64 77 79
104 117 9 22 24 37 50 63 76 78 91
116 8 21 23 36 49 62 75 88 90 103
7 20 33 35 48 61 74 87 89 102 115
19 32 34 47 60 73 86 99 101 114 6
31 44 46 59 72 85 98 100 113 5 18
43 45 58 71 84 97 110 112 4 17 30
55 57 70 83 96 109 111 3 16 29 42
56 69 82 95 108 121 2 15 28 41 54
Magic number: 671
Fortran
Works with: Fortran version 95 and later
program Magic_Square
implicit none
integer, parameter :: order = 15
integer :: i, j
write(, "(a, i0)") "Magic Square Order: ", order
write(, "(a)") "----------------------"
do i = 1, order
do j = 1, order
write(, "(i4)", advance = "no") f1(order, i, j)
end do
write(,)
end do
write(, "(a, i0)") "Magic number = ", f2(order)
contains
integer function f1(n, x, y)
integer, intent(in) :: n, x, y
f1 = n mod(x + y - 1 + n/2, n) + mod(x + 2y - 2, n) + 1
end function
integer function f2(n)
integer, intent(in) :: n
f2 = n (1 + n n) / 2
end function
end program
Output:
Magic Square Order: 15
122 139 156 173 190 207 224 1 18 35 52 69 86 103 120
138 155 172 189 206 223 15 17 34 51 68 85 102 119 121
154 171 188 205 222 14 16 33 50 67 84 101 118 135 137
170 187 204 221 13 30 32 49 66 83 100 117 134 136 153
186 203 220 12 29 31 48 65 82 99 116 133 150 152 169
202 219 11 28 45 47 64 81 98 115 132 149 151 168 185
218 10 27 44 46 63 80 97 114 131 148 165 167 184 201
9 26 43 60 62 79 96 113 130 147 164 166 183 200 217
25 42 59 61 78 95 112 129 146 163 180 182 199 216 8
41 58 75 77 94 111 128 145 162 179 181 198 215 7 24
57 74 76 93 110 127 144 161 178 195 197 214 6 23 40
73 90 92 109 126 143 160 177 194 196 213 5 22 39 56
89 91 108 125 142 159 176 193 210 212 4 21 38 55 72
105 107 124 141 158 175 192 209 211 3 20 37 54 71 88
106 123 140 157 174 191 208 225 2 19 36 53 70 87 104
Magic number = 1695
Frink
This program takes an order from command-line or requests an odd order from the user. It uses an algorithm from Dr. Crypton's column in Science Digest in the 1980s which the developer of Frink remembered and used to use by hand to create giant magic squares until his English teacher told him "don't do that in class."
order = length[ARGS] > 0 ? eval[ARGS@0] : undef
until isInteger[order] and order mod 2 == 1
order = eval[input["Enter order (must be odd): ", 3]]
a = new array[[order, order], undef]
x = order div 2
y = 0
for i = 1 to order^2
{
ny = (y - 1) mod order
nx = (x + 1) mod order
if a@ny@nx != undef
{
nx = x
ny = (y + 1) mod order
}
a@y@x = i
y = ny
x = nx
}
println[formatTable[a]]
println["Magic number is " + sum[a@0]]
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic number is 65
Go
Translation of: C
package main
import (
"fmt"
"log"
)
func ms(n int) (int, []int) {
M := func(x int) int { return (x + n - 1) % n }
if n <= 0 || n&1 == 0 {
n = 5
log.Println("forcing size", n)
}
m := make([]int, nn)
i, j := 0, n/2
for k := 1; k <= nn; k++ {
m[in+j] = k
if m[M(i)n+M(j)] != 0 {
i = (i + 1) % n
} else {
i, j = M(i), M(j)
}
}
return n, m
}
func main() {
n, m := ms(5)
i := 2
for j := 1; j <= nn; j = 10 {
i++
}
f := fmt.Sprintf("%%%dd", i)
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
fmt.Printf(f, m[in+j])
}
fmt.Println()
}
}
Output:
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
Haskell
Translating imperative code
Translation of: cpp
-- as a translation from imperative code, this is probably not a "good" implementation
import Data.List
type Var = (Int, Int, Int, Int) -- sx sy sz c
magicSum :: Int -> Int
magicSum x = ((x x + 1) div 2) x
wrapInc :: Int -> Int -> Int
wrapInc max x
| x + 1 == max = 0
| otherwise = x + 1
wrapDec :: Int -> Int -> Int
wrapDec max x
| x == 0 = max - 1
| otherwise = x - 1
isZero :: -> Int -> Int -> Bool
isZero m x y = m !! x !! y == 0
setAt :: (Int,Int) -> Int -> ->
setAt (x, y) val table
| (upper, current : lower) <- splitAt x table,
(left, this : right) <- splitAt y current
= upper ++ (left ++ val : right) : lower
| otherwise = error "Outside"
create :: Int ->
create x = replicate x $ replicate x 0
cells :: -> Int
cells m = xx where x = length m
fill :: Var -> ->
fill (sx, sy, sz, c) m
| c < cells m =
if isZero m sx sy
then fill ((wrapInc sz sx), (wrapDec sz sy), sz, c + 1) (setAt (sx, sy) (c + 1) m)
else fill ((wrapDec sz sx), (wrapInc sz(wrapInc sz sy)), sz, c) m
| otherwise = m
magicNumber :: Int ->
magicNumber d = transpose $ fill (d div 2, 0, d, 0) (create d)
display :: -> String
display (x:xs)
| null xs = vdisplay x
| otherwise = vdisplay x ++ ('\n' : display xs)
vdisplay :: [Int] -> String
vdisplay (x:xs)
| null xs = show x
| otherwise = show x ++ " " ++ vdisplay xs
magicSquare x = do
putStr "Magic Square of "
putStr $ show x
putStr " = "
putStrLn $ show $ magicSum x
putStrLn $ display $ magicNumber x
Transpose . cycled
Defining the magic square as two applications of (transpose . cycled) to a simply ordered square.
import Control.Monad (join)
import Data.List (maximumBy, transpose)
import Data.List.Split (chunksOf)
import Data.Ord (comparing)
magicSquare :: Int ->
magicSquare n
| 1 == mod n 2 =
applyN 2 (transpose . cycled) $
plainSquare n
| otherwise = []
plainSquare :: Int ->
plainSquare = chunksOf <> enumFromTo 1 . (^ 2)
-------------------------- TEST ---------------------------
main :: IO ()
main =
mapM_ putStrLn $
showSquare . magicSquare <$> [3, 5, 7]
------------------------- GENERIC -------------------------
applyN :: Int -> (a -> a) -> a -> a
applyN n f = foldr (.) id (replicate n f)
cycled :: ->
cycled rows =
let n = length rows
d = quot n 2
in zipWith
(\d xs -> take n $ drop (n - d) (cycle xs))
[d, subtract 1 d .. - d]
rows
------------------------ FORMATTING ----------------------
justifyRight :: Int -> a -> [a] -> [a]
justifyRight n c = (drop . length) <> (replicate n c <>)
showSquare :: Show a => -> String
showSquare rows =
( (\xs w -> unlines ((justifyRight w ' ' =<<) <$> xs))
<> succ . maximum . fmap length . join
)
$ fmap show <$> rows
Output:
8 1 6
3 5 7
4 9 2
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
Siamese method
Encoding the traditional 'Siamese' method
{-# LANGUAGE TupleSections #-}
import Control.Monad (forM_)
import Data.List (intercalate, transpose)
import qualified Data.Map.Strict as M
import Data.Maybe (fromJust, isJust)
magic :: Int ->
magic = mapAsTable <> siamMap
----------------- SIAMESE METHOD FUNCTIONS ---------------
-- Highest zero-based index of grid ->
-- 'Siamese' indices keyed by coordinates
siamMap :: Int -> M.Map (Int, Int) Int
siamMap n
| odd n = go n
| otherwise = M.fromList []
where
go n = sPath uBound (M.fromList []) (quot uBound 2, 0) 1
where
h = quot n 2
uBound = n - 1
sPath uBound sMap (x, y) h =
let newMap = M.insert (x, y) h sMap
in if y == uBound && x == quot uBound 2
then newMap
else
sPath
uBound
newMap
(nextSiam uBound sMap (x, y))
(succ h)
-- Highest index of square -> Siam xys so far -> xy ->
-- next xy coordinate
nextSiam :: Int -> M.Map (Int, Int) Int -> (Int, Int) -> (Int, Int)
nextSiam uBound sMap (x, y) =
let alt (a, b)
-- Top right corner?
| a > uBound && b < 0 = (uBound, 1)
-- beyond right edge?
| a > uBound = (0, b)
-- above top edge?
| b < 0 = (a, uBound)
-- already filled?
| isJust (M.lookup (a, b) sMap) = (a - 1, b + 2)
| otherwise = (a, b) -- Up one, right one.
in alt (x + 1, y - 1)
---------------- DISPLAY AND TEST FUNCTIONS --------------
-- Size of square -> integers keyed by coordinates
-- -> rows of integers
mapAsTable :: Int -> M.Map (Int, Int) Int ->
mapAsTable nCols xyMap =
let axis = [0 .. nCols - 1]
in fmap (fromJust . flip M.lookup xyMap)
<$> (axis >>= \y -> [(,y) <$> axis])
checked :: -> (Int, Bool)
checked square =
let diagonals =
fmap (flip (zipWith (!!)) [0 ..])
. ( (:)
<> (return . reverse)
)
h : t =
sum <$> square
<> transpose square
<> diagonals square
in (h, all (h ==) t)
table :: String -> -> [String]
table delim rows =
let justifyRight c n s =
drop
(length s)
(replicate n c <> s)
in intercalate delim
<$> transpose
( (fmap =<< justifyRight ' ' . maximum . fmap length)
<$> transpose rows
)
main :: IO ()
main =
forM_ [3, 5, 7] $
\n -> do
let test = magic n
putStrLn $ unlines (table " " (fmap show <$> test))
print $ checked test
putStrLn ""
Output:
8 1 6
3 5 7
4 9 2
(15,True)
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
(65,True)
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
(175,True)
Icon and Unicon
This is a Unicon-specific solution because of the use of the [: ...:] construct.
procedure main(A)
n := integer(!A) | 3
write("Magic number: ",n(nn+1)/2)
sq := buildSquare(n)
showSquare(sq)
end
procedure buildSquare(n)
sq := [: |list(n)\n :]
r := 0
c := n/2
every i := !(nn) do {
/sq[r+1,c+1] := i
nr := (n+r-1)%n
nc := (c+1)%n
if /sq[nr+1,nc+1] then (r := nr,c := nc) else r := (r+1)%n
}
return sq
end
procedure showSquare(sq)
n := sq
s := (nn)+2
every r := !sq do every writes(right(!r,s)|"\n")
end
Output:
->ms 5
Magic number: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
->
J
Based on
ms=: i:@<.@-: |."_1&|:^:2 >:@i.@,~
In other words, generate a square of counting integers, like this:
:@i.@,~ 3
1 2 3
4 5 6
7 8 9
Then generate a list of integers centering on 0 up to half of that value, like this:
i:@<.@-: 3
_1 0 1
Finally, rotate each corresponding row and column of the table by the corresponding value in the list. We can use the same instructions to rotate both rows and columns if we transpose the matrix before rotating (and perform this transpose+rotate twice).
Example use:
ms 5
9 15 16 22 3
20 21 2 8 14
1 7 13 19 25
12 18 24 5 6
23 4 10 11 17
~.+/ms 5
65
~.+/ms 101
515201
Note also that an important feature of magic squares is that their diagonals sum the same way:
9+21+13+5+17
65
3+8+13+18+23
65
Java
public class MagicSquare {
public static void main(String[] args) {
int n = 5;
for (int[] row : magicSquareOdd(n)) {
for (int x : row)
System.out.format("%2s ", x);
System.out.println();
}
System.out.printf("\nMagic constant: %d ", (n n + 1) n / 2);
}
public static int[][] magicSquareOdd(final int base) {
if (base % 2 == 0 || base < 3)
throw new IllegalArgumentException("base must be odd and > 2");
int[][] grid = new int[base][base];
int r = 0, number = 0;
int size = base base;
int c = base / 2;
while (number++ < size) {
grid[r][c] = number;
if (r == 0) {
if (c == base - 1) {
r++;
} else {
r = base - 1;
c++;
}
} else {
if (c == base - 1) {
r--;
c = 0;
} else {
if (grid[r - 1][c + 1] == 0) {
r--;
c++;
} else {
r++;
}
}
}
}
return grid;
}
}
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic constant: 65
JavaScript
ES5
Translation of: Mathematica
( and referring to )
(function () {
// n ->
function magic(n) {
return n % 2 ? rotation(
transposed(
rotation(
table(n)
)
)
) : null;
}
// ->
function rotation(lst) {
return lst.map(function (row, i) {
return rotated(
row, ((row.length + 1) / 2) - (i + 1)
);
})
}
// ->
function transposed(lst) {
return lst.map(function (col, i) {
return lst.map(function (row) {
return row[i];
})
});
}
// [a] -> n -> [a]
function rotated(lst, n) {
var lng = lst.length,
m = (typeof n === 'undefined') ? 1 : (
n < 0 ? lng + n : (n > lng ? n % lng : n)
);
return m ? (
lst.slice(-m).concat(lst.slice(0, lng - m))
) : lst;
}
// n ->
function table(n) {
var rngTop = rng(1, n);
return rng(0, n - 1).map(function (row) {
return rngTop.map(function (x) {
return row n + x;
});
});
}
// [m..n]
function rng(m, n) {
return Array.apply(null, Array(n - m + 1)).map(
function (x, i) {
return m + i;
});
}
/ TEST WITH 3, 5, 11 /
// Results as right-aligned wiki tables
function wikiTable(lstRows, blnHeaderRow, strStyle) {
var css = strStyle ? 'style="' + strStyle + '"' : '';
return '{| class="wikitable" ' + css + lstRows.map(
function (lstRow, iRow) {
var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|'),
strDbl = strDelim + strDelim;
return '\n|-\n' + strDelim + ' ' + lstRow.join(' ' + strDbl + ' ');
}).join('') + '\n|}';
}
return [3, 5, 11].map(
function (n) {
var w = 2.5 n;
return 'magic(' + n + ')\n\n' + wikiTable(
magic(n), false, 'text-align:center;width:' + w + 'em;height:' + w + 'em;table-layout:fixed;'
)
}
).join('\n\n')
})();
Output:
magic(3)
8 3 4
1 5 9
6 7 2
magic(5)
17 23 4 10 11
24 5 6 12 18
1 7 13 19 25
8 14 20 21 2
15 16 22 3 9
magic(11)
68 80 92 104 116 7 19 31 43 55 56
81 93 105 117 8 20 32 44 45 57 69
94 106 118 9 21 33 34 46 58 70 82
107 119 10 22 23 35 47 59 71 83 95
120 11 12 24 36 48 60 72 84 96 108
1 13 25 37 49 61 73 85 97 109 121
14 26 38 50 62 74 86 98 110 111 2
27 39 51 63 75 87 99 100 112 3 15
40 52 64 76 88 89 101 113 4 16 28
53 65 77 78 90 102 114 5 17 29 41
66 67 79 91 103 115 6 18 30 42 54
ES6
Cycled . transposed . cycled
Translation of: Haskell
(2nd Haskell version: cycledRows . transpose . cycledRows)
(() => {
// magicSquare:: Int ->
const magicSquare = n =>
n % 2 !== 0 ? (
compose([transpose, cycled, transpose, cycled, enumSquare])(n)
) : [];
// Size of square -> rows containing integers [1..]
// enumSquare:: Int ->
const enumSquare = n =>
chunksOf(n, enumFromTo(1, n n));
// Table of integers -> Table with rows rotated by descending deltas
// cycled:: ->
const cycled = rows => {
const d = Math.floor(rows.length / 2);
return zipWith(listCycle, enumFromTo(d, -d), rows)
};
// Number of positions to shift to right -> List -> Wrap-cycled list
// listCycle:: Int -> [a] -> [a]
const listCycle = (n, xs) => {
const d = -(n % xs.length);
return (d !== 0 ? xs.slice(d)
.concat(xs.slice(0, d)) : xs);
};
// GENERIC FUNCTIONS ------------------------------------------------------
// chunksOf:: Int -> [a] ->
const chunksOf = (n, xs) =>
xs.reduce((a, _, i, xs) =>
i % n ? a : a.concat([xs.slice(i, i + n)]), []);
// compose:: [(a -> a)] -> (a -> a)
const compose = fs => x => fs.reduceRight((a, f) => f(a), x);
// enumFromTo:: Int -> Int -> Maybe Int -> [Int]
const enumFromTo = (m, n, step) => {
const d = (step || 1) (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, (_, i) => m + (i d));
};
// intercalate:: String -> [a] -> String
const intercalate = (s, xs) => xs.join(s);
// min:: Ord a => a -> a -> a
const min = (a, b) => b < a ? b : a;
// show:: a -> String
const show = JSON.stringify;
// transpose:: ->
const transpose = xs =>
xs.map((_, iCol) => xs.map(row => row[iCol]));
// unlines:: [String] -> String
const unlines = xs => xs.join('\n');
// zipWith:: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) =>
Array.from({
length: min(xs.length, ys.length)
}, (_, i) => f(xs[i], ys[i]));
// TEST -------------------------------------------------------------------
return intercalate('\n\n', [3, 5, 7]
.map(magicSquare)
.map(xs => unlines(xs.map(show))));
})();
Output:
[8,1,6]
[3,5,7]
[4,9,2]
[17,24,1,8,15]
[23,5,7,14,16]
[4,6,13,20,22]
[10,12,19,21,3]
[11,18,25,2,9]
[30,39,48,1,10,19,28]
[38,47,7,9,18,27,29]
[46,6,8,17,26,35,37]
[5,14,16,25,34,36,45]
[13,15,24,33,42,44,4]
[21,23,32,41,43,3,12]
[22,31,40,49,2,11,20]
Traditional 'Siamese' method
Encoding the traditional 'Siamese' method
Translation of: Haskell
(() => {
// Number of rows -> n rows of integers
// oddMagicTable:: Int ->
const oddMagicTable = n =>
mapAsTable(n, siamMap(quot(n, 2)));
// Highest index of square -> Siam xys so far -> xy -> next xy coordinate
// nextSiam:: Int -> M.Map (Int, Int) Int -> (Int, Int) -> (Int, Int)
const nextSiam = (uBound, sMap, [x, y]) => {
const [a, b] = [x + 1, y - 1];
return (a > uBound && b < 0) ? (
[uBound, 1] // Move down if obstructed by corner
) : a > uBound ? (
[0, b] // Wrap at right edge
) : b < 0 ? (
[a, uBound] // Wrap at upper edge
) : mapLookup(sMap, [a, b])
.nothing ? ( // Unimpeded default: one up one right
[a, b]
) : [a - 1, b + 2]; // Position occupied: move down
};
// Order of table -> Siamese indices keyed by coordinates
// siamMap:: Int -> M.Map (Int, Int) Int
const siamMap = n => {
const
uBound = 2 n,
sPath = (uBound, sMap, xy, n) => {
const [x, y] = xy,
newMap = mapInsert(sMap, xy, n);
return (y == uBound && x == quot(uBound, 2) ? (
newMap
) : sPath(
uBound, newMap, nextSiam(uBound, newMap, [x, y]), n + 1));
};
return sPath(uBound, {}, [n, 0], 1);
};
// Size of square -> integers keyed by coordinates -> rows of integers
// mapAsTable:: Int -> M.Map (Int, Int) Int ->
const mapAsTable = (nCols, dct) => {
const axis = enumFromTo(0, nCols - 1);
return map(row => map(k => fromJust(mapLookup(dct, k)), row),
bind(axis, y => [bind(axis, x => [
[x, y]
])]));
};
// GENERIC FUNCTIONS ------------------------------------------------------
// bind:: [a] -> (a -> [b]) -> [b]
const bind = (xs, f) => [].concat.apply([], xs.map(f));
// curry:: Function -> Function
const curry = (f, ...args) => {
const go = xs => xs.length >= f.length ? (f.apply(null, xs)) :
function () {
return go(xs.concat(Array.from(arguments)));
};
return go([].slice.call(args, 1));
};
// enumFromTo:: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// fromJust:: M a -> a
const fromJust = m => m.nothing ? {} : m.just;
// fst:: [a, b] -> a
const fst = pair => pair.length === 2 ? pair : undefined;
// intercalate:: String -> [a] -> String
const intercalate = (s, xs) => xs.join(s);
// justifyRight:: Int -> Char -> Text -> Text
const justifyRight = (n, cFiller, strText) =>
n > strText.length ? (
(cFiller.repeat(n) + strText)
.slice(-n)
) : strText;
// length:: [a] -> Int
const length = xs => xs.length;
// log:: a -> IO ()
const log = (...args) =>
console.log(
args
.map(show)
.join(' -> ')
);
// map:: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
// mapInsert:: Dictionary -> k -> v -> Dictionary
const mapInsert = (dct, k, v) =>
(dct[(typeof k === 'string' && k) || show(k)] = v, dct);
// mapKeys:: Map k a -> [k]
const mapKeys = dct =>
sortBy(mappendComparing([snd, fst]),
map(JSON.parse, Object.keys(dct)));
// mapLookup:: Dictionary -> k -> Maybe v
const mapLookup = (dct, k) => {
const
v = dct[(typeof k === 'string' && k) || show(k)],
blnJust = (typeof v !== 'undefined');
return {
nothing: !blnJust,
just: v
};
};
// mappendComparing:: [(a -> b)] -> (a -> a -> Ordering)
const mappendComparing = fs => (x, y) =>
fs.reduce((ord, f) => {
if (ord !== 0) return ord;
const
a = f(x),
b = f(y);
return a < b ? -1 : a > b ? 1 : 0
}, 0);
// maximum:: [a] -> a
const maximum = xs =>
xs.reduce((a, x) => (x > a || a === undefined ? x : a), undefined);
// Integral a => a -> a -> a
const quot = (n, m) => Math.floor(n / m);
// show:: a -> String
const show = x => JSON.stringify(x);
//
// snd:: (a, b) -> b
const snd = tpl => Array.isArray(tpl) ? tpl : undefined;
//
// sortBy:: (a -> a -> Ordering) -> [a] -> [a]
const sortBy = (f, xs) => xs.slice()
.sort(f);
// table:: String -> -> [String]
const table = (delim, rows) =>
map(curry(intercalate)(delim),
transpose(map(col =>
map(curry(justifyRight)(maximum(map(length, col)))(' '), col),
transpose(rows))));
// transpose:: ->
const transpose = xs =>
xs.map((_, col) => xs.map(row => row[col]));
// unlines:: [String] -> String
const unlines = xs => xs.join('\n');
// TEST -------------------------------------------------------------------
return intercalate('\n\n',
bind([3, 5, 7],
n => unlines(table(" ",
map(xs => map(show, xs), oddMagicTable(n))))));
})();
Output:
8 1 6
3 5 7
4 9 2
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
jq
Adapted from #AWK
def odd_magic_square:
if type != "number" or . % 2 == 0 or . <= 0
then error("odd_magic_square requires an odd positive integer")
else
. as $n
| reduce range(1; 1 + ($n$n)) as $i
( [0, (($n-1)/2), []];
. as $x | . as $y
| .
| setpath([$x, $y]; $i )
| if getpath([(($x+$n-1) % $n), (($y+$n+1) % $n)])
then [(($x+$n+1) % $n), $y, .]
else [ (($x+$n-1) % $n), (($y+$n+1) % $n), .]
end ) | .
end ;
Examples
def task:
def pp: if length == 0 then empty
else "(.)", (.[1:] | pp )
end;
"The magic sum for a square of size (.) is ( (.. + 1)./2 ):",
(odd_magic_square | pp)
;
(3, 5, 9) | task
Output:
$ jq -n -r -M -c -f odd_magic_square.jq
The magic sum for a square of size 3 is 15:
[8,1,6]
[3,5,7]
[4,9,2]
The magic sum for a square of size 5 is 65:
[17,24,1,8,15]
[23,5,7,14,16]
[4,6,13,20,22]
[10,12,19,21,3]
[11,18,25,2,9]
The magic sum for a square of size 9 is 369:
[47,58,69,80,1,12,23,34,45]
[57,68,79,9,11,22,33,44,46]
[67,78,8,10,21,32,43,54,56]
[77,7,18,20,31,42,53,55,66]
[6,17,19,30,41,52,63,65,76]
[16,27,29,40,51,62,64,75,5]
[26,28,39,50,61,72,74,4,15]
[36,38,49,60,71,73,3,14,25]
[37,48,59,70,81,2,13,24,35]
Julia
v0.6.0
function magicsquareodd(base::Int)
if base & 1 == 0 || base < 3; error("base must be odd and >3") end
square = fill(0, base, base)
r, number = 1, 1
size = base base
c = div(base, 2) + 1
while number ≤ size
square[r, c] = number
fr = r == 1 ? base : r - 1
fc = c == base ? 1 : c + 1
if square[fr, fc] != 0
fr = r == base ? 1 : r + 1
fc = c
end
r, c = fr, fc
number += 1
end
return square
end
for n in 3:2:7
println("Magic square with size $n - magic constant = ", div(n ^ 3 + n, 2))
println("----------------------------------------------------")
square = magicsquareodd(n)
for i in 1:n
println(square[i, :])
end
println()
end
Output:
Magic square with size 3 - magic constant = 15
[8, 1, 6]
[3, 5, 7]
[4, 9, 2]
Magic square with size 5 - magic constant = 65
[17, 24, 1, 8, 15]
[23, 5, 7, 14, 16]
[4, 6, 13, 20, 22]
[10, 12, 19, 21, 3]
[11, 18, 25, 2, 9]
Magic square with size 7 - magic constant = 175
[30, 39, 48, 1, 10, 19, 28]
[38, 47, 7, 9, 18, 27, 29]
[46, 6, 8, 17, 26, 35, 37]
[5, 14, 16, 25, 34, 36, 45]
[13, 15, 24, 33, 42, 44, 4]
[21, 23, 32, 41, 43, 3, 12]
[22, 31, 40, 49, 2, 11, 20]
Kotlin
Translation of: C
// version 1.0.6
fun f(n: Int, x: Int, y: Int) = (x + y 2 + 1) % n
fun main(args: Array) {
var n: Int
while (true) {
print("Enter the order of the magic square: ")
n = readLine()!!.toInt()
if (n < 1 || n % 2 == 0) println("Must be odd and >= 1, try again")
else break
}
println()
for (i in 0 until n) {
for (j in 0 until n) print("%4d".format(f(n, n - j - 1, i) n + f(n, j, i) + 1))
println()
}
println("\nThe magic constant is ${(n n + 1) / 2 n}")
}
Sample input/output:
Output:
Enter the order of the magic square : 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic constant is 369
Lua
For all three kinds of Magic Squares(Odd, singly and doubly even)
See Magic_squares/Lua.
Mathematica /Wolfram Language
Rotate rows and columns of the initial matrix with rows filled in order 1 2 3 .... N^2 Method from
rp[v_, pos_] := RotateRight[v, (Length[v] + 1)/2 - pos];
rho[m_] := MapIndexed[rp, m];
magic[n_] :=
rho[Transpose[rho[Table[in + j, {i, 0, n - 1}, {j, 1, n}]]]];
square = magic // Grid
Print["Magic number is ", Total[square]]
Output:
(alignment lost in translation to text):
{68, 80, 92, 104, 116, 7, 19, 31, 43, 55, 56},
{81, 93, 105, 117, 8, 20, 32, 44, 45, 57, 69},
{94, 106, 118, 9, 21, 33, 34, 46, 58, 70, 82},
{107, 119, 10, 22, 23, 35, 47, 59, 71, 83, 95},
{120, 11, 12, 24, 36, 48, 60, 72, 84, 96, 108},
{1, 13, 25, 37, 49, 61, 73, 85, 97, 109, 121},
{14, 26, 38, 50, 62, 74, 86, 98, 110, 111, 2},
{27, 39, 51, 63, 75, 87, 99, 100, 112, 3, 15},
{40, 52, 64, 76, 88, 89, 101, 113, 4, 16, 28},
{53, 65, 77, 78, 90, 102, 114, 5, 17, 29, 41},
{66, 67, 79, 91, 103, 115, 6, 18, 30, 42, 54}
Magic number is 671
Output from code that checks the results Rows
{671,671,671,671,671,671,671,671,671,671,671}
Columns
{671,671,671,671,671,671,671,671,671,671,671}
Diagonals
671
671
Maxima
wrap1(i):= if i>%n% then 1 else if i<1 then %n% else i;
wrap(P):=maplist('wrap1, P);
uprigth(P):= wrap(P + [-1, 1]);
down(P):= wrap(P + [1, 0]);
magic(n):=block([%n%: n,
M: zeromatrix (n, n),
P: [1, (n + 1)/2],
m: 1, Pc],
do (
M[P,P]: m,
m: m + 1,
if m>n^2 then return(M),
Pc: uprigth(P),
if M[Pc,Pc]=0 then P: Pc
else while(M[P,P]#0) do P: down(P)));
Usage:
(%i6) magic(3);
[ 8 1 6 ]
[ ]
(%o6) [ 3 5 7 ]
[ ]
[ 4 9 2 ]
(%i7) magic(5);
[ 17 24 1 8 15 ]
[ ]
[ 23 5 7 14 16 ]
[ ]
(%o7) [ 4 6 13 20 22 ]
[ ]
[ 10 12 19 21 3 ]
[ ]
[ 11 18 25 2 9 ]
(%i8) magic(7);
[ 30 39 48 1 10 19 28 ]
[ ]
[ 38 47 7 9 18 27 29 ]
[ ]
[ 46 6 8 17 26 35 37 ]
[ ]
(%o8) [ 5 14 16 25 34 36 45 ]
[ ]
[ 13 15 24 33 42 44 4 ]
[ ]
[ 21 23 32 41 43 3 12 ]
[ ]
[ 22 31 40 49 2 11 20 ]
/ magic number for n=7 /
(%i9) lsum(q, q, first(magic(7)));
(%o9) 175
Nim
Translation of: Python
import strutils
proc magic(n: int) =
let length = len($(n n))
for row in 1 .. n:
for col in 1 .. n:
let cell = (n ((row + col - 1 + n div 2) mod n) +
((row + 2 col - 2) mod n) + 1)
stdout.write ($cell).align(length), ' '
echo ""
echo "\nAll sum to magic number ", (n n + 1) n div 2
for n in [3, 5, 7]:
echo "\nOrder ", n, "\n======="
magic(n)
Output:
Order 3
8 1 6
3 5 7
4 9 2
All sum to magic number 15
Order 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
All sum to magic number 65
Order 7
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
All sum to magic number 175
Oforth
: magicSquare(n)
| i j wd |
n sq log asInteger 1+ ->wd
n loop: i [
n loop: j [
i j + 1- n 2 / + n mod n
i j + j + 2 - n mod 1 + +
System.Out swap <<w(wd) " " << drop
]
printcr
]
System.Out "Magic constant is : " << n sq 1 + 2 / n << cr ;
Output:
5 magicSquare
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic constant is : 65
PARI/GP
Translation of: Perl
The index-fiddling differs from Perl since GP vectors start at 1.
magicSquare(n)={
my(M=matrix(n,n),j=n\2+1,i=1);
for(l=1,n^2,
M[i,j]=l;
if(M[(i-2)%n+1,j%n+1],
i=i%n+1
,
i=(i-2)%n+1;
j=j%n+1
)
);
M;
}
magicSquare(7)
Output:
[30 39 48 1 10 19 28]
[38 47 7 9 18 27 29]
[46 6 8 17 26 35 37]
[ 5 14 16 25 34 36 45]
[13 15 24 33 42 44 4]
[21 23 32 41 43 3 12]
[22 31 40 49 2 11 20]
Pascal
Works with: Free Pascal version 1.0
Translation of: C
PROGRAM magic;
( Magic squares of odd order )
CONST
n=9;
VAR
i,j :INTEGER;
BEGIN (magic)
WRITELN('The square order is: ',n);
FOR i:=1 TO n DO
BEGIN
FOR j:=1 TO n DO
WRITE((i2-j+n-1) MOD nn + (i2+j-2) MOD n+1:5);
WRITELN
END;
WRITELN('The magic number is: ',n(nn+1) DIV 2)
END (magic).
Output:
The square order is: 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
The magic number is: 369
improved
shuffles columns and rows and changed col<-> row to get different looks. n! x n! 2 different arrangements. See last column of version before moved to the top row.
PROGRAM magic;
{$IFDEF FPC }{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
( Magic squares of odd order )
type
tsquare = array of array of LongInt;
trowcol = array of NativeInt;
function GenShuffleRowCol(n: nativeInt):trowcol;
var
i,j,tmp: NativeInt;
begin
setlength(result,0);
IF n > 0 then
Begin
setlength(result,n);
For i := 0 to n-1 do
result[i] := i;
//shuffle
For i := n-1 downto 1 do
Begin
j := random(i+1);//j == [0..i]
tmp := result[i];result[i]:= result[j];result[j]:= tmp;
end;
end;
end;
function MagicSqrOdd(n:nativeInt;SwapColRoW:boolean):tsquare;
VAR
rowIdx,colIdx,row,col,num :NativeInt;
cols,rows :trowcol;
BEGIN
rows:= GenShuffleRowCol(n);
cols:= GenShuffleRowCol(n);
setlength(result,n,n);
FOR rowIdx:= 0 TO n-1 DO
BEGIN
row := rows[rowIdx];
FOR colIdx:=0 TO n-1 DO
Begin
col := cols[colIdx];
//corrected formula cause row :0..n1-> corrected to 1..n
num := (row2-col+n+2) MOD nn + (row2+col+1) MOD n+1;
IF SwapColRoW then
result[colIdx,rowIdx] := num
else
result[rowIdx,colIdx] := num;
end;
END;
END;
function MagicSqrCheck(const Mq:tsquare):boolean;
var
row,col,rowsum,mn,n,itm: NativeInt;
colSum:trowcol;
begin
n := length(Mq);
mn := n(nn+1) DIV 2;
setlength(colsum,n);//automatic initialised to zero
For row := n-1 downto 0 do
Begin
//check one row
rowsum := 0;
For col := n-1 downto 0 do
Begin
itm := Mq[row,col];
write(itm:4);
inc(rowsum,itm);
//sum up the columns too, for I'm just here
inc(colSum[col],itm);
end;
writeln;
result := (rowsum=mn);
IF Not(result) then begin writeln(row:4,col:4,rowsum:10);EXIT;end;
end;
//check columns
For col := n-1 downto 0 do
Begin
result := (colSum[col]=mn);
IF Not(result) then begin writeln(col:4,colSum[col]:10);EXIT;end;
end;
writeln;
end;
var
n,mn : nativeInt;
Mq : tsquare;
Begin
randomize;
n := 9;
mn := n(nn+1) DIV 2;
WRITELN('The square order is: ',n);
WRITELN('The magic number is: ',mn);
Mq := MagicSqrOdd(n,random(2)=0);
writeln(MagicSqrCheck(Mq));
end.
Output:
The square order is: 9
The magic number is: 369
70 30 20 9 40 50 10 80 60
13 63 53 33 64 74 43 23 3
54 14 4 65 24 34 75 55 44
5 46 45 25 56 66 35 15 76
37 6 77 57 16 26 67 47 36
29 79 69 49 8 18 59 39 19
78 38 28 17 48 58 27 7 68
62 22 12 73 32 42 2 72 52
21 71 61 41 81 1 51 31 11
TRUE
PascalABC.NET
See Pascal.
Perl
See Magic squares/Perl for a general magic square generator.
Phix
with javascript_semantics
function magic_square(integer n)
if mod(n,2)!=1 or n<1 then return false end if
sequence square = repeat(repeat(0,n),n)
for i=1 to n do
for j=1 to n do
square[i,j] = nmod(2i-j+n-1,n) + mod(2i+j-2,n) + 1
end for
end for
return square
end function
procedure check(sequence sq)
integer n = length(sq)
integer magic = n(nn+1)/2
integer bd=0, fd=0
for i=1 to length(sq) do
if sum(sq[i])!=magic then ?9/0 end if
if sum(columnize(sq,i))!=magic then ?9/0 end if
bd += sq[i,i]
fd += sq[n-i+1,n-i+1]
end for
if bd!=magic or fd!=magic then ?9/0 end if
end procedure
for i=1 to 7 by 2 do
sequence square = magic_square(i)
printf(1,"maqic square of order %d, sum: %d\n", {i,sum(square[i])})
string fmt = sprintf("%%%dd",length(sprintf("%d",ii)))
pp(square,{pp_Nest,1,pp_IntFmt,fmt,pp_StrFmt,3,pp_IntCh,false,pp_Pause,0})
check(square)
end for
Output:
maqic square of order 1, sum: 1
{{1}}
maqic square of order 3, sum: 15
{{2,9,4},
{7,5,3},
{6,1,8}}
maqic square of order 5, sum: 65
{{ 2,23,19,15, 6},
{14,10, 1,22,18},
{21,17,13, 9, 5},
{ 8, 4,25,16,12},
{20,11, 7, 3,24}}
maqic square of order 7, sum: 175
{{ 2,45,39,33,27,21, 8},
{18,12, 6,49,36,30,24},
{34,28,15, 9, 3,46,40},
{43,37,31,25,19,13, 7},
{10, 4,47,41,35,22,16},
{26,20,14, 1,44,38,32},
{42,29,23,17,11, 5,48}}
Picat
Translation of: J
import util.
go =>
foreach(N in [3,5,17])
M=magic_square(N),
print_matrix(M),
check(M),
nl
end,
nl.
%
% Not as nice as the J solution.
% But I like the chaining of the functions.
%
magic_square(N) = MS =>
if N mod 2 = 0 then
printf("N (%d) is not odd!\n", N),
halt
end,
R = make_rotate_list(N), % the rotate indices
MS = make_square(N).transpose().rotate_matrix(R).transpose().rotate_matrix(R).
%
% make a square matrix of size N (containing the numbers 1..NN)
%
make_square(N) = [[IN+J : J in 1..N]: I in 0..N-1].
%
% rotate list:
% rotate_list(11) = [-5,-4,-3,-2,-1,0,1,2,3,4,5]
%
make_rotate_list(N) = [I - ceiling(N / 2) : I in 1..N].
%
% rotate the matrix M according to rotate list R
%
rotate_matrix(M, R) = [rotate_n(Row,N) : {Row,N} in zip(M,R)].
%
% Rotate the list L N steps (either positive or negative N)
% rotate(1..10,3) -> [4,5,6,7,8,9,10,1,2,3]
% rotate(1..10,-3) -> [8,9,10,1,2,3,4,5,6,7]
%
rotate_n(L,N) = Rot =>
Len = L.length,
R = cond(N < 0, Len + N, N),
Rot = [L[I] : I in (R+1..Len) ++ 1..R].
%
% Check if M is a magic square
%
check(M) =>
N = M.length,
Sum = N(NN+1) // 2, % The correct sum.
println(sum=Sum),
Rows = [sum(Row) : Row in M],
Cols = [sum(Col) : Col in M.transpose()],
Diag1 = sum([M[I,I] : I in 1..N]),
Diag2 = sum([M[I,N-I+1] : I in 1..N]),
All = Rows ++ Cols ++ [Diag1, Diag2],
OK = true,
foreach(X in All)
if X != Sum then
printf("%d != %d\n", X, Sum),
OK := false
end
end,
if OK then
println(ok)
else
println(not_ok)
end,
nl.
% Print the matrix
print_matrix(M) =>
N = M.len,
printf("N=%d\n",N),
Format = to_fstring("%%%dd",max(flatten(M)).to_string().length+1),
foreach(Row in M)
foreach(X in Row)
printf(Format, X)
end,
nl
end,
nl.
Output:
N=3
6 7 2
1 5 9
8 3 4
sum = 15
ok
N=5
9 15 16 22 3
20 21 2 8 14
1 7 13 19 25
12 18 24 5 6
23 4 10 11 17
sum = 65
ok
N=17
27 45 63 81 99 117 135 153 154 172 190 208 226 244 262 280 9
62 80 98 116 134 152 170 171 189 207 225 243 261 279 8 26 44
97 115 133 151 169 187 188 206 224 242 260 278 7 25 43 61 79
132 150 168 186 204 205 223 241 259 277 6 24 42 60 78 96 114
167 185 203 221 222 240 258 276 5 23 41 59 77 95 113 131 149
202 220 238 239 257 275 4 22 40 58 76 94 112 130 148 166 184
237 255 256 274 3 21 39 57 75 93 111 129 147 165 183 201 219
272 273 2 20 38 56 74 92 110 128 146 164 182 200 218 236 254
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289
36 54 72 90 108 126 144 162 180 198 216 234 252 270 288 17 18
71 89 107 125 143 161 179 197 215 233 251 269 287 16 34 35 53
106 124 142 160 178 196 214 232 250 268 286 15 33 51 52 70 88
141 159 177 195 213 231 249 267 285 14 32 50 68 69 87 105 123
176 194 212 230 248 266 284 13 31 49 67 85 86 104 122 140 158
211 229 247 265 283 12 30 48 66 84 102 103 121 139 157 175 193
246 264 282 11 29 47 65 83 101 119 120 138 156 174 192 210 228
281 10 28 46 64 82 100 118 136 137 155 173 191 209 227 245 263
sum = 2465
ok
Testing a larger instance
go2 =>
N = 313,
M = magic_square(N),
check(M),
nl.
Output:
sum = 15332305
ok
PicoLisp
(load "@lib/simul.l")
(de magic (A)
(let
(Grid (grid A A T T)
Sum (/ ( A (inc ( A 2))) 2))
(println 'N A 'Sum Sum)
# cut one important edge
(with (last (last Grid)) (con (: 0 1)))
(with (get Grid (inc (/ A 2)) A)
(for N ( A A)
(=: V N)
(setq This
(if
(with (setq @@ (north (east This)))
(not (: V)) )
@@
(south This) ) ) ) )
# display
(mapc
'((L)
(for This L
(prin (align 4 (: V))) )
(prinl) )
Grid )
# clean
(mapc '((L) (mapc zap L)) Grid) ) )
(magic 5)
(prinl)
(magic 7)
Output:
N 5 Sum 65
11 10 4 23 17
18 12 6 5 24
25 19 13 7 1
2 21 20 14 8
9 3 22 16 15
N 7 Sum 175
22 21 13 5 46 38 30
31 23 15 14 6 47 39
40 32 24 16 8 7 48
49 41 33 25 17 9 1
2 43 42 34 26 18 10
11 3 44 36 35 27 19
20 12 4 45 37 29 28
PL/I
magic: procedure options (main); / 18 April 2014 /
declare n fixed binary;
put skip list ('What is the order of the magic square?');
get list (n);
if n < 3 | iand(n, 1) = 0 then
do; put skip list ('The value is out of range'); stop; end;
put skip list ('The order is ' || trim(n));
begin;
declare m(n, n) fixed, (i, j, k) fixed binary;
on subrg snap put data (i, j, k);
m = 0;
i = 1; j = (n+1)/2;
do k = 1 to nn;
if m(i,j) = 0 then
m(i,j) = k;
else
do;
i = i + 2; j = j + 1;
if i > n then i = mod(i,n);
if j > n then j = 1;
m(i,j) = k;
end;
i = i - 1; j = j - 1;
if i < 1 then i = n;
if j < 1 then j = n;
end;
do i = 1 to n;
put skip edit (m(i, )) (f(4));
end;
put skip list ('The magic number is' || sum(m(1,)));
end;
end magic;
Output:
What is the order of the magic square?
The order is 5
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
The magic number is 65
What is the order of the magic square?
The order is 7
28 19 10 1 48 39 30
29 27 18 9 7 47 38
37 35 26 17 8 6 46
45 36 34 25 16 14 5
4 44 42 33 24 15 13
12 3 43 41 32 23 21
20 11 2 49 40 31 22
The magic number is 175
Pluto
Translation of: Wren
local function ms(n)
local M = |x| -> (x + n - 1) % n
if n <= 0 or n & 1 == 0 then
n = 5
print($"forcing size {n}")
end
local m = {}
for i = 1, n n do m[i] = 0 end
local i = 0
local j = n // 2
for k = 1, n n do
m[i n + j + 1] = k
if m[M(i) n + M(j) + 1] != 0 then
i = (i + 1) % n
else
i = M(i)
j = M(j)
end
end
return {n, m}
end
local [n, m] = ms(5)
for i = 1, n do
for j = 1, n do io.write(string.format("%4d", m[(i - 1) n + j])) end
print()
end
print($"\nMagic number: {(n n + 1) // 2 n}")
Output:
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
Magic number : 65
Python
Procedural
def magic(n):
for row in range(1, n + 1):
print(' '.join('%i' % (len(str(n2)), cell) for cell in
(n ((row + col - 1 + n // 2) % n) +
((row + 2 col - 2) % n) + 1
for col in range(1, n + 1))))
print('\nAll sum to magic number %i' % ((n n + 1) n // 2))
for n in (5, 3, 7):
print('\nOrder %i\n=======' % n)
magic(n)
Order 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
All sum to magic number 65
Order 3
8 1 6
3 5 7
4 9 2
All sum to magic number 15
Order 7
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
All sum to magic number 175
Composition of pure functions
Two applications of (transposed . cycled) to a sequentially ordered square:
Translation of: Haskell
Works with: Python version 3.7
'''Magic squares of odd order N'''
from itertools import cycle, islice, repeat
from functools import reduce
magicSquare:: Int ->
def magicSquare(n):
'''Magic square of odd order n.'''
return applyN(2)(
compose(transposed)(cycled)
)(plainSquare(n)) if 1 == n % 2 else []
plainSquare:: Int ->
def plainSquare(n):
'''The sequence of integers from 1 to N^2,
subdivided into N sub-lists of equal length,
forming N rows, each of N integers.
'''
return chunksOf(n)(
enumFromTo(1)(n 2)
)
cycled:: ->
def cycled(rows):
'''A table in which the rows are
rotated by descending deltas.
'''
n = len(rows)
d = n // 2
return list(map(
lambda d, xs: take(n)(
drop(n - d)(cycle(xs))
),
enumFromThenTo(d)(d - 1)(-d),
rows
))
TEST ----------------------------------------------------
main:: IO ()
def main():
'''Magic squares of order 3, 5, 7'''
print(
fTable( doc + ':')(lambda x: '\n' + repr(x))(
showSquare
)(magicSquare)([3, 5, 7])
)
GENERIC -------------------------------------------------
applyN:: Int -> (a -> a) -> a -> a
def applyN(n):
'''n applications of f.
(Church numeral n).
'''
def go(f):
return lambda x: reduce(
lambda a, g: g(a), repeat(f, n), x
)
return lambda f: go(f)
chunksOf:: Int -> [a] ->
def chunksOf(n):
'''A series of lists of length n,
subdividing the contents of xs.
Where the length of xs is not evenly divible,
the final list will be shorter than n.'''
return lambda xs: reduce(
lambda a, i: a + [xs[i:n + i]],
range(0, len(xs), n), []
) if 0 < n else []
compose (<<<):: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))
drop:: Int -> [a] -> [a]
drop:: Int -> String -> String
def drop(n):
'''The sublist of xs beginning at
(zero-based) index n.'''
def go(xs):
if isinstance(xs, (list, tuple, str)):
return xs[n:]
else:
take(n)(xs)
return xs
return lambda xs: go(xs)
enumFromThenTo:: Int -> Int -> Int -> [Int]
def enumFromThenTo(m):
'''Integer values enumerated from m to n
with a step defined by nxt-m.
'''
def go(nxt, n):
d = nxt - m
return range(m, n - 1 if d < 0 else 1 + n, d)
return lambda nxt: lambda n: list(go(nxt, n))
enumFromTo:: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
take:: Int -> [a] -> [a]
take:: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
return lambda xs: (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
transposed:: Matrix a -> Matrix a
def transposed(m):
'''The rows and columns of the argument transposed.
(The matrix containers and rows can be lists or tuples).
'''
if m:
inner = type(m)
z = zip(m)
return (type(m))(
map(inner, z) if tuple != inner else z
)
else:
return m
DISPLAY -------------------------------------------------
fTable:: String -> (a -> String) ->
(b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
indented:: Int -> String -> String
def indented(n):
'''String indented by n multiples
of four spaces
'''
return lambda s: (n 4 ' ') + s
showSquare:: -> String
def showSquare(rows):
'''Lines representing rows of lists.'''
w = 1 + len(str(reduce(max, map(max, rows), 0)))
return '\n' + '\n'.join(
map(
lambda row: indented(1)(''.join(
map(lambda x: str(x).rjust(w, ' '), row)
)),
rows
)
)
MAIN ---
if name == 'main':
main()
Output:
Magic squares of odd order N:
3 ->
8 1 6
3 5 7
4 9 2
5 ->
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
7 ->
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
R
See here for the solution for all three cases.
Example
magic(5)
[,1] [,2] [,3] [,4] [,5]
[1,] 17 24 1 8 15
[2,] 23 5 7 14 16
[3,] 4 6 13 20 22
[4,] 10 12 19 21 3
[5,] 11 18 25 2 9
Racket
lang racket
;; Using "helpful formulae" in:
;;
(define (squares n) n)
(define (last-no n) (sqr n))
(define (middle-no n) (/ (add1 (sqr n)) 2))
(define (M n) ( n (middle-no n)))
(define ((Ith-row-Jth-col n) I J)
(+ ( (modulo (+ I J -1 (exact-floor (/ n 2))) n) n)
(modulo (+ I ( 2 J) -2) n)
1))
(define (magic-square n)
(define IrJc (Ith-row-Jth-col n))
(for/list ((I (in-range 1 (add1 n)))) (for/list ((J (in-range 1 (add1 n)))) (IrJc I J))))
(define (fmt-list-of-lists l-o-l width)
(string-join
(for/list ((row l-o-l))
(string-join (map (λ (x) (~a #:align 'right #:width width x)) row) " "))
"\n"))
(define (show-magic-square n)
(format "MAGIC SQUARE ORDER:~a~%~a~%MAGIC NUMBER:~a~%"
n (fmt-list-of-lists (magic-square n) (+ (order-of-magnitude (last-no n)) 1)) (M n)))
(displayln (show-magic-square 3))
(displayln (show-magic-square 5))
(displayln (show-magic-square 9))
Output:
MAGIC SQUARE ORDER:3
8 1 6
3 5 7
4 9 2
Magic Number:15
MAGIC SQUARE ORDER:5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Magic Number:65
MAGIC SQUARE ORDER:9
47 58 69 80 1 12 23 34 45
57 68 79 9 11 22 33 44 46
67 78 8 10 21 32 43 54 56
77 7 18 20 31 42 53 55 66
6 17 19 30 41 52 63 65 76
16 27 29 40 51 62 64 75 5
26 28 39 50 61 72 74 4 15
36 38 49 60 71 73 3 14 25
37 48 59 70 81 2 13 24 35
Magic Number:369
Raku
(formerly Perl 6)
See Magic squares/Raku for a general magic square generator.
Output:
With a parameter of 5:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
The magic number is 65
With a parameter of 19:
192 213 234 255 276 297 318 339 360 1 22 43 64 85 106 127 148 169 190
212 233 254 275 296 317 338 359 19 21 42 63 84 105 126 147 168 189 191
232 253 274 295 316 337 358 18 20 41 62 83 104 125 146 167 188 209 211
252 273 294 315 336 357 17 38 40 61 82 103 124 145 166 187 208 210 231
272 293 314 335 356 16 37 39 60 81 102 123 144 165 186 207 228 230 251
292 313 334 355 15 36 57 59 80 101 122 143 164 185 206 227 229 250 271
312 333 354 14 35 56 58 79 100 121 142 163 184 205 226 247 249 270 291
332 353 13 34 55 76 78 99 120 141 162 183 204 225 246 248 269 290 311
352 12 33 54 75 77 98 119 140 161 182 203 224 245 266 268 289 310 331
11 32 53 74 95 97 118 139 160 181 202 223 244 265 267 288 309 330 351
31 52 73 94 96 117 138 159 180 201 222 243 264 285 287 308 329 350 10
51 72 93 114 116 137 158 179 200 221 242 263 284 286 307 328 349 9 30
71 92 113 115 136 157 178 199 220 241 262 283 304 306 327 348 8 29 50
91 112 133 135 156 177 198 219 240 261 282 303 305 326 347 7 28 49 70
111 132 134 155 176 197 218 239 260 281 302 323 325 346 6 27 48 69 90
131 152 154 175 196 217 238 259 280 301 322 324 345 5 26 47 68 89 110
151 153 174 195 216 237 258 279 300 321 342 344 4 25 46 67 88 109 130
171 173 194 215 236 257 278 299 320 341 343 3 24 45 66 87 108 129 150
172 193 214 235 256 277 298 319 340 361 2 23 44 65 86 107 128 149 170
The magic number is 3439
REXX
This REXX version will also generate a square of an even order, but it'll not be a magic square.
/REXX program generates and displays magic squares (odd N will be a true magic square)./
parse arg N . /obtain the optional argument from CL./
if N=='' | N=="," then N=5 /Not specified? Then use the default./
NN=NN; w=length(NN) /W: width of largest number (output)./
r=1; c=(n+1) % 2 /define the initial row and column./
@.=. /assign a default value for entire @./
do j=1 for NN / [↓] filling uses the Siamese method/
if r<1 & c>N then do; r=r+2; c=c-1; end /the row is under, column is over./
if r<1 then r=N / " " " " make row=last. /
if r>N then r=1 / " " " over, " " first./
if c>N then c=1 / " column " over, " col=first./
if @.r.c\==. then do; r=min(N,r+2); c=max(1,c-1); end /at the previous cell? /
@.r.c=j; r=r-1; c=c+1 /assign # ───► cell; next row & column/
end /j/
/ [↓] display square with aligned #'s/
do r=1 for N; _= /display one matrix row at a time. /
do c=1 for N; = right(@.r.c, w) /construct a row of the magic square. /
end /c/
say substr(_, 2) /display a row of the magic square. /
end /r/
say / [↓] If an odd square, show magic #./
if N//2 then say 'The magic number (or magic constant is): ' N (NN+1) % 2
/stick a fork in it, we're all done. /
output when using the default input of: 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
The magic number (or magic constant is): 65
output when using the input of: 3
8 1 6
3 5 7
4 9 2
The magic number (or magic constant is): 15
output when using the input of: 19
192 213 234 255 276 297 318 339 360 1 22 43 64 85 106 127 148 169 190
212 233 254 275 296 317 338 359 19 21 42 63 84 105 126 147 168 189 191
232 253 274 295 316 337 358 18 20 41 62 83 104 125 146 167 188 209 211
252 273 294 315 336 357 17 38 40 61 82 103 124 145 166 187 208 210 231
272 293 314 335 356 16 37 39 60 81 102 123 144 165 186 207 228 230 251
292 313 334 355 15 36 57 59 80 101 122 143 164 185 206 227 229 250 271
312 333 354 14 35 56 58 79 100 121 142 163 184 205 226 247 249 270 291
332 353 13 34 55 76 78 99 120 141 162 183 204 225 246 248 269 290 311
352 12 33 54 75 77 98 119 140 161 182 203 224 245 266 268 289 310 331
11 32 53 74 95 97 118 139 160 181 202 223 244 265 267 288 309 330 351
31 52 73 94 96 117 138 159 180 201 222 243 264 285 287 308 329 350 10
51 72 93 114 116 137 158 179 200 221 242 263 284 286 307 328 349 9 30
71 92 113 115 136 157 178 199 220 241 262 283 304 306 327 348 8 29 50
91 112 133 135 156 177 198 219 240 261 282 303 305 326 347 7 28 49 70
111 132 134 155 176 197 218 239 260 281 302 323 325 346 6 27 48 69 90
131 152 154 175 196 217 238 259 280 301 322 324 345 5 26 47 68 89 110
151 153 174 195 216 237 258 279 300 321 342 344 4 25 46 67 88 109 130
171 173 194 215 236 257 278 299 320 341 343 3 24 45 66 87 108 129 150
172 193 214 235 256 277 298 319 340 361 2 23 44 65 86 107 128 149 170
The magic number (or magic constant is): 3439
Ring
n=9
see "the square order is : " + n + nl
for i=1 to n
for j = 1 to n
x = (i2-j+n-1) % nn + (i2+j-2) % n + 1
see "" + x + " "
next
see nl
next
see "the magic number is : " + n(nn+1) / 2 + nl
Output:
the square order is : 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
the magic number is : 369
RPL
Translation of: IS-BASIC
≪ → n
≪ n DUP 2 →LIST 0 CON
1 n FOR j
1 n FOR k
j k 2 →LIST
j 2 k - n + 1 - n MOD n
j 2 k + 2 - n MOD 1 +
+ PUT
NEXT NEXT
n DUP SQ 1 + 2 /
≫ ≫ 'ODDMAGIC' STO
5 ODDMAGIC
Output:
2: [[2 23 19 15 6]
[14 10 1 22 18]
[21 17 13 9 5]
[8 4 25 16 12]
[20 11 7 3 24]]
1: 65
Ruby
def odd_magic_square(n)
raise ArgumentError "Need odd positive number" if n.even? || n <= 0
n.times.map{|i| n.times.map{|j| n((i+j+1+n/2)%n) + ((i+2j-5)%n) + 1} }
end
[3, 5, 9].each do |n|
puts "\nSize #{n}, magic sum #{(nn+1)/2n}"
fmt = "%#{(nn).to_s.size + 1}d" n
odd_magic_square(n).each{|row| puts fmt % row}
end
Output:
Size 3, magic sum 15
8 1 6
3 5 7
4 9 2
Size 5, magic sum 65
16 23 5 7 14
22 4 6 13 20
3 10 12 19 21
9 11 18 25 2
15 17 24 1 8
Size 9, magic sum 369
50 61 72 74 4 15 26 28 39
60 71 73 3 14 25 36 38 49
70 81 2 13 24 35 37 48 59
80 1 12 23 34 45 47 58 69
9 11 22 33 44 46 57 68 79
10 21 32 43 54 56 67 78 8
20 31 42 53 55 66 77 7 18
30 41 52 63 65 76 6 17 19
40 51 62 64 75 5 16 27 29
Rust
fn main() {
let n = 9;
let mut square = vec![vec![0; n]; n];
for (i, row) in square.iter_mut().enumerate() {
for (j, e) in row.iter_mut().enumerate() {
e = n (((i + 1) + (j + 1) - 1 + (n >> 1)) % n) + (((i + 1) + (2 (j + 1)) - 2) % n) + 1;
print!("{:3} ", e);
}
println!("");
}
let sum = n (((n n) + 1) / 2);
println!("The sum of the square is {}.", sum);
}
Output:
47 58 69 80 1 12 23 34 45
57 68 79 9 11 22 33 44 46
67 78 8 10 21 32 43 54 56
77 7 18 20 31 42 53 55 66
6 17 19 30 41 52 63 65 76
16 27 29 40 51 62 64 75 5
26 28 39 50 61 72 74 4 15
36 38 49 60 71 73 3 14 25
37 48 59 70 81 2 13 24 35
The sum of the square is 369.
Scala
def magicSquare( n:Int ) : Option[Array[Array[Int]]] = {
require(n % 2 != 0, "n must be an odd number")
val a = Array.ofDimInt
// Make the horizontal by starting in the middle of the row and then taking a step back every n steps
val ii = Iterator.continually(0 to n-1).flatten.drop(n/2).sliding(n,n-1).take(nn2).toList.flatten
// Make the vertical component by moving up (subtracting 1) but every n-th step, step down (add 1)
val jj = Iterator.continually(n-1 to 0 by -1).flatten.drop(n-1).sliding(n,n-2).take(nn2).toList.flatten
// Combine the horizontal and vertical components to create the path
val path = (ii zip jj) take (nn)
// Fill the array by following the path
for( i<-1 to (nn); p=path(i-1) ) { a(p._1)(p._2) = i }
Some(a)
}
def output() : Unit = {
def printMagicSquare(n: Int): Unit = {
val ms = magicSquare(n)
val magicsum = (n n + 1) / 2
assert(
if( ms.isDefined ) {
val a = ms.get
a.forall(.sum == magicsum) &&
a.transpose.forall(.sum == magicsum) &&
(for(i<-0 until n) yield { a(i)(i) }).sum == magicsum
}
else { false }
)
if( ms.isDefined ) {
val a = ms.get
for (y <- 0 to n 2; x <- 0 until n) (x, y) match {
case (0, 0) => print("╔════╤")
case (i, 0) if i == n - 1 => print("════╗\n")
case (i, 0) => print("════╤")
case (0, j) if j % 2 != 0 => print("║ " + f"${ a(0)((j - 1) / 2) }%2d" + " │")
case (i, j) if j % 2 != 0 && i == n - 1 => print(" " + f"${ a(i)((j - 1) / 2) }%2d" + " ║\n")
case (i, j) if j % 2 != 0 => print(" " + f"${ a(i)((j - 1) / 2) }%2d" + " │")
case (0, j) if j == (n 2) => print("╚════╧")
case (i, j) if j == (n 2) && i == n - 1 => print("════╝\n")
case (i, j) if j == (n 2) => print("════╧")
case (0, ) => print("╟────┼")
case (i, ) if i == n - 1 => print("────╢\n")
case (i, _) => print("────┼")
}
}
}
printMagicSquare(7)
}
Output:
╔════╤════╤════╤════╤════╤════╤════╗
║ 30 │ 39 │ 48 │ 1 │ 10 │ 19 │ 28 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 38 │ 47 │ 7 │ 9 │ 18 │ 27 │ 29 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 46 │ 6 │ 8 │ 17 │ 26 │ 35 │ 37 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 5 │ 14 │ 16 │ 25 │ 34 │ 36 │ 45 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 13 │ 15 │ 24 │ 33 │ 42 │ 44 │ 4 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 21 │ 23 │ 32 │ 41 │ 43 │ 3 │ 12 ║
╟────┼────┼────┼────┼────┼────┼────╢
║ 22 │ 31 │ 40 │ 49 │ 2 │ 11 │ 20 ║
╚════╧════╧════╧════╧════╧════╧════╝
Seed7
$ include "seed7_05.s7i";
const func integer: succ (in integer: num, in integer: max) is
return succ(num mod max);
const func integer: pred (in integer: num, in integer: max) is
return succ((num - 2) mod max);
const proc: main is func
local
var integer: size is 3;
var array array integer: magic is 0 times 0 times 0;
var integer: row is 1;
var integer: column is 1;
var integer: number is 0;
begin
if length(argv(PROGRAM)) >= 1 then
size := integer parse (argv(PROGRAM));
end if;
magic := size times size times 0;
column := succ(size div 2);
for number range 1 to size 2 do
magic[row][column] := number;
if magic[pred(row, size)][succ(column, size)] = 0 then
row := pred(row, size);
column := succ(column, size);
else
row := succ(row, size);
end if;
end for;
for key row range magic do
for key column range magic[row] do
write(magic[row][column] lpad 4);
end for;
writeln;
end for;
end func;
Output:
s7 magicSquaresOfOddOrder 7
SEED7 INTERPRETER Version 5.0.5203 Copyright (c) 1990-2014 Thomas Mertes
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
Sidef
func magic_square(n {.is_pos && .is_odd}) {
var i = 0
var j = idiv(n, 2)
var magic_square = []
for l in (1 .. n2) {
magic_square[i][j] = l
if (magic_square[i.dec % n][j.inc % n]) {
i = (i.inc % n)
}
else {
i = (i.dec % n)
j = (j.inc % n)
}
}
return magic_square
}
func print_square(sq) {
var f = "%#{(sq.len2).len}d";
for row in sq {
say row.map{ f % _ }.join(' ')
}
}
var(n=5) = ARGV»to_i()»...
var sq = magic_square(n)
print_square(sq)
say "\nThe magic number is: #{sq.sum}"
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
The magic number is: 65
Stata
See here for all three cases.
. mata magic(5)
1 2 3 4 5
+--------------------------+
1 | 17 24 1 8 15 |
2 | 23 5 7 14 16 |
3 | 4 6 13 20 22 |
4 | 10 12 19 21 3 |
5 | 11 18 25 2 9 |
+--------------------------+
Swift
extension String: Error {}
struct Point: CustomStringConvertible {
var x: Int
var y: Int
init(_ _x: Int,
_ _y: Int) {
self.x = _x
self.y = _y
}
var description: String {
return "(\(x), \(y))\n"
}
}
extension Point: Equatable,Comparable {
static func == (lhs: Point, rhs: Point) -> Bool {
return lhs.x == rhs.x && lhs.y == rhs.y
}
static func < (lhs: Point, rhs: Point) -> Bool {
return lhs.y != rhs.y ? lhs.y < rhs.y : lhs.x < rhs.x
}
}
class MagicSquare: CustomStringConvertible {
var grid:[Int:Point] = [:]
var number: Int = 1
init(base n:Int) {
grid = [:]
number = n
}
func createOdd() throws -> MagicSquare {
guard number < 1 || number % 2 != 0 else {
throw "Must be odd and >= 1, try again"
return self
}
var x = 0
var y = 0
let middle = Int(number/2)
x = middle
grid = Point(x,y)
for i in 2 ... numbernumber {
let oldXY = Point(x,y)
x += 1
y -= 1
if x >= number {x -= number}
if y < 0 {y += number}
var tempCoord = Point(x,y)
if let _ = grid.firstIndex(where: { (k,v) -> Bool in
v == tempCoord
})
{
x = oldXY.x
y = oldXY.y + 1
if y >= number {y -= number}
tempCoord = Point(x,y)
}
grid[i] = tempCoord
}
print(self)
return self
}
fileprivate func gridToText(_ result: inout String) {
let sorted = sortedGrid()
let sc = sorted.count
var i = 0
for c in sorted {
result += " \(c.key)"
if c.key < 10 && sc > 10 { result += " "}
if c.key < 100 && sc > 100 { result += " "}
if c.key < 1000 && sc > 1000 { result += " "}
if i%number==(number-1) { result += "\n"}
i += 1
}
result += "\nThe magic number is \(number (number number + 1) / 2)"
result += "\nRows and Columns are "
result += checkRows() == checkColumns() ? "Equal" : " Not Equal!"
result += "\nRows and Columns and Diagonals are "
let allEqual = (checkDiagonals() == checkColumns() && checkDiagonals() == checkRows())
result += allEqual ? "Equal" : " Not Equal!"
result += "\n"
}
var description: String {
var result = "base \(number)\n"
gridToText(&result)
return result
}
}
extension MagicSquare {
private func sortedGrid()->[(key:Int,value:Point)] {
return grid.sorted(by: {$0.1 < $1.1})
}
private func checkRows() -> (Bool, Int?)
{
var result = Set<Int>()
var index = 0
var rowtotal = 0
for (cell, _) in sortedGrid()
{
rowtotal += cell
if index%number==(number-1)
{
result.insert(rowtotal)
rowtotal = 0
}
index += 1
}
return (result.count == 1, result.first ?? nil)
}
private func checkColumns() -> (Bool, Int?)
{
var result = Set<Int>()
var sorted = sortedGrid()
for i in 0 ..< number {
var rowtotal = 0
for cell in stride(from: i, to: sorted.count, by: number) {
rowtotal += sorted[cell].key
}
result.insert(rowtotal)
}
return (result.count == 1, result.first)
}
private func checkDiagonals() -> (Bool, Int?)
{
var result = Set<Int>()
var sorted = sortedGrid()
var rowtotal = 0
for cell in stride(from: 0, to: sorted.count, by: number+1) {
rowtotal += sorted[cell].key
}
result.insert(rowtotal)
rowtotal = 0
for cell in stride(from: number-1, to: sorted.count-(number-1), by: number-1) {
rowtotal += sorted[cell].key
}
result.insert(rowtotal)
return (result.count == 1, result.first)
}
}
try MagicSquare(base: 3).createOdd()
try MagicSquare(base: 5).createOdd()
try MagicSquare(base: 7).createOdd()
Demonstrating:
Works with: Swift 5
Output:
base 3
8 1 6
3 5 7
4 9 2
The magic number is 15
Rows and Columns are Equal
Rows and Columns and Diagonals are Equal
base 5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
The magic number is 65
Rows and Columns are Equal
Rows and Columns and Diagonals are Equal
base 7
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
The magic number is 175
Rows and Columns are Equal
Rows and Columns and Diagonals are Equal
Tcl
proc magicSquare {order} {
if {!($order & 1) || $order < 0} {
error "order must be odd and positive"
}
set s [lrepeat $order [lrepeat $order 0]]
set x [expr {$order / 2}]
set y 0
for {set i 1} {$i <= $order2} {incr i} {
lset s $y $x $i
set x [expr {($x + 1) % $order}]
set y [expr {($y - 1) % $order}]
if {[lindex $s $y $x]} {
set x [expr {($x - 1) % $order}]
set y [expr {($y + 2) % $order}]
}
}
return $s
}
Demonstrating:
Works with: Tcl version 8.6
package require Tcl 8.6
set square [magicSquare 5]
puts [join [lmap row $square {join [lmap n $row {format "%2s" $n}]}] "\n"]
puts "magic number = [tcl::mathop::+ {}[lindex $square 0]]"
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
magic number = 65
TI-83 BASIC
Translation of: C
Works with: TI-83 BASIC version TI-84Plus 2.55MP
9→N
DelVar [A]:{N,N}→dim([A])
For(I,1,N)
For(J,1,N)
Remainder(I2-J+N-1,N)N+Remainder(I2+J-2,N)+1→A
End
End
[A]
Output:
[[2 75 67 59 51 43 35 27 10]
[22 14 6 79 71 63 46 38 30]
[42 34 26 18 1 74 66 58 50]
[62 54 37 29 21 13 5 78 70]
[73 65 57 49 41 33 25 17 9 ]
[12 4 77 69 61 53 45 28 20]
[32 24 16 8 81 64 56 48 40]
[52 44 36 19 11 3 76 68 60]
[72 55 47 39 31 23 15 7 80]]
VBScript
Translation of: Liberty BASIC
Sub magic_square(n)
Dim ms()
ReDim ms(n-1,n-1)
inc = 0
count = 1
row = 0
col = Int(n/2)
Do While count <= nn
ms(row,col) = count
count = count + 1
If inc < n-1 Then
inc = inc + 1
row = row - 1
col = col + 1
If row >= 0 Then
If col > n-1 Then
col = 0
End If
Else
row = n-1
End If
Else
inc = 0
row = row + 1
End If
Loop
For i = 0 To n-1
For j = 0 To n-1
If j = n-1 Then
WScript.StdOut.Write ms(i,j)
Else
WScript.StdOut.Write ms(i,j) & vbTab
End If
Next
WScript.StdOut.WriteLine
Next
End Sub
magic_square(5)
Output:
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
V (Vlang)
Translation of: Ring
fn main() {
mut n, mut x := 9, 0
println("the square order is: ${n}" + "\n")
for i in 1 .. n + 1 {
for j in 1 .. n + 1 {
x = (i 2 - j + n - 1) % n n + (i 2 + j - 2) % n + 1
print(" ${x:2} ")
}
println("")
}
println("\n" + "the magic number is: ${n (n n+1) / 2}")
}
Output:
the square order is : 9
2 75 67 59 51 43 35 27 10
22 14 6 79 71 63 46 38 30
42 34 26 18 1 74 66 58 50
62 54 37 29 21 13 5 78 70
73 65 57 49 41 33 25 17 9
12 4 77 69 61 53 45 28 20
32 24 16 8 81 64 56 48 40
52 44 36 19 11 3 76 68 60
72 55 47 39 31 23 15 7 80
the magic number is : 369
VTL-2
Translation of: C
10 N=1
20 ?="Magic square of order ";
30 ?=N
40 ?=" with constant ";
50 ?=NN+1/2N
60 ?=":"
70 Y=0
80 X=0
90 ?=Y2+N-X/N0+%N+(Y2+X+1/N0+%+1
100 $=9
110 X=X+1
120 #=X<N90
130 ?=""
140 Y=Y+1
150 #=YN20
Output:
Magic square of order 1 with constant 1:
1
Magic square of order 3 with constant 15:
2 9 4
7 5 3
6 1 8
Magic square of order 5 with constant 65:
2 23 19 15 6
14 10 1 22 18
21 17 13 9 5
8 4 25 16 12
20 11 7 3 24
Magic square of order 7 with constant 175:
2 45 39 33 27 21 8
18 12 6 49 36 30 24
34 28 15 9 3 46 40
43 37 31 25 19 13 7
10 4 47 41 35 22 16
26 20 14 1 44 38 32
42 29 23 17 11 5 48
Wren
Translation of: Go
Library:Wren-fmt
import "./fmt" for Fmt
var ms = Fn.new { |n|
var M = Fn.new { |x| (x + n - 1) % n }
if (n <= 0 || n&1 == 0) {
n = 5
System.print("forcing size %(n)")
}
var m = List.filled(n n, 0)
var i = 0
var j = (n/2).floor
for (k in 1..nn) {
m[in + j] = k
if (m[M.call(i)n + M.call(j)] != 0) {
i = (i + 1) % n
} else {
i = M.call(i)
j = M.call(j)
}
}
return [n, m]
}
var res = ms.call(5)
var n = res
var m = res
for (i in 0...n) {
for (j in 0...n) Fmt.write("$4d", m[in+j])
System.print()
}
System.print("\nMagic number: %(((nn + 1)/2).floor n)")
Output:
15 8 1 24 17
16 14 7 5 23
22 20 13 6 4
3 21 19 12 10
9 2 25 18 11
Magic number : 65
XPL0
Translation of: ALGOL W
\Construct a magic square of odd order - as a procedure can't return an
\ array, the caller must supply one that is big enough.
function MagicSquare( Square, Order );
integer Square, Order;
integer Row, Col, I, J;
\Ensure a row/col position is on the square
function InSquare; int Pos ;
return if Pos < 1 then Order else if Pos > Order then 1 else Pos;
\move "up" a row in the square
function Up; int Row; return InSquare( Row - 1 );
\move "across right" in the square
function Right; int Col ; return InSquare( Col + 1 );
if (Order&1) = 0 or Order < 1 then begin
\can't make a magic square of the specified order
return false
end
else begin
\Order is OK - construct the square using de la Loubere's
\ algorithm as in the Wikipedia page
\initialise square
for I := 1 to Order do for J := 1 to Order do Square( I, J ) := 0;
\initial position is the middle of the top row
Col := ( Order + 1 ) / 2;
Row := 1;
\construct square
for I := 1 to ( Order Order ) do begin
Square( Row, Col ) := I;
if Square( Up( Row ), Right( Col ) ) # 0 then begin
\the up/right position is already taken, move down
Row := Row + 1;
end
else begin
\can move up/right
Row := Up( Row );
Col := Right( Col );
end
end; \for_i
\sucessful result
return true
end; \magicSquare
\prints the magic square
procedure PrintSquare( Square, Order );
integer Square, Order;
integer Sum, W, I_W, I, J;
begin
\set integer width to accomodate the largest number in the square
W := ( Order Order ) / 10;
I_W := 1;
while W > 0 do begin I_W := I_W + 1; W := W / 10 end;
Format(I_W+1, 0);
Sum:= 0;
for I := 1 to Order do Sum := Sum + Square( 1, I );
Text(0, "maqic square of order "); IntOut(0, Order);
Text(0, " : Sum: "); IntOut(0, Sum );
for I := 1 to Order do begin
CrLf(0);
RlOut(0, float(Square( I, 1 )) );
for J := 2 to Order do RlOut(0, float(Square( I, J )) )
end; \for_I
CrLf(0);
end; \printSquare
\test the magic square generation
integer Sq ( 1+11, 1+11 ), L, I;
begin
L:= [1, 3, 5, 7];
for I := 0 to 3 do begin
if MagicSquare( Sq, L(I) ) then PrintSquare( Sq, L(I) )
else Text(0, "can't generate square^m^j" );
end \for_I
end]
Output:
maqic square of order 1 : Sum: 1
1
maqic square of order 3 : Sum: 15
8 1 6
3 5 7
4 9 2
maqic square of order 5 : Sum: 65
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
maqic square of order 7 : Sum: 175
30 39 48 1 10 19 28
38 47 7 9 18 27 29
46 6 8 17 26 35 37
5 14 16 25 34 36 45
13 15 24 33 42 44 4
21 23 32 41 43 3 12
22 31 40 49 2 11 20
zkl
Translation of: Ruby
fcn rmod(n,m){ n=n%m; if (n<0) n+=m; n } // Ruby: -5%3-->1
fcn odd_magic_square(n){ //-->list of nn numbers, row order
if (n.isEven or n <= 0) throw(Exception.ValueError("Need odd positive number"));
}
T(3, 5, 9).pump(Void,fcn(n){
"\nSize %d, magic sum %d".fmt(n,(nn+1)/2n).println();
fmt := "%%%dd".fmt((nn).toString().len() + 1) n;
odd_magic_square(n).pump(Console.println,T(Void.Read,n-1),fmt.fmt);
});
Output:
Size 3, magic sum 15
8 1 6
3 5 7
4 9 2
Size 5, magic sum 65
16 23 5 7 14
22 4 6 13 20
3 10 12 19 21
9 11 18 25 2
15 17 24 1 8
Size 9, magic sum 369
50 61 72 74 4 15 26 28 39
60 71 73 3 14 25 36 38 49
70 81 2 13 24 35 37 48 59
80 1 12 23 34 45 47 58 69
9 11 22 33 44 46 57 68 79
10 21 32 43 54 56 67 78 8
20 31 42 53 55 66 77 7 18
30 41 52 63 65 76 6 17 19
40 51 62 64 75 5 16 27 29
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189717 | https://www.ahdictionary.com/word/search.html?q=SYNTAX | American Heritage Dictionary Entry: SYNTAX
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syn·tax (s ĭ nt ă ks′)
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n.
1.
a.The study of the rules whereby words or other elements of sentence structure are combined to form grammatical sentences.
b.A publication, such as a book, that presents such rules.
c.The pattern of formation of sentences or phrases in a language.
d.Such a pattern in a particular sentence or discourse.
2.Computers The rules governing the formation of statements in a programming language.
3.A systematic, orderly arrangement.
[French syntaxe, from Late Latin syntaxis, from Greek suntaxis, from suntassein, to put in order : sun-, syn- + tassein , tag-, to arrange.]
The American Heritage® Dictionary of the English Language, Fifth Edition copyright ©2022 by HarperCollins Publishers. All rights reserved.
Indo-European & Semitic Roots Appendices
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The Indo-European appendix covers nearly half of the Indo-European roots that have left their mark on English words. A more complete treatment of Indo-European roots and the English words derived from them is available in our Dictionary of Indo-European Roots.
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189718 | https://www.academia.edu/104714200/Legal_Transplants | (PDF) Legal Transplants
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Legal Transplants
Нина Кршљанин
2023, Springer eBooks
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Abstract
A legal transplant is a legal norm or legal institution (or, indeed, a group of institutions, entire normative act, etc.) that has been transferred to one legal system from another. The system from which the transplant originates is often referred to as the "donor system" or "model system," while the receiving system is sometimes named "donee" or "borrowing system." Transplantation includes anything from a norm being copied ad verbum, to extensive modifications being performed by the receiving system, while still keeping some core semblance of the original idea. The term, while older, has been popularized by Alan Watson's theory of legal transplants, which underlines the significance that non-legal factors, including pure chance, often play in the process of transplantation. Many approximate synonyms for the process of transplantation exist (e.g., one can also talk of the reception or borrowing of foreign law), but few provide suitable nouns for the law being transferred, which could partially explain the popularity of the term "transplant." Origins of the Term The earliest known use of the term appears in Jeremy Bentham's essay "Of the Influence of Time and Place in Matters of Legislation" (drafted in 1782, published in 1802), where he speaks of "transplanting laws" (Bentham 1962), though he often uses the term "transferring" as well. The purpose of the essay was not purely theoretical: it was also meant as practical advice for the English government of Bengal, and as a call for reform of English law at home (Huxley 2007). Bentham's view is that laws that are perfect in one country would not necessarily be such for another where different circumstances (both natural and social) reign. For a transplanting to succeed, he believes a detailed insight into such circumstances is required, and he provides a number of examples of differences that cause the same act or situation to have different legal implications in various countries. However, Bentham's advice for transplanters is mostly focused on causes, methods, and the advised (gradual) pace of the process, and the essay ends on a quite different note than it began, asserting that the best laws (and that would be those following utilitarian principles) would be the best for any time and place. The main fault with transplantation is, then, the faultiness of the transplanted laws (Bentham 1962). The term does not seem to have found widespread use in the nineteenth century. Only about a century and a half later, on August 1, 1927, Frederick P. Walton held a lecture "The Historical
... Read more
Related papers
A new theoretical framework for Legal Transplants? (Working Paper)
Hjalmar Newmark Diaz
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The Impossibility of a Grand Transplant Theory
Ilaria Minucci
This paper examines legal transplant theories in comparative law and in particular, the Turkish legal transplant experience through the example of reception of a civil law system in Turkey in the 1920s. Although not looking for an answer that finds either culturalists or transferists right, the author argues that a comparativist should avoid falling into the trap of either/or binary thinking about legal transplants and avoid being necessarily either transferist or culturalist, or a middle ground theorist. The author argues that comparativists need a separate interpretation for each individual legal transplant case that they study by applying social science methodology without providing a "grand theory of transplantation". This methodology requires a move away from simplicity to the complexities of real experiences situated in historical, social and cultural contexts.
download Download free PDFView PDF chevron_right
Legal Transplant of Undue Influence
Mindy ChenWishart
International Comparative Law Quarterly, 2013
Is legal transplant possible? The stark bipolarity of a 'yes' or 'no' answer attracted by such a question is much less interesting and revealing than the question: what shapes the life of legal transplants? The answer to the latter question is contingent on a wide range of variables triggered by the particular transplant; the result can occupy any point along the spectrum from faithful replication to outright rejection. This case study of the transplant of the English doctrine of undue influence into Singaporean law asks why the Singaporean courts have applied the doctrine in family guarantee cases to such divergent effect, when they profess to apply the same law. The answer owes less to grand theories than to a careful examination of the nature of the transplanted law and the relationship between the formal and informal legal orders of the originating and the recipient society raised by the particular transplant.
download Download free PDFView PDF chevron_right
Reverse Legal Transplants
Sital Kalantry
2020
In early modern history, laws often moved in one direction—from imperial nations to their colonies. In the contemporary era, we would expect legal solutions to move in many directions, including from economically weaker countries (“Global South” countries) to economically more developed ones (“Global North” countries). The legal transplant literature, however, documents relatively few transplants from Global South countries to Global North countries. There is also little critical analysis in the literature about why there is a paucity of “reverse legal transplants”—transplants from Global South countries to Global North countries. This Article presents a case study of a reverse legal transplant—the movement of restrictions on abortion from India to the United States. U.S. statutes restricting women from terminating their pregnancies on the basis of the predicted sex of the fetus have been enacted in nine states and introduced in over half of all state legislatures and the U.S. Congr...
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Dynamics in (Auto)-acculturation of Legal Transplants
Christo Grozev
Legal transplants were first conceptualized in the 1790s by W.A.J Wilson 1 , who pinpointed the cross-border and cross-cultural genesis of a substantial body of national law. Watson explained it largely with the simple efficiency of borrowing -generally universal -legal concepts, and maintained that transplantation was the most fertile source of legal development.
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Legal Transplants and the Frontiers of Legal Knowledge
michele graziadei
Theoretical Inquiries in Law, 2009
The study of legal transplants provides a vital critical supplement to mainstream theories about legal change. Legal transplants are not exceptional or isolated occurrences, despite the economic, social, political and cultural barriers that separate the world's legal systems. This Article goes beyond traditional approaches to the study of transplants by substituting the figurative language of transplants with explicit theory about how legal change is produced. It first provides a brief account of what the literature on legal transplants has achieved so far in terms of "macro" explanations of legal change currently available. It then argues that legal transplants as social acts performed by individuals call for a study of the "micro" level of engagement with legal change by individuals. The key notion that is advanced to explore this dimension is the notion of mediated action, which denotes action that is performed by individuals making use of features of the environment as tools to interact in a specific setting. The notion of mediated action was first introduced in cultural-historical psychological investigations of the social formation of the mind. As social acts, legal transplants represent instances of mediated action. The last part of this Article highlights how legal transplants raise questions of justice and discusses briefly how the new approach to the study of transplants advocated here relates to them.
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Comparative Law as the Study of Transplants and Receptions, The Oxford Handbook of Comparative Law (2006)
michele graziadei
The comparative study of transplants and receptions investigates contacts of legal cultures and explores the complex patterns of change triggered by them. The study of legal transfers offers considerable intellectual rewards. It shows that the law is a complex phenomenon and corrects simplistic views regarding what law is and how it develops. The spread of legal institutions, ideals, ideologies, doctrines, rules, and so on, is often in the hands of professional elites. The study of transplants and receptions demonstrates that the knowledge and standing of those elites comes from interactions between the local and non-local dimensions of the law, that is, between the national and international spheres. This picture is true in Berlin and in New York, in London and in Lima, but it is also true in less cosmopolitan environments.
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The challenges of legal transplants in a globalized context: A case study on ‘working’ examples
Maria P . Reyes
Globalization has led legal systems to influence each other throughout history and legal transplants have not been immune to that process either. Nowadays we do not just see examples of legal transplants but come across cases of 'cross-fertilization' of case law. An emergence of unstudied trends that have spread around the world in a voluntary and involuntary way across legal cultures can also be witnessed. However it seems that legal transplants have been largely misunderstood and have been condemned as mere copy-paste exercises due to their history. This paper studies the different theories of legal transplants in comparative law and also analyses some particular cases that have eventually ‘worked’ efficiently both in the country of origin and in the country where those laws were adopted. We can no longer say that legal transplants are impossible; they are a reality we cannot run away from. Amongst the many challenges faced by this phenomenon, the primary one is to overcome all the obstacles imposed by a ‘legal globalization’ that is threatening developing countries by focusing more on the liberalization of markets, rather than on policies supporting social welfare. Despite this fact, there is still hope for a new concept of ‘legal globalization’, one that includes executing a serious study of the trends that not only developing countries are in need of, but also is a need of the developed countries, which is the adoption of policies towards social welfare, one of the biggest challenges that face modern legal comparatists today.
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By Chance and Prestige: Legal Transplants in Russia and Eastern Europe
GIANMARIA AJANI
The American Journal of Comparative Law, 1995
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A strategic interpretation of legal transplants
Nuno Garoupa
DISCUSSION PAPER SERIES-CENTRE …, 2003
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189719 | https://mathuniversecompass.com/sumIndexShift.html | Typesetting math: 100%
Summation index shift
The summation index 'i=1' indicates that the summation begins at one,
with subsequent values plugged into 'i' starting from 1 and incrementing by one each time. Make sure you understand the basics - Summation Notation first.
In some examples, we may want to change the limits (boundaries) of summation.
The terms that we sum must stay the same, so the operation is equivalence.
To better understand, let's look at an example where we will change the boundaries:
here we start from 1. Let's say we want to start from 2:
we must keep the summation terms the same as in the first summation!
But now each summation term is greater by one in comparison to
the initial sum. So, we will subtract 1 as follows:
because we shifted the lower bound from 1 to 2, we also had to shift the upper bound
by one, up from 3 to 4 — so we sum three numbers as before. Done.
From the above, we can conclude that the following equality holds:
And in general, to shift up by 1:
And to shift by any whole number p:
where m is any whole number - the lower bound we start from
Let's do more similar examples together, and then you will do a couple on your own.
Example 1: shift summation index up by 1 for:
Solution: . Now, let's shift the lower and upper bounds by
one according to the formula above: Great, everything works.
Final solution:
Now it is your turn!
"Student, Example 1":
a) shift the summation index up by 1 for:
b) shift the summation index up by 1 for:
you can check solutions at the bottom of this page
Let's continue, Example 2: shift the summation index up by 2 for:
Solution:. Now let's shift lower and upper bounds by
two. Then subtract two from summation variable: Good.
Final solution:
Example 3: shift the summation index up by 3 for:
Solution: we have:
shifting the boundaries by 3 we get:
Final solution:
Now it is your turn!
Solve summation index shift examples (you can check sollutions on the buttom of this page):
Shift the summation index up by 1:
3.
5.
Shift the summation index up by 5:
6.
7.
Shift the summation index down by one:
8.
9.
Solutions are at the bottom on purpose. First, do them yourself :)) and then scroll down.
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Clément, P. 1985. The relationships of rotifers. In: S. Conway Morris et al. (eds.), The origins and relationships of lower invertebrates. Clarendon Press, Oxford.
Clément, P. and E. Wurdak. 1991. Rotifera. Pages 219–297 in: Microscopic Anatomy of Invertebrates, Vol. 4. F. W. Harrison and E. E. Ruppert, eds. Wiley-Liss, New York.
Ferraguti, M. and G. Melone. 1999. Spermiogenesis in Seison nebaliae (Rotifera, Seisonidea): further evidence of a rotifer—acanthocephalan relationship. Tissue Cell 31:428–440.
Funch, P., M.V. Sørensen and M. Obst. 2005. On the phylogenetic position of Rotifera—have we come any further? Hydrobiologia 546:11–28.
García-Varela, M. and S. A. Nadler. 2006. Phylogenetic relationships among Syndermata inferred from nuclear and mitochondrial gene sequences. Molecular Phylogenetics and Evolution 40 (1):61-72.
Garey, J. R., T. J. Near, M. R. Nonnemacher, and S. A. Nadler. 1996. Molecular evidence for Acanthocephala as a subtaxon of Rotifera. J. Mol. Evol. 43:287-292.
Garey, J. R., A. Schmidt-Rhaesa, T. J. Near, and S. A. Nadler. 1998. The evolutionary relationships of rotifers and acanthocephalans. Hydrobiologia 387:83-91.
Herlyn, H., O. Piskurek, J. Schmitz, U. Ehlers, and H. Zischler. 2003. The syndermatan phylogeny and the evolution of acanthocephalan endoparasitism as inferred from 18S rDNA sequences. Molecular Phylogenetics and Evolution 26:155–164.
Kutikova, L. 1983. Parallelism in the evolution of rotifers. Hydrobiologia 104:3-7.
Lorenzen, S. 1985. Phylogeny of pseudocoelomate evolution. In: S. Conway Morris et al. (eds), The origins and relationships of lower invertebrates. Clarendon Press, Oxford.
Mark Welch, D. B. 2000. Evidence from a protein-coding gene that acanthocephalans are rotifers. Invertebrate Biology 119:17-26.
Mark Welch, D. B. 2001. Early contributions of molecular phylogenetics to understanding the evolution of Rotifera. Hydrobiologia 446/447:315-322.
Melone, G., C. Ricci, H. Segers, and R. L. Wallace. 1998. Phylogenetic relationships of phylum Rotifera with emphasis on the families of Bdelloidea. Hydrobiologia 387/388:101-107.
Miquelis, A., J.-F. Martin, E. W. Carson, G. Brun, and André Gilles. 2000. Performance of 18S rDNA helix E23 for phylogenetic relationships within and between the Rotifera–Acanthocephala clades. C.R. Acad. Sci. Paris, Sciences de la vie/Life Sciences 323:925–941.
Nogrady, T., R. L. Wallace, and T. W. Snell. 1993. Rotifera, Vol. 1: Biology, Ecology and Systematics. Guides to the Identification of the Microinvertebrates of the Continental Waters of the World. H. J. Dumont, ed. SPB Academic Publishers bv., The Hague.
Segers, H. 2002. The nomenclature of the Rotifera: annotated checklist of valid family- and genus-group names. Journal of Natural History 36:631-640.
Sørensen, M. V. 2002. On the evolution and morphology of the rotiferan trophi, with a cladistic analysis of Rotifera. Journal of Zoological Systematics and Evolutionary Research 40:129-154.
Sørensen, M. V. and G. Giribet 2006. A modern approach to rotiferan phylogeny: Combining morphological and molecular data. Molecular Phylogenetics and Evolution 40(2):585-608.
Sudzuki, M. 1964. New systematical approach to the Japanese planktonic Rotatoria. Hydrobiologia 23:1-124.
Wallace, R.L. and T. W. Snell. 1991. Rotifera. In: Ecology and Classification of North American Freshwater Invertebrates. J. H. Thorpe & A. P. Covich, eds. Academic Press, New York.
Witek, A., H. Herlyn, A. Meyer, L. Boell, G. Bucher and T. Hankeln. 2008. EST based phylogenomics of Syndermata questions monophyly of Eurotatoria. BMC Evolutionary Biology 2008, 8:345doi:10.1186/1471-2148-8-345
Zrzavy, J. 2001. The interrelationships of metazoan parasites: a review of phylum- and higher-level hypotheses from recent morphological and molecular phylogenetic analyses. Folia Parasitologica 48:81-103.
Information on the Internet
Rotifer Systematic Database. Elizabeth Walsh.
Micscape magazine:
The Wonderfully Weird World of Rotifers. A article written by Richard L. Howey and illustrated by Wim van Egmond.
Rotifers by Wim van Egmond.
Gallery of Rotifers by Wim van Egmond.
Rotifers and how to find them by Roy Winsby.
Title Illustrations
Click on an image to view larger version & data in a new window
| Scientific Name | Acanthocephala |
| Comments | with proboscis extended |
| Creator | Houseman |
| Specimen Condition | Dead Specimen |
| Sex | Female |
| Copyright | © BIODIDAC |
| Scientific Name | Bdelloidea |
| Creator | J. M. Cavanihac |
| Copyright | © BIODIDAC |
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189721 | https://books.google.mw/books?id=ZZ8kAAAAQBAJ&printsec=copyright&hl=pt-BR&source=gbs_pub_info_r | Quantitative Chemical Analysis - Daniel C. Harris - Google Livros
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Quantitative Chemical Analysis
Daniel C. Harris
W. H. Freeman, 27 de jul. de 2010 - 1008 páginas
Dan Harris's Quantitative Chemical Analysis continues to be the most widely used textbook for analytical chemistry. It offers consistently modern portrait of the tools and techniques of chemical analysis, incorporating real data, spreadsheets, and a wealth of applications, all presented in a witty, personable style that engages students without compromising the principles and depth necessary for a thorough and practical understanding.
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Sobre o autor(2010)
Biographical Statement for Nomination of Daniel C. Harris for
J. Calvin Giddings Award for Excellence in Analytical Chemical EducationI was born in 1948 in Brooklyn, New York. As a teenager, I enjoyed a science program on Saturdays at Columbia University, where I took note of especially good teaching by astronomy professor Lloyd Motz. In my freshman year at Massachusetts Institute of Technology, excellent teaching of organic chemistry by Daniel S. Kemp diverted me from biochemistry to chemistry. A spectroscopy class from George F. Whitesides led me to Whitesides and his student Chuck Casey (later President of the American Chemical Society) for senior thesis research. I developed a strong consciousness for high quality teaching. Two other classes with noteworthy teaching quality were quantum mechanics from John S. Waugh and group theory from F. A. Cotton.
After graduating from MIT shortly before my 20th birthday, I headed to Caltech where I joined the research group of Harry B. Gray—an exceptional lecturer. After a year as a teaching assistant in organic chemistry, George S. Hammond and Harry Gray recognized a spark for teaching and offered me the opportunity to team teach an advanced freshman course. My graduate student partner, Michael D. Bertolucci, and I were given carte blanche to develop an interesting course for freshman that would not overlap other courses in the curriculum. We chose an overview of general chemistry for one term, followed by two terms of introduction to group theory and spectroscopy. We conducted a critique of each other’s lecture immediately after every class. I placed highest value in interest, content, clarity, and physical understanding, which became main goals in my textbook writing. At the age of 21, I found myself driven to write lecture notes which, upon the recommendation of Harry Gray, evolved into the book Symmetry and Spectroscopy. I team-taught the freshman course with other graduate students and had the academic rank of Instructor during my last year of graduate studies. For part of that year, I was a postdoc in the fledgling field of 13C-NMR spectroscopy with John D. Roberts.
After two years as a postdoc at the Albert Einstein College of Medicine in New York City with Philip Aisen—an exemplary mentor—I started my first faculty position at the University of California at Davis in 1975. I was assigned to teach analytical chemistry for sophomores and accelerated freshmen. This assignment was interesting because I had never taken a course in analytical chemistry. I arrived at MIT after analytical chemistry became an elective and flew through MIT too quickly to partake in the analytical course. I had practical analytical experience from undergraduate, graduate, and postdoctoral research. My source of instruction in chemical equilibrium was the graduate course “Aquatic Chemistry” taught by J. J. Morgan at Caltech. At Davis, I sat in on an analytical courses taught by a senior member of the department to “learn the ropes” before being thrust before my first students in analytical chemistry.
My burning desire at Davis was to be the best teacher I could be. I was known for being available at all hours for student questions, for circulating through laboratories every day, and for memorizing the names and faces of every student. It became apparent to students that sitting in the back row of a 300-seat lecture hall did not offer immunity from being called upon by name to answer a question during lecture. I brought a demonstration into almost every lecture and eachterm ended with a series of explosions. The last class each term attracted far more students than were enrolled in the course. The majority of my students at Davis were life science majors whose interests resonated with my research interest in metalloproteins.
I surveyed every analytical textbook I could find and taught from several. I found the more thorough books to be dull and the more interesting books to be less thorough. After two years, I decided to write text to accompany my lectures. My goal was to be interesting and thorough in the selected topics. Publisher’s representatives saw my notes in the bookstore and soon there were five offers for publication. I visited each publisher and unashamedly adopted the best suggestions from each editor. In 1978, I signed with W. H. Freeman as the publisher I thought would produce the nicest book. After two more years of writing, a year of revision, and a year of production, the first edition of Quantitative Chemical Analysis was born in 1982.
By this time, I had not been offered tenure at Davis or at Franklin and Marshall College. I loved teaching, but decided to try a different career. In 1983, I moved to the U.S. Navy’s Michelson Laboratory at China Lake, California, where my present title is Senior Scientist. In the course of 25 years with the Navy, I was elected an Esteemed Fellow and received a Top Navy Scientist award. My research concerns transparent ceramic sensor windows. I have been teaching a professional course in this subject several times each year since 1990 and wrote the monograph Materials for Infrared Windows and Domes, which is the standard reference in its field.
Meanwhile, Quantitative Chemical Analysis sold well enough for the publisher to invite me to prepare a 2nd edition. I found myself with two full-time jobs—one for the Navy and a second as a textbook writer. My wife Sally has been editorial assistant and proofreader on every book. She produced all of the illustrations for Symmetry and Spectroscopy with a one-year-old watching over her shoulder. Thirty years after signing our contract with Freeman, we are working on the 8th edition. The book has had 12 foreign translations.
Our chief competitor, Doug Skoog (with coauthors West, Holler, and Crouch) had “big” and “little” books to serve two market levels. Freeman asked me to write a small book to complement Quantitative Chemical Analysis, but I hesitated to go into competition with myself. By 1995, we no longer had children in the house and the time was ripe for a “small” book. My priorities for Exploring Chemical Analysis were to be (1) short, (2) interesting, and (3) elementary—in that order. This book has now gone through 4 editions and 3 foreign translations.
A survey published in 2002 found that my two books were used in over half of the analytical chemistry courses in the United States. [P. A. Mabrouk, Anal. Chem. 2002, 74, 269A.] In 2008, Quantitative Chemical Analysis received the McGuffey Longevity Award from the Textbook and Academic Authors Association.
In my writing, I try to catch the reader’s attention and to convey excitement by illustrating each topic with interesting real-world examples. I try to get to the heart of a topic with the minimum number of words. It is good pedagogy to explain everything and not to assume prior knowledge on the part of the reader. Heavy use of illustrations makes ideas more understandable and memorable. Chapters are broken into short sections which are more digestable than long sections. Recalling my own student days, I include answers to all problems at the back of the book. Some teachers would rather have a set of problems without answers, but I have never heard a student complain about immediate feedback after working a problem. An informal writing style and a little humor provide a relaxed tone.Quantitative Chemical Analysis evolved over 30-years. Spectrophotometry grew from one to three chapters as it moved from the middle of the book to the front and then to the middle again. Chromatography expanded from two to four chapters as its importance grew. Electrophoresis and mass spectrometry were added. Quality assurance, sampling, and sample preparation were added and quality assurance increased in importance. Computer programming projects were introduced in the second edition. Spreadsheets appeared in the fourth edition and increased in each subsequent edition. A spreadsheet-oriented chapter on advanced chemical equilibrium appeared in the seventh edition. Uniform, high-interest opening vignettes appeared in the fourth edition. Chapter 0 on the “analytical process” describing an actual student analysis of caffeine in chocolate appeared in the fifth edition. Gravimetric analysis was demoted to the back of the book. Electroanalytical chemistry decreased from five to four chapters. Instructions for experiments moved to the web in the sixth edition to make room for growth in other subjects.
Exploring Chemical Analysis began with brevity as the first goal. User feedback directed me to add several topics that had been rejected for the first edition. These topics included activity coefficients, systematic treatment of equilibrium, EDTA and redox titration curve calculations, and an expanded discussion of spectrophotometry. Placement of spectrophotometry early in the book did not fit well with many curricula, so the subject was moved back in the second edition. The third edition increased emphasis on quality assurance, integrated mass spectrometry with chromatography, and introduced inductively coupled plasma-mass spectrometry. Spreadsheets gradually increased in every edition. A short “ask yourself” question with an answer at the end of every worked example appeared in the fourth edition.
The most common comment I receive from teachers can be paraphrased as “I love your book and I wish it weren’t so long. And please add more on (fill in favorite topic).” Kolthoff, Sandell, Mehan, and Bruckenstein wrote in the preface of what was perhaps the most venerable analytical textbook of the 20th century, “as much as anyone, we regret the length of this revised edition ” (1170 pages) and “it is a very hard undertaking to seek to please everybody.”
A good textbook has the attributes of a good teacher. The best description I have seen for a good teacher is a person with a “deep understanding of the subject, unbounded enthusiasm, humor, and the ability to communicate excitement, clarity and precision of thought and word, and the ability to put oneself in the mind of a student new to the subject.” [C. Thyagaraja, Caltech News, 2000, 34, 11.] To these I would add the ability to convey the significance and applications of the subject. I strive toward these ends in my writing.
Informações bibliográficas
Título Quantitative Chemical Analysis
AutorDaniel C. Harris
Edição 7, revisada
Editora W. H. Freeman, 2010
ISBN 1429277882, 9781429277884
Num. págs.1008 páginas
Exportar citaçãoBiBTeXEndNoteRefMan
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189722 | https://m.youtube.com/watch?v=E5HkIX9dGaU&t=0s | HOW TO FIND LOCAL MAXIMUM AND MINIMUM - Apply the first derivative test to find local max and min
Jake's Math Lessons
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Posted: 10 Jun 2020
In this video I'll show you how to find local maximum and minimum values of a function. You can find the local maximum and minimum values by using the first derivative test. Local extrema and critical points go hand in hand. The process of finding the local extrema using first derivative test is closely related to finding the critical values and the increasing and decreasing intervals of a function.
You can find the local extrema with the first derivative test with a few simple steps:
1. Find the critical numbers of the function - 1:00
2. Determine which intervals are increasing and decreasing - 3:52
3. Use the first derivative test to classify critical numbers as local max or min - 7:31
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+ Increasing and Decreasing Intervals -
+ How to Find Global Maximums and Minimums -
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Transcript:
hey guys jake here coming at you with another math problem today today i'm going to be going over how to find the local maximum and minimum of a function for the maximum and minimum values if there's multiple and i'm going to be showing you how to do that with this example here so this is kind of building off of finding the increasing and decreasing intervals of a function which i talked about a couple days ago i'll put a link to that up here so you can check that out because that is a big part of the process for finding the local maximum and minimum values of a function so really when you're trying to find these local or relative maximum and minimum values you're going to start by figuring out the increasing and the decreasing intervals just like we did in the previous video and then you're going to kind of take it a step further by applying the first derivative test since essentially to figure out if each of these critical numbers within the function is a local maximum or a local minimum so let me show you what i mean by that like i said it's going to be very similar to finding increasing and decreasing intervals the first step is going to be to find your critical numbers the critical numbers are the places where the local maximum and minimum values may occur so that's where you want to start so to find the critical numbers i have a video on that too i'll put a link here so you can check that out if you want some more detail because i'm just going to kind of go over this a little quickly until we get to the first derivative portion but to find the critical numbers we just need to find the derivative of our function so to find the derivative of this function we're going to be able to do that with just using power rule so power rule says we'll bring the 3 down in front that'll give us 6x and then we'll lower the power by one six x squared bring this power down in front will give us minus six x and then lowering the power by one will just leave us with an x and then minus twelve the x just kind of falls off for the derivative so now to find the critical numbers we need to figure out the places where f prime equals zero and we need to figure out the places where f prime does not exist this is a polynomial polynomials are defined everywhere so as a result there are no x values where this does not exist so all we need to do to find our critical numbers is set our derivative equal to zero and solve for x so the first thing we can do notice all these terms have a six in it so we can pull a six out factor that out that'll just leave us with x squared minus x minus two and then we can factor out this uh what's left in the parentheses here so x squared minus x minus two would factor out to x minus two times x plus one and to kind of double check that negative two and positive 1 would add up to negative 1 which is the coefficient of our x term here and then negative 2 times positive 1 gives us negative 2. so that's how we can kind of double check so now let's just uh kind of erase up here and move up towards the top here to figure out where this equals zero we can take each of these factors individually and set any of them set each of them equal to zero and then solve each factor on its own so first of all we have six six will never equal zero so this won't give us any any critical numbers then x minus two if we set that equal to zero we can add two to both sides and that'll just give us x equals two so that's one of our critical numbers and then lastly x plus one equals zero we can subtract one from both sides and that'll give us x equals negative one so x equals negative one and x equals two are the only two critical numbers for this function so now what we want to do is figure out basically which intervals our function is increasing and decreasing and then we will kind of use this with the first derivative test so all i've done here is kind of moved up the factored version of this f prime so that we can use that to figure out our intervals and the reason why is it's going to be a little bit easier to test where our function is increasing and decreasing so remember when we're applying the first derivative test and figuring out the intervals where our function is increasing and decreasing the only points that we want to put on our number line are the critical numbers one thing that i see a lot of people do is put 0 in your number line you don't want to do that 0 is not a critical number in this case so you don't want to just automatically put 0 in your number line you only want the critical numbers of course if zero is a critical number you do want it there but don't put zero on there every single time if it's not a critical number we only want our critical numbers on our number line so now what we can do is plug in some values or one value from each of these three intervals right we have all the numbers over here all the numbers between these two and then all the numbers over here we just want to plug in one number from each of these intervals into our derivative to figure out if our function is increasing or decreasing on each of those intervals so let's take negative two for example is less than negative one that would represent this interval so if we plug negative 2 into our derivative we're going to get a positive number for this factor we're going to get negative 2 minus negative 2 is a negative number for that factor and then negative 2 plus 1 is negative 1 which is still a negative number so we're going to get a positive number times a negative number times a negative number which will give us a positive number so what that means is on this interval over here for all x values less than negative 1 our function has a positive slope because the output of f prime is positive so that means our function f is increasing for all those values a positive slope a positive output in your f prime just means that your function is increasing at that point so now let's plug in some number in here so let's say x equals 0 because that's between negative 1 and 2 we'll plug that into our f prime and figure out if f is increasing or decreasing there so if we plug in 0 6 is still going to be a positive number 0 minus 2 is going to be a negative 2 that's a negative number and then 0 plus 1 is positive 1. so we're going to have a positive number times a negative number times a positive number which is going to give us a negative number so what that tells us is f prime is negative between x equals negative 1 and x equals 2. so therefore the slope of f is negative which means f is decreasing it's going down at that point as we go from left to right our function is going to be going down in this interval and it's going to be going up in this interval now we just need to figure out this third integral over here take some x value greater than 2 so let's just plug in x equals 3 for example 6 will still be positive x minus two three minus two would be positive one that's a positive number and three plus one would be positive four so we're gonna get a positive number times a positive number times a positive number gives us a positive slope so f is increasing for all x is greater than 2. so now what the first derivative test says is if we have a critical number where our function is increasing to the left of that critical number and it's decreasing to the right of that critical number then this critical number will be a local max or a local maximum so x equals negative 1 is a local maximum the first derivative test also tells us if we have a critical number on our function where f is decreasing to the left of that critical number and then it's increasing to the right of that critical number then that critical number is a local minimum or a relative minimum so at x equals 2 that critical number is a local minimum the last thing that the first derivative tells us is if it's doing the same thing on both sides of the critical number so we don't have an example of that in this problem but let's just say that we had some critical number where when we drew our number line out we got some critical number here where our function was increasing on both sides it was increasing to the left and increasing to the right that would not be a local maximum or a local minimum it just means that the slope is zero there so basically if you had something like this your function would be increasing and then it would flatten out at that critical number and then it would start increasing again so this is you could see that the slope is zero there so it would be a critical number but it's not a local maximum or a local minimum and then same thing if it was decreasing to both sides if it was decreasing to the left and to the right we would get a function that would be like this right if we had a flat slope right here so it was a critical number but it was decreasing to the left and decreasing to the right this is not a local maximum or minimum so that's really all the first derivative test says i'll do another example of this and that video will be coming out tomorrow so be sure to subscribe and hit that bell icon so you're notified when i come out with that video and you can keep getting some practice with the first derivative test |
189723 | https://artofproblemsolving.com/wiki/index.php/Sum_and_difference_of_powers?srsltid=AfmBOooReNsQ86UqjOZ27uY5q1LLYu3QALwvvktj5-bXxgu24qGniBkk | Art of Problem Solving
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Sum and difference of powers
The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
Contents
[hide]
1 Sums of Odd Powers
2 Differences of Powers
3 Sum of Cubes
4 Factorizations of Sums of Powers
5 See Also
Sums of Odd Powers
Differences of Powers
If is a positive integer and and are real numbers,
For example:
Note that the number of terms in the second factor is equal to the exponent in the expression being factored.
An amazing thing happens when and differ by , say, . Then and
.
For example:
If we also know that then:
Sum of Cubes
Factorizations of Sums of Powers
Note that all these sums of powers can be factorized as follows:
If we have a difference of powers of degree , then
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial , except for the fact that the coefficient on each of the terms is . This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
icecreamrolls8
See Also
Factoring
Difference of squares, an extremely common specific case of this.
Binomial Theorem
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¿El «agua» es femenino o masculino?
El sustantivo "agua" es de género femenino, pero al comenzar por "a" tónica -en la que recae el acento- se utiliza la forma del artículo en singular por razones de fonética histórica, según la Real Academia de la Lengua.
¿La palabra «agua» es masculino o femenino? Y es que a veces la lengua resulta confusa, ya que la mayoría de palabras de género femenino en español vienen precedidas por «la». Sin embargo hay excepciones, y esta es posiblemente más importante de nuestro idioma. Y aunque es una palabra tan importante y usada, no solo en el español, en general, resulta curioso que muchas personas cometan este error.
Existe un caso similar, pero a la inversa, se trata de la palabra «moto», nadie dice «el moto». Sin duda es un misterio sin resolver. A veces, el género de las palabras puede resultar confuso, por lo que en este artículo responderemos a algunas de las dudas más frecuentes sobre palabras relacionadas con el mundo del agua.
Entonces, ¿»agua» es masculino o femenino?
El sustantivo «agua» es de género femenino, pero al comenzar por «a» tónica -en la que recae el acento- se utiliza la forma del artículo en singular por razones de fonética histórica, según la Real Academia de la Lengua (RAE). Esta regla solo opera cuando el artículo antecede inmediatamente al sustantivo. De ahí que digamos «el agua«. Pero si entre el artículo y el sustantivo se interpone otra palabra, la regla queda sin efecto. De ahí que se deba decir «la misma agua».
Cuando la palabra agua está precedida del indefinido una toma generalmente la forma «un». Asimismo, los indefinidos alguna y ninguna pueden adoptar en estos casos las formas apocopadas (algún agua, ningún agua) o mantener las formas plenas (alguna agua, ninguna agua).
Si ponemos la palabra en plural, si identificaremos la marca femenina cuando aparecen en plural. Este es un fenómeno que no es exclusivo a la palabra «agua», puesto que muchas otras palabras del español se rigen por a misma norma. Por ejemplo, «águila» o «área» experimentan los mismos cambios.
En ese sentido, hablaríamos de las águilas, las áreas, las aves, unas hadas, unas almas, etc. Por otro lado, si interponemos algún tipo de palabra entre el artículo y el sustantivo, volverá a aparecer la marca del femenino: Por ejemplo, si hablamos: el hada/la mágica hada, el área/las peligrosas áreas. Este uso se da, asimismo, con algún y ningún:algún alma, algún hada, algún aula, algún hacha, ningún alma, ningún hada, ningún aula, ningún hacha.
Por su parte, el indefinido una toma generalmente la forma un cuando antecede inmediatamente a sustantivos femeninos que comienzan por /a/ tónica:un área, un hacha, un águila(si bien no es incorrecto, aunque sí poco frecuente, utilizar la forma plena una: una área, una hacha, una águila).
Concordancia de la palabra
Puesto que estas palabras son femeninas, los adjetivos deben concordar siempre en femenino: el agua clara, el agua limpia, el agua dulce. Con los demostrativos este, ese, aquel o con cualquier otro adjetivo determinativo, como todo, mucho, poco, otro, etc., deben usarse las formas femeninas correspondientes: esta agua, aquella misma agua, toda el agua, mucha agua… Por eso hay que decir, cuando se hable de nosotros mismos, «La Fundación del Agua» y no «La Fundación de la Agua».
Existen otros casos en que la lengua española muestra una gran flexibilidad a la hora de referirse a las palabras relacionadas con el agua. Piensa en la palabra «mar»… ¿dirías que es una palabra masculina o femenina? En este caso, la respuesta es ambas.
El uso de los dos géneros está aceptado por la RAE, por lo que la palabra «mar» es de género ambiguo. Lo normal hoy es usar mar como masculino (el mar). Sin embargo, también es un uso extendido y considerado como válido hacer uso como femenino en algunos contextos y expresiones: entrar en alta mar, etc.
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Islas de calor: el desafío de enfriar las ciudades
Las grandes ciudades experimentan temperaturas superiores a las de su entorno debido a que las superficies construidas capturan y retienen el calor. Esta situación se ve agravada por el cambio climático. Para reducir este efecto se plantea un urbanismo que, entre otras medidas, introduzca más zonas de vegetación y agua en el diseño urbano
Las ciudades concentran hoy más de la mitad de la población mundial y, según estimaciones de Naciones Unidas, en 2050 esta cifra crecerá hasta el 70%. El acelerado crecimiento de los entornos urbanos en todo el planeta plantea numerosos desafíos para conseguir que sean inclusivos, seguros, resilientes y sostenibles, tal y como propone el ODS 11.
Entre los muchos retos ligados al bienestar humano que abordan las ciudades está el fenómeno de las islas de calor, una situación que provoca que muchas ciudades experimenten temperaturas significativamente más altas que las de las zonas de alrededor que no están urbanizadas.
La diferencia térmica puede oscilar entre 1 °C y 3 °C durante el día, pero en episodios extremos y, sobre todo, durante las noches de verano, puede alcanzar hasta 10 grados adicionales.
Este exceso térmico afecta al bienestar de las personas y es un un desafío de salud pública. Y más si se tiene en cuenta que el cambio climático trae aparejado un aumento creciente de las temperaturas globales y olas de calor más frecuentes y acusadas. Esto provoca que, en muchas ciudades del mundo, la sensación térmica se haga insoportable en las épocas del año con más calor.
Por qué se forman las islas de calor urbanas
El origen de las islas de calor urbanas es complejo y responde a la interacción de múltiplesfactores físicos, ambientalesy de planificación del territorio.
En primer lugar hay que tener en cuenta que los materiales predominantes en las ciudades, como el asfalto, el hormigón o las cubiertas oscuras de los edificios, tienen una elevada capacidad de absorción térmica. Durante el día capturan energía solar y, por la noche, la liberan lentamente, provocando que las temperaturas nocturnas se mantengan elevadas y dificulten el descanso.
Pero a esta acumulación térmica por los materiales empleados hay que añadir el efecto que tiene la propia geometría urbana: las calles estrechas, los edificios altos y los espacios poco ventilados atrapan el calor y reducen la circulación del aire.
Además, hay que tener en cuenta que los vehículos a motor, los sistemas de climatización, las industrias y otras actividades liberan calor e incrementan la temperatura ambiental en la ciudad.
La falta de vegetación suficiente en los entornos urbanos agrava esta situación. Las plantas regulan la temperatura de forma natural mediante dos mecanismos esenciales: la sombra y la evapotranspiración, un proceso mediante el cual el agua absorbida por raíces y hojas se libera a la atmósfera, enfriando el aire. Además de la vegetación, es importante la presencia de zonas con fuentes y láminas de agua en la ciudad, ya que tienen la capacidad de refrescar el entorno cercano.
Cualquier persona puede apreciar la diferencia de sensación térmica que se experimenta al pasar de una calle soleada a un parque cubierto de árboles y regado. Sin árboles, parques ni zonas húmedas, las ciudades pierden este sistema de refrigeración natural. El resultado es un círculo vicioso: días más cálidos, noches más sofocantes, menor confort térmico y un aumento del consumo energético, ya que las personas dependen más de sistemas de refrigeración que, paradójicamente, generan aún más calor y emisiones de gases de efecto invernadero.
Consecuencias para la salud
Las consecuencias para la salud y el bienestar son considerables. La Organización Mundial de la Salud (OMS) advierte que las olas de calor son responsables de un aumento preocupante de la mortalidad. Añade, además, que el número de personas expuestas al calor extremo está aumentando exponencialmente debido al cambio climático en todas las regiones del mundo
Por ejemplo, según la OMS, sólo en Europa, en 2023 fallecieron 70.000 personas como consecuencia del calor sufrido entre junio y agosto. Los colectivos más vulnerables son las personas mayores, los niños pequeños y quienes padecen enfermedades cardiovasculares, respiratorias o metabólicas, aunque el impacto alcanza a toda la población.
El calor extremo, señala la OMS, también afecta al bienestar psicológico: interfiere con el sueño, reduce la concentración y eleva los niveles de ansiedad e irritabilidad.
El calor extremo también afecta al mundo laboral. Según un informe conjunto publicado en 2025 por la OMS y la Organización Meteorológica Mundial (OMM), la productividad de las personas trabajadoras disminuye entre un 2% y un 3% por cada grado por encima de 20 °C, especialmente cuando a la temperatura alta se suman condiciones de humedad elevada y potente radiación solar. Es algo que afecta especialmente alos trabajos desarrollados al aire libre.
En ese mismo sentido, un informe publicado en enero de 2025 por el World Economic Forum, o Foro de Davos, en colaboración con la compañía de seguros Allianz, señala que que el calor extremo podría ocasionar pérdidas de productividad por valor de 2,4 billones de dólares anuales para 2030.
Existen otros riesgos: por ejemplo, según el Instituto Nacional de Seguridad y Salud en el Trabajo (INSST), del Gobierno de España, los accidentes en el entorno laboral aumentan un 17 % durante los episodios de calor extremo.
Infraestructuras verdes y azules
Los expertos coinciden en que la solución al fenómeno de las islas de calor urbanas pasa por repensar la planificación y el diseño urbano. Una de las estrategias más efectivas es apostar por las infraestructuras verdes y azules, un enfoque que integra naturaleza y agua en la ciudad.
Las infraestructuras verdes abarcan parques, corredores ecológicos, jardines verticales, techos ajardinados y arbolado urbano, mientras que las infraestructuras azules incorporan elementos hídricos como fuentes, estanques, canales o humedales artificiales. La combinación de ambas permite reducir las temperaturas mediante sombra y evapotranspiración, además de mejorar la calidad del aire, aumentar la biodiversidad y favorecer el bienestar psicológico de la población.
El papel del agua en la mitigación de las islas de calores esencial. Los cuerpos de agua estratégicamente diseñados generan microclimas más frescos gracias a la evaporación, especialmente cuando se integran con vegetación circundante.
El empleo de sistemas de drenaje urbano sostenible, como pavimentos permeables, jardines de lluvia y sistemas de retención contribuye a reducir la escorrentía superficial y a permitir que el agua se integre en el suelo de las ciudades y no pase por encima de ella a toda velocidad.
Aplicar estas técnicas evita la saturación de los sistemas de drenaje durante episodios de lluvias torrenciales, cada vez más frecuentes debido al cambio climático y permite que pueda ser contenida más tiempo para aportarle la depuración adecuada antes de devolverla al medio ambiente. También se puede emplear para generar zonas húmedas estacionales, espacios que potencian la adaptación al cambio climático y crean zonas de enfriamiento de la urbe
Los techos verdes y las fachadas vegetalesrepresentan otra solución de gran potencial. Además de mejorar el aislamiento térmico de los edificios y reducir el consumo energético, ayudan a rebajar la temperatura superficial de las cubiertas en varios grados.
También cabe emplear pavimentos reflectantes, también llamados “pavimentos fríos”, que incrementan la capacidad de las superficies urbanas para reflejar la radiación solar y evitar la acumulación de calor.
El futuro de las ciudades requiere una planificación que las haga más resilientes y sostenibles. Apostar por infraestructuras verdes y azules, aumentar el arbolado, rehabilitar espacios públicos mediante soluciones basadas en la naturaleza y repensar los materiales constructivos son pasos para construir ciudades más frescas, habitables y resistentes frente al cambio climático.
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Praderas marinas y carbono azul: conoce los bosques del mar
En las costas de todo el planeta prosperan ecosistemas esenciales para la salud de los océanos. Son las praderas marinas, formadas por vegetales adaptados a vivir en el agua salada. Son esenciales para la salud oceánica, ya que limpian el agua, producen oxígeno, fijan sedimentos y carbono y ofrecen un nicho ecológico a cientos de especies
Bajo la superficie del mar, en aguas poco profundas y luminosas, crecen lentamente los pastos o praderas marinas, uno de los ecosistemas más productivos y vitales del planeta. Aunque a menudo pasan inadvertidos, los hábitats formados por praderas marinas tienen un papel crucial en la salud de los océanos, en la biodiversidad y en la lucha contra el cambio climático, según señala el programa de las Naciones Unidas para el Medio Ambiente (UNEP), en su informe Out of The Blue, publicado en 2020.
Las praderas marinas están compuestas por plantas, emparentadas con las que vemos en tierra firme. Son vegetales que han evolucionado a partir de plantas terrestres que regresaron al mar hace millones de años y se adaptaron al agua salada.
Técnicamente, las especies que forman las praderas marinas son fanerógamas, es decir, vegetales con raíces, hojas, flores, frutos y semillas. Se las suele confundir con algas, pero no lo son, ya que las algas son un tipo de organismo evolutivamente más antiguo que carece de esos elementos.
Las fanerógamas marinas se desarrollan en fondos arenosos poco profundos y con buena luminosidad que les permita realizar el proceso de fotosíntesis. Forman praderas sumergidas que se extienden por cientos de kilómetros cuadrados junto a las costas. Dependiendo de la claridad de las aguas, pueden prosperar incluso a 40 metros de profundidad, por lo que tienen un amplio rango de litoral para ocupar.
¿Qué especies forman las praderas marinas?
Existen unas 60 especies reconocidas en todo el mundo de fanerógamas marinas, agrupadas en géneros como Zostera, Halophila, Thalassia, Cymodocea y el de Posidonia, que incluye especies como la emblemática Posidonia oceanica.
Esta última es una especie que solo se encuentra en el mar Mediterráneo, donde forma praderas densas y longevas, algunas con miles de años de antigüedad. Su crecimiento es muy lento, pero persistente, y acaba generando un rico ecosistema que transforma los arenales costeros en un punto caliente de biodiversidad.
En cierto sentido, ofrecen el mismo servicio que los arrecifes de coral en aguas más cálidas: transforman el entorno creando nichos ecológicos para multitud de especies de peces, moluscos y crustáceos y todo tipo de fauna marina.
La posidonia se puede reproducir por semillas, que generan nuevas plantas. Pero habitualmente prosperapor medio de sus raíces o rizomas, que, van extendiéndose por el fondo marino y creando nuevos brotes de hojas, colonizando poco a poco el espacio. Diversos estudios científicos han demostrado que algunos tramos de praderas marinas están compuestos en realidad por un único ejemplar de planta, que ha ido colonizando decenas de metros con clones de sí misma nacidos desde la raíz.
Importancia ecológica de los pastos marinos
Los pastos marinos son auténticos ingenieros del ecosistema. Entre sus funciones más destacadas están:
Producción de oxígeno: al realizar la fotosíntesis, liberan oxígeno al agua, beneficiando la vida marina.
Hábitat y refugio: proporcionan alimento, refugio y zona de cría para miles de especies, incluyendo capturas de interés comercial de peces, moluscos y crustáceos, por lo que ofrecen servicios ambientales y también económicos directos al ser humano.
Protección costera: sus raíces y rizomas fijan los sedimentos del fondo marino, reduciendo la erosión y protegiendo playas y costas. Ayudan a mantener y regenerar las playas: los temporales arrastran hasta las costas enormes cantidades de hojas secas de plantas marinas. Esta masa vegetal ayuda a retener la arena y, a medida que se degrada, va generando una masa que ayuda a proteger las playas de la erosión de las olas.
Captura de carbono azul: los pastos marinos almacenan grandes cantidades de carbono en sus hojas, raíces y sedimentos. Durante su crecimiento generan grandes cantidades de material orgánico que queda enterrado en suelo marino en condiciones de falta de oxígeno, lo que permite la captura a largo plazo del carbono presente en esa materia orgánica. Este carbono capturado en ambientes costeros se conoce como carbono azul.
Plantas de ‘Posidonia oceanica’ en las costas del Mar Mediterráneo. | FOTO: Albert Kok vía Wikimedia Commons
El valor del carbono azul
Los pastos marinos, junto con los manglares y las marismas, forman los principales reservorios de carbono azul. A pesar de ocupar solo un pequeño porcentaje de la superficie oceánica, estas praderas pueden capturar y almacenar hasta 35 veces más carbono por hectárea que un bosque tropical, según indican diversos estudios.
Este carbono queda enterrado durante siglos en los sedimentos marinos, lo que los convierte en una herramienta natural y poderosa contra el cambio climático. La Posidonia oceanica, en particular, es una de las plantas más eficaces en la fijación de carbono.
Praderas marinas: hábitat amenazado
A pesar de su importancia, los pastos marinos están en declive, principalmente por la contaminación del agua, el fondeo de embarcaciones que destruyen físicamente las praderas, la urbanización costera, el dragado de fondos, el aumento de temperaturas, la acidificación de los océanos, y a invasión de especies exóticas.
Es una suma de impactos que han hecho que en las últimas décadas estos ecosistemas pierdan terreno y vitalidad en todo el mundo.
En el caso de Posidonia oceanica, las praderas han disminuido drásticamente amplias zonas costeras del Mediterráneo. Disminuir los impactos es clave, ya que debido a su lento crecimiento, una vez dañadas, estas plantas tardan décadas o incluso siglos en recuperarse, si es que lo hacen.
Conservación y protección
Frente a esa situación, se están desarrollando múltiples estrategias para conservar y restaurar los pastos marinos:
Áreas marinas protegidas: que limitan el acceso o las actividades humanas en zonas clave.
Sistemas de fondeo ecológico: boyas que evitan el anclaje directo sobre las praderas.
Monitorización científica: mediante satélites, buceo y sensores submarinos para evaluar la salud de las praderas.
Proyectos de restauración: plantación de brotes de pasto marino en zonas degradadas.
Educación ambiental: para sensibilizar a la ciudadanía, pescadores y turistas sobre su valor.
Las praderas marinas son mucho más que “plantas bajo el mar”. Son infraestructuras naturalesesenciales para la biodiversidad, el clima y la protección costera. Cuidarlas ayuda a proteger los mares, frenar el cambio climático y asegurar el futuro de la vida en la Tierra.
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189726 | https://themathbehindthemagic.wordpress.com/2015/05/13/nim-part-1/ | the Math behind the Magic
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May 13, 2015 · 6:00 AM
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Nim Part 1
According to Wikipedia, “Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps.” We are going to simplify the game slightly and only have one heap of objects with players removing up to a maximum number per turn. The winner of Nim is the person who removes the last object from the pile. The two players are Alice and Bob. Alice always starts. So let us get started with a simple example. Say you have a pile of 10 stones and each player can remove up to 2 stones, what could a game look like? Here we go:
Alice removes 1 stone, 9 stones remaining.
Bob removes 2 stones, 7 stones remaining.
Alice removes 1 stone, 6 stones remaining.
Bob removes 1 stone, 5 stones remaining.
Alice removes 2 stone, 3 stones remaining.
Bob removes 2 stones, 1 stone remaining.
Alice removes 1 stone, 0 stones remaining. Alice wins!
There you have it: a simple game of Nim with Alice as the winner. However, a perceptive individual might wonder, “Is there a pattern to this game? Is there a winning strategy?”
Rewinding a bit, what if Alice were to take the 4th last stone, leaving 3 stones in the pile? Then Bob could either take 1 or 2 stones. Using a diagram, we can see what happens.
From the above flow chart, it is clear that no matter what action Bob makes, Alice can counter it and come out victorious. Therefore, as long as Alice is able to leave a pile of 3 stones for Bob, Alice will always win. Facing down a pile of 3 stones is bad news for Bob.
Rewinding another step, what if Bob was facing a pile of 6 stones?
Again we see that from 6 stones, no matter what move Bob makes, Alice will be able to counter him and leave him with a pile of 3 stones. Further, we already know that leaving Bob with 3 stones assures her a victory. Poor Bob loses when he is facing down a pile of 6 stones as well. The same logic can be applied if Bob is staring at a pile of 9 stones.
Since the original pile starts with 10 stones, if Alice takes 1 stone to start the game off, leaving a pile of 9 stones, she will win 100% of the time. What do 3, 6 and 9 all have in common? They are all multiples of 3!
So there you have it: if you can leave your opponent with a pile of stones that is a multiple of 3 then you will win. Another way of thinking about this is that 3, 6 and 9 all have remainder 0 when divided by 3. The strategy employed by Alice is to remove enough stones each round so that she leaves Bob on a multiple of 3. As seen in the flow charts above, if Bob removes 1 stone, Alice should remove 2 stones. Similarly, if Bob removes 2 stones, Alice should remove 1 stone. And her victory is certain.
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189727 | http://physics.bu.edu/~duffy/java/RelV2.html | Relative velocity
The river is 50 meters wide. The boat is the red circle. Although the boat points in the direction of the red arrow, it will actually travel along the black line, because the water carries the boat downstream. The black line represents the velocity of the boat with respect to the shore. Note that this is the sum of the velocity of the boat with respect to the water (in red) + the velocity of the water with respect to the shore (in blue).
The positive x direction is to the right. The positive y direction is up.
Here are a few things to think about:
Adjust the speed of the river. Does it affect the time it takes the boat to cross?
With the river speed at 2 m/s, can you find a way to make the boat go straight across?
What determines whether the boat goes upstream, downstream, or straight across?
Created by Andrew Duffy, Boston University Physics Department.
Last update September 9, 1998 |
189728 | https://artofproblemsolving.com/wiki/index.php/Rotation?srsltid=AfmBOor6HqLRLJ-nD3L0G7npcwB4rr4_1AyV6jJKgIEElZlekN7o42By | Art of Problem Solving
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Rotation
A rotation of a planar figure is a transformation that preserves area and angles, but not orientation. The resulting figure is congruent to the first.
Suppose we wish to rotate triangle clockwise around a point , also known as the center of rotation.
We would first draw segment . Then, we would draw a new segment, such that the angle formed is , and . Do this for points and , to get the new triangle
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189729 | https://www.khanacademy.org/math/up-class-8/xc58caf62fe43a614:circle-and-cyclic-quadrilateral/xc58caf62fe43a614:circle-and-cyclic-quadrilateral-13-b/v/inscribed-angles-indian-accent | Inscribed angles (video) | Khan Academy
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UP Class 8th
Course: UP Class 8th>Unit 11
Lesson 2: Angles subtended by chords of a circle
Inscribed angles
Inscribed angle theorem proof
Angle subtended by chords of a circle
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Angles subtended by chords of a circle
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Inscribed angles
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Sal finds a missing inscribed angle using the inscribed angle theorem.Created by Kalakrit (Dubbing).
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189730 | https://www.askiitians.com/forums/Mechanics/the-density-of-a-thin-rod-of-length-l-varies-with_258085.htm | The density of a thin rod of length l varies with the distance x from - askIITians
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Mechanics>The density of a thin rod of length l var...
The density of a thin rod of length l varies with the distance x from one end as rho is equal to rho zero x square by l square find the position of centre of mass of rod
Tapan Swain , 5 Years ago
Grade 11
1 Answers
Arun
Last Activity: 5 Years ago
Dear student
Please attach an image as question is not clear.
I will be happy to help you.
Thanks and regards
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189731 | https://www.youtube.com/watch?v=tedzsRH0Jas | Piecewise function formula from graph | Functions and their graphs | Algebra II | Khan Academy
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[Voiceover] By now we're used to seeing functions defined like h(y)=y^2 or f(x)= to the square root of x. But what we're now going to explore is functions that are
defined piece by piece over different intervals
and functions like this you'll sometimes
view them as a piecewise, or these types of function definitions they might be called a
piecewise function definition. Let's take a look at this
graph right over here. This graph, you can see that the function is constant over this interval, 4x. And then it jumps up
in this interval for x, and then it jumps back down
for this interval for x. Let's think about how we would write this using our function notation. If we say that this right
over here is the x-axis and this is the y=f(x) axis. Then, let's see, our function
f(x) is going to be equal to, there's three different intervals. So let me give myself some space for the three different intervals. Now this first interval
is from, not including -9, and I have this open circle here. Not a closed in circle. So not including -9 but
x being greater than -9 and all the way up to and including -5. I could write that as -9 is less than x, less than or equal to -5. That's this interval, and what is the value of the function
over this interval? Well we see, the value
of the function is -9. It's a constant -9 over that interval. It's a little confusing because the value of the function is actually also the value of the lower bound on this
interval right over here. It's very important to look at
this says, -9 is less than x, not less than or equal. If it was less than or
equal, then the function would have been defined at
x equals -9, but it's not. We have an open circle right over there. But now let's look at the next interval. The next interval is
from -5 is less than x, which is less than or equal to -1. Over that interval, the
function is equal to, the function is a constant 6. It jumps up here. Sometimes people call this a
step function, it steps up. It looks like stairs to some degree. Now it's very important
here, that at x equals -5, for it to be defined only one place. Here it's defined by this part. It's only defined over here. So that's why it's
important that this isn't a -5 is less than or equal to. Because then if you put
-5 into the function, this thing would be filled in, and then the function would
be defined both places and that's not cool for a function, it wouldn't be a function anymore. So it's very important that when you input - 5 in here, you know which
of these intervals you are in. You can't be in two of these intervals. If you are in two of these intervals, the intervals should
give you the same values so that the function maps, from one input to the same output. Now let's keep going. We have this last
interval where we're going from -1 to 9. >From -1 to +9. And x starts off with -1 less than x, because you have an open
circle right over here and that's good because X equals -1 is defined up here, all the way to x is
less than or equal to 9. Over that interval, what is
the value of our function? Well you see, the value of
our function is a constant -7. A constant -7 and we're done. We have just constructed a piece by piece definition
of this function. Actually, when you see this
type of function notation, it becomes a lot clearer why function notation is useful even. Hopefully you enjoyed that. I always find these piecewise
functions a lot of fun. |
189732 | https://www.ncbi.nlm.nih.gov/books/NBK190419/ | Enterococcal Bacteriophages and Genome Defense - Enterococci - NCBI Bookshelf
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Gilmore MS, Clewell DB, Ike Y, et al., editors. Enterococci: From Commensals to Leading Causes of Drug Resistant Infection [Internet]. Boston: Massachusetts Eye and Ear Infirmary; 2014-.
Enterococci: From Commensals to Leading Causes of Drug Resistant Infection [Internet].
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Gilmore MS, Clewell DB, Ike Y, et al., editors.
Boston: Massachusetts Eye and Ear Infirmary; 2014-.
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Enterococcal Bacteriophages and Genome Defense
Breck A. Duerkop, Kelli L. Palmer, and Malcolm J. Horsburgh.
Author Information and Affiliations
Authors
Breck A. Duerkop,1 Kelli L. Palmer,2 and Malcolm J. Horsburgh 3.
Affiliations
1 Department of Immunology, University of Texas Southwestern Medical Center, Dallas, Texas 75390, USA
Email: ude.nretsewhtuostu@pokreuD.kcerB
2 Department of Molecular and Cell Biology, University of Texas at Dallas, Richardson, TX 75080, USA
3 Institute of Integrative Biology, University of Liverpool, Liverpool, L69 7ZB, United Kingdom
Corresponding author.
Created: February 11, 2014.
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Enterococcal Bacteriophages
A brief overview of bacteriophages: Bacteriophages (phages) are viruses that infect bacteria. Similar to the viruses of plants and animals, phages are inert and are unable to propagate themselves in the absence of a host. Phages depend on host metabolism to provide the organic material and machinery necessary for their replication and for the subsequent packaging of the viral genetic material during phage particle biosynthesis. Phages are associated with nearly all known bacterial taxa and, as a result, are found in diverse environments that range from soil to oceans and even in deserts (Prestel, Salamitou, & DuBow, 2008; Prigent, Leroy, Confalonieri, Dutertre, & DuBow, 2005; Srinivasiah, Bhavsar, Thapar, Liles, Schoenfeld, & Wommack, 2008; Wommack & Colwell, 2000). Phages are found either directly associated with their bacterial hosts or in large numbers as free virions in the environment. Since there is a vast distribution of phages across the globe, it is possible to theorize that phages constitute the most abundant biological entities on earth. Their numbers have been estimated to reach as high as 10 31 particles with the potential for 10 25 phage infections occurring every second (Pedulla, et al., 2003; Wommack & Colwell, 2000). As many more phage genome sequences have become available in recent years, it is obvious that phages are extremely incongruent at the genomic level. This diversity in genetic makeup is proposed to result from the fastidious replication of phage particles during the infection of highly permissive hosts. During these infections, phages are able to exchange DNA within host genomes through recombination, and continually generate diversity as a result (Hendrix, Smith, Burns, Ford, & Hatfull, 1999).
The vast majority of phages belong to the order of Caudovirales, which are tailed phages that have dsDNA and an isometric capsid. Caudovirales is comprised of three phylogenetically-related families that are discriminated by tail morphology: Myoviridae (long contractile tails), Siphoviridae (long non-contractile tails), and Podoviridae (short tails) (Ackermann, 2007; Krupovic, Prangishvili, Hendrix, & Bamford, 2011). The most well-studied tailed phages are the coliphages ʎ (Siphoviridae), T4 (Myoviridae), and T7 (Podoviridae) which infect Escherichia coli and which have served as workhorses for elucidating the mechanisms of modern molecular genetics and biochemistry (Johnson, Poteete, Lauer, Sauer, Ackers, & Ptashne, 1981; Miller, Kutter, Mosiq, Arisaka, Kunisawa, & Rüger, 2003; Ptashne, et al., 1980; Tabor & Richardson, 1985). Far less abundant are the non-tailed phages, which encompass numerous families with great morphological distinction; these include phages that are filamentous (long filaments to short rods), polyhedral (vesicular and envelope-like), and pleomorphic (including those that are lemon, droplet, and ampule shaped) (Ackermann, 2007). The nucleic acid content of phage genomes is either DNA or RNA and both double and single stranded DNA and RNA phages have been identified. In addition, the size of the phage genome can range from under ten kilobases to several hundred kilobases.
Phages have evolved replication strategies that can be lytic, lysogenic (temperate), or chronic. Chronic replication results in the continual, non-lethal shedding of virions by protrusion through the membrane. All phages have common life-cycle stages of adsorption, DNA injection and replication, virion production, and release. Tailed phages mediate host cell lysis through the combined action of a holin, which perforates the membrane, and an endolysin (lysin), which hydrolyses cell wall peptidoglycan. Lytic phages are restricted to a life-cycle that results in the lysis of their host. Temperate phages have two possible life-cycles: lysis, or the recombination of their genome at a chromosomal attachment site using a phage-encoded integrase. Temperate phages are maintained within the host chromosome by transcriptional repressors that determine when the phage undergoes an infectious or lytic switch. The lytic switch occurs when conditions within their host promote excision. Excision usually proceeds during times of hardship when host health is threatened, either by physical stress or by chemical stress, such as antibiotics, ultraviolet (UV) light, or reactive oxygen species (Allen, et al., 2011; DeMarini & Lawrence, 1992; Little & Mount, 1982). Temperate phages provide key insights into the evolution of bacterial pathogenesis, since many temperate phages encode virulence factors used by pathogenic bacteria during both human and animal infections (Bensing, Siboo, & Sullam, 2001; Brüssow, Canchaya, & Hardt, 2004; Novick, Christie, & Penadés, 2010).
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Distribution of phages across the enterococci
The first documented reports of enterococcal phages were published over 70 years ago (Clark & Clark, 1927; Evans, 1934). It was not until the early 1960s that more comprehensive analyses of enterococcal phages began to take shape (Brock, 1964; Rogers & Sarles, 1963). At the time, which was prior to the advent of modern molecular phylogenetics, the enterococci were characterized as group D streptococci. Rogers and Sarles (Rogers & Sarles, 1963) isolated phage-like particles from the intestinal tracts of Sprague-Dawley rats, and identified two phages that were capable of forming plaques on Streptococcus faecalis var. zymogenes. After careful analysis of host range using intestinal isolates of S. faecalis, followed by phage antibody serotyping, Rogers and Sarles captured some of the first images of enterococcal phages by using electron microscopy. They determined that the enterococcal phages appeared to have icosahedral heads and long non-contractile tails (Rogers & Sarles, 1963). One year later, Thomas Brock published a detailed survey of the host range of several enterococcal phages and identified enterococcal lysogens for the first time (Brock, 1964). Furthermore, Brock made the distinction that when testing for plaque formation on S. faecalis var. zymogenes, Streptococcus faecalis var. liquefaciens, and Streptococcus faecium, a degree of cross-reactivity of the phages for different host strains was present. Interestingly, this promiscuity was unique within S. faecalis and S. faecium, and phages specific to one species did not otherwise infect the other. When this type of phage resistance does occur, it is often due to inherent differences between species, with respect to their cell wall structure, mechanisms of protection against the acquisition of foreign DNA, and immunity due to lysogeny (superinfection exclusion).
Currently the well-studied enterococcal phages are those that infect and lysogenize Enterococcus faecalis and Enterococcus faecium. Numerous lytic phages that infect E. faecalis and E. faecium have been isolated from diverse environments, including sewage and wastewater sites, livestock runoff, and the intestinal tract (Horiuchi, Sakka, Hayashi, Shimada, Kimura, & Sakka, 2012; Lee & Park, 2012; Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010; Otawa, Hirakata, Kaku, & Nakai, 2012; Parasion, Kwiatek, Mizak, Gryko, Bartoszcze, & Kocik, 2012; Rogers & Sarles, 1963). To achieve host lysis phages encode cell membrane holins and peptidoglycan hydrolases. These cell wall hydrolases, most commonly from lytic phages, have great potential as novel therapeutics that can target pathogenic strains of E. faecalis and E. faecium.
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Known phage families found among Enterococci
Until recently, all characterized enterococcal phages belonged to the Podoviridae, Siphoviridae, or Myoviridae families. These phages exhibit considerable genetic and morphological diversity. For example, the genome sequence of E. faecalis strain 62, which was isolated from a healthy infant, revealed the presence of a Podoviridae phage, EF62Φ, which has a linear extrachromasomal genome. The maintenance of EF62Φ in E. faecalis is proposed to be the result of an encoded RepB and a toxin-antitoxin system (Brede, Snipen, Ussery, Nederbragt, & Nes, 2011). Mazaheri Nezhad Fard et al. isolated the first non-tailed enterococcal phages that belong to the polyhedral, filamentous, and pleomorphic (PFP) phages. In this study, the authors isolated phages from E. faecalis, E. faecium, and Enterococcus gallinarum strains found in piggery effluent. These phages, which were similar to PFPs, included a filamentous Inoviridae family phage, a polyhedral phage of the Leviviridae family, and several abnormally shaped pleomorphic phages that resembled droplet or lemon-like structures and which belong to the Guttaviridae and Fuselloviridae families (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010). Little is known about the genetic organization of these diverse phages, and it is also unclear whether these phages have lysogenic life-cycles.
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Environments where enterococcal phages are found
Enterococcal species reside in the oral cavity and urogenital tract of mammals as well as the digestive systems of mammals and insects. Enterococci are also frequently found in fecal waste sites, such as sewage treatment plants and uncontained areas of fecal contaminated groundwater. Furthermore, enterococci have evolved to become opportunistic pathogens that cause life-threatening infections, including endocarditis, septic bacteremia, and hospital-acquired wound infections. It is quite likely that the phages associated with different enterococcal species are also located within these diverse environments.
Enterococcal phages have been isolated from human and animal origins where the enterococci live as commensal bacteria (Caprioli, Zaccour, & Kasatiya, 1975; Nigutová, Styriak, Javorský, & Pristas, 2008; Rogers & Sarles, 1963). A comprehensive study from the 1970s isolated numerous enterococcal phages from the human urogenital tract, including phages found at urethral, endocervical, and ano-rectal body sites (Caprioli, Zaccour, & Kasatiya, 1975). These phages were highly successful at infecting numerous E. faecalis and E. faecium strains. Based on their ability to infect and lyse these enterococci, the phages were used to classify isolates of E. faecalis and E. faecium into 27 and 22 distinct strain types, respectively (Caprioli, Zaccour, & Kasatiya, 1975). This study highlights the importance of environments where enterococci are commensal bacteria as sources of phages. Further studies of these types of phages both within and outside their natural habitats may shed light on the ecology and community dynamics of commensal enterococci.
Enterococcal phages isolated from the intestinal tracts of rodents have been characterized (Rogers & Sarles, 1963); however, studies on enterococcal phage particles from human intestinal contents are limited. One particular study that addresses the composition of intestinal bacterial communities associated with a premature low-weight infant discovered that this infant maintained an amplified clonal population of Enterococcus in its intestines (Morowitz, et al., 2010). It was determined that the 16S rRNA sequence of this enterococcal strain, UC1ENC, was identical to the 16S rRNA sequences of several E. faecalis strains. Using total bacterial DNA sequences from the infant gut, the authors were able to reassemble the UC1ENC genome. UC1ENC shared ~81% of its protein coding sequence with E. faecalis V583. Two UC1ENC chromosomal prophage elements shared DNA sequence similarity with two phages (phage02 and phage04) found in the E. faecalis V583 chromosome. A second study further substantiated this evidence, and showed that the rise of Enterococcus within premature low-weight infants was not exclusive to the individual infant of the Morowitz et al. study. These data showed that in the case of eleven different premature low-weight infants, 75% of the individuals screened were highly populated with Enterococcus species within their intestinal bacterial communities (LaTuga, et al., 2011). In this study, numerous phage DNA sequences were also identified from the intestinal contents of the premature low-weight infants; however, this analysis did not elaborate as to whether any of these phages may be associated with the colonizing enterococcal strains. As an effort to better understand the microbial communities of the intestinal tract, the National Institutes of Health–funded Human Microbiome Project has begun to sequence numerous bacterial isolates from the intestine, including many species of enterococci (Proctor, 2011). These enterococcal genome sequences have revealed a large number of putative prophages that may be important to the commensal biology of the enterococci in the intestine.
The enterococci are also commensals of the mammalian oral cavity, and in some instances, have been associated with periodontal disease (Kayaoglu & Ørstavik, 2004). Enterococcal infections of the root canal, especially those caused by E. faecalis, are extremely resistant to current therapies (Stuart, Schwartz, Beeson, & Owatz, 2006). Unlike the intestinal tract, E. faecalis phages from the human oral cavity have been isolated and studied in detail. E. faecalis isolates have been recovered from the infected root canals of humans for whom therapeutic interventions were unsuccessful (Stevens, Ektefaie, & Fouts, 2011; Stevens, Porras, & Delisle, 2009). Four out of ten of these E. faecalis isolates could be induced to produce lytic phages through mitomycin C treatment. Three of these phages resembled Siphoviridae phages, and were characterized by long non-contractile tails and spherical heads. The fourth phage identified resembled a phage that was more closely related to the Myoviridae, with a contractile tail with tail fibers and an icosahedral head structure. This study was the first to show that enterococcal strains living within the oral cavity can be induced to produce phages. Enterococcal phages have been used to successfully reduce the ability of E. faecalis to grow on the surface of human dental roots. A phage multiplicity of infection of 0.1 was sufficient to minimize the ability of E. faecalis to colonize dental roots (Paisano, Spira, Cai, & Bombana, 2004). Treatment of E. faecalis endontic infection is difficult and recurrent infection is a concern. It has been proposed that phages that infect these bacteria may prove useful as an alternative to current treatment options (Paisano, Spira, Cai, & Bombana, 2004; Stevens, Porras, & Delisle, 2009).
Being members of the intestinal microbiota, Enterococcus species are shed from humans and animals in fecal waste. Phages have been instrumental in determining the existence of enterococci within contaminated water environments and are currently being used to monitor fecal contaminated water sources for the presence of enterococcal strains. One method relies on Enterococcus isolates from diverse contaminated water sources including waste-water run-off areas of grazing cattle, pigs, or sheep, and municipal waste-water sites (Räisänen, et al., 2007). These enterococcal isolates, which include multiple strains from the species E. faecalis, E. faecium, E. gallinarum, and Enterococcus casseliflavus, are used to survey water samples where enterococcal contamination is unknown. Enterococcal contamination within these water samples is indicated by the presence of phages capable of infecting the collection of enterococcal strains. This method, referred to as microbial source tracking, relies on using indicator strains of enterococci to identify infectious phages within water samples as a metric of enterococcal fecal contamination (Purnell, Ebdon, & Tayor, 2011). A second method uses a similar approach; however, a bank of known enterococcal phages termed “enterophages” are used to screen water samples for fecal E. faecalis contamination (Bonilla, Santiago, Marcos, Urdaneta, Domingo, & Toranzos, 2010; Santiago-Rodriguez, et al., 2010). Enterophages exclusively infect strains of E. faecalis and are present at levels of ~10 2 phages per 100 ml of domestic raw sewage (Bonilla, Santiago, Marcos, Urdaneta, Domingo, & Toranzos, 2010). Enterophages are diverse and include phages similar to the Siphoviridae as well as non-tailed phages with icosahedral shaped capsids. It is also thought that enterophages may be indicative of water sources contaminated with human fecal waste, although, enterophages similar to those isolated from contaminated water sources have yet to be identified in human stool samples (Santiago-Rodriguez, et al., 2010).
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Enterococcal Temperate Phages and the Impact of Lysogeny
Temperate phages possess an alternative life-cycle that is absent from the reproduction of lytic bacteriophages, whereby the bacterial host harbors the phage genome and then replicates it during cell division. This transmits the phage genome vertically to daughter cells, which subsequently propagate the phage. Lysogens—bacteria acting as phage genome hosts—express resistance to superinfection by the same phage, but not to superinfection by heterologous phages (Birge, 1994). Multiple lysogenic infections, over time and with heterologous temperate bacteriophages, produce polylysogens. Comparative genome analyses of several low G+C genera of the Bacilli subbranch, such as Enterococcus, Streptococcus, Staphylococcus, and Listeria reveal that polylysogeny is common. During lysogeny, which follows recombination of the phage genome into the host chromosome; most of the phage genes are repressed. Those genes that are expressed are mostly involved in the maintenance of lysogeny, and the expressed proteins have regulatory functions that prevent the transcription of genes encoding replication, morphogenesis, and lytic components responsible for the lytic life-cycle of the phage (Ptashne, 2004). Phage conversion genes that are present on some prophages are often expressed during lysogeny. In numerous prophages of the low G+C Firmicutes (e.g. Staphylococcus aureus), phage conversion genes encode known or proposed virulence and fitness genes (Desiere, Lucchini, Canchaya, Ventura, & Brüssow, 2002; Prévost, et al., 1995; Tormo, et al., 2008; van Wamel, Rooijakkers, Ruyken, van Kessel, & van Strijp, 2006). Lysogeny is not a permanent state and during bacterial growth infective virions arise due to spontaneous prophage induction. The rate at which prophages enter the lytic cycle is phage and host-specific and, in addition, chemical or physical agents that damage DNA, including oxidants, some antibiotics like mitomycin C, and UV radiation, can all induce prophage entry into the lytic cycle.
Temperate phage genomics
The 3.2 Mb chromosome of the vancomycin-resistant E. faecalis strain V583 revealed that lysogeny contributed the largest component of horizontally-acquired DNA in this clinical isolate. Seven potential integrated phage–derived sequences were identified that comprised close to 10% of the host cell DNA (Paulsen, et al., 2003). phage01 (EF0303-55), phage03 (EF1417-89), phage04 (EF1988-2043), phage05 (EF2084-145), and phage06 (EF2798-855) have sufficient composition for integration/excision, DNA replication, and capsid/tail morphogenesis to generate functional virions, either alone or synergistically with other prophages. The phage02 region (EF1276-93) identified in the V583 genome, was later described to form part of the core genome (McBride, Fischetti, Leblanc, Moellering, Jr., & Gilmore, 2007). This region appears to be the remnant of a prophage and this cryptic phage likely retains no capacity for induction into the lytic life-cycle. Similarly, the phage07 region (EF2936-55) of strain V583 appeared to be a cryptic prophage but was recently shown to produce infectious virions (Duerkop, Clements, Rollins, Rodrigues, & Hooper, 2012; Matos, et al., 2013). Phage07 is similar to the phage-related chromosomal islands of Gram-positive bacteria which are mobile elements that utilize the packaging and structural elements of a helper phage for dissemination (Novick, Christie, & Penadés, 2010). Phage07 hijacks phage01 particles in this manner upon excision from the chromosome and has been re-named E. faecalis chromosomal island of V583 (EfCIV583) (Matos, et al., 2013). Sequencing of the 2.8 Mb genome of strain OG1RF, the parent of which was originally isolated from the human oral cavity, provided a stark contrast to strain V583, since only one phage element was present (the cryptic phage02). Multiple sequenced E. faecalis genomes are deposited in publicly-available databases and the presence or absence of prophages across many of these different genomes has been determined by using DNA microarray-based comparative genomic hybridization and comparative genomic analyses (Lepage, et al., 2006; McBride, Fischetti, Leblanc, Moellering, Jr., & Gilmore, 2007; Solheim, Brekke, Snipen, Willems, Nes, & Brede, 2011). These studies confirm the variable presence of prophage elements integrated at six regions identified in the V583 genome (phage01, phages 03–06, and EfCIV583).
Lysogenic conversion
In addition to bacterial genome comparison approaches, studies have described the genomics of purified virions liberated by phage-generated lysis after induction (Stevens, Porras, & Delisle, 2009; Yasmin, et al., 2010). The benefit of these studies lies in their capacity to determine the infectious temperate phages present within discrete strain sets, as well as a description of the laterally transferred DNA sequences. In the study by Yasmin et al. (Yasmin, et al., 2010), a collection of 47 clinical E. faecalis bacteremia isolates were screened for prophage induction using norfloxacin, mitomycin C, and UV light as stressors. Thirty-four unique phages were induced from these strains, as determined by host range and restriction fragment length polymorphisms (RFLP). Twelve strains in the study (26%) were polylysogens that contained up to five inducible prophages, based upon their host range and the restriction digest of DNA (Yasmin, et al., 2010). Of the phages that were identified, eight were confirmed as Siphoviridae by their morphology using electron microscopy indicating their long, non-contractile tails (~200 nm) and isometric capsids (~50 nm diameter) (Figure 1). The genome sequences of these phages were determined using DNA pyrosequencing.
Figure 1.
Electron micrograph of induced Siphoviridae prophage (ΦFL1A) from a clinical isolate of E. faecalis. The long, non-contractile tail (~200 nM in length) enables adsorption for delivery of phage DNA from the isometric capsid (~50 nm in diameter). (more...)
Comparative genomic analyses grouped the eight Siphoviridae phages into four phage sequence types (ΦFL1A, B. and C; ΦFL2A and B; ΦFL3A and B; and ΦFL4A). Of these, ΦFL4A shares a high nucleotide identity with the phage01 region of strain V583, and the integrase proteins of these two phages have 99% amino acid identity (Yasmin, et al., 2010). The ΦFL1, ΦFL2, and ΦFL3 phages all have identical integrase proteins, and two of these phages, ΦFL2B and ΦFL3A, were induced from the same polylysogen host. This finding indicates that polylysogenized genomes are likely to exhibit diversity in the order that phages lysogenize their host. The DNA replication and packaging regions of the ΦFL1 and ΦFL2 phages share sequence identity with the phage03 and phage05 regions of strain V583, but appear to be otherwise distinct (Yasmin, et al., 2010). The study of Stevens et al. (Stevens, Porras, & Delisle, 2009) identified a single Siphoviridae phage, ΦEF11, from the lysogen host strain TUSoD11, following induction with mitomycin C, that was morphologically similar with those described by Yasmin et al. (Yasmin, et al., 2010). From these two studies, it appears that the genome organization of the Siphoviridae phages of E. faecalis is similar to that of many of the Siphoviridae phages that infect low G+C Gram-positive bacteria (Stevens, Ektefaie, & Fouts, 2011; Yasmin, et al., 2010). Phage genomes have been described as being modular in their organization (Desiere, Lucchini, Canchaya, Ventura, & Brüssow, 2002; Hatfull, Cresawn, & Hendrix, 2008) and the genomes of the described E. faecalis Siphoviridae phages are similarly modular within the three transcriptional units (Figure 2). The first unit of the prophage (i.e., as it appears on the host chromosome) is the leftward-transcribed integrase/cI region for the maintenance of lysogeny. A large rightward-transcribed region encoding proteins for the lytic pathway, replication, packaging, head/tail morphogenesis, and lysis functions is followed by a variable leftward-transcribed region. This terminal leftward-transcribed region between the lysin and the right-hand phage attachment site in many temperate phages of the low G+C Gram-positive bacteria often contains phage conversion genes. For example, the genes located in these regions resemble those that encode innate immune evasion proteins of S. aureus (Desiere, Lucchini, Canchaya, Ventura, & Brüssow, 2002; Prévost, et al., 1995; Tormo, et al., 2008; van Wamel, Rooijakkers, Ruyken, van Kessel, & van Strijp, 2006).
Figure 2.
Schematic of E. faecalis temperate Siphoviridae genomes. Modular organization based upon the different ΦFL1, ΦFL2, ΦFL3, ΦFL4 and ΦEF11 phages (Stevens, Ektefaie, & Fouts, 2011; Yasmin, et al., 2010). The (more...)
Analysis of the ΦEF11 phage genome and the genomes of the ΦFL1, ΦFL2, ΦFL3, ΦFL4 phages identified several potential phage conversion genes, also called cargo (Stevens, Ektefaie, & Fouts, 2011; Yasmin, et al., 2010). The ΦEF11 phage appears to have an extended set of lysins immediately followed by three leftward-transcribed genes, two of which encode a putative membrane protein and a lysM domain-containing protein (Stevens, Ektefaie, & Fouts, 2011). Based on their predicted function, these proteins are likely to be located on the host cell surface, and if expressed during growth, represent bona fide phage conversion genes. These proteins could also be involved with lysogeny or immunity to phages—or, as proposed by Stevens et al (Stevens, Porras, & Delisle, 2009), they may facilitate host cell lysis. The ΦFL1, ΦFL2, and ΦFL3 phages contain one or more small open reading frames within their cargo regions that have inferred amino acid sequence identity with control proteins of streptococcal phages, which suggests a possible role in the maintenance of lysogeny (Yasmin, et al., 2010). The ΦFL4 phage encodes no potential phage conversion genes in the terminal region of its genome. Further studies remain to be performed to investigate the carriage of phage conversion genes in enterococcal temperate phages.
The initial description of the E. faecalis strain V583 genome proposed that phage04 contains a ferrochelatase-encoding gene (EF1989) that could function in heme biosynthesis (Paulsen, et al., 2003). Within phage04, the gene cspA (EF1991) encodes a cold shock family protein homologous to that of E. coli (Lee, Xie, Jiang, Etchegaray, Jones, & Inouye, 1994). CspA homologs are encoded within a prophage genome of the Lactococcus lactis strain ll1403 and the Streptococcus phage bIL312 genome. A second copy of cspA (EF0781), with a high-sequence identity to the cspA homolog on phage04, is found elsewhere in the genome of its strain V583 host. Comparative prophage analyses have revealed that there are clear similarities between many modular phage genes present in Enterococcus, Lactococcus, Streptococcus, and Staphylococcus species (Paulsen, et al., 2003; Stevens, Ektefaie, & Fouts, 2011; Villion, Chopin, Deveau, Ehrlich, Moineau, & Chopin, 2009; Yasmin, et al., 2010). For example, the ΦFL3A and ΦFL3B phages share 34% and 31% sequence identity, respectively, with prophages of L. lactis subsp. cremoris SK11 and MG1363 (Yasmin, et al., 2010). As described earlier in this chapter, these bacteria often reside together in similar host environments, foodstuffs, or fecal-contaminated water sources, which may potentiate the proximal lateral transfer of DNA by phages and other mobile DNA elements.
E. faecium temperate phages
Phages that infect E. faecium and E. faecium genome–encoded prophages have been described in several studies (Galloway-Peña, Roh, Latorre, Qin, & Murray, 2012; Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010; van Schaik, et al., 2010). The ability to induce prophages from a group of genome-sequenced E. faecium strains has recently been demonstrated (van Schaik, et al., 2010). These induced prophages were all Siphoviridae, and were morphologically identical to prophages from E. faecalis. Within the genome of these seven prophages, 3-5% of the coding sequence was determined to originate from phage DNA. This suggests that the majority of the E. faecium prophage sequence contributes potentially novel DNA, which drives the genomic diversity of these strains (van Schaik, et al., 2010).
Other enterococcal temperate phages
Far less is known about the temperate phages of other enterococci; however, current sequencing efforts will facilitate future studies. Non-faecalis and non-faecium enterococcal species , including E. casseliflavus and E. gallinarum, respectively, are likely to carry prophage elements, as determined by the presence of phage integrase genes in genome sequences from those strains (Palmer, et al., 2010). The phage genome integrity, boundaries, and organization of these elements are unclear and await annotation. Moreover, induction experiments are required to determine whether potential prophages can be induced from these species to produce infectious phage particles. Recent studies have also identified a number of novel enterococcal phages from diverse environments that can infect E. gallinarum and E. casseliflavus (discussed above) (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010; Purnell, Ebdon, & Tayor, 2011).
Role of phages in virulence
Comparative genome analyses of the E. faecalis V583 strain and of the ΦFL1, ΦFL2, ΦFL3, and ΦFL4 groups of phages revealed that they encoded multiple homologs of the Streptococcus mitis phage M1 PblA platelet-binding protein (Bensing, Siboo, & Sullam, 2001; Mitchell & Sullam, 2009). The PblA and PblB phage tail proteins of S. mitis were clearly demonstrated to contribute to platelet adhesion via interactions with the α2-8-linked sialic acid residues on gangliosides of platelet membranes (Mitchell & Sullam, 2009). V583 strains harboring singly lysogenized pbl gene containing phages (phage01, 04, or 06) adhered to human platelet cells, whereas lysogenized strains carrying only phage03 and 05 or EfCIV583 alone were unable to bind platelets (Matos, et al., 2013). The contribution to phage encoded Pbl proteins during E. faeclais diseases such as endocarditis has not yet been tested.
There have been no systematic studies that characterize the role of prophages and cryptic phages in the pathogenesis of E. faecalis. A study by Yasmin et al. (Yasmin, et al., 2010) revealed that ΦFL3A and ΦFL3B phage lysogens of E. faecalis JH2-2 reduced the survival of Galleria mellonella caterpillars, as compared to their non-lysogen parent strain. No clear effects were observed in this model for lysogens of the other phages identified. However, this preliminary investigation, using only one insect model of infection, did not study lysogen survival within the Galleria caterpillars, or the rates of spontaneous lysis for the various lysogens during growth. Moreover, it will be important to determine the transcription and replication of each of the phage elements during growth in the infection models. In contrast to E. faecalis, no comparative genomic analysis of E. faecium phages has been undertaken to date, and as a result, their potential contribution to pathogenesis is therefore unknown. The sequential deletion of prophage elements from a lysogenized enterococcal chromosome, or the generation of allelic replacement mutants of phage genes prior to comparative virulence studies, represents a more rigorous approach for future analysis. These types of studies have been performed in S. aureus (Bae, Baba, Hiramatsu, & Schneewind, 2006) and E. coli (Wang, et al., 2010), and in each case, multiple phage-encoded genes were found to positively contribute to changes in either virulence potential or environmental survival (such as antibiotic resistance).
More recently, a comparative genomic analysis of hospital-associated clonal complex 2 (CC2) strains of E. faecalis revealed that there was enrichment of multiple lateral transfer elements in these strains (Solheim, Brekke, Snipen, Willems, Nes, & Brede, 2011). Among 252 genes that were statistically enriched in CC2, as compared to non-CC2 strains, the majority of such genes were found in known lateral transfer elements. This group included 51 genes (nearly the complete genome) of phage03 and several genes of phage04 (Solheim, Brekke, Snipen, Willems, Nes, & Brede, 2011). Most of these genes were expressed based upon transcription analyses; however, no account was recorded for spontaneous lysis from prophages entering the lytic cycle within the cultures, which could account for the observed transcription. Aside from this finding, the determination that phage elements are positively correlated with virulence enhancement supports the future study of phage deletion strains, as described above.
Phage transduction
Temperate phages are well-recognized for their lateral gene transfer capabilities. The bacteriophage-mediated transfer of DNA from a donor cell to a recipient cell is termed transduction, and can be mediated by either lytic or temperate phages. Transducing particles occur during the lytic stage of phage life-cycles and can be of two types: either generalized or specialized transducing phage particles. Generalized transducing particles arise from the packaging of any host bacterial genome sequence of the requisite size with the absence of phage DNA. Specialized transducing phage particles are generated only by a temperate phage that excise and package DNA immediately adjacent to one end of the integrated prophage through covalent attachment (Birge, 1994). Transduction has the capacity to facilitate lateral gene transfer and was demonstrated in the enterococci by using the lytic phage EFRM1 (discussed below).
It was recently shown that several E. faecalis temperate phages were capable of generalized transduction (Yasmin, et al., 2010). Temperate phage transduction in this study was demonstrated by the transfer and recombination of DNA encoding antibiotic resistance in E. faecalis. This termperate phage transduction property has implications for the evolution of multi-drug resistant enterococci, as well as being a generally useful tool for laboratory-mediated genetic transfer. The capacity of enterococcal phages to facilitate generalized transduction was extended in a later study, which proposed that phages could mediate inter-species transfer of antibiotic resistance genes between different enterococci, including E. faecalis, E. faecium, E. gallinarum, Enterococcus hirae, and E. casseliflavus (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2011).
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Enterococcal Lytic Phages
By definition, a lytic phage is one that can infect a bacterial host strain, but does not possess the ability to lysogenize. Lytic phages are considered predators of bacteria, as their purpose is to infect, replicate, and lyse the host for transmission. This purpose opposes to that of temperate phages, which can enter a life-cycle of parasitism maintaining chronic infection within their host (Weinbauer, 2004). Of course, prophages themselves can enter the lytic life-cycle under specific inducing conditions and at host strain-specific frequencies. Lytic phages are often devoid of an integrase gene and lack the genes for the maintenance of lysogeny, which is why these phages exist in a perpetually predatory state. Lytic phages and the enzymes that code for host cell wall destruction, including the membrane-permeating holins and peptidoglycan hydrolytic enzymes, called lysins, are debated as alternative therapies against bacteria resistant to multiple antibiotics (Chhibber, Kaur, & Kumari, 2008; Matsuzaki, et al., 2003; McVay, Velásquez, & Fralick, 2007). Several obligate lytic enterococcal phages have been identified (Horiuchi, Sakka, Hayashi, Shimada, Kimura, & Sakka, 2012; Lee & Park, 2012; Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010; Otawa, Hirakata, Kaku, & Nakai, 2012; Parasion, Kwiatek, Mizak, Gryko, Bartoszcze, & Kocik, 2012; Parasion, Kwiatek, Mizak, Gryko, Bartoszcze, & Kocik, 2012); however, due to space restrictions only a few will be discussed in detail.
ϕEF24C
The lytic phage ϕEF24C was isolated from a water channel in Kochi City, Japan after passage in culture with E. faecalis strain EF24 (Uchiyama, et al., 2008). Further analysis revealed that ϕEF24C was able to infect and proliferate in multiple strains of E. faecalis, including several vancomycin-resistant isolates; however, ϕEF24C was unable to infect or lyse any of the ten E. faecium strains that were tested. ϕEF24C shows broad activity for E. faecalis, but appears to lack the ability of cross-strain infectivity and lysis. The phylogeny of ϕEF24C places this phage in the Myoviridae family. ϕEF24C has an icosahedral head of ~93 nm in diameter and a contractile tail that when fully extended is ~204 nm in length (Uchiyama, et al., 2008). N-terminal degradation sequencing of many of the ϕEF24C structural proteins, including tail and capsid proteins, suggests that ϕEF24C is related to the SPO1 phage genus, which includes members such as Staphylococcus phages K and 812, Lactobacillus phage LP65, and Listeria phage P100, all of which are known lytic phages (Chibani-Chennoufi, Dillmann, Marvin-Guy, Rami-Shojaei, & Brüssow, 2004; Uchiyama, et al., 2008). The genome of ϕEF24C is large, with ~142 kbp of circular double stranded DNA. The ϕEF24C genome houses 221 open reading frames and five tRNA synthetase genes (Uchiyama, Rashel, Takemura, Wakiguchi, & Matsuzaki, 2008). Just fewer than 50% of the open reading frames found in the ϕEF24C are predicted to encode proteins with known functional domains. The ϕEF24C genome is organized into three distinct modules. The first module is structural and includes the genes involved in capsid and tail biosynthesis. The second module contains genes that encode replication proteins. The third module is a short structural module that includes a gene with an immunoglobulin protein motif, which is thought to be structurally associated with the phage particle. Outside of these modular domains are genes that encode putative nucleic acid precursor biosynthesis enzymes, which may be involved in the de novo synthesis of modified nucleotides. Furthermore, the ϕEF24C genome contains no genes homologous to a site specific integrase, which verifies the strictly lytic life-cycle of ϕEF24C (Uchiyama, Rashel, Takemura, Wakiguchi, & Matsuzaki, 2008).
EFRM31
EFRM31 is a double-stranded DNA phage that was originally isolated from runoff water from a pig farm, and was amplified in E. faecalis isolates found within piggery waste water (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010). Electron microscopy revealed that EFRM31 is a member of the Siphoviridae family and has a long non-contractile tail (206 nm, length) and a spherical capsid (55 nm, width). The EFRM31 genome is a circular ~17 kbp genome and is characterized by low G+C content. There are 87 open reading frames that are predicted to encode proteins, and 17 of these open reading frames are putative morphogenesis genes that encode structural and accessory polypeptides for tail and capsid assembly. EFRM31 also encodes numerous proteins of unknown function (Fard, Barton, Arthur, & Heuzenroeder, 2010). The EFRM31 genome is homologous to the EFAP-1 bacteriophage genome, a lytic phage of E. faecalis (Son, et al., 2010).
Outside of encoding proteins for phage particle biosynthesis and phage genome replication, EFRM31 also encodes a putative virulence factor homologous to a zinc metalloprotease, a DNA primase gene, an antibiotic resistance membrane pump, and a polyamine transporter. These genes are clustered together and reside outside of the organized modules for phage particle assembly and genome replication. It is likely that these genes have been acquired from a bacterial host chromosome through recombination (Fard, Barton, Arthur, & Heuzenroeder, 2010). EFRM31 is proficient in packaging DNA, aside from its own chromosome, during phage particle assembly. This can be concluded from studies that measured the ability of EFRM31 to transduce antibiotic resistance markers from E. faecalis host strain chromosomal or plasmid DNA to enterococcal host strains where sensitivity to a particular antibiotic had been previously determined (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2011). EFRM31 was highly capable of transducing gentamicin resistance to multiple enterococcal species, and as a result, EFRM31 was also able to undergo inter-species transduction of gentamicin resistance to two E. faecium strains and an Enterococcus isolate most closely related to the durans/hirae species (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2011). In some, but not all, of these transfer events, there was an association of an ant2-I gene (which codes for an aminoglycoside nucleotidyltransferase) cassette acquisition that is known to confer low level gentamicin resistance in the transduced strain. Since transducing phages are mostly species-specific, EFRM31 is the first example of an enterococcal phage that is capable of inter-species host range generalized transduction.
EFAP-1
EFAP-1 is a lytic phage that was isolated from environmental samples taken from a variety of sources within a cowshed. EFAP-1 infects E. faecalis. This phage is a member of the Siphoviridae family , and by electron microscopy, was determined to have a non-contractile tail and an isometric capsid. EFAP-1 is most closely related to Bacillus phage ϕ105 and Staphylococcus prophage ϕPV83 (Son, et al., 2010). The EFAP-1 genome is composed of 24 open reading frames and similar to other related phages, the open reading frames are modular and consist of clusters of genes annotated to be involved in phage structure, host cell lysis, phage genome replication, and electron transport. Several open reading frames within these modules share homology with proteins that are found on bacterial chromosomes. Although these open reading frames may originate from bacterial chromosomes, they may perform functions for the EFAP-1 phage. For example, the gene ORF1 encodes a putative thioredoxin-like superfamily protein that may be involved in phage DNA replication and may also be used during bacterial cell electron transport (Son, et al., 2010). Several open reading frames found in the EFAP-1 genome show no sequence similarity to those found in the current public sequence databases, and as a result, have been delineated as genes of unknown function. This result suggests that, like many other phages, EFAP-1 encodes potentially novel genes that, if expressed, may uniquely aid in determining the biology of this lytic phage.
Many more enterococcal lytic phages have been identified including at least one that is known to infect E. faecium as well as several recently identified environmental phages of diverse morphologies that infect E. faecalis (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010; Otawa, Hirakata, Kaku, & Nakai, 2012).
Enterococcal phage lysins and therapeutic applications
After replication and packaging of phage particles within a bacterial cell, lytic phages must exit the host cell in order to disseminate to a new host bacterium. To achieve such an exodus, non-filamentous double stranded DNA phage genomes encode enzymes, typically referred to as lysins, that accumulate in the cytoplasm of the host cell and target the destruction of its cell wall (Fischetti, 2005). Lysins act on the peptidoglycan of the bacterial cell wall by binding to polysaccharide or peptide moieties and subsequently cleaving these macromolecules, which aids in the release of phage progeny. Lysins are diverse hydrolases, including endo-β-N-acetylglucosaminidase or N-acetylmuramidase, that bind to sugar residues of peptidoglycan and endopeptidases that target a peptide ligand of the peptidoglycan cell wall. A fourth type of lysin, N-acetylmuramoyl-L-alanine amidase (amidase), hydrolyzes the amide bond between the peptide moiety and the sugar moiety of peptidoglycan (Young R. , 1992). The activity of a lysin is coupled to a second phage protein known as a holin—a transmembrane protein that inserts into the bacterial cell membrane causing disruption (Wang, Smith, & Young, 2000). Following holin insertion, the lysin can then traverse the cell membrane and gain access to the cell wall peptidoglycan to cause lysis. Holins are instrumental in allowing lysins to target the cell wall, because most lysins lack N-terminal signal sequences and cannot independently gain access to the cell wall (Young, Wang, & Roof, 2000).
It has been postulated that phage lysins could be developed into an alternative to antimicrobials, as there is a great need for new antimicrobial agents to fight the steady rise in multi-drug resistant bacteria that has taken place in recent decades. Many phage lysins have been purified and these enzymes have been used in proof of principal experiments to kill pathogenic bacteria both in vitro and in vivo (Fischetti, 2005; Fischetti, 2003). Lysins as therapeutics have mostly been applied to Gram-positive bacteria, where the lysin has easy access to the peptidoglycan on the outside of the bacterial cell. This process is in direct contrast to Gram-negative bacteria, which usually resist killing by exogenous lysins, due to the peptidoglycan of these cells being protected by the bacterial outer membrane. The enterococci are notorious for their antimicrobial resistance phenotypes (Palmer, Kos, & Gilmore, 2010; Willems & Bonten, 2007). In some cases, enterococcal strains can be resistant to a broad range of antibiotics, including some recently developed antibiotics (Arias, et al., 2011; Palmer, Daniel, Hardy, Silverman, & Gilmore, 2011). Due to their intrinsic antibiotic resistance, the efficacy of enterococcal phage lysins as antimicrobial agents has been tested in various studies.
PlyV12 is a lysin that was identified from a genomic DNA library of the ϕ1 lytic Myoviridae family enterococcal phage. This DNA library was screened for lytic factors that had killing activity against the E. faecalis strain V12 found in the human urogenital tract (Caprioli, Zaccour, & Kasatiya, 1975; Yoong, Schuch, Nelson, & Fischetti, 2004). The identification of a bacteriolytic polypeptide, PlyV12, that shares homology with N-acetylmuramoyl-L-alanine amidase type lysin enzymes, had a high specific lytic activity against both E. faecalis and E. faecium strains, including vancomycin-resistant strains. Interestingly, PlyV12 was also active in killing strains of S. aureus, Staphylococcus epidermidis, and multiple groups of streptococci. This established the PlyV12 lysin as a broad spectrum bacteriolysin, which is uncharacteristic of these types of enzymes, as most kill only the species that the parental phage can infect (Fischetti, 2005). Furthermore, it is proposed that PlyV12 may target a cell wall moiety that is conserved within many Gram-positive genera. The amino acid sequence of PlyV12 revealed a protein of ~32 kDa that has an N-terminal catalytic domain, and that shares sequence similarity to N-acetylmuramoyl-L-alanine amidase -like lysins from the group A and B streptococci Streptococcus pyogenes and Streptococcus agalactiae, as well as Streptococcus pneumoniae and Streptococcus mitis (Yoong, Schuch, Nelson, & Fischetti, 2004). This similarity in the lysin catalytic domain may account for the broad-spectrum killing ability of PlyV12, although the C-terminal region of PlyV12 where the predicted lysin epitope binding site resides is highly divergent from these streptococcal lysins (Yoong, Schuch, Nelson, & Fischetti, 2004).
A second study focused on a lysin enzyme from the lytic phage EFAP-1. This lysin, designated as EFAL-1, is also predicted to be a cell wall hydrolytic N-acetylmuramoyl-L-alanine amidase. EFAL-1 was purified after overexpression in E. coli cells, and was found to have broad spectrum killing activity (Son, et al., 2010). EFAL-1 was found to be lytic against 13 E. faecalis strains, seven E. faecium strains, two strains of S. agalactiae, and two strains of Streptococcus uberis. This potent activity of EFAL-1 is impressive, considering the parental lytic phage that encodes this lysin, EFAP-1, is only capable of lysing a single E. faecalis strain from this group of susceptible bacteria.
A third study characterized the lysin encoded by orf9 from the lytic enterococcal phage ϕEF24C. Using MALDI-TOF mass spectrometry, Orf9 was shown to specifically cleave peptidoglycan at the peptide sugar linkage, which confirms that Orf9 is also a hydrolytic N-acetylmuramoyl-L-alanine amidase (Uchiyama, et al., 2011). Site-specific truncations of the orf9 coding region confirmed that the ability of Orf9 to lyse E. faecalis depended on an intact C-terminus. An extensive analysis of the N-terminal region of Orf9 was not possible, as truncations within this region rendered the protein insoluble (Uchiyama, et al., 2011). Unlike the EFAL-1 and PlyV12 lysins, Orf9 was shown to have lytic activity against only E. faecalis and E. faecium. However, this analysis was only extended to 10 Staphylococcus strains, and further screening of other Gram-positive bacteria may reveal similar broad activity.
It is worth noting that phage lysins from non-enterococcal bacteria have been shown to be effective at killing Enterococcus species. The temperate phage ϕ-0303 of Lactobacillus helveticus encodes a 40 kDa lysin called Mur-LH that functions as a cell wall–degrading N-acetylmuramidase (Deutsch, Guezenec, Piot, Foster, & Lortal, 2004). Purified Mur-LH can lyse a diverse array of Gram-positive bacteria, including E. faecium. Mur-LH was unable to lyse E. faecalis, suggesting that Mur-LH binds to a cell surface epitope that is not shared between these two enterococcal species. Lactobacillus gasseri phage ϕgaY encodes an N-acetylmuramidase. This lysin, Lys gaY, exhibits broad bacteriolytic activity and can lyse over 20 different species of Gram-positive bacteria, including enterococci (Sugahara, et al., 2007). Purified Lys gaY also causes logarithmically growing Gram-positive bacterial cells to aggregate and form long chains. This phenotype is mostly likely caused by Lys gaY binding to the site of cell wall peptidoglycan synthesis during cell division, which results in the inability of daughter cells to completely partition.
These data emphasize two important points. First, contrary to popular belief, many lysins, specifically those of enterococcal lytic phages, have broad tropism and are capable of intra- and inter-generic killing. Second, bactericidal therapeutics based on highly promiscuous lysins may be powerful tools for combating diverse Gram-positive pathogens. Lysins have been shown to act synergistically with some antibiotics (Loeffler & Fischetti, 2003) and specifically when used in combination with antibiotics that target cell wall synthesis, like penicillin. Bacterial resistance to lysins appears to be a rare event. It is thought that low incidence resistance to phage lysins is caused by phages evolving their lysins to target essential molecules in the bacterial cell wall (Fischetti, 2008). Therapies that combine phage lysins and antibiotics may also decrease the incidence of bacterial resistance through synergistic effects. However, one concern with this course of action is that phage lysins are enzymes, and therapeutic treatments using these types of proteins would result in the production of neutralizing antibodies and the elimination of the lysins (Fischetti, 2005). The efficacy of the bacteriolytic activity of lysins in the presence of neutralizing antibodies has been addressed. A study by Loeffler et al showed that a streptococcal phage lysin still maintained killing in the presence of specific neutralizing antibodies, although it was reduced in activity (Loeffler, Djurkovic, & Fischetti, 2003). Furthermore, mice pre-immunized with a streptococcal lysin prior to receiving an intranasal S. pneumoniae challenge produced lysin-specific neutralizing antibodies, yet were still protected upon administration of the lysin following infection (Loeffler, Djurkovic, & Fischetti, 2003). These data suggest that immunogenic phage lysins are still capable of acting against their target in the presence of neutralizing antibodies, and may warrant consideration for development into alternative antimicrobial agents.
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Enterococcal Genome Defense Mechanisms
Prokaryotes possess multiple mechanisms that can interfere with phage infection, including receptor/adsorption blocking; abortive infection; clustered, regularly interspaced short palindromic repeats (CRISPR) with CRISPR-associated (Cas) proteins (referred to here as CRISPR-Cas); and restriction modification (RM) (51). CRISPR-Cas and RM can also limit uptake of other mobile elements, such as plasmids (discussed below). Despite clear differences in the abundance of prophages, plasmids, and other mobile elements in enterococcal genomes (Bourgogne, et al., 2008; Leavis, Willems, van Wamel, Schuren, Caspers, & Bonten, 2007; Palmer, et al., 2012; Solheim, Brekke, Snipen, Willems, Nes, & Brede, 2011), little is known about the mechanisms by which some strains appear more susceptible to phage infection and plasmid uptake than others. Recent studies have highlighted potential roles for genome defense systems in the evolution of multidrug resistance and hospital adaptation in the enterococci (Lindenstrauss, Pavlovic, Bringmann, Behr, Ehrmann, & Vogel, 2011; Palmer & Gilmore, 2010; Solheim, Brekke, Snipen, Willems, Nes, & Brede, 2011) and in the transfer of antibiotic resistance genes from enterococci to S. aureus (Monk, Shah, Xu, Tan, & Foster, 2012). This section introduces CRISPR-Cas proteins and restriction modification defense, and reviews evidence for these systems in the enterococci. Methods to identify CRISPR-cas loci in newly sequenced enterococcal genomes are also discussed. Because no phage receptors have been identified for the enterococci, nor have any phage abortive infection mechanisms been reported, those genome defense strategies are not discussed here.
CRISPR-Cas, a prokaryotic acquired immune system
Clusters of roughly palindromic repeat sequences, referred to as CRISPR, were identified in bacterial and archaeal genomes to be sequenced (Jansen, Embden, Gasstra, & Schouls, 2002; Mojica F. J., Díez-Villaseñor, Soria, & Juez, 2000). The repeats, which can vary in length (typically ~24-48 bp) and palindromic structure (Haft, Selengut, Mongodin, & Nelson, 2005; Kunin, Sorek, & Hugenholtz, 2007), were detected in 83% of archaeal and 56% of bacterial species analyzed by the CRISPI database as of February 2012 (103 archaeal and 956 bacterial species with available genome data) (Rousseau, Gonnet, Le Romancer, & Nicolas, 2009). Thus, CRISPR appear to be both ancient and widespread among prokaryotes. Genes encoding nucleases and other proteins involved in DNA and RNA processing are typically associated with CRISPR, and are called CRISPR-associated genes (cas genes) (Jansen, Embden, Gasstra, & Schouls, 2002). The cas genes encode Cas proteins. CRISPR-cas loci have diverse cas gene cohorts and repeat structures, and several classification systems have been reported (Haft, Selengut, Mongodin, & Nelson, 2005; Kunin, Sorek, & Hugenholtz, 2007; Makarova, Grishin, Shabalina, Wolf, & Koonin, 2006; Makarova, et al., 2011). Recently, a unifying nomenclature was established that groups CRISPR-Cas systems into three broad types (I, II, and III) defined by type-specific cas genes (Makarova, et al., 2011). Only two conserved cas genes, cas1 and cas2, are associated with all CRISPR-cas loci that are predicted to be functional (Makarova, et al., 2011). Exemplifying the diversity and complexity of these loci, the three broad CRISPR-Cas types function by distinct molecular mechanisms (Wiedenheft, Sternberg, & Doudna, 2012), and prokaryotes can harbor multiple CRISPR-cas loci of varying types in their genomes. The CRISPR-Cas type that appears to be most relevant to the enterococci is Type II, and this chapter focuses mainly on Type II systems. Readers should refer to an excellent and recent review for more information on Type I and Type III systems (Wiedenheft, Sternberg, & Doudna, 2012). Note that the Type II nomenclature was established in 2011 (Makarova, et al., 2011), and that some prior literature referred to these loci as Nmeni (Haft, Selengut, Mongodin, & Nelson, 2005) or Cas4 (Makarova, Grishin, Shabalina, Wolf, & Koonin, 2006) systems.
A representative Type II CRISPR-cas locus is shown in Figure 3A. Repeat sequences in Type II loci are typically 36 bp in length, and the unique sequences that occur between repeats, called spacers, are typically 30 bp (Haft, Selengut, Mongodin, & Nelson, 2005). The final repeat of the array is sometimes degenerate in sequence. A conserved leader sequence located 5ʹ to the repeat-spacer array drives the expression of the array (Deltcheva, et al., 2011) and likely contains motifs that are important for the addition of new spacers (Yosef, Goren, & Qimron, 2012). Type II CRISPR-cas loci typically possess 4 cas genes: the type-specific gene cas9, the core genes cas1 and cas2, and either csn2 or cas4 (Makarova, et al., 2011), although variations exist (Deltcheva, et al., 2011; Palmer & Gilmore, 2010).The csn2 gene appears to be specific to Type II-A systems, while cas4 has a wider distribution, being present in both Type I and Type II-B systems (Makarova, et al., 2011).
Figure 3.
A generalized model of a Type II CRISPR-Cas defense system. (A) A typical Type II CRISPR-cas locus. Spacers are shown as black diamonds and repeats are not shown. Either csn2 or cas4 may be present as the fourth gene in the cas cluster. (B) Upon plasmid (more...)
Catalytic mechanisms and crystal structures for some Cas proteins are known. Cas1 from the Type I CRISPR-Cas system of Pseudomonas aeruginosa is a metal-dependent deoxyribonuclease that generates 80 bp double-stranded DNA fragments (note that spacer lengths for this CRISPR-cas locus are 32 bp) (Wiedenheft, Zhou, Jinek, Coyle, Ma, & Doudna, 2009). Cas2 proteins from multiple prokaryotes have been characterized, and all are endoribonucleases that preferentially cleave single-stranded RNA targets in U-rich regions (Beloglazova, et al., 2008). Csn2 from E. faecalis ATCC 4200 has been purified and its structure characterized; the protein oligomerizes to form a tetrameric ring that binds double-stranded DNA (Nam, Kurinov, & Ke, 2011). The Type II-specific Cas9 proteins are large (~1000 amino acids) proteins that possess predicted RuvC-like nuclease and HNH restriction endonuclease-like domains (Makarova, Aravind, Wolf, & Koonin, 2011). No structural data are currently available for these proteins. Similarly, Cas4 is a predicted nuclease for which no structural data are available.
Genome analysis provided the initial evidence that CRISPR-Cas systems interact with mobile elements. Studies published in 2005 reported that some CRISPR spacers are identical to plasmid and phage sequences; a result that suggests that a prior encounter with mobile elements resulted in spacer acquisition (Figure 3B) (Bolotin, Quinquis, Sorokin, & Ehrlich, 2005; Mojica F. J., Díez-Villaseñor, García-Martínez, & Soria, 2005; Pourcel, Salvignol, & Vergnaud, 2005). A region of a phage, plasmid, or other genome with sequence identity to a spacer sequence is called a protospacer (Deveau, et al., 2008). A relationship between the presence of certain CRISPR spacers and the absence of prophage possessing corresponding protospacers was noted for S. pyogenes (Pourcel, Salvignol, & Vergnaud, 2005). These observations contributed to the hypothesis that spacer-derived small RNAs could block phage infection by interfering with phage gene expression via an RNA interference (RNAi)-like mechanism (Bolotin, Quinquis, Sorokin, & Ehrlich, 2005; Makarova, Grishin, Shabalina, Wolf, & Koonin, 2006; Mojica F. J., Díez-Villaseñor, García-Martínez, & Soria, 2005).
In 2007, a role for the Type II CRISPR-Cas as an anti-phage defense system of S. thermophilus was experimentally demonstrated (Barrangou, et al., 2007). The S. thermophilus type II CRISPR-cas locus possesses cas9, cas1, cas2, and cas4 genes, and for the purposes of this chapter, is referred to as StCRISPR1-Cas system. S. thermophilus DGCC7710 was challenged in vitro with lytic phages. The authors sequenced the StCRISPR1 of 9 phage-resistant mutants recovered from the phage challenge experiments and found that the loci had acquired new spacer sequences. The new spacer sequences were similar to coding and non-coding genomic sequences from the lytic phages used in the infection challenge. Novel spacer sequences were added only to the leader end of the CRISPR. Most notably, S. thermophilus mutant strains that were engineered to lack spacers became phage sensitive; while strains engineered to possess spacers matching phage sequence gained phage resistance.
These experiments confirmed that the newly acquired spacers contributed to phage resistance. One of two cas genes queried, cas9 (formerly cas5) (Makarova, et al., 2011) was also required for phage resistance. When taken together, these data demonstrated that the StCRISPR1-Cas system protected a subpopulation of S. thermophilus from phage attack through the acquisition of novel, heritable spacer sequences, which in turn provided protection from subsequent attacks by phages with similar sequences by a Cas9-dependent mechanism (Figure 3B and 3D). Additional work has found that StCRISPR1 can acquire spacers from a plasmid resident in S. thermophilus, which apparently promotes loss of that plasmid, also by a Cas9-dependent mechanism (Garneau, et al., 2010).
An additional Type II CRISPR-cas locus is present in S. thermophilus DGCC7710, called CRISPR3-Cas (referred to as StCRISPR3-Cas) (Horvath, et al., 2008). This locus possesses cas9, cas1, cas2, and csn2 genes, as well as a repeat sequence that is similar (but not identical) to the repeats of StCRISPR1 (Horvath, et al., 2008). Both StCRISPR1 and StCRISPR3 of S. thermophilus DGCC7710 obtain new spacers in response to phage challenges (Barrangou, et al., 2007; Horvath, et al., 2008); thus, each of these loci are active in S. thermophilus DGCC7710. Despite the presence and activity of StCRISPR3-Cas in S. thermophilus DGCC7710, that loci's cas9 is apparently unable to complement a StCRISPR1-Cas cas9 mutant (Barrangou, et al., 2007; Garneau, et al., 2010; Horvath, et al., 2008). This indicates that the Cas9 proteins are highly specific to certain repeat sequences and/or other CRISPR structures.
While spacer acquisition has been experimentally demonstrated for the Type II CRISPR-Cas system of S. thermophilus (Barrangou, et al., 2007; Deveau, et al., 2008; Garneau, et al., 2010), little of the mechanistic detail is known. It has been proposed that cas1 and cas2 are important for spacer addition regardless of CRISPR type, based on their biochemical activities and because of their ubiquity among CRISPR-cas loci (Makarova, Aravind, Wolf, & Koonin, 2011). In support of this theory, novel spacer addition occurs when both cas1 and cas2 are overexpressed in an E. coli strain that possesses a Type I CRISPR but otherwise lacks chromosomally encoded cas genes (Yosef, Goren, & Qimron, 2012). In the initial demonstration of StCRISPR1-Cas as an anti-phage system, the authors noted that spacer addition did not occur in strains in which the cas7 gene (now cas4) was insertionally inactivated (Barrangou, et al., 2007). A similar observation was made for anti-plasmid spacer acquisition experiments (Garneau, et al., 2010). Thus it appears that this gene may also play an important role in spacer acquisition.
The selection of novel spacer sequences does not appear to be random. Conserved sequence motifs have been identified immediately adjacent to protospacer sequences targeted by StCRISPR1-Cas (NNAGAAW) and STCRISPR3-Cas (NGGNG) (Deveau, et al., 2008; Horvath, et al., 2008), which suggests that these motifs play a role in spacer selection. In addition to a putative role in spacer selection, these protospacer-adjacent motifs (PAMs) (Mojica F. J., Díez-Villaseñor, García-Martínez, & Almendros, 2009) are important for the interference of StCRISPR-Cas with mobile elements. Mutations in the PAM allow both phages and plasmids to avoid Type II CRISPR interference (Deveau, et al., 2008; Garneau, et al., 2010; Sapranauskas, Gasiunas, Fremaux, Barrangou, Horvath, & Siksnys, 2011), although it appears that the StCRISPR1-Cas system tolerates non-consensus PAM sequences in plasmid, but not in phage targets (Garneau, et al., 2010). More work is required to determine the precise roles of PAMs in CRISPR spacer selection and mobile element interference, as well as why PAMs in the hostʹs own genome are not readily utilized for spacer selection.
What is the precise role of the CRISPR spacer in mobile element interference? It is now known that Type I, Type II, and Type III CRISPR repeat-spacer arrays are transcribed and processed to generate small guide RNAs called CRISPR RNAs (crRNAs) (Brouns, et al., 2008), which vary in structure depending on their CRISPR type (Wiedenheft, Sternberg, & Doudna, 2012). For Type II systems, transcription of the repeat-spacer array generates a long, pre-crRNA which is processed to short, 39-42 ribonucleotide (nt) crRNAs (Deltcheva, et al., 2011) (Figure 3C). This processing has been observed for Type II loci in S. thermophilus, S. pyogenes, Streptococcus mutans, and Listeria monocytogenes, among others (Deltcheva, et al., 2011). A mature Type II crRNA consists of 20 3ʹ nt of a spacer and 19-22 5ʹ nt of the following repeat. Thus, each Type II crRNA is a unique guide molecule that consists of a memory of a previous mobile element exposure (a partial spacer) and a conserved handle (a partial repeat) that likely associates with cellular machinery and prevents unproductive self-interference. Based on the structure of mature crRNAs, it appears that the 5ʹ end of the spacer sequence is less important than the 3ʹ end for crRNA-mediated interference. This is supported by the observation that when the Type II StCRISPR3-cas locus is heterologously expressed in E. coli, spacer-protospacer mismatches are tolerated at the 5ʹ spacer end and the central region (positions 2, 11, 18, and 23 of a 30 nt spacer), but not at the 3ʹ spacer end (positions 25 and 28) (Sapranauskas, Gasiunas, Fremaux, Barrangou, Horvath, & Siksnys, 2011). Additionally, StCRISPR1-Cas tolerates a spacer-plasmid protospacer mismatch at the first 5ʹ position of the spacer sequence (Garneau, et al., 2010).
Processing of the Type II pre-crRNA proceeds via a unique mechanism that is distinct from pre-crRNA processing in Type I and Type III systems (Deltcheva, et al., 2011; Wiedenheft, Sternberg, & Doudna, 2012). A second non-coding RNA associated with the Type II CRISPR-cas locus, called the trans-activating CRISPR RNA (tracrRNA), is required for pre-crRNA processing (Deltcheva, et al., 2011). For the type II CRISPR-cas locus of S. pyogenes, tracrRNA is encoded 5ʹ to csn1; however, the location of tracrRNA varies for different Type II loci. tracrRNA contains a repeat-like sequence oriented antisense to the repeats of pre-crRNA. tracrRNA acts as a guide in pre-crRNA processing and forms an RNA-RNA duplex with pre-crRNA repeats, thereby facilitating cleavage within the crRNA repeat/tracrRNA anti-repeat by the housekeeping RNase III. Interestingly, a role for RNase III in Type II crRNA maturation is reminiscent of small interfering RNA and microRNA processing in eukaryotes (Deltcheva, et al., 2011). Certain aspects of the Type II processing mechanism remain to be elucidated, including the factor that is responsible for cleavage within the spacer sequence to liberate a mature crRNA (Deltcheva, et al., 2011). While the cas9 gene is required for pre-crRNA and tracrRNA processing (Deltcheva, et al., 2011), the precise role of the Cas9 protein in processing is unknown. Cas9 may carry out cleavage within the spacer sequence to liberate a mature crRNA, and/or may stabilize and facilitate base pairing of pre-crRNA and tracrRNA (Deltcheva, et al., 2011). As expected based on their roles in pre-crRNA processing, RNase III, tracrRNA, and cas9 are each required for S. pyogenes CRISPR spacer-dependent interference with plasmid uptake (Deltcheva, et al., 2011).
There is evidence for crRNA-directed cleavage of both RNA and DNA targets in Type III CRISPR-Cas systems (Hale, et al., 2009; Marrafini & Sontheimer, 2008; Zhang, et al., 2012), but there is none thus far for Type II CRISPR-Cas systems. It is clear that DNA is a target of the StCRISPR1-Cas system (Garneau, et al., 2010). Blunt-end cleavage occurs within the target protospacer of double-stranded plasmid DNA, 3 bp 5ʹ to the PAM (after the 27th nucleotide of a 30 nucleotide protospacer) (Garneau, et al., 2010). The same cleavage site is observed for phage DNA targets (Garneau, et al., 2010). Cleavage occurs 3 bp 5ʹ to the PAM even if the spacer is < 30 bp, indicating that the cleavage proceeds by a 3ʹ anchored mechanism (Garneau, et al., 2010). An additional cleavage event occurs within protospacers when spacers target the positive strand of a phage genome. This second cleavage occurs within the protospacer, 19 or 20 bp 5ʹ to the PAM (35), which corresponds to the 5ʹ end of mature crRNAs (Deltcheva, et al., 2011). As a result, it appears that the Type II CRISPR-Cas interference machinery can discriminate between DNA strands in the target DNA. Because only a negative strand protospacer has been investigated for plasmid targets (Garneau, et al., 2010), it is unclear whether this strand discrimination is specific to phage targets or is a more general property of Type II CRISPR-Cas interference.
To summarize, Type II CRISPR-Cas are genome defense systems that evolve in response to plasmid and phage attacks and that confer immunity to progeny cells through a genetically programmed memory of these attacks (Figure 3). Despite rapid advances in this field, much remains to be learned about Type II CRISPR-Cas systems. Arguably the most important aspect are the mechanisms by which CRISPR-Cas exclude their own genome's potential spacers from inclusion in CRISPR arrays, and the precise roles of each of the Cas proteins in acquiring new spacers, processing crRNAs, and providing defense. While structural data are available for Type I (Wiedenheft, et al., 2011) and Type III (Zhang, et al., 2012) crRNA interference complexes (i.e. crRNA with its associated protein effectors), this is not the case for Type II. Determining a crystal structure for Cas9 will be especially informative, as this protein acts at both the crRNA processing and interference stages of Type II CRISPR defense. An open question remains as to why a 30 bp spacer is incorporated into a CRISPR array if only 2/3 of that sequence is used in a mature crRNA; perhaps different ruler mechanisms guide spacer addition and crRNA processing.
Finally, while many CRISPR spacers have sequence identity to plasmid and phage sequences, some have identity to the host's own genome These spacers are referred to as self-targeting spacers. Self-targeting spacers were identified in StCRISPR1 and StCRISPR3 loci of various S. thermophilus strains (Horvath, et al., 2008). The role of self-targeting spacers is unclear, although it has been hypothesized that CRISPR systems can erroneously incorporate "self" genome into the spacer array, leading to selection for progeny with degenerate (broken) CRISPR-cas loci (Stern, Keren, Wurtzel, Amitai, & Sorek, 2010). An alternative and not necessarily mutually exclusive explanation, depending on the CRISPR type in question, is that self-targeting spacers regulate the expression of chromosomal genes through an RNAi-like mechanism (Horvath, et al., 2008). Despite sequence identities of some CRISPR spacers to known mobile elements and host genomes, most spacers analyzed do not match known sequences—a finding that indicates that prokaryotes are exposed to a diverse variety of uncharacterized phages and plasmids that have not been sampled by genomic and metagenomic (total DNA retrieved from an environmental sample) studies to date.
Enterococcal CRISPR-Cas systems
Enterococcal CRISPR-cas was first identified in the E. faecalis OG1RF genome (Bourgogne, et al., 2008). A Type II-A locus possessing a CRISPR and cas9, cas1, cas2, and csn1 genes was identified in OG1RF between homologues of the E. faecalis V583 ORFs EF0672 and EF0673, which the authors named CRISPR1 (referred to as the EfsCRISPR1-cas locus for the purposes of this chapter, shown in Figure 4). An additional gene of unknown function is encoded 3ʹ to the CRISPR1 array in OG1RF. A second CRISPR locus that lacks cas genes was identified between V583 homologues of EF2062 and EF2063, which the authors named CRISPR2 (EfsCRISPR2 for the purposes of this chapter, and also shown in Figure 4). The consensus repeat sequences for EfsCRISPR1 and EfsCRISPR2 are identical, suggesting that the two loci are functionally linked (Bourgogne, et al., 2008; Horvath, Coûté-Monvoisin, Romero, Boyaval, Fremaux, & Barrangou, 2009). An EfsCRISPR2 locus of identical repeat sequence and different spacer content occurring between EF2062 and EF2063 was subsequently identified in the V583 genome (Horvath, Coûté-Monvoisin, Romero, Boyaval, Fremaux, & Barrangou, 2009; Palmer & Gilmore, 2010). The presence of EfsCRISPR1-cas in E. faecalis OG1RF was proposed to account for the low prophage content of this strain (OG1RF possesses only the cryptic phage02 which is part of the E. faecalis core genome) (Bourgogne, et al., 2008; McBride, Fischetti, Leblanc, Moellering, Jr., & Gilmore, 2007; Palmer, et al., 2012). This is in contrast to E. faecalis V583, which possesses seven prophage elements (Paulsen, et al., 2003) and lacks EfsCRISPR1-Cas (Bourgogne, et al., 2008). PCR-based screening for the presence of EfsCRISPR1, cas1, and csn1 in 14 additional E. faecalis isolates indicated that this locus is variable in the faecalis species (Bourgogne, et al., 2008).
Figure 4.
E. faecalis CRISPR-cas loci. Representative EfsCRISPR1-cas, EfsCRISPR2, and EfsCRISPR3-cas loci are shown. E. faecalis V583 open reading frames and their homologues in other strains are shown in grey. cas genes are shown in white. CRISPR spacers (30 nucleotides) (more...)
The availability of genome sequence data for an additional 16 E. faecalis genomes (Palmer, et al., 2010) allowed for a more comprehensive analysis of CRISPR distribution in the faecalis species (Palmer & Gilmore, 2010). As expected, EfsCRISPR1-cas was variably distributed among the 16 genomes (present in 5/16), and when present occurred invariably between homologues of EF0672 and EF0673. The number of spacers in the loci varied, as did the presence of the gene of unknown function occurring 3ʹ to the CRISPR array in E. faecalis OG1RF. A second CRISPR-cas locus, named CRISPR3-cas (referred to as the EfsCRISPR3-cas locus for the purposes of this chapter), was identified in two additional faecalis genomes between homologues of the V583 ORFs EF1759 and EF1760 (Figure 4). EfsCRISPR3-cas encodes a putative Type II CRISPR-Cas system. Its cohort of cas genes contains csn1, cas1, cas2, and a gene of unknown function. EfsCRISPR3 repeats are distinct from EfsCRISPR1/CRISPR2 repeats, sharing only 42% nucleotide sequence identity. EfsCRISPR1-cas and EfsCRISPR3-cas loci did not co-occur in the same strain; thus, 7 of the 16 genomes analyzed possessed a Type II CRISPR-Cas system.
Interestingly, an EfsCRISPR2 locus of conserved repeat sequence and varying spacer content was identified between EF2063 and EF2061 homologues in all 16 E. faecalis genomes analyzed (Palmer & Gilmore, 2010). Note that EF2062 homologues are not annotated in these genomes, despite the nucleotide sequence being ≥ 98% conserved. As a result, it appears that an orphan CRISPR locus of varying spacer content is core to the faecalis genome, and in 9 of 16 genomes, this locus is maintained in the absence of cas genes. EfsCRISPR2 does not appear to confer defense from mobile elements, as genomes possessing only EfsCRISPR2 are significantly larger and encode more protein domains associated with mobile elements as compared to genomes possessing an EfsCRISPR1-cas or EfsCRISPR3 -cas locus (Palmer, et al., 2012). Further, the absence of cas9 in EfsCRISPR2-only strains indicates that pre-crRNA generated from EfsCRISPR2 would not be processed to mature crRNAs.
Analysis of 140 spacer sequences extracted from the EfsCRISPR loci of 16 E. faecalis genomes revealed identities to phage and plasmid sequences (Palmer & Gilmore, 2010). It appears that the E. faecalis Type II CRISPR-Cas systems provide defense from attack by both lytic and temperate phage. Spacers in two strains were similar to sequences from the lytic enterococcal phage ΦEF24C (Uchiyama, et al., 2008), while spacers from seven strains were similar to sequences from the temperate phage ΦFL2B (Yasmin, et al., 2010), or prophages present in the V583 genome (Paulsen, et al., 2003). Spacers from six strains were similar to sequences from pheromone-responsive plasmids and plasmids integrated into the V583 genome, suggesting that E. faecalis Type II CRISPR-Cas systems also confer defense against these elements.
Because plasmids are common disseminators of antibiotic resistance genes in E. faecalis (Palmer, Kos, & Gilmore, 2010), it is likely that CRISPR-Cas acts as a barrier to the acquisition of these genes. This is supported by the finding that the absence of either Type II EfsCRISPR-cas locus is significantly associated with antibiotic resistance acquired by horizontal gene transfer in a collection of 48 E. faecalis strains (Palmer & Gilmore, 2010). Further, high-risk E. faecalis multilocus sequence typing (MLST) lineages lack EfsCRISPR1-cas and EfsCRISPR3-cas (Palmer & Gilmore, 2010), and the presence of certain virulence factor genes is associated with EfsCRISPR-cas absence in E. faecalis (Lindenstrauss, Pavlovic, Bringmann, Behr, Ehrmann, & Vogel, 2011). It appears that the absence of Type II CRISPR-Cas defense in certain enterococcal lineages explains the preponderance of acquired antibiotic resistance genes, prophage, plasmids, and other mobile element traits observed for these strains.
A Type II-A CRISPR-cas locus has also been identified in E. faecium, occurring in 3 of 8 draft genomes analyzed (Palmer & Gilmore, 2010). This locus, EfmCRISPR1-cas, possesses a CRISPR with repeat sequences that are distinct from the E. faecalis CRISPR loci, although EfmCRISPR1 appears to be more closely related to EfsCRISPR1 than EfsCRISPR1 is to EfsCRISPR3. EfmCRISPR1-cas possesses csn1, cas1, cas2, and csn2, as well as an additional gene of unknown function encoded 5ʹ to csn1. Spacers in one of the three strains are similar to predicted phage sequences from Clostridium novyi and L. lactis, and could be derived from an uncharacterized enterococcal phage. The EfmCRISPR-cas locus was absent from three vancomycin-resistant E. faecium strains and from three of four multidrug resistant strains, a finding that is consistent with a role for the system in limiting acquisition of antibiotic resistance genes by horizontal gene transfer.
Have Type II CRISPR-Cas systems been lost from certain E. faecalis and E. faecium lineages, and if so, by what mechanism? It was postulated that S. thermophilus Type II CRISPR-cas loci could be lost by recombination occurring between a CRISPR repeat and a partial repeat sequence occurring 5ʹ to the cas genes (likely the anti-repeat of tracrRNA (Deltcheva, et al., 2011; Horvath, et al., 2008). This implies the deletion of the CRISPR-cas locus, which does not appear to be the mechanism underlying the presence of variable Type II CRISPR-cas loci in E. faecium or E. faecalis. For E. faecium, genomic analyses have revealed that what is clinically classified as E. faecium is actually composed of two distinct phylogenetic clades (Galloway-Peña, Roh, Latorre, Qin, & Murray, 2012; Palmer, et al., 2012) that possess ~ 4-6% divergence in shared gene sequences (Palmer, et al., 2012). Hospital adaptation appears to have originated from only one of these clades, while the other clade consists of fecal commensal or otherwise non-hospital adapted strains. Recombination can occur between strains from these two clades, generating new strains with hybrid genomes (Palmer, et al., 2012). Strikingly, the EfmCRISPR1-cas locus has been identified only in strains from the non-hospital adapted clade, as well as in a hybrid strain that apparently acquired the locus by recombination (Palmer, et al., 2012). This suggests that the ancestral absence of CRISPR-Cas defense in one E. faecium clade facilitated the eventual emergence of acquired multidrug resistance and hospital adaptation specifically from that clade.
Unlike E. faecium, many E. faecalis strains are so closely related in core genome sequence that their relationships to each other cannot be cleanly resolved by whole genome analysis (Palmer, et al., 2012). It is therefore more difficult to postulate that ancestral E. faecalis was composed of distinct clades heterogeneous for CRISPR-Cas defense. In the place of the EfsCRISPR1-cas locus, strains such as E. faecalis V583 possess a novel, ~200 nucleotide sequence found only in strains that lack that locus (Palmer & Gilmore, 2010). Similarly, strains lacking EfsCRISPR3-cas possess a novel, ~50 nucleotide sequence found only in strains that lack that locus. The origin and possible function of these novel sequences is unclear. What is clear is that recombination between E. faecalis strains can cause the loss of EfsCRISPR1-Cas.
Displacement of the OG1RF EfsCRISPR1-cas locus by incoming V583 genomic DNA has been observed for in vitro conjugation experiments (Manson, Hancock, & Gilmore, 2010; Palmer & Gilmore, 2010). Pheromone-responsive plasmids resident in the V583 cell can integrate into the chromosome, and from there initiate plasmid transfer functions, ultimately mobilizing large regions of the V583 chromosome into OG1RF recipients (Manson, Hancock, & Gilmore, 2010). These high frequency recombination-like transfers generate transconjugant strains with hybrid V583-OG1RF genomes that lack EfsCRISPR1-cas and concomitantly possess acquired antibiotic resistance genes and other clinically relevant traits originating from V583 (Manson, Hancock, & Gilmore, 2010; Palmer & Gilmore, 2010). Conceivably, antibiotic resistant, CRISPR-cas deficient strains such as these hybrid strains, as well as high-risk strains such as V583, could be specifically selected for during antibiotic therapy, and could ultimately become the dominant enterococcal lineages in antibiotic-treated patients. Collectively, data from E. faecium and E. faecalis genome analysis and plasmid transfer experiments indicate that enterococcal CRISPR-cas loci can be both disseminated and lost via recombination.
Methods to identify CRISPR-cas loci in enterococcal genomes
Several methods are available to search for CRISPR and cas genes in enterococcal genomes. Because the chromosomal locations of EfsCRISPR1-cas, EfsCRISPR2, EfsCRISPR3-cas, and EfmCRISPR1-cas are conserved (Palmer & Gilmore, 2010), those locations can be specifically interrogated for the presence or absence of these loci, either by genomic analysis or by PCR-based screening. To identify novel CRISPR loci, repeat finder programs can be used. A web-based program, CRISPRFinder (Grissa, Vergnaud, & Pourcel, 2007), is useful for the identification of CRISPR candidates. A genome sequence, either draft or complete, can be loaded into the CRISPRFinder web server and analyzed for potential CRISPR content with a short (< 1 minute) turnaround time. Candidates can then be mapped back to the genome sequence and compared to the structure of known CRISPR loci. As a caveat to this approach, small CRISPR loci like EfsCRISPR2 (consisting of 1 repeat, 1 spacer, and a degenerate repeat) are not identified by CRISPRFinder, and require annotation by analysis of a conserved genomic location (Palmer & Gilmore, 2010). If present, cas genes may be identified in the vicinity of CRISPR candidates, and can be easily found using genome annotations or by searching for genes encoding conserved protein domains (for example, TIGRFAM domains) that are diagnostic for CRISPR-Cas systems (Makarova, et al., 2011).
To illustrate this approach, analysis of the sequenced genome of Enterococcus italicus DSM 15952 (GenBank accession number AEPV00000000) for potential CRISPR-cas loci using CRISPRFinder, identified three potential CRISPR arrays: two located on genome contig 74 and one located on contig 75. The CRISPR on contig 75 possesses a 36 nucleotide repeat sequence and 23 spacer sequences with an average size of 30 nucleotides, and is associated with a set of cas genes including csn1, cas1, cas2, and csn2. Thus, E. italicus possesses a Type II-A CRISPR-cas locus. The two CRISPR on contig 74 flank a set of cas genes that include cas1, cas2, cas10, csm2, csm3, csm4, csm5, csm6, and cas6. cas10 is the signature gene for Type III CRISPR-cas loci, and csm2 is the signature gene for Type III-A CRISPR-cas loci (Makarova, et al., 2011). It appears that in addition to a Type II-A CRISPR-cas locus, E. italicus also possesses a Type III-A CRISPR-cas locus. Whether both the Type II-A and Type III-A CRISPR-Cas systems are active and whether they have overlapping or specialized functions in E. italicus genome defense remains to be determined.
Restriction-modification (RM)
There is a small amount of evidence for RM occurring in enterococci. A Type II restriction endonuclease, SfaI, was purified from Streptococcus faecalis var. zymogenes strain TR (likely an E. faecalis isolate) (Wu, King, & Jay, 1978). The recognition sequence of SfaI is GGCC, and cleavage occurs between the G and C (Wu, King, & Jay, 1978). SfaI activity was not identified in other S. faecalis and S. faecium strains, and based on its variable presence, the authors postulated that the gene encoding SfaI was plasmid-based (Wu, King, & Jay, 1978). Two other S. faecalis enzymes, SfaGU and SfaNI, were reported in a survey of RM enzymes, published in 1982 (Roberts R. J., 1984). A RM system (M.SfeI and R.SfeI) encoded by a Streptococcus faecalis SE72 plasmid has also been characterized (Okhapkina, et al., 2002). Finally, a methyltransferase with Dam-like activity encoded by the enterococcal VanB-type vancomycin resistance transposon Tn 1549 has been identified (Radlińska, Piekarowicz, Galimand, & Bujnicki, 2005). Available literature suggests that RM does occur in the enterococci, although the systems identified to date are associated with mobile elements or are otherwise strain-specific. It remains to be determined whether these systems act as barriers to the uptake of additional mobile elements.
New England Biolabs maintains a database of computationally predicted and biochemically verified RM proteins, called REBASE (Roberts, Vincze, Posfai, & Macelis, 2010). A search for "Enterococcus" in REBASE yields 147 hits (as of March 2012), encompassing putative restriction, modification, and specificity subunits encoded by enterococcal chromosomes, plasmids, transposons, and temperate phages, including ΦFL1, ΦFL2, and ΦFL3. The majority of these RM candidates (100/147) are predicted Type II system components (as compared to 29 hits for Type I, 2 hits for Type III, and 15 hits for Type IV). For some of these proteins, recognition and/or cleavage sites have been identified. In the future, it will be of interest to integrate biochemical data from REBASE with genome analyses to better understand the distribution of these systems across the genus, and how their distributions might relate to prophage abundance, acquired antibiotic resistance, and hospital adaptation.
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Perspectives and Future Directions
Enterococcal phage receptors
The receptors utilized by enterococcal phages and the host cell ligands that these receptors bind are mostly uncharacterized. To better understand the biology of enterococcal phages, it will be essential to identify the receptors they use to interact with their host. Peptidoglycan, repeating glycan strands of N-acetyl-D-glucosamine and N-acetylmuramic acid linked to small tetra-peptides, as well as cell wall teichoic acids, covalently linked to peptidoglycan disaccharides, are likely phage binding site candidates (Osborn, 1969; Tipper & Strominger, 1968). Additional teichoic acids called lipoteichoic acids are also anchored in the phospholipid membrane of enterococci. Teichoic acids in the cell wall of S. aureus and Bacillus subtilis have been determined to be phage receptors (Lindberg, 1973) and for S. aureus, alterations in wall teichoic acids results in phage resistance (Wolin, Archibald, & Baddiley, 1966). Lipoteichoic acids of the Lactobacillus delbrueckii cell membrane are also bound by phages. Substitutions in the glycerol backbone of the lipoteichoic acid render L. delbrueckii phage incapable of binding to the host cell surface (Räisänen, et al., 2007). Enterococcal teichoic acid structures vary. E. faecium U0317 produces a wall teichoic acid polymer of 2-acetamido-2-deoxy-D-galactose, glycerol, and phosphate (Bychowska, et al., 2011), while both poly(glycerol-phosphate) and poly(ribitol-phosphate) wall teichoic acid polymers have been detected in E. hirae ATCC 9790 (Armstrong, Baddiley, Buchanan, Davison, Kelemen, & Neuhaus, 1959) E. faecalis produces a 1,3 poly(glycerol-phosphate) polymer attached to either D-alanine, kojibiose (α-D-glucopyranosyl-(1,2)-α-D-glucose), or 6,6′-di-alanyl-α-kojibiose (Hogendorf, Bos, Overkleeft, Codée, & Marel, 2010). The sugars and peptides that compose the cell wall teichoic acids and lipoteichoic acids of the enterococci are strong candidates for enterococcal phage attachment structures. Previous work has also shown that the extracts of enterococcal cell walls can inactivate staphylococcal phages, a finding that suggests that cell wall sugars and/or peptides may be natural ligands for enterococcal phage binding (Rakieten & Tiffany, 1938).
Surface proteins are also likely receptor candidates for enterococcal phages. Numerous types of cell surface proteins have been identified, mainly in E. coli, as receptors for phages. These cell surface proteins represent a diverse array of protein types and include peptidoglycan-associated proteins, membrane transport proteins or channel porins, enzymes, substrate receptors used to transport metabolites, and secretion systems (Rakhuba, Kolomiets, Dey, & Novik, 2010). Enterococcal genomes contain many genes whose products are thought to be associated with the bacterial cell surface (Palmer, et al., 2012; Paulsen, et al., 2003). For instance, the E. faecalis V583 genome sequence contains greater than 50 putative lipoproteins with predicted signal sequences that may be associated with the E. faecalis membrane; approximately 20 different carbohydrate utilization pathways which have identifiable membrane transport proteins; and a large number of virulence related genes that are predicted to be exposed at the bacterial cell surface (Paulsen, et al., 2003). Many of these proteins may be used by enterococcal phages for the transmission of their nucleic acid during infections.
Other cell surface structures may be used for enterococcal phage attachment to the host cell, including capsular polysaccharides and pili. Capsular polysaccharides are secreted from bacterial cells and deposited at the cell surface, creating a layer of carbohydrate slime that covers the cell. In some instances, polysaccharide capsules are protective against phage adsorption; however, other phages utilize cell surface capsular polysaccharide for initial adsorption, a process which is often reversible (Lindberg, 1973; Rakhuba, Kolomiets, Dey, & Novik, 2010). E. faecalis encodes two clusters of genes that are known to contribute to capsule biosynthesis: the epa and cps genes. The epa genes are thought to contribute to the production of a rhamnose containing polysaccharide capsule, whereas the cps gene cluster polysaccharides are composed of glucose and galactose and are sometimes referred to as diheteroglycan (Hancock & Gilmore, 2002; Theilacker, et al., 2011; Thurlow, Thomas, & Hancock, 2009). The production of the cps polysaccharides is variable in E. faecalis and is the basis of the group A-D serotyping.
Little is known about capsule biogenesis in other enterococcal species, although E. faecium, E. casseliflavus, and E. gallinarum possess a putative capsule biosynthesis system that is distinct from the cps system of E. faecalis and is more similar to the capsule biosynthesis system of S. pneumoniae (Palmer, et al., 2012). It is possible that enterococcal phages bind to these polysaccharides in a cell trophic manner, due to the variable possession and expression of the capsular polysaccharide synthesis loci between enterococcal strains. If so, this may be a useful way to determine host specificity for particular enterococcal phage groups.
Some phages have also evolved to bind to the bacterial cell surface through interactions with surface appendages, such as flagella and pili (Lindberg, 1973; Rakhuba, Kolomiets, Dey, & Novik, 2010). Enterococci such as E. gallinarum and E. casseliflavus are flagellated, and many species produce pili (Nallapareddy, et al., 2006; Sillanpää, Prakash, Nallapareddy, & Murray, 2009). Pili receptors for phage adsorption have been identified for pleomorphic-RNA–containing phages and filamentous phages (Rakhuba, Kolomiets, Dey, & Novik, 2010). The recent identification of numerous novel enterococcal phages belonging to the polyhedral, filamentous, and pleomorphic phage group suggests that pili could be cell surface targets for phage adsorption (Mazaheri Nezhad Fard, Barton, & Heuzenroeder, 2010). Future studies on phage receptors of enterococci will be needed to better understand the mechanisms used by enterococcal phages to interact with their hosts, and may yield insight into how these phages influence the community structures of enterococci within natural habitats.
Role of enterococcal phages in mammalian systems
Studies have begun to elucidate the human virome, which is the type and abundance of viral particles found at distinct anatomical sites where commensal bacteria reside, which includes the intestinal tract, oral cavity, and the lungs (Pride, et al., 2012; Reyes, et al., 2010; Willner, et al., 2009). It is clear that the areas of the human body colonized by commensal bacteria are highly populated with phages. It is also evident that these phage populations fluctuate when changes in bacterial community structure occur; for instance, when the human diet is altered (Minot, et al., 2011). So far, our knowledge of the identity of these phages is limited to metagenomic DNA libraries of phage DNA sequences (i.e. integrases and phage structural genes) and their comparison to known phage and prophage sequences that have been deposited in public nucleotide databases. From these data, it is possible to obtain an idea of the diversity of natural phage populations associated with humans, and, to an extent, determine some information on the relative abundance of certain types of phages (i.e. double- versus single-stranded DNA phages) (Reyes, et al., 2010).
Analysis of metagenomic DNA sequences from the intestinal tract has revealed sequences with similarity to enterococcal phages, specifically those of E. faecalis (Duerkop, Clements, Rollins, Rodrigues, & Hooper, 2012; Morowitz, et al., 2010). Studying enterococcal species in the intestinal tract, especially those that are polylysogenic, may provide critical information about the biological significance of enterococcal phages in the intestines. It is possible that phages that infect the enterococci may influence genetic exchange between Enterococcus species within the intestine. Phage-mediated exchange of DNA may be another mechanism that enterococcal species use for generating genetic variation to evade the immune system at the intestinal mucosa. Enterococcal phages are a likely source of novel genes. Therefore, phage-mediated transfer of genetic traits may result in the acquisition of genes that contribute to hypervariable phenotypic traits for immune evasion—or, conversely, phages may transfer virulence factors that aid in the transition of enterococci from commensals to pathogens. This type of phage-bacterial interaction could have far-reaching implications for the evolution of pathogenic enterococci, as multiple species have evolved to become successful opportunistic pathogens. Recently, Enterococcal phages have been proposed to play an ecological role in the intestine by aiding in niche competition. In the intestines of gnotobiotic mice, prophage01 and EfCIV583 aided E. faecalis V583 during competition with closely related E. faecalis strains (Duerkop, Clements, Rollins, Rodrigues, & Hooper, 2012). This suggests that prophage induction could be a means by which certain communities of enterococci compete with neighboring, genetically similar communities for nutrients in the intestine by restricting the invasion of neighboring enterococcal strains through phage-mediated killing.
The use of in vitro co-culture systems and mouse models of enterococcal intestinal colonization, including the use of gnotobiotic mice, will be instrumental toward further elucidating the genetic and ecological roles of enterococcal phages within the intestinal tract. Furthermore, these types of studies are not limited to the intestinal tract and can be applied to other environments where enterococci reside, including the oral and vaginal microbiota, as well as endocarditis and bacteremia infections. It is likely that enterococcal phages will profoundly influence enterococcal biology within a multitude of different environments where both commensal and pathogenic enterococci are found, ultimately leading to discoveries relevant to the ways in which commensal and pathogenic enterococci interact with their mammalian hosts.
Enterococcal CRISPR-Cas and RM Systems
Most sequence analysis of enterococcal CRISPR was reported in 2010 (Palmer & Gilmore, 2010), prior to the discovery of tracrRNA and the structure of mature Type II CRISPR crRNAs (Deltcheva, et al., 2011). A strict sequence identity cut-off was used to identify enterococcal protospacer-spacer sequence matches (27/30 identical residues required) (Palmer & Gilmore, 2010), and it is likely that potential targets of the enterococcal CRISPR-Cas systems were missed. Based on current literature on Type II CRISPR-Cas systems, it now appears that 20/30 identical resides at the 3ʹ spacer end would be a stringent cut-off for spacer analysis. While informative for identifying crRNA targets (especially for E. faecium, for which few were identified), additional protospacers would facilitate the identification of consensus PAM sequences for each of the three enterococcal CRISPR-Cas systems. PAM sequences will be important to consider in designing experimental systems interrogating the specific roles of enterococcal CRISPR and CRISPR-Cas in conferring defense against phages and plasmids.
One attribute that distinguishes enterococcal Type II systems from those studied in other bacteria is the EfsCRISPR2 of E. faecalis. It is unclear how prevalent orphan CRISPR loci are in other prokaryotes. An orphan pre-crRNA in L. monocytogenes EGD-e appears to stabilize a transcript that encodes iron acquisition proteins (Mandin, Repoila, Vergassola, Geissmann, & Cossart, 2007). This result suggests that EfsCRISPR2 may have a secondary function in E. faecalis. Whether EfsCRISPR2 is expressed and processed remains to be determined, although the absence of cas9 suggests that the generation of mature crRNAs will not occur. A long transcript encoded antisense to the V583 ORF EF2062 and EfsCRISPR2 region has also been detected, which may complicate the analysis of EfsCRISPR2 expression and processing (d’Hérouel, et al., 2011). The significance of this antisense transcript is unknown, although presumably it could contribute to processing and/or stability of the EfsCRISPR2 transcript.
Beyond the CRISPR-Cas defense, it will be important to determine the way in which differential genome modifications and restriction enzyme cohorts affects phage susceptibilities, as well as the frequency of horizontal gene transfer in the enterococci. Restriction modification appears to be a major barrier to gene transfer in S. aureus (Monk, Shah, Xu, Tan, & Foster, 2012), and the same may be true for the enterococci. Bioinformatic identification and analysis of candidate restriction-modification and abortive infection systems from available enterococcal genomes will be informative, as very little is known about genome defense in the enterococci. Some therapeutic approaches have already started to explore the use of CRISPR and RM systems (Marrafini & Sontheimer, 2008; Torres, Jaenecke, Timmis, García, & Díaz, 2000).
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Concluding Remarks
Enterococcal genomes harbor a diversity of integrated prophage and other mobile elements. Some strains are also susceptible to infection by non-integrating lytic phages. The study of phages and mobile element biology has revealed several exciting areas of enterococcal research. First, enterococcal phage lytic proteins, the lysins, have prospects as novel therapeutics that can target antibiotic resistant strains of enterococci and other Gram-positive bacteria that resist broad-spectrum antibiotic treatment. However, not only lysins, but also CRISPR, RM systems, and bacteriophages have the potential to be used as modern therapeutics. Second, enterococcal phages have great potential as important mediators of genetic exchange through transduction during intra- and inter-species infections. It is these phage-host interactions, along with other mobile elements like plasmids, which facilitate the acquisition of novel traits (i.e. antibiotic resistance determinants and virulence factors). The attainment of genes from various mobile elements helps to uniquely distinguish clonal populations of enterococci. Prophages and plasmids are abundant in the enterococci, even in strains thought to live a strictly commensal lifestyle. Therefore, the contribution of phage and plasmids to the biology of enterococci may provide insight into what role, if any, these elements play in the transition of enterococci from commensals to pathogens. That being said, one idea has surfaced. Many enterococci have evolved mechanisms to protect their genomes from the assault of plasmids and phage by employing CRISPR-Cas regulatory systems. These systems restrict host genome acquisition of foreign mobile elements through their destruction. Notably, those enterococcal strains possessing one or more complete CRISPR-Cas system are parasitized less by phage and other mobile elements which include antibiotic resistance genes. Those strains that lack functional CRISPR-Cas systems retain many more foreign DNA elements. This dichotomy may be one way in which enterococci have evolved to become successful opportunistic pathogens. If true, the lack of CRISPR-Cas systems would represent a legitimate mechanism for the promotion of the commensal to pathogen transition in the enterococci. These studies on enterococcal phage biology, mobile elements, and CRISPR-Cas have just begun to scratch the surface of an emerging field, wide open for new discoveries.
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Enterococcal Bacteriophages
Distribution of phages across the enterococci
Known phage families found among Enterococci
Environments where enterococcal phages are found
Enterococcal Temperate Phages and the Impact of Lysogeny
Enterococcal Lytic Phages
Enterococcal Genome Defense Mechanisms
Perspectives and Future Directions
Concluding Remarks
References
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Enterococcal Bacteriophages
Distribution of phages across the enterococci
Known phage families found among Enterococci
Environments where enterococcal phages are found
Enterococcal Temperate Phages and the Impact of Lysogeny
Enterococcal Lytic Phages
Enterococcal Genome Defense Mechanisms
Perspectives and Future Directions
Concluding Remarks
References
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189733 | https://www.sciencedirect.com/science/article/pii/S2772736X2200041X | Skip to article
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Human Pathology Reports
Volume 28, June 2022, 300629
Case Report Disseminated opportunistic infections masquerading as central nervous system malignancies
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Abstract
Central nervous system manifestations are not only rare but extremely challenging to diagnose in patients with disseminated opportunistic infections. These CNS presentations can be very non-specific. On rare occasions, they can present as space-occupying lesions raising suspicions of primary or metastatic tumors. In this manuscript, we report a series of three cases of disseminated opportunistic infections with space-occupying lesions in the brains that are extremely challenging to diagnose. We are reporting a series of three cases of disseminated opportunistic infections with intracranial space-occupying lesions that were extremely challenging to diagnose; two patients with disseminated toxoplasmosis and one with disseminated nocardiosis. The clinicopathologic findings, radiological studies, and important diagnostic tests are discussed along with a review of the literature.
Keywords
Toxoplasmosis
Nocardiosis
Encephalitis
Brain tumor
Immunocompromised
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© 2022 The Author(s). Published by Elsevier Inc. |
189734 | https://www.hilarispublisher.com/open-access/what-is-the-retrospective-correlation-between-high-grade-squamous-intraepithelial-lesion-hsil-on-cytology-and-the-histol.pdf | Open Access ISSN: 2157-7099 Journal of Cytology & Histology Research Article Volume 11:1, 2020 DOI: 10.37421/jch.2020.11.550 What is the Retrospective Correlation between High-Grade Squamous Intraepithelial Lesion (HSIL) on Cytology and the Histological Diagnosis of Cervical Intraepithelial Neoplasia 2 (CIN2) or More in AML, Antwerp, Belgium?
Abstract Introduction: Cervical cancer is a major worldwide health problem. Therefore, regular cervical screening in order to make an early diagnosis can help to prevent cervical cancer. The aim of this retrospective study is to evaluate the correlation between HSIL on cytology and histological CIN2+ in AML, Antwerp and to compare two liquid-based cytological techniques ThinPrep® LBC (TP) and SurePath™ LBC (SP).
Methods: 120 women with a HSIL positive cytological smear from 2014 (ThinPrep® LBC) and another 120 from 2010 (SurePath™ LBC) were anonymously randomised out of the AML database, according to predefined in- and exclusion criteria. The Belgian Cancer Registry (CIB and CHP) and the AML database were consulted for histological and cytological data and the researched variables (doctor’s speciality, age, HPV status, -genotypes and -persistence) of these 240 women. 184 women, with histological follow-up within one year, out of 240 were included. Statistical analysis was performed using Stata 15.1 (StataCorp, USA). P-values and Odds-ratios were calculated.
Results: The CIN2+/HSIL ratio of all included 184 subjects was 75.5% (95%CI=69.3-81.8). The found CIN2+ percentages for TP and SP, were 75.8% (95%CI 67.0-84.6) and 75.3% (95%CI 66.5-84.1) respectively. For all included subjects the variables hrHPV infection (p=0,008; OR=6.97) and HPV16 infection (p=0.004; OR=2.79) were statistically significant for having CIN2+ on histology.
Conclusions: The found CIN2+/HSIL ratio of 75.5% in AML, Antwerp is similar to the percentages found in worldwide laboratories. HSIL positive women who are HPV16+ or hrHPV+ are at significant higher risk for invasive cervical disease. No statistically significant difference in CIN2+% was found between the two LBC techniques TP and SP.
Keywords: High-grade squamous intraepithelial lesion • Cervical intraepithelial neoplasia grade 2 or more • Cervical cytology • Cervical histology • Cytohistological correlation Nina Karia1#, Alison Van Loon1#, Ina Benoy1,2,3 and Johannes Bogers1,2,3,4 1AMBIOR, Laboratory for Cell Biology & Histology, University of Antwerp, Universiteitsplein 1, 2610 Antwerp, Belgium 2Laboratory of Molecular Pathology, AML, Emiel Vloorsstraat 9, Antwerp, Belgium 3National Reference Centre for HPV, Juliette Wytsmanstraat 14, Brussels, Belgium 4International Centre for Reproductive Health, Ghent University, De Pintelaan 185, Ghent, Belgium #Both authors contributed equally to this research article Address for Correspondence: Nina Karia, University of Antwerp, Universiteitsplein 1, Antwerp, 2020, Belgium, Tel: 0032499416423, E-mail: Nina.
Karia@student.uantwerpen.be Copyright: © 2020 Karia N, 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.
Received 19 January 2020; Accepted 03 February 2020; Published 10 February 2020 Introduction As cervical cancer is a major health problem, regular cervical screening to make an early diagnosis can help prevent cervical cancer, through identifying and treating pre-invasive cervical lesions. The yearly number of new cervical cancer diagnoses in Belgium was 640 in 2016 . Concerning the prevalence of High-grade Squamous Intraepithelial Lesion (HSIL) cytology in 2017, 0.4% of cervical smears in Flanders (North Belgium) has a HSIL outcome . As found in our systematic review , the worldwide percentage of HSIL-positive women who have a histological result of Cervical Intraepithelial Neoplasia 2 or more (CIN2+) is 77.5%. In AML (Algemeen Medisch Laboratorium), Antwerp, Belgium no research has been done on the correlation between cytological HSIL and the diagnosis of CIN2+. Therefore, in this retrospective study this relationship is evaluated for AML collecting samples from Flanders, Belgium. The aim of our study is to evaluate the correlation between the cytological screening and histological outcome in the diagnosis of cervical cancer, more specifically the correlation between HSIL on cytology and histological CIN2+.
AML used to work exclusively with the BD SurePath™ (SP) liquid-based cytology (LBC) until June 2013, thereafter ThinPrep® (TP) LBC was used exclusively. In previous studies little research has been done regarding the differences of effectiveness between these two LBC tests. Therefore, in this study a comparison between these methods is made, using two cohorts.
It is relevant to know if the CIN2+/HSIL ratio for AML is high or low, as this would guide further policy concerning treatment and follow-up. If it is low, regarding the practical relevance for individuals, patients will be confronted with unnecessary physical complications, for example future pregnancy problems . Looking at the practical relevance in the field of medicine, in this setting the focus is wrongly on a “healthy” population who do not need treatment. This conducts a faulty expenditure of medical sources. Concerning the theoretical relevance, this study would contribute to a more up-to-date scientific knowledge in the field of cervical screening. Finally mentioning relevance in public health, the overtreatment does not increase the “healthy life years” and brings more costs to the community .
The goal of our study was to evaluate the correlation between HSIL on cytology and histological CIN2+ in AML, Antwerp and to analyse the contribution of several variables (doctor’s speciality, age, HPV status, -genotypes and -persistence) on the outcome (CIN2+/HSIL ratio). Females living in Flanders with HSIL on LBC from AML were enrolled. A similar value to the value found in our literature search, was expected in AML . Next to the main aim of our study, a second research question concerned the comparison between the two liquid-based cytological techniques (SurePath™ and ThinPrep®), for the CIN2+/HSIL ratio in AML.
J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 2 of 11 Materials and Methods Study design and study subjects The study design used is a retrospective observational cohort study. Two cohorts of HSIL positive women, retrieved from the anonymised AML database, were analysed. No control group was composed. For cohort one, women of 18 years and older with a cytological diagnosis of HSIL made by ThinPrep® LBC were enrolled. Conventional cytology and SurePath™ LBC test were excluded. For cohort two, women of 18 years and older with a cytological diagnosis of HSIL made by SurePath™ LBC test were enrolled. So ThinPrep® LBC and conventional cytology were excluded. The full overview for the criteria of enrolment in a chronological order can be found in Figure 1. For both cohorts cytological samples from which the identity could not be retrieved, were excluded, for example anonymous samples from “Gezondheidszorg en hulpverlening aan prostituees” (GHAPRO) and anonymous samples within the framework of participation in scientific research. Also, women based abroad, for example women living in the Netherlands visiting a Belgian doctor, were excluded.
Concerning the number of study subjects, 120 females with HSIL were enrolled for each cohort. This number was obtained by the Survey System Sample Size Calculator [6,7]. For the sample size calculation, a confidence level of 95% and a confidence interval margin of +/- 10% were used. After applying the predefined exclusion criteria, a study population of 797 women for cohort one and 681 women for cohort two was obtained (Figure 1). After calculation, the needed sample size was at least 86 women for cohort one and 84 for cohort two. Note that following definition for inclusion was used: of the enrolled women, only those with a histological follow-up within one year were included. To make sure that ample patients could be included, 120 females with HSIL were enrolled for both cohorts (240 enrolled women in total). In our study, final exclusion or “lost in follow-up” was thus defined as the following: all enrolled women with cytological follow-up within one year or any kind of follow-up after more than one year. These defined criteria were used because, as defined in our research question, the outcome of this study is the correlation between HSIL on cytology and CIN2+ on histology. CIN2+ is a tissue diagnosis, and thus needs to be made on a histological specimen. When follow-up only consists of a repeat cytology, this does not give us any information about any possible histological tissue diagnosis.
Diagnostic tests For this study, no additional interventions were conducted. Although, included patients underwent two interventions in the past, more specifically a cytological test (with HPV co-testing) and a histological test.
Concerning the cytological test, cervical cells were collected using the Cervex-Brush® Combi (Rovers). After collection, cells were immediately transferred into a vial containing a fixative fluid and transported in room temperature to AML for further processing. Samples collected in 2010 were processed according to the BD SurePath™ method which used an ethanol-based fluid as preservative. Cervical cells of the women who were analysed in 2014, were processed according to the Hologic ThinPrep® method, a methanol based fixative fluid. Upon arrival in the lab, the samples were split into two parts. One part was used for HPV genotyping with the RIATOL qPCR test, the other part for the preparation and staining of the thin-layer slides. All slides were screened microscopically, using Focal Point Guided Screening (FPGS), by well-trained cytotechnicians and classified according to the Bethesda classification. This method implies a first cytological examination with automated computer guided microscopes, which analyses the PAP smear slides and selects 22 focal points based on segmentation, feature extraction and object classification. These focal points are provided for the second cytological examination .
All lab results were entered in a database which also contains patient demographics (age) and sample collection information (sampling date, specialization smear taker and collection medium).
The AML database was linked with two Belgian Cancer Registry databases, to find histological and cytological follow-up data from study participants who underwent a first cytological test at AML. The Belgian Cancer Registry provides two separate registers containing histological and/or cytological information. The first register is called the “Cancer In Belgium register” (CIB), which contains only histological information about all cervical cancers registered in Belgium. The second register is called the “CytoHistoPathological register” (CHP), which contains all results of all cytological and histological cervical samples examined in Belgium. We consulted both registers in the given order. Consulting the CHP was necessary because the CIB database only reports new cancer diagnoses. CHP in the contrary, contains all cytological and histological cervix sample Serial co-testing : a unique algorithm for AML that positively influences the sensitivity of cytological screening, from 59,14% to 74,42%. This means that at the time of cytological examination, there is foreknowledge of the HPV-status.
INSZ number = Identification number of social security in Belgium a. Cohort 1 (ThinPrep® LBC, 2014) b. Cohort 2 (SurePath™ LBC, 2010) n = 94.697 •Serial co-testing performed in AML •Cervical smear taken in 2014 •INSZ number available (for linking with the Belgian Cancer Registry) n = 94.027 •Age at time of cervical smear restricted to >17,9 years old n = 92.728 •Cells preserved in ThinPrep® LBC n = 797 •Confirmed cytological diagnosis of HSIL n = 120 •Randomisation of 120 women n = 89.761 •Serial co-testing performed in AML •Cervical smear taken in 2010 •INSZ number available (for linking with the Belgian Cancer Registry) n = 88.897 •Age at time of cervical smear restricted to >17,9 years old n = 88.368 •Cells preserved in SurePath™ LBC n = 681 •Confirmed cytological diagnosis of HSIL n = 120 •Randomisation of 120 women Figure 1. Flowcharts enrolment of subjects.
J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 3 of 11 results taken in Flanders. This means that cancer relapses will be described in CHP, but not in CIB. Thus, by solely consulting the CIB these would be missed as histological CIN2+ follow-up.
In practice, several histological methods are used after a cytological HSIL diagnosis, including colposcopically directed biopsy (CDB) from the most abnormal appearing area, large loop excision of the transformation zone (LLETZ), a large endocervical excision procedure (LEEP) and conisation . In the consulted registers the used histological method was not mentioned.
In this study no extra intervention was conducted in addition to the abovementioned interventions. To conclude, we worked with the anonymised database of AML completed with the databases of the Belgian Cancer Registry, more specifically for Flanders the “Centrum voor Kankeropsporing” (CvKO).
Respecting the ethical considerations, approval of the Commission of Medical Ethics (UA-UZA) was obtained. The processing of personal data in pursuit of this study was limited to those data that are reasonably necessary to investigate the correlation between HSIL on cytology and histological CIN2+, in AML. These data were processed anonymously before statistical analysis with adequate precautions to ensure confidentiality. Researchers had no access to patient identification.
Data collection Two random patient cohorts of 120 women each were enrolled using the excel randomisation function (Figure 1). For these 240 enrolled women, all data was retrieved from the AML database and the Belgian Cancer Registry CIB and CHP databases. The CIB database was always consulted first. After this, the CHP database and AML database were only consulted for those women who were not found in the CIB database. All data was anonymised during analysis, using a unique identification number. For both cohorts the information listed in Table 1 was retrieved in an anonymised manner from the AML database and Belgian Cancer Registry (CIB and CHP).
All the gathered information was collected into an overview excel spreadsheet. The elements histologic outcome, type of diagnostic intervention during follow-up and source of information were coded to make a statistical analysis, as further mentioned in Table 2a. Relevant variables for further statistical analysis were determined. The data concerning these variables in Table 2b were coded. The used codes are mentioned in the last column of Table 2b. Before the statistical analysis was performed, the calculation of the proportion CIN2+/HSIL for cohort one, cohort two and the entire group of included subjects was made. 95% confidence intervals (95%CI) were calculated for these proportions, which equal β +/- z , where β is the calculated proportion, n is the sample size and z is the appropriate value from the standard normal distribution for a 95%-confidence level (equalling 1.96).
Statistics What concerns the statistical analysis of this study, a logistic regression was performed to evaluate the effect of different variables on the outcome (CIN2+ yes/no) and to investigate the effect of selection bias. Also, a multivariate regression was performed on the HPV types 16, 31 and 67, to investigate if one of these HPV types is a confounder. Statistical analysis was performed using Stata 15.1 (StataCorp, USA). P-values and Odds-ratios with 95%CI were calculated. The obtained information was processed anonymously.
Results In the following sections the found outcomes of each defined cohort are described separately, starting with cohort one (2014), followed by cohort two (2010). Afterwards, both cohorts are looked at together as one group and calculations are made for each of the abovementioned variables (see Methods). In the final section the results of the statistical analysis are described.
Follow-up of the 2014 ThinPrep® cohort Analysing the 120 women from cohort one who had a TP test in 2014, we find that 64 women are found with follow-up in the CIB database, 44 women are found in the CHP and AML database and 12 women had no histological or cytological follow-up. From the 108 women that had undergone any follow-up, 102 had this follow-up within one year after their HSIL diagnosis. The other 6 women were excluded. 91 women of the remaining 102 women who had follow-up within 1 year, had histological follow-up and thus fit the inclusion criteria by our definition. When only looking at the women who had histological follow-up within 1 year after the cytological HSIL diagnosis, 91 of the 120 women remain. Identifying the CIN2+ results gives a number of 69 women who had a diagnosis of CIN2 or more on histology and 22 women who had CIN1 or less. Calculating the CIN2+ percentage gives a ratio of 69 over 91 with an outcome of 75.8%. In Figure 2 an illustration can be found of the cohort from 2014, using TP test.
Follow-up of the 2010 SurePath™ cohort Analysing the 120 women from cohort two who had a SP test in 2010, we find that 49 women are found with follow-up in the CIB database and 71 women are found in the CHP database. From the total of 120 women, 107 had follow-up within one year after their HSIL diagnosis. The other 13 women were excluded. 93 women of the remaining 107 women who had follow-up within 1 year, had histological follow-up and fit thus the inclusion criteria by our definition. So, when only looking at the women who had histological follow-up within 1 year after the cytological HSIL diagnosis, 93 of the 120 women remain. Identifying the CIN2+ results gives a number of 70 women who had a diagnosis of CIN2 or more on histology and 23 women who had CIN1 or less. Thus, calculating the CIN2+ percentage gives a ratio of 70 over 93 with an outcome of 75.3%. In Figure 3 an illustration can be found of the cohort from 2010, using SP test.
Outcome of cohort one and two The found CIN2+ percentages for cohort one and two were 75.8% and 75.3% respectively. 91 out of 120 and 93 out of 120 women with a HSIL-diagnosis on cytology from cohort one and two had histological follow-up within one year. This is 184 out of 240 women. So, 76.7% of the HSIL positive women went for a biopsy within one year of their abnormal smear, and 23.3% were lost in follow-up.
AML database CIB registry CHP registry Date of HSIL diagnosis Date of cervical histology Date of cervical histology or cytology Age at the time of HSIL diagnosis Outcome of histology according to the CIN classification Type of diagnostic intervention carried out (cytology or histology) HPV status and genotypes at the time of HSIL diagnosis Country code Outcome of the carried out histological or cytological test Duration of the HPV infection Organ code Organ code Specialty of the doctor who did the smear test Collection medium Table 1. Retrieved data per source.
J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 4 of 11 Outcome of all included subjects applying the variables Opposed to the above paragraph, in this section we combine both cohorts to consider the entire group of included subjects, not separating the women in the two cohorts of 2014 and 2010. Thus, the 91 included women from 2014 and the 93 included women from 2010 are looked at together. For this entire group of 184 women a global CIN2+ percentage was calculated, what represents the main aim of our study.
After this, we identified diverse subgroups. We further calculated the percentages of CIN2+ for these different variable subgroups, next to the global CIN2 or more percentage.
For this entire group of 184 women we found that 75.5% have CIN2+ on histology within one year, with a range of 66.5%-78.5% and a 95%CI of 69.29%-81.72% (Table 3).
To point out, some women with a HSIL positive cervical smear do not undergo a biopsy but have a repeat cytology as follow-up. In the Methods section the inclusion criteria are mentioned which state that these women are excluded because cytology cannot give certainty about tissue diagnosis. In order to tackle this uncertainty a range was established, as seen in Table 3. The “range” consists of a lower limit and an upper limit. The lower limit is defined when hypothetically cytological follow-up within one year is also included as a valid follow-up and these cytological follow-up results are automatically identified as CIN2 negative on biopsy. The upper limit is defined when hypothetically cytological follow-up within one year is included as a valid follow-up and these cytological follow-up results are automatically identified as CIN2 positive on biopsy.
In Table 4 an overview of the calculation of the percentages of CIN2+ for the analyzed variables can be found. When looking at the two subgroups a. Elements Element Possible responses Coded response Outcome of histology according to the CIN classification CIN2-0 CIN2+ 1 Lost in follow-up according to definition 9 Type of diagnostic intervention carried out Histology 1 Cytology 2 No follow-up 3 Source of information CIB 1 CHP 2 AML 3 Variable Type Possible responses Coded response Cervical histology or cytology result available within one year after HSIL diagnosis Dichotomous No follow-up within one year 0 Follow-up within one year 1 Collection medium Categorical BD SurePath™ liquid-based Pap test 1 ThinPrep® LBC 2 Doctor’s specialty who did the smear test Categorical Gynaecologist 1 General practitioner 2 Age at the time of HSIL diagnosis Dichotomous <30 years old 0 30 years old or more 1 HPV status at the time of HSIL diagnosis Dichotomous HPV -0 HPV + 1 hrHPV status at the time of HSIL diagnosis Dichotomous hrHPV -0 hrHPV + 1 lrHPV status at the time of HSIL diagnosis Dichotomous lrHPV -0 lrHPV + 1 Number of different hrHPV types at the time of HSIL diagnosis Categorical 0 0 1 1 2 2 more than 2 3 HPV type 16 status at the time of HSIL diagnosis Dichotomous HPV16 -0 HPV16 + 1 Analogous strategy as described above for HPV type 16 for following HPV types: 18, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 67, 68 Persistence of HPV infection at the time of HSIL diagnosis Categorical The present one and the same HPV type infection has not been present for at least two years or absence of HPV infection 0 The present one and the same HPV type infection has been present for at least two years 1 Unknown 9 b. Variables Table 2. Description and coding of elements and variables, Lost in follow-up = all women with HSIL cytology and cytological follow-up within one year or any kind of follow-up after more than one year. No follow-up = all women with HSIL cytology with no cytological nor histological follow-up at any given time.
hrHPV = high risk HPV types, these are the following: 16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, 66 and 68; lrHPV = low risk HPV types, these are the following 6, 11, 53 and 67 J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 5 of 11 of collection medium, there is a difference of 0.5% in CIN2+ percentage between both subgroups. There is a difference of 0.8% in CIN2+ percentage between the subgroups of doctor specialty. In the subgroups of age, there is a difference of 2.5% in CIN2+%.
A difference of 44.4% in CIN2+% is seen between the subgroups of hrHPV status. 95.1% of all included women are hrHPV+. When looking at the subgroups of number of different HPV types, following percentages are found for respectively 0, 1, 2, more than 2 number of hrHPV types: 33.3%, 78.7%, 75.5%, 78.1%. For the subgroups of lrHPV, a difference of 13.6% is found, with 12% of all included women who are lrHPV+. Concerning the HPV types 16, 18, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 67, and 68, a difference of respectively 18.5%, 7.4%, 8.4%, 9.5%, 9.5%, 6.8%, 24.9%, 8.9%, 3.6%, 15.3%, 17.4%, 17.3%, 6.7%, 4.3%, 33.9% and 26.1% is noted in CIN2% between the genotype positive and negative subgroups. When looking at the subgroups of persistence of HPV infection, this was unknown in 77.2% of the women. Between the other two subgroups (persistent Diagnosis Cytology or histology Follow-up period HSIL diagnosis in 2014 120 Follow-up within 1 year: 102 Histological : 91 CIN2+ : 69 - SQCA : 6 - SQIS : 60 - CIN2 : 3 CIN2- : 22 - CIN1 : 13 - ABST : 9 Cytological : 11 - HSIL : 1 - LSIL : 6 - NILM : 3 -ASC-H: 1 Follow-up over 1 year: 6 Cytological : 6 - HSIL: 1 - LSIL : 2 - NILM: 3 No follow-up: 12 Figure 2. Visual analytics cohort 1.
SQCA= Invasive squamous carcinoma, SQIS= Squamous carcinoma in situ, ABST= No dysplasia, nor tumor, HSIL= High-grade squamous intraepithelial lesion, LSIL= Low-grade squamous intraepithelial lesion, NILM= negative for intraepithelial lesion or malignacy, ASC-H= Atypical squamous cells, cannot exclude HSIL Diagnosis Cytology or histology Follow-up period HSIL diagnosis in 2010 120 Follow-up within 1 year: 107 Histological : 93 CIN2+ : 70 - SQCA : 5 - ADCA: 2 - SQIS : 47 - ADIS: 2 - CIN2 : 13 - OTHMAL: 1 CIN2- : 23 - CIN1 : 7 - ABST : 15 - CGIN: 1 Cytological : 14 - HSIL : 2 - LSIL : 2 - NILM : 8 -ASC-H: 1 - SQCA: 1 Follow-up over 1 year: 13 Histological: 8 CIN2+: 6 - SQIS: 4 - CIN2: 2 CIN2-: 2 (ABST) Cytological : 5 - NILM: 4 - ATYP: 1 Figure 3. Visual analytics cohort 2.
ADCA = Invasive adenocarcinoma; ADIS = Adenocarcinoma in situ; OTHMAL = Other malignancies, specified: tumors including metastasis and invasion of the cervix, other than ADIS, SQIS, ADSQIS, ADCA, SQCA, ADSQC; CGIN = Cervical Glandular Intraepithelial Neoplasia; ATYP = Atypia, not further specified J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 6 of 11 infection for at least two years and no persistent infection) there is a difference of 16% in CIN2+%.
Statistical analysis In this section the outcome of the statistical analysis with Stata 15.1 is described. It is divided into three segments: statistical analysis of the whole group of 184 included women, the first cohort from 2014 and the second cohort from 2010 consecutively.
Previous to these analyses on the 184 included women, a statistical analysis was performed on the 240 women enrolled after randomisation. No statistically significant variables were found to enhance the risk of being lost in follow-up (as defined in Table 2). These 56 women (23.3%) had no significant predicting features for not having histological follow-up within one year. In particular nor age (p=0.609), nor doctor’s specialty (p=0.615), nor hrHPV positivity (p=0.889) influenced having histological follow-up within one year after HSIL diagnosis as shown in Figure 4.
Statistical analysis of all 184 included women Of all 184 included women, those with a hrHPV infection at the time of HSIL diagnosis had 6.97 more chance (i.e. Odd’s ratio, OR) to be CIN2+ on histology than those who were hrHPV negative at that moment, with a 95%-confidence interval (95%CI) of 1.67-29.17. This is statistically significant (p=0.008) as in Table 5 and Figure 5.
A second statistically significant variable found for the whole group was HPV16 infection at the time of HSIL diagnosis. More specifically, of all 184 included women, those with a HPV16 infection had 2.79 more chance (OR) to be CIN2+ on histology than those who were HPV16 negative, with a 95%CI of 1.38-5.66 (p=0.004) (Table 5 and Figure 5).
Following variables were proven not to be statistically significant: the other researched HPV types (18, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 67 and 68), persistent HPV-infection as per definition, number of hrHPV types, lrHPV type, age and doctor specialty. Also, the variable “collection medium”, the comparison between cohort one (TP) and two (SP), was again proven to be a not statistically significant variable (Table 3) (p=0.930, OR =1.03).
The multivariate regression performed on the HPV types 16, 31 and 67, showed that there is no confounding as the statistical significance of the different HPV types stays unchanged with this analysis. HPV type 16 continues to be statistically significant with p=0.006 and OR 2.76. HPV type 31 and 67 are still not significant (HPV31: p=0.201 and OR=1.90; HPV67: p=0.157 and OR=0.32).
included CIN2-CIN2+ CIN2+/ HSIL (%) 95%CI (%) Range (%) P-value All included subjects 184 45 139 75.5 69.3- 81.8 66.5-78.5 Collection medium: TP (cohort 1) 91 22 69 75.8 67.0- 84.6 67.7-78.4 0.930 Collection medium: SP (cohort 2) 93 23 70 75.3 66.5- 84.1 65.4-78.5 ADCA = Invasive adenocarcinoma; ADIS = Adenocarcinoma in situ; OTHMAL = Other malignancies, specified: tumors including metastasis and invasion of the cervix, other than ADIS, SQIS, ADSQIS, ADCA, SQCA, ADSQC; CGIN = Cervical Glandular Intraepithelial Neoplasia; ATYP = Atypia, not further specified Table 3. Main outcome. a. Effect of age for included/LFU b. Effect of doctor specialty for included/LFU c. Effect of hrHPV status for included/LFU 21.7 25 78.3 75 0% 20% 40% 60% 80% 100% included LFU <30y ≥30y p = 0.609 77.2 80.4 22.8 19.6 0% 20% 40% 60% 80% 100% included LFU GYN GP p = 0.615 4.9 5.4 95.1 94.6 0% 20% 40% 60% 80% 100% included LFU hrHPV-hrHPV+ p = 0.889 Figure 4. Examples of nonsignificant variables (lost in follow-up).
LFU = lost in follow-up; GYN = gynaecologist; GP = general practitioner J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 7 of 11 Variable # included (%) CIN2-CIN2+ CIN2+/ HSIL (%) P-value Doctor specialty: gynaecologist 142 (77.2) 35 107 75.4 0.912 Doctor specialty: general practitioner 42 (22.8) 10 32 76.2 Age at the time of HSIL diagnosis: <30y 40 (21.7) 9 31 77.5 0.745 Age at the time of HSIL diagnosis: 30y or more 144 (78.3) 36 108 75 hrHPV status at the time of HSIL diagnosis: -9 (4.9) 6 3 33.3 0.008 hrHPV status at the time of HSIL diagnosis: + 175 (95.1) 39 136 77.7 lrHPV status at the time of HSIL diagnosis: -162 (88) 37 125 77.2 0.172 lrHPV status at the time of HSIL diagnosis: + 22 (12) 8 14 63.6 Number of different hrHPV types at the time of HSIL diagnosis: 0 9 (4.9) 6 3 33.3 0.264 Number of different hrHPV types at the time of HSIL diagnosis: 1 94 (51) 20 74 78.7 Number of different hrHPV types at the time of HSIL diagnosis: 2 49 (26.6) 12 37 75.5 Number of different hrHPV types at the time of HSIL diagnosis: more than 2 32 (17.5) 7 25 78.1 HPV type 16 status at the time of HSIL diagnosis: -88 (47.8) 30 58 65.9 0.004 HPV type 16 status at the time of HSIL diagnosis: + 96 (52.2) 15 81 84.4 HPV type 18 status at the time of HSIL diagnosis: -168 (91.3) 40 128 76.2 0.510 HPV type 18 status at the time of HSIL diagnosis: + 16 (8.7) 5 11 68.8 HPV type 31 status at the time of HSIL diagnosis: -150 (81.5) 39 111 74 0.310 HPV type 31 status at the time of HSIL diagnosis: + 34 (18.5) 6 28 82.4 HPV type 33 status at the time of HSIL diagnosis: -172 (93.5) 41 131 76.2 0.463 HPV type 33 status at the time of HSIL diagnosis: + 12 (6.5) 4 8 66.7 HPV type 35 status at the time of HSIL diagnosis: -172 (93.5) 41 131 76.2 0.463 HPV type 35 status at the time of HSIL diagnosis: + 12 (6.5) 4 8 66.7 HPV type 39 status at the time of HSIL diagnosis: -161 (87.5) 38 123 76.4 0.477 HPV type 39 status at the time of HSIL diagnosis: + 23 (12.5) 7 16 69.6 HPV type 45 status at the time of HSIL diagnosis: -181 (98.4) 45 136 75.1 omitted HPV type 45 status at the time of HSIL diagnosis: + 3 (1.6) 0 3 100 HPV type 51 status at the time of HSIL diagnosis: -160 (87) 41 119 74.4 0.346 HPV type 51 status at the time of HSIL diagnosis: + 24 (13) 4 20 83.3 HPV type 52 status at the time of HSIL diagnosis: -156 (84.8) 39 117 75 0.686 HPV type 52 status at the time of HSIL diagnosis: + 28 (15.2) 6 22 78.6 HPV type 53 status at the time of HSIL diagnosis: -174 (94.6) 44 130 74.7 0.297 HPV type 53 status at the time of HSIL diagnosis: + 10 (5.4) 1 9 90 HPV type 56 status at the time of HSIL diagnosis: -164 (89.1) 37 127 77.4 0.093 HPV type 56 status at the time of HSIL diagnosis: + 20 (10.9) 8 12 60 HPV type 58 status at the time of HSIL diagnosis: -172 (93.5) 44 128 74.4 0.209 HPV type 58 status at the time of HSIL diagnosis: + 12 (6.5) 1 11 91.7 HPV type 59 status at the time of HSIL diagnosis: -173 (94) 43 130 75.1 0.620 HPV type 59 status at the time of HSIL diagnosis: + 11 (6) 2 9 81.8 HPV type 66 status at the time of HSIL diagnosis: -177 (96.2) 43 134 75.7 0.797 HPV type 66 status at the time of HSIL diagnosis: + 7 (3.8) 2 5 71.4 HPV type 67 status at the time of HSIL diagnosis: -177 (96.2) 41 136 76.8 0.058 HPV type 67 status at the time of HSIL diagnosis: + 7 (3.8) 4 3 42.9 HPV type 68 status at the time of HSIL diagnosis: -180 (97.8) 43 137 76.1 0.254 HPV type 68 status at the time of HSIL diagnosis: + 4 (2.2) 2 2 50 Non-persistence of HPV infection at the time of HSIL diagnosis: The present one and the same HPV type infection has not been present for at least two years or absence of HPV infection 19 (10.3) 8 11 57.9 0.113 Persistence of HPV infection at the time of HSIL diagnosis: The present one and the same HPV type infection has been present for at least two years 23 (12.5) 6 17 73.9 Persistence of HPV infection at the time of HSIL diagnosis: unknown 142 (77.2) 31 111 78.2 Table 4: Overview of the calculation of the percentages of CIN2+ for the analyzed variables J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 8 of 11 Next, a logistic regression was performed on the subgroup of HPV16 positive women, to investigate the effect of more than one different hrHPV type infections. This showed to have no statistically significant effect (p=0.5, OR=1.26).
Statistical analysis of cohort one Analogous to the analysis of the whole group (5.5.1.), the variables hrHPV infection and HPV16 infection were statistically significant for the 91 included women from cohort one (TP, 2014) (Table 5). Those with a hrHPV infection at the time of HSIL diagnosis had 6.47 more chance (OR) to be CIN2+ on histology than those who were hrHPV negative at that moment, with a 95%CI of 1.40-29.80. This is statistically significant (p=0.017). Those with a HPV16 infection had 3.54 more chance to be CIN2+ than those who were HPV16 negative, with a 95%CI of 1.28-9.84 (p=0.015).
For this cohort an OR of 0.093 was found for the variable HPV67 infection, which would imply that a HPV67 infection at the time of HSIL diagnosis reduces the chance to be CIN2+. Note that the p-value is 0.045 and the 95%CI is large (0.009-0.947) as in Table 5.
The other analyses variables were not statistically significant for cohort one.
For cohort one alone, the multivariate regression performed on the HPV types 16, 31 and 67, again showed that there is no confounding.
Statistical analysis of cohort two For the 93 included women from cohort two (SP, 2010), the only variable found to be statistically significant (p=0.029) was persistent HPV infection (Table 5). Women from cohort two had 1.17 more chance to be CIN2+ when having a persistent HPV infection at the time of HSIL diagnosis (95%CI 1.02-1.35).
The other analysed variables were not statistically significant for cohort two. The multivariate regression performed on the HPV types 16, 31 and 67, again, showed no confounding.
Discussion The percentage of HSIL-positive women with a histological CIN2+ result within one year is 75.5% (95%CI: 69.33-81.75) in AML, Antwerp. Comparing this to the result of 77.5% of our previously conducted systematic review , we find that there is only a difference of 2.0% between these two outcomes and that this difference is not significant as 77.5% lies within the 95%Cl. However, some notable differences are found between this study and the cited systematic review. A major distinction is the worldwide extent of the systematic review, in contrast to this study where only Flemish (North Belgium) women were included. Further, the review only discussed screening cytological samples, whereas this study discusses all collected samples (excluding GHAPRO and women without INSZ number) in AML without distinguishing between screening and diagnostic samples. Thus, in this study there is no knowledge of symptomatology of the included women. Also, the systematic review did not exclude women with histological follow-up after more than one year, in contrary to our defined inclusion criteria. Several articles that were included in the review have a study period of 5 years or more, which could lead to a higher detection rate of CIN2+. This could be a contributing factor for the slightly higher percentage found in the systematic review . a. hrHPV proportions for CIN2-/CIN2+ b. HPV16 proportions for CIN2-/CIN2+ 13.3 2.2 86.7 97.8 0% 20% 40% 60% 80% 100% CIN 2-CIN 2+ hrHPV-hrHPV+ p = 0.008 66.6 41.7 33.3 58.3 0% 20% 40% 60% 80% 100% CIN 2-CIN 2+ HPV16-HPV16+ p = 0.004 Figure 5. Significant variables (all included subjects).
Odds-ratio P value 95% Confidence interval All subjects hrHPV 6.97 0.008 1.67-29.17 HPV16 2.79 0.004 1.38-5.66 Cohort one hrHPV 6.47 0.017 1.40-29.80 HPV16 3.54 0.015 1.28-9.84 HPV67 0.093 0.045 0.009-0.947 Cohort two Persistent HPV infection 1.17 0.029 1.02-1.35 Table 5. Statistical data of significant variables.
J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 9 of 11 An important reason appointed in literature for the found false positivity rate of 24.5%, is the use of CDB. Multiple studies find that a higher detection rate of CIN2+ is observed when women undergo a LEEP or conisation, which are more reliable specimens [9-14]. Other variables like parity, menopausal status and use of oral contraceptive pills, are reasons for false positive results. Particularly HSIL positive nulliparous women, post-menopausal women and women who do not use oral contraceptive pills are found to have more CIN2- results on histology [9,14].
Next to the main aim of our study, it also seemed interesting to make an extra feature on the difference between the two cytological collection media used in AML over the past years (SurePath™ LBC and ThinPrep® LBC). The found CIN2+ percentages for TP and SP, were 75.8% (95%CI 67.0- 84.6) and 75.3% (95%CI 66.5-84.1) respectively. It is demonstrated that there is no significant difference between these two cytological techniques concerning outcome (p-value=0.930, OR=1.03). Still, to be precise, the statistical analyses were also made on both cohorts separately. In literature, we find similar results. Cox et al. indirectly compared the two liquid based cytology methods and did not detect a difference in sensitivity . Also, Zhao et al. concluded that SurePath™ and ThinPrep® samples yielded analogous validity in detecting significant cervical lesions . Yet, the rate of satisfactory slides was higher with use of SP, as the cell yield is larger because the brush of the collecting device is retained, contrary to TP, where the brush is rinsed and discarded [16,17]. Also, residual volumes for Hybrid Capture 2 (HC2) HPV DNA testing were larger when using SP . However, we note that little literature was found specifically researching the difference between SP and TP LBC. More often, the focus lies on the comparison between conventional cytology and LBC.
Regarding the applied variables on the total group of 184 included women, the statistical analysis demonstrates that for women with a HSIL-positive smear the variables hrHPV positivity and HPV16 positivity are significant for having a histological CIN2+ outcome within one year after cytological testing. The highest OR (6.97) is found for hrHPV-positivity, which implies that a HSIL positive woman who is also hrHPV positive has around seven times more risk for developing a histological CIN2+ lesion within one year than a HSIL positive woman who is hrHPV negative. For HPV16 positive women this risk is around three times higher (OR=2.79). HPV16 is the most carcinogenic type. This corresponds with literature, as Smelov et al. systematically reviewed risks of different HPV types for CIN2+ and found that HPV genotype 16 contributed the greatest proportion of CIN2+ . They also found that all hrHPV types together contributed 86.9% of CIN2+ lesions, which is analogous to our findings .
Persistence of HPV-infection was found to be a nonsignificant variable. There was a nonsignificant difference of 16% in CIN2+ percentage between the two subgroups “persistent infection for at least two years” and “no persistent infection”. This is different from what is found in several articles [19,20]. One of the largest prospective screening trials with a follow-up period of over 13 years, Elfgren et al., found that all (40 of 40, 100% [95%CI, 91-100%]) patients who have persistent HPV-infection eventually will be diagnosed with CIN2+, many in as few as two years. In this study a persistent HPV-infection is defined as an infection persisting over 12 months . In our study we found that 73,9% (17 of 23) of women with a persistent HPV-infection, defined as an infection persisting over 2 years, developed a CIN2+ lesion. A reason for the lower percentage found in our study could be the small group of women who had one of the same HPV type infection for at least two years. It is important to highlight that of 77.2% of women included in our study the persistence of HPV-infection is unknown. The two main reasons for unknown persistence status are the following: the longest follow-up interval is less than two years (if the specific HPV-type was persistent at that moment), the shortest follow-up interval is more than two years (if the specific HPV-type was cleared at that moment). The large amount of 142 women (out of 184 women) with an unknown HPV-persistence can be explained by our set definition for persistence of 2 years. Across studies, it is clear that the definition of HPV persistence varies considerably . Koshiol et al. conducted a systematic review regarding persistent HPV infections. We cite: “The minimum duration of HPV persistence was less than 6 months for 30 percent of studies, 6-12 months for 45%, and more than 12 months for 25%” . Another example is Sand et al. who defined persistence as having the same genotype twice 1-4.5 years apart .
The number of different hrHPV-types at the time of HSIL diagnosis was found to be a nonsignificant variable. There is little difference between the presence of one (78.7%), two (75.5%), or more than two (78.1%) types. In literature we find that consensus is pending regarding the role of multiple hrHPV-strains in developing CIN2+ histology . In Chaturvedi et al. co-infection with multiple high-risk species (more specifically HPV 16, 31, 33, 35, 52 and 58) resulted in a significantly increased risk of CIN2+ (OR=2.2; 95% CI=1.1-4.6), compared with single HPV-infection . In Li et al. on the other hand, they found that patients with single HPV16-infection had higher incidence of CIN2+ (62.2%) than patients with HPV16 infection mixed with other hrHPV-types (52.4%) .
The age at time of HSIL-diagnosis was also found to be a nonsignificant variable. In our study we found that 77.5% of HSIL-positive women aged <30 years and 75% of HSIL-positive women aged 30 years or older have a histological CIN2+ lesion. This slightly higher percentage of younger women does not entirely correspond with the found literature on this subject. Einstein et al. states that cervical lesions found in women younger than 30 are rarely clinically significant . Amongst young females who have experienced their first sexual intercourse there is a high HPV-infection rate. Winer et al. reports a 29% one-year cumulative incidence of HPV-infection after first sexual introduction, increasing to 50% by three years . In young women these infections are often transient, resulting in transient cytological lesions. A possible explanation for the higher percentage found in our study could be the fact that in this study no exclusive screening population was used and thus diagnostic smears could have been included.
It is noticeable that 100% of the HPV45+ women are CIN2+, keeping in mind that this regards a very small group of only 3 women. Only 1.7% of the included HPV+ women harbored the HPV45 genotype. During statistical analysis no p-value could be calculated thus no conclusions could be made regarding the significance of this variable. In literature, a low prevalence of HPV45 infection is indeed found. In Peralta-Rodriguez et al., 0.82% of HSIL+ women (3 of 364) were HPV45 positive .
Regarding the applied variables on the two cohorts of 91 and 93 included women respectively, the statistical analysis demonstrates that for women from cohort one (TP, 2014) with a HSIL-positive smear the variables hrHPV-positivity and HPV16-positivity are significant for having a histological CIN2+ outcome within one year after cytological testing. On the contrary, HPV67-positivity appears to be a protective factor with an OR of 0.093 and p-value of 0.045. It must be noted that the 95%CI of this low risk-HPV type is very large (0.009-0.947) which implies that this is a coincidental finding. Also, no literature could be found that demonstrates a possible protective role of HPV67-infection for the development of high-grade cervical lesions.
For cohort two (SP, 2010) the only statistically significant variable found, was persistent HPV infection (OR 1.17). As mentioned above, this is congruent with the existing literature [19,20].
In our study 76.7% had histological follow-up within one year. This percentage is lower than the percentages found in the 2018 report of the Flemish government concerning the cervical screening program “Jaarrapport 2018 Bevolkingsonderzoek baarmoederhalskanker” . In this report only 5-10% of HSIL positive women have no histological follow-up within twelve months. However, it must be noted that this report concerns samples dated from 2016, whereas our included samples date from 2010 and 2014. As stated in the report, the follow-up rate has inclined from 73.0% in 2013 to 80.2% in 2016 (when looking at all abnormal cervical cytology results as one group, including HSIL) , which could explain the higher follow-up percentage found in 2016. In our study we found no statistically significant variable that influenced the follow-up rate.
Looking at the relevance of our study, theoretically it contributes to a J Cytol Histol, Volume 11:1, 2020 Karia N, et al.
Page 10 of 11 more up-to-date scientific knowledge in de field of cervical screening in Flanders. Regarding the external validity of this study, we can point out that the found CIN2+% in AML, Antwerp is similar to the percentages found in worldwide laboratories. Thus, AML performs equally as good as other studied laboratories worldwide. Practically, based on our findings we could advice that in HSIL positive women who are HPV16 or hrHPV positive, extra attention should be paid to adequate anamnesis for alarm symptoms and timely histologic follow-up. Although, the results of our study are mostly practically relevant for cytopathologists, not general practitioners, because we only looked at HSIL+ women. Regarding consequences for public health, we should still be aware of the factor of overtreatment and the costs this brings to society.
Conclusion The found CIN2+/HSIL ratio of 75.5% in AML, Antwerp is similar to the percentages found in worldwide laboratories. HSIL positive women who are HPV16+ or hrHPV+ are at significant higher risk for invasive cervical disease. No statistically significant difference in CIN2+% was found between the two LBC techniques TP and SP.
Limitations and Recommendations for Future Studies A limitation of our study is the fact that we solely researched a homogenous group of HSIL positive women. No control group of HSIL negative women was enrolled, thus no sensitivity and specificity of cytology could be calculated. Also, the initial HSIL diagnosis was not reviewed by a second cytopathologist. A third limitation is the retrospective character of our study, as no fixed follow-up period could be set in the initial outline of the study. In a prospective cohort study loss to follow-up including missing values could be reduced.
However, with our study design we mimic a real-life situation yielding realistic data (e.g. non-compliance due to personal reasons), opposed to an artificially set up study. Other strengths of our study are working with the accredited laboratory “AML” with a large dataset (94.697 analysed cervical smears in 2014 and 89.761 in 2010), the comparison between TP and SP and the interpretation of these cohorts by the same pathologist. A last strength is the practice of HPV informed guided screening, a combination of serial co-testing and FPGS [6,8,29].
More grounded clinical relevance will need to be proved in future larger studies. Another suggestion for further research is an adaptation of the study design with implementation of a control group (of HSIL negative women) and a prospective approach.
Acknowledgements This study was initiated by Johannes Bogers, Professor in the Faculty of Cell Biology and Histology at the University of Antwerp and Medical Director of Pathology in AML, Laboratory of Molecular Pathology, Antwerp. We are grateful for the help of Ina Benoy, Medical Director of the Department of Molecular Diagnostics in AML.
Conflict of Interest Authors NK and AVL declare that they have no conflict of interest. Authors IB and JB are both working in a laboratory processing cytological and molecular biology specimens (AML BVBA).
References 1. Cancer in an Ageing Population in Belgium 2004-2016, Belgian Cancer Registry, Brussels, 2018. Available from:
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4. Agramunt, Sílvia, Miguel Ángel Checa, Mireia González and Larrazabal Fernando, et al. “High-grade squamous intraepithelial lesion could be managed conservatively in women up to 25 years: results from a retrospective cohort study.” Journal of Lower Genital Tract Disease 17 (2013):459-462.
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How to cite this article: Nina Karia, Alison Van Loon, Ina Benoy and Johannes Bogers. What is the Retrospective Correlation between High-Grade Squamous Intraepithelial Lesion (HSIL) on Cytology and the Histological Diagnosis of Cervical Intraepithelial Neoplasia 2 (CIN2) or More in AML, Antwerp, Belgium? J Cytol Histol 10 (2020) doi: 10.37421/jch.2020.11.550 |
189735 | https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s11-06-end-of-chapter-material.html | End-of-Chapter Material
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7.6 End-of-Chapter Material
Application Problems
Problems marked with a ♦ involve multiple concepts.
♦ Most plants, animals, and bacteria use oxygen as the terminal oxidant in their respiration process. In a few locations, however, whole ecosystems have developed in the absence of oxygen, containing creatures that can use sulfur compounds instead of oxygen. List the common oxidation states of sulfur, provide an example of any oxoanions containing sulfur in these oxidation states, and name the ions. Which ion would be the best oxidant?
Titanium is currently used in the aircraft industry and is now used in ships, which operate in a highly corrosive environment. Interest in this metal is due to the fact that titanium is strong, light, and corrosion resistant. The densities of selected elements are given in the following table. Why can an element with an even lower density such as calcium not be used to produce an even lighter structural material?
| Element | Density (g/cm 3) | Element | Density (g/cm 3) |
:---: :---: |
| K | 0.865 | Cr | 7.140 |
| Ca | 1.550 | Mn | 7.470 |
| Sc | 2.985 | Fe | 7.874 |
| Ti | 4.507 | Co | 8.900 |
| V | 6.110 | Ni | 8.908 |
3. ♦ The compound Fe 3 O 4 was called lodestone in ancient times because it responds to Earth’s magnetic field and can be used to construct a primitive compass. Today Fe 3 O 4 is commonly called magnetite because it contains both Fe 2+ and Fe 3+, and the unpaired electrons on these ions align to form tiny magnets. How many unpaired electrons does each ion have? Would you expect to observe magnetic behavior in compounds containing Zn 2+? Why or why not? Would you expect Fe or Zn to have the lower third ionization energy? Why?
4. ♦ Understanding trends in periodic properties allows us to predict the properties of individual elements. For example, if we need to know whether francium is a liquid at room temperature (approximately 20°C), we could obtain this information by plotting the melting points of the other alkali metals versus atomic number. Based on the data in the following table, would you predict francium to be a solid, a liquid, or a gas at 20°C?
| | Li | Na | K | Rb | Cs |
:---: :---: :---: |
| Melting Point (°C) | 180 | 97.8 | 63.7 | 39.0 | 28.5 |
Francium is found in minute traces in uranium ores. Is this consistent with your conclusion? Why or why not? Why would francium be found in these ores, but only in small quantities?
Answers
Due to its 3 s 2 3 p 4 electron configuration, sulfur has three common oxidation states: +6, +4, and −2. Examples of each are: −2 oxidation state, the sulfide anion, S 2− or hydrogen sulfide, H 2 S; +4 oxidation state, the sulfite ion, SO 3 2−; +6 oxidation state, the sulfate ion, SO 4 2−. The sulfate ion would be the best biological oxidant, because it can accept the greatest number of electrons.
Iron(II) has four unpaired electrons, and iron(III) has five unpaired electrons. Compounds of Zn 2+ do not exhibit magnetic behavior, because the Zn 2+ ion has no unpaired electrons. The third ionization potential of zinc is larger than that of iron, because removing a third electron from zinc requires breaking into the closed 3 d 10 subshell.
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Understanding Theory Of Congruence and Number Theory
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This is about understanding Congruence, I started with (Elementary Number Theory by David M. Burton), I am studying Chapter 4, Theory Of Congruence and the hard part is understanding the proof of Chinese Remainder Theorem and a Theorem related to solving system of linear congruences.
Further, please suggest If there is a better book for holding grip on solving problems related to Congruences and insightful reading on Number Theory somewhat written in Soviet-Mathematician-style, like no need to have a teacher.
Last, how should one approach Number Theory while learning it like Solving Problems, Understanding Proofs and Writing Proofs and Applying it in real life like Cryptography, programming problems.
Please help.
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edited May 5, 2020 at 12:35
Shantanu Sharma
asked May 5, 2020 at 11:36
Shantanu SharmaShantanu Sharma
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There are many good book recommendations for self learning number theory on this site, e.g., here. The best method is to do examples and counterexamples. Then you can make every definition and every statement explicit, until you understand it.
Dietrich Burde
–
Dietrich Burde
05/05/2020 11:38:56
Commented
May 5, 2020 at 11:38
@DietrichBurde need something more subtle readings.
Shantanu Sharma
–
Shantanu Sharma
05/05/2020 12:00:11
Commented
May 5, 2020 at 12:00
@DietrichBurde Already mentioned that something in soviet style. I know Mathstackexcahnge is great. Those are good recommendations which you shared but I am looking for something different.
Shantanu Sharma
–
Shantanu Sharma
05/05/2020 12:16:04
Commented
May 5, 2020 at 12:16
1
Are you at (or near) a university? Is there a faculty member you could discuss this with? you'll get better advice from someone you can sit down and talk to, than you will from random people on a website.
Gerry Myerson
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Gerry Myerson
05/05/2020 12:36:27
Commented
May 5, 2020 at 12:36
You need to be more specific about precisely what you have difficulty understanding in order for us to best help you. Section 4.4 contains a common (unmotivated) proof of CRT. See here for a motivated conceptual derivation of that CRT formula. The section ends with a theorem showing how to solve a system of two equations in two variables modn by elimination (or, equivalently, by Cramer's rule, e.g. see here and here).
Bill Dubuque
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Bill Dubuque
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I read a lot of Joe Silverman's A Friendly Introduction to Number Theory. He actually says that he intended it for non-math majors. But don't let that deter you.
It's a fun book, well-written, thorough, covering lots of the standard number theory topics.
It's very accessible.
The section on the Chinese remainder theorem is well done. And it's noted that this implies Euler's phi function is multiplicative.
Let's see. Twin primes, Goldbach's conjecture, Fermat primes, Mersenne primes, Fibonacci numbers, big and little oh, cryptography, primality testing, Fermat and Euler's theorems, Carmichael numbers, congruence and modular arithmetic. Not in that order of course. The list goes on. Oh, and, big oversight on my part, the prime number theorem.
The last part covers Fermat's Last Theorem via elliptic curves and other trickery.
I have also heard good things about Baker's book.
And Davenport's The Higher Arithmetic sounds good.
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189737 | https://www.uptodate.com/contents/role-of-echocardiography-in-atrial-fibrillation | Role of echocardiography in atrial fibrillation - UpToDate
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Role of echocardiography in atrial fibrillation
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SUMMARY AND RECOMMENDATIONS
INTRODUCTION
OBJECTIVES
INDICATIONS
TRANSTHORACIC ECHOCARDIOGRAPHY
Left atrial size
Mitral valve function
Left ventricular function
TRANSESOPHAGEAL ECHOCARDIOGRAPHY
Left atrial thrombi
Spontaneous echocardiographic contrast
Blood flow velocity
Right atrial thrombi
SOCIETY GUIDELINE LINKS
SUMMARY AND RECOMMENDATIONS
REFERENCES
GRAPHICS
Movies
- TEE LA appendage thrombus
- TEE left atrial appendage thrombus and spontaneous echo contrast
- TEE LA thrombus and spontaneous echo contrast
- TEE LA and prosthetic MV thrombus
Figures
- LV function stroke risk in AF
- TEE abnormalities stroke risk
Diagnostic Images
- Apical 4 chamber echo right atrial thrombus
RELATED TOPICS
Antiarrhythmic drugs to maintain sinus rhythm in patients with atrial fibrillation: Recommendations
Atrial fibrillation in adults: Selection of candidates for long-term anticoagulation
Atrial fibrillation in adults: Use of oral anticoagulants
Atrial fibrillation: Cardioversion
Atrial fibrillation: Left atrial appendage occlusion
Atrial fibrillation: Overview and management of new-onset atrial fibrillation
Control of ventricular rate in patients with atrial fibrillation who do not have heart failure: Pharmacologic therapy
Echocardiographic evaluation of the atria and appendages
Epidemiology, risk factors, and prevention of atrial fibrillation
Mechanisms of thrombogenesis in atrial fibrillation
Medical management of asymptomatic aortic stenosis in adults
Mitral stenosis: Surgical interventions and investigational approaches in adults
Percutaneous mitral balloon commissurotomy in adults
Prevention of embolization prior to and after restoration of sinus rhythm in atrial fibrillation
Society guideline links: Atrial fibrillation
Role of echocardiography in atrial fibrillation
Author:Warren J Manning, MDSection Editor:Bradley P Knight, MD, FACCDeputy Editors:Han Li, MDSusan B Yeon, MD, JD
Literature review current through:Aug 2025.
This topic last updated:Nov 22, 2024.
INTRODUCTION
Atrial fibrillation (AF) is the most common treated arrhythmia. Echocardiography plays a key role in evaluation and management of patients with AF.
The topic will review the use of echocardiography in evaluating patients with AF. An overview of AF is presented separately. (See "Atrial fibrillation: Overview and management of new-onset atrial fibrillation".)
OBJECTIVES
The role of echocardiographic imaging among patients with AF can be divided into two main categories:
●Assessment of cardiac chamber sizes and function, the atrial contribution to left ventricular filling, the pericardium, and valvular function. This information may be helpful in determining the conditions associated with AF, the risk for recurrent AF following cardioversion, and the hemodynamic benefit of maintaining sinus rhythm. This information is generally obtained from transthoracic echocardiography (TTE), with moderately invasive transesophageal echocardiography (TEE) generally reserved for assessment of the left atrial appendage for thrombus prior to cardioversion. (See "Epidemiology, risk factors, and prevention of atrial fibrillation" and "Antiarrhythmic drugs to maintain sinus rhythm in patients with atrial fibrillation: Recommendations".)
●Identification of patients at increased risk for thromboembolic complications of AF before cardioversion and in patients with chronic AF. (See "Prevention of embolization prior to and after restoration of sinus rhythm in atrial fibrillation" and "Atrial fibrillation in adults: Selection of candidates for long-term anticoagulation".)
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Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. It does NOT include all information about conditions, treatments, medications, side effects, or risks that may apply to a specific patient. It is not intended to be medical advice or a substitute for the medical advice, diagnosis, or treatment of a health care provider based on the health care provider's examination and assessment of a patient's specific and unique circumstances. Patients must speak with a health care provider for complete information about their health, medical questions, and treatment options, including any risks or benefits regarding use of medications. This information does not endorse any treatments or medications as safe, effective, or approved for treating a specific patient. UpToDate, Inc. and its affiliates disclaim any warranty or liability relating to this information or the use thereof. The use of this information is governed by the Terms of Use, available at 2025© UpToDate, Inc. and its affiliates and/or licensors. All rights reserved.
Topic Feedback
Movies
Left atrial appendage thrombus seen on TEETransesophageal echocardiography (TEE) showing left atrial appendage thrombus and spontaneous echo contrastTransesophageal echocardiogram (TEE) showing left atrial (LA) thrombus and spontaneous echo contrastLeft atrial thrombus and thrombus adjacent to a prosthetic mitral valve seen on TEE
Left atrial appendage thrombus seen on TEETransesophageal echocardiography (TEE) showing left atrial appendage thrombus and spontaneous echo contrastTransesophageal echocardiogram (TEE) showing left atrial (LA) thrombus and spontaneous echo contrastLeft atrial thrombus and thrombus adjacent to a prosthetic mitral valve seen on TEE
Figures
Significant left ventricular dysfunction predicts stroke in AFLeft atrial abnormalities and complex aortic plaque correlate with the risk of thromboembolism in atrial fibrillation
Significant left ventricular dysfunction predicts stroke in AFLeft atrial abnormalities and complex aortic plaque correlate with the risk of thromboembolism in atrial fibrillation
Diagnostic Images
Apical 4 chamber echocardiogram showing right atrial thrombus
Apical 4 chamber echocardiogram showing right atrial thrombus
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189738 | https://www.quora.com/Why-is-the-vector-cross-product-equal-to-zero-whenever-the-two-vectors-are-perpendicular | Something went wrong. Wait a moment and try again.
Cross Product
Perpendicular Vectors
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5
Why is the vector cross product equal to zero whenever the two vectors are perpendicular?
Alexander Farrugia
frequent user of linear algebra.
·
Upvoted by
Max Gretinski
, PhD Mathematics · Author has 3.2K answers and 27.5M answer views
·
6y
It’s the
scalar
, or
dot
, product of vectors that is equal to zero whenever the two vectors are perpendicular. The vector, or cross, product does not have this property. Indeed, the vector product is equal to zero (to be precise, to the zero vector) if and only if the two vectors are parallel.
Alexandru Carausu
Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010)
·
Author has 3K answers and 875.1K answer views
·
Feb 13
Another (the fifth) wrong / improper question received in the last two days !
If the cross product of two free vectors is u x v = 0 , then (at least) one of the two vectors is = 0 , or they are collinear, but not perpendicular / orthogonal.
The magnitude (or length) of u x v is
| u x v | = u v sin θ where u = | u | , v = | v | & θ = ∢ ( u , v ) ∈ [ 0 , π ] . (1)
A (free) vector v is = 0 <==> | v | = 0 . Thus, it follows from (1) that
u x v = 0 <==> | u x v | = 0 <==> sin θ = 0 <==> θ ∈ { 0 , π } <==> u || v . (2)
Final Comment. The sender of this question had the possibility to find the correct answ
Another (the fifth) wrong / improper question received in the last two days !
If the cross product of two free vectors is u x v = 0 , then (at least) one of the two vectors is = 0 , or they are collinear, but not perpendicular / orthogonal.
The magnitude (or length) of u x v is
| u x v | = u v sin θ where u = | u | , v = | v | & θ = ∢ ( u , v ) ∈ [ 0 , π ] . (1)
A (free) vector v is = 0 <==> | v | = 0 . Thus, it follows from (1) that
u x v = 0 <==> | u x v | = 0 <==> sin θ = 0 <==> θ ∈ { 0 , π } <==> u || v . (2)
Final Comment. The sender of this question had the possibility to find the correct answer from the chapter of Vector Algebra which (usually) opens any textbook of ANALYTIC GEOMETRY. He/she would have learned that
u ⊥ v if u • v = 0 because u • v = u v cos θ with θ as in (1) ==> θ = π/2 . (3)
The product-type operation in (3) is just called the dot product of the two vectors ; the trivial case when u v = 0 ==> u = 0 and/or v = 0 is excluded since the zero vectors don’t have any direction.
Md. Sadman Sakib
Science Student
·
Author has 162 answers and 198.2K answer views
·
4y
Whenever two vectors are perpendicular,
the angle between these vector in 90 degree
Suppose , A & B are two vectors.
The dot product of two vectors is a scalar which’s value is the multiplication of the two vectors’ value and the cosine of the angle between two vectors.
A.B = A×B×Cos90 =
:> A.B = A×B×0
:> A.B = 0
The cross product of two vectors that are perpendicular , means cosine90 is the minimum answer which is zero.
Now,
The cross product of two vectors is a vector which’s value is the multiplication of the values and the sine of the angle between two vectors. and its direction is ( where η is a u
Whenever two vectors are perpendicular,
the angle between these vector in 90 degree
Suppose , A & B are two vectors.
The dot product of two vectors is a scalar which’s value is the multiplication of the two vectors’ value and the cosine of the angle between two vectors.
A.B = A×B×Cos90 =
:> A.B = A×B×0
:> A.B = 0
The cross product of two vectors that are perpendicular , means cosine90 is the minimum answer which is zero.
Now,
The cross product of two vectors is a vector which’s value is the multiplication of the values and the sine of the angle between two vectors. and its direction is ( where η is a unit vector which is perpendicular to the dimension of the two vector)
So,
A×B= A×B×sin90×η
:> A×B= A×B×1×η
:> A×B= AB ×η
( where η is a unit vector which is perpendicular to the dimension of A& B )
So,
If Angle between A & B is 90 , so from cross product, sine 90 is the maximum value, so in 90 degree , the value of cross product is maximum.
Related questions
Why is the cross product of two vectors perpendicular to the plane they reside in?
What is the cross product if two vectors, A and B, are perpendicular to each other?
The resultant of two vectors is perpendicular to one of the vectors. What is the angle between the two vectors?
How does a cross product of two vectors give a perpendicular vector?
Why does the cross product of two vectors always result in a zero vector when the vectors are subtracted from one another?
Deb P. Choudhury
Former Professor at University of Allahabad
·
Author has 10K answers and 8M answer views
·
2y
Originally Answered: Why does the cross product equal to zero if both vectors are parallel?
·
For the same reason as sin(0) = 0, or a parallelogram whose adjacent sides are along the same line, has zero area.
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1y
Actually, the statement that the vector cross product is equal to zero whenever the two vectors are perpendicular is incorrect. In fact, the cross product of two vectors is non-zero when the vectors are perpendicular.
Here's a brief explanation:
Definition of the Cross Product: The cross product of two vectors A and B is given by:
A×B=|A||B|sin(θ)n
where:
|A| and |B| are the magnitudes of the vectors,
θ is the angle between the two vectors,
n is a unit vector perpendicular to t
Actually, the statement that the vector cross product is equal to zero whenever the two vectors are perpendicular is incorrect. In fact, the cross product of two vectors is non-zero when the vectors are perpendicular.
Here's a brief explanation:
Definition of the Cross Product: The cross product of two vectors A and B is given by:
A×B=|A||B|sin(θ)n
where:
|A| and |B| are the magnitudes of the vectors,
θ is the angle between the two vectors,
n is a unit vector perpendicular to the plane formed by A and B.
Perpendicular Vectors: When two vectors are perpendicular, the angle θ between them is 90∘ (or π2 radians). The sine of 90∘ is 1, so the formula becomes:
A×B=|A||B|⋅1⋅n=|A||B|n
This result is a vector with a magnitude equal to the product of the magnitudes of A and B, and it points in the direction given by the right-hand rule.
When is the Cross Product Zero?: The cross product is equal to zero when:
The vectors are parallel (i.e., θ=0∘ or θ=180∘), where sin(0)=0 and sin(180)=0.
At least one of the vectors is the zero vector.
In summary, the cross product of two perpendicular vectors is not zero; rather, it has a magnitude equal to the product of their magnitudes and is directed perpendicular to both vectors.
Igor
PhD in Mechanical Engineering, Technion - Israel Institute of Technology (Graduated 1997)
·
Author has 3.1K answers and 1.3M answer views
·
4y
Originally Answered: What if two vectors are orthogonal and cross product is zero?
·
This means that the length of one of them (or even both of them) is zero. In this case, however, we can hardly talk about the orthogonality
Related questions
What is the resultant vector when two perpendicular vectors are added together? Is it equal to zero or not?
Why cross product is zero if two vectors are in same direction?
If two vectors are perpendicular, what is their cross product?
If the magnitude of the cross product of three vectors is 0 are they perpendicular?
Why is the cross product of two vectors perpendicular?
Karan Hemdev
I believe imagination is more than knowledge...
·
Author has 51 answers and 217.1K answer views
·
9y
Related
Why cross product is zero if two vectors are in same direction?
This question goes in the core of Cross product and I have to define cross product to answer this
Cross product of two vectors is magnitudewise equal to the product of their magnitude with their sine of that angle between them.
By this defination
R = A X B
R = ABSin(Φ)
it's clear that if A and B are parallel then Φ =0
so,
R = 0
This is just a defination and it's corresponding answer but it's I find its very theoretical..
Here is it's physical interpretation….
If You have two vectors then they can be assumed as the adjacent sides of an parallelogram then their Cross product gives you the area of that p
This question goes in the core of Cross product and I have to define cross product to answer this
Cross product of two vectors is magnitudewise equal to the product of their magnitude with their sine of that angle between them.
By this defination
R = A X B
R = ABSin(Φ)
it's clear that if A and B are parallel then Φ =0
so,
R = 0
This is just a defination and it's corresponding answer but it's I find its very theoretical..
Here is it's physical interpretation….
If You have two vectors then they can be assumed as the adjacent sides of an parallelogram then their Cross product gives you the area of that parallelogram
Now Your question is just a logical one…
If You have two vectors which are parallel or in one line so.... what's the area of parallelogram?
it's simply zerooo since there's no height of that parallelogram….
I hope uh got this one..
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Charmi Jain
BFM from K.P.B. Hinduja College of Commerce (Graduated 2021)
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8y
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Why is the product of perpendicular vectors 0?
It is because:
Dot product of two vectors is :a.b = abcosø
And angle between perpendicular vectors is 90°
Hence , when we substitute ø =90°
a.b becomes zero. [ cos90°=0]
Hence, dot product of perpendicular vectors is zero.
Drake Way
Advocate of science and critical thinking.
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Author has 4.2K answers and 3.3M answer views
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6y
It isn't for two reasons.
The cross product is a vector. It wouldn't be equal to 0, it would be equal to
→
0
, the zero vector.
→
a
×
→
b
=
|
→
a
|
|
→
b
|
(
sin
θ
)
^
n
;
sin
θ
is maximized when the vectors are perpendicular.
It's the dot-product that is 0 for perpendicular vectors.
Sponsored by Grammarly
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Work faster with AI, while ensuring your writing always makes the right impression.
Lionel Brits
Minor in math, graduate studies in physics.
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Author has 97 answers and 187.4K answer views
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5y
Related
Why is the cross product of two vectors perpendicular?
This is by design. Suppose you wanted to create a linear operation to test whether two (or more) vectors are linearly dependent. Why linear? Glad you asked. Let's move along. So let's call this operation the wedge, or ^ for short. Given vectors A and B, we would like this operation to fail (return zero) if A is proportional to B, and by linearity this demands that we define our operation via the following rule:
A ^ A = 0 for every A
This immediately gives us A ^ B = - B ^ A, which you can prove by applying the previous rule to A - B.
Now this operation we have created is nice, and we can extend i
This is by design. Suppose you wanted to create a linear operation to test whether two (or more) vectors are linearly dependent. Why linear? Glad you asked. Let's move along. So let's call this operation the wedge, or ^ for short. Given vectors A and B, we would like this operation to fail (return zero) if A is proportional to B, and by linearity this demands that we define our operation via the following rule:
A ^ A = 0 for every A
This immediately gives us A ^ B = - B ^ A, which you can prove by applying the previous rule to A - B.
Now this operation we have created is nice, and we can extend it to more than two vectors. Let's see what happens if we write A ^ B ^ C, assuming that such an operation can be defined. At the very least, we'll assume associativity because why not? So we define A ^ B ^ C = A ^ (B ^ C) which by associativity also equals (A ^ B) ^ C.
Note we have yet to figure out how to calculate with ^, we are only defining its properties according to our needs. Let's check if it fits the bill. Obviously if A and B are linearly dependent, then A ^ B ^ C = 0, because associativity and linearity. But what about A and C? Observe:
A ^ B ^ C = - A ^ (C ^ B) = - (A ^ C) ^ B = 0.
So our operation faithfully tells us if any pair of vectors in a set are linearly dependent, and does so for any number of vectors (though in 3 dimensions you'll get zero if you try to wedge more than 3 vectors, since the 4th vector must be a linear combination of the other 3).
Ok, so how do we actually compute with the wedge? Easy. Knowing it's action on any orthogonal basis, we can figure the rest out by linearity. So given A and B in terms of basis vectors e1, e2, e3, we can express A ^ B in terms of some linear combination of e1^e2, e2^e3, and e1^e3. We don't actually need to know what those objects are, only that they aren't equal. Furthermore, we can show that they also form a basis of a 3 dimensional vector space. Similarly e1^e2^e^3 forms a basis for a vector space of dimension one.
Congratulations, you've just discovered the determinant!
Ok, so before the sheer power of this new operation goes to your head, you notice that the vector space defined by e1^e2, e2^e3, and e1^e3 is the same dimensionality that you started with. Maybe we can identify elements between these two vector spaces. The most natural way of doing this actually turns out to be by recognizing that since A^B^C is a one-dimensional vector space, we can trivially identify e1^e2^e3 with the number 1 (that is, they are dual to one another). But this immediately turns the operation A^B^C into an inner product (dot product) between vectors in the (e1^e2, e2^e3, e1^e3) vector space and vectors in the (e1, e2, e3) vector space. That is, it takes one of each type, and returns a real number, satisfying all the usual properties of an inner product.
So we define < A^B, C > = dual(A^B^C).
This immediately requires that, for example, = 1, so it is natural to define dual(e1^e2) = e3.
So not only do we have a powerful new operation, a determinant, a new inner product and an associated duality, but we've got yet another new object:
dual ( A ^ B )
defines a linear operation that maps from VxV to V, that is, it takes two vectors and produces another vector, which, as we've seen, is orthogonal to the first two. Why? Because
dual ( A ^ B ) dot C = dual(A^B^C)
which you can see by expanding A^B and X in terms of a basis. This new object, dual ( A ^ B ), is called the cross product, and the fact that it is a 3 dimensional vector is just a happenstance. In 4 dimensions, A^B defines a 6-dimensional vector space!
Khenan Mak
Former Lecturer/Researcher at MMU Cyberjaya (2004–2020)
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Author has 2.6K answers and 1.2M answer views
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4y
It is the dot, or scalar, product that has that property. When the vectors are perpendicular, the magnitude of the cross product is at its maximum.
Stephen Miller
PhD in algebraic topology
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 380 answers and 2.1M answer views
·
6y
Related
Two vectors that are orthogonal have a dot product of zero. Does this mean two parallel vectors will always have a cross product of zero?
Yes and no. You seem to have two questions here:
Will two parallel vectors always have a cross product of zero? Yes.
Does this follow from the fact that two orthogonal vectors have a dot product of zero? Not really. There may be a convoluted way to deduce one fact from the other using relationships between dot products and vector products, but the fact that two parallel vectors have a vector product of zero follows from basic properties of vector products. Namely:
Vector product commutes with scalar multiplication i.e. (aU) x V = a (U x V) = U x (aV).
Vector product is anticommutative, i.e. U x V
Yes and no. You seem to have two questions here:
Will two parallel vectors always have a cross product of zero? Yes.
Does this follow from the fact that two orthogonal vectors have a dot product of zero? Not really. There may be a convoluted way to deduce one fact from the other using relationships between dot products and vector products, but the fact that two parallel vectors have a vector product of zero follows from basic properties of vector products. Namely:
Vector product commutes with scalar multiplication i.e. (aU) x V = a (U x V) = U x (aV).
Vector product is anticommutative, i.e. U x V = - V x U.
Applying those two rules to parallel vectors gives you the result.
Inayat Ullah
Studied Mathematics (Graduated 1997)
·
Author has 129 answers and 496.3K answer views
·
5y
Related
Why is the cross product of two vectors perpendicular?
Let us do it in a simple manner. Take two three dimensional vectors →A=a^i+b^j+c^k and →B=e^i+f^j+g^k
As we know that two vectors →p and →q are perpendicular if and only →p.→q=0
we shall use his fact to show that the cross product of two vectors →A and →Bis perpendicular to both vectors , precisely →A×→Bis perpendicular to he plane containing both→A and →B.
So we first find →A×→Band then take dot product of→A×→Bwith →A and →B.
→A×→B=\b
Let us do it in a simple manner. Take two three dimensional vectors →A=a^i+b^j+c^k and →B=e^i+f^j+g^k
As we know that two vectors →p and →q are perpendicular if and only →p.→q=0
we shall use his fact to show that the cross product of two vectors →A and →Bis perpendicular to both vectors , precisely →A×→Bis perpendicular to he plane containing both→A and →B.
So we first find →A×→Band then take dot product of→A×→Bwith →A and →B.
Now we take dot product of with
So is perpendicular to . likewise we can show that is perpendicular to
You can also recall to right hand rule to examine
Related questions
Why is the cross product of two vectors perpendicular to the plane they reside in?
What is the cross product if two vectors, A and B, are perpendicular to each other?
The resultant of two vectors is perpendicular to one of the vectors. What is the angle between the two vectors?
How does a cross product of two vectors give a perpendicular vector?
Why does the cross product of two vectors always result in a zero vector when the vectors are subtracted from one another?
What is the resultant vector when two perpendicular vectors are added together? Is it equal to zero or not?
Why cross product is zero if two vectors are in same direction?
If two vectors are perpendicular, what is their cross product?
If the magnitude of the cross product of three vectors is 0 are they perpendicular?
Why is the cross product of two vectors perpendicular?
Why is the resultant vector zero when two vectors have equal magnitude but opposite directions?
What is the dot product and why does it equal to zero if two vectors are orthogonal (perpendicular)?
Why is the product of perpendicular vectors 0?
Why does the cross product equal to zero if both vectors are parallel?
Why is the unit vector of the product in cross product perpendicular to the plane where two vectors are present?
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189739 | https://dspace.mit.edu/bitstream/handle/1721.1/104426/6-042j-spring-2010/contents/lecture-notes/MIT6_042JS10_lec31.pdf | 1 Albert R Meyer, April 23, 2010 lec 11F.1 Mathematics for Computer Science MIT 6.042J/18.062J Generating Functions Albert R Meyer, April 23, 2010 Infinite Geometric Sum S(x) ::= 1+ x + x2 ++ xn + lec 11F.2 xS(x) = x + x2 ++ xn + Albert R Meyer, April 23, 2010 Infinite Geometric Sum S(x) ::= 1+ x + x2 ++ xn + S(x)xS(x) = lec 11F.3 1 xS(x) = x + x2 ++ xn + Albert R Meyer, April 23, 2010 Ordinary Generating Functions The ordinary generating function for the infinite sequence g0, g1, g2, , gn, is the power series: G(x) = g0+g1x+g2x2++gnxn+ lec 11F.5 gn, gn Albert R Meyer, April 23, 2010 lec 11F.6 “corresponds to” Infinite Geometric Sum = 1 1 x Albert R Meyer, April 23, 2010 Bags of fruit lec 11F.8 In how many ways can we fill a bag with n fruits given the following constraints?
•At most 2 oranges.
•Any number of apples.
•Any number of bananas that only come in bunches of 3. 2 Albert R Meyer, April 23, 2010 Bags with n = 4 fruits •0 oranges, 1 apple, 3 bananas •0 oranges, 4 apples, 0 bananas •1 orange, 0 apples, 3 bananas •1 orange, 3 apples, 0 bananas •2 oranges, 2 apples, 0 bananas lec 11F.9 Number of 4-fruit bags: 5 Albert R Meyer, April 23, 2010 At most 2 oranges lec 11F.10 # ways to pick k oranges Albert R Meyer, April 23, 2010 There is only 1 way to pick a bag of k apples: ak = 1 Any number of apples lec 11F.11 Albert R Meyer, April 23, 2010 Substituting xk for x lec 11F.12 Albert R Meyer, April 23, 2010 Bananas in bunches of 3 lec 11F.13 Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.14 We can use the individual generating functions to solve original fruit problem! 3 Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.15 Ways to pick 12 apples & bananas: • 0 apples and 12 bananas • 1 apple and 11 bananas # ways • 12 apples and 0 bananas Total=5 1 0 1 Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.16 = # ways to pick k bananas bk = # ways to pick j apples aj = # ways to pick j apples and rest bananas ajb12-j Ways to pick 12 apples & bananas: Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.17 # ways to pick 12 apples & bananas = a 0b12 + a1b11 +…+ a11b1 + a12b0 But this is the coefficient of x12 in A(x)·B(x) Albert R Meyer, April 23, 2010 lec 11F.18 Convolution Rule The coefficient of x12 in the product A(x)·B(x): Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.19 # ways to pick 12 apples & bananas is the coefficient of x12 in A(x)·B(x) the generating function for picking apples & bananas Albert R Meyer, April 23, 2010 Convolution Rule lec 11F.20 The gen func for choosing from a union of disjoint sets is the product of the gen funcs for choosing from each set. 4 Albert R Meyer, April 23, 2010 Bags of Fruit lec 11F.21 Gen func for the bags of fruit: = 1 1 x ( ) 2 Albert R Meyer, April 23, 2010 A Familiar Generating Function? lec 11F.22 1 / 1 x ( ) 2 so # of our bags with n fruits is the coefficient of xn in We can easily relate 1/(1-x)2 to something we already know how to count! Albert R Meyer, April 23, 2010 A Familiar Generating Function? lec 11F.24 The gen func for selecting n donuts of a given flavor: Albert R Meyer, April 23, 2010 A Familiar Generating Function? lec 11F.25 The gen func for selecting n donuts using both flavors: Albert R Meyer, April 23, 2010 A Familiar Generating Function? lec 11F.26 The gen func for selecting n donuts among k flavors: Albert R Meyer, April 23, 2010 The Donut Number! lec 11F.27 Using k different flavors, how many ways are there to form a bag of n donuts? (You already know the answer to this one.) 5 Albert R Meyer, April 23, 2010 The Donut Number! lec 11F.28 Using k different flavors, how many ways are there to form a bag of n donuts? n + k 1 n Albert R Meyer, April 23, 2010 The Donut Number! lec 11F.29 n + k 1 n so coeff of xn in Albert R Meyer, April 23, 2010 Conclusion: Bags of Fruit lec 11F.30 In how many ways can we fill a bag with n of our fruits?
F(x) = 1 1 x ( ) 2 fn = n + 2 1 n = n + 1 Albert R Meyer, April 23, 2010 Finding coefficients lec 11F.34 If a generating function H(x) is a rational function there is a simple way to find the nth coefficient hn quotient of polynomials Albert R Meyer, April 23, 2010 Partial Fraction Expansions H(x) ::= x 2x2 3x + 1 Express as sum Factor denominator lec 11F.35 Albert R Meyer, April 23, 2010 Partial Fractions for H(x) H(x) = A 1 1 1 2x + A 2 1 1 x h n = A 1 2n + A 2 1 TO DO: find A1 and A2.
lec 11F.36 6 Albert R Meyer, April 23, 2010 Solve for A1 and A2 Multiply both sides by denom of LHS.
lec 11F.37 Albert R Meyer, April 23, 2010 Solve for A1 and A2 Substitute in values for x. lec 11F.38 Albert R Meyer, April 23, 2010 Finding the coefficients the partial fraction expansion lec 11F.39 = 1 1 2x 1 1 x hn = 2n 1 Albert R Meyer, April 23, 2010 In General… The partial fraction expansion of P(x)/Q(x) contains terms of the form …+ A 1 x ( ) k +… where 1/ is a root of Q(x).
We know the nth coeff of this! lec 11F.40 A n + k -1 n n Albert R Meyer, April 23, 2010 Partial Fractions Caveat #1 For roots with multiplicity k>1 in factored denominator of gen func 1 x ( ) k need k partial fractions: A 1 1 x ( ) 1 + A 2 1 x ( ) 2 + + A k 1 x ( ) k + lec 11F.41 Albert R Meyer, April 23, 2010 Partial Fractions Caveat #2 use polynomial long division to find Q(x) and R(x) such that If deg(N) > deg(D)… and deg(R) < deg(D).
lec 11F.42 7 Albert R Meyer, April 23, 2010 Team Problems Problems 1 & 2 lec 11F.44 MIT OpenCourseWare 6.042J / 18.062J Mathematics for Computer Science Spring 2010 For information about citing these materials or our Terms of Use, visit: |
189740 | https://elsevier-elibrary.com/contents/fullcontent/58090/epubcontent_v2/OEBPS/B9780443069529000072.htm | Gray’s Anatomy for Students
Page 54
2 Back
Conceptual overview 56
General description 56
Functions 57
Support 57
Movement 57
Protection of the nervous system 58
Component parts 58
Bones 58
Muscles 60
Vertebral canal 62
Spinal nerves 63
Relationship to other regions 64
Head 64
Thorax, abdomen, and pelvis 65
Limbs 65
Key features 65
Long vertebral column and short spinal cord 65
Intervertebral foramina and spinal nerves 66
Innervation of the back 66
Regional anatomy 67
Skeletal framework 67
Vertebrae 67
Intervertebral foramina 75
Posterior spaces between vertebral arches 75
Joints 79
Joints between vertebrae in the back 79
Ligaments 82
Anterior and posterior longitudinal ligaments 82
Ligamenta flava 82
Supraspinous ligament and ligamentum nuchae 83
Interspinous ligaments 84
Back musculature 86
Superficial group of back muscles 86
Intermediate group of back muscles 92
Deep group of back muscles 93
Suboccipital muscles 99
Spinal cord 101
Vasculature 102
Meninges 104
Arrangement of structures in the vertebral canal 106
Spinal nerves 107
Surface anatomy 112
Back surface anatomy 112
Absence of lateral curvatures 112
Primary and secondary curvatures in the sagittal plane 112
Useful nonvertebral skeletal landmarks 112
How to identify specific vertebral spinous processes 114
Visualizing the inferior ends of the spinal cord and subarachnoid space 115
Identifying major muscles 116
Clinical cases 118
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Page 56
Conceptual overview
GENERAL DESCRIPTION
The back consists of the posterior aspect of the body and provides the musculoskeletal axis of support for the trunk. Bony elements consist mainly of the vertebrae, although proximal elements of the ribs, superior aspects of the pelvic bones, and posterior basal regions of the skull contribute to the back’s skeletal framework (Fig. 2.1).
Fig. 2.1Skeletal framework of the back.
Associated muscles interconnect the vertebrae and ribs with each other and with the pelvis and skull. The back contains the spinal cord and proximal parts of the spinal nerves, which send and receive information to and from most of the body.
Page 57
FUNCTIONS
Support
The skeletal and muscular elements of the back support the body’s weight, transmit forces through the pelvis to the lower limbs, carry and position the head, and brace and help maneuver the upper limbs. The vertebral column is positioned posteriorly in the body at the midline. When viewed laterally, it has a number of curvatures (Fig. 2.2):
the primary curvature of the vertebral column is concave anteriorly, reflecting the original shape of the embryo, and is retained in the thoracic and sacral regions in adults;
secondary curvatures, which are concave posteriorly, form in the cervical and lumbar regions and bring the center of gravity into a vertical line, which allows the body’s weight to be balanced on the vertebral column in a way that expends the least amount of muscular energy to maintain an upright bipedal stance.
Fig. 2.2Curvatures of the vertebral column.
As stresses on the back increase from the cervical to lumbar regions, lower back problems are common.
Movement
Muscles of the back consist of extrinsic and intrinsic groups:
the extrinsic muscles of the back move the upper limbs and the ribs;
the intrinsic muscles of the back maintain posture and move the vertebral column; these movements include flexion (anterior bending), extension, lateral flexion, and rotation (Fig. 2.3).
Fig. 2.3Back movements.
Although the amount of movement between any two vertebrae is limited, the effects between vertebrae are additive along the length of the vertebral column. Also, freedom of movement and extension are limited in the thoracic region relative to the lumbar part of the vertebral column. Muscles in more anterior regions flex the vertebral column.
In the cervical region, the first two vertebrae and associated muscles are specifically modified to support and position the head. The head flexes and extends, in the nodding motion, on vertebra CI, and rotation of the head occurs as vertebra CI moves on vertebra CII (Fig. 2.3).
Page 58
Protection of the nervous system
The vertebral column and associated soft tissues of the back contain the spinal cord and proximal parts of the spinal nerves (Fig. 2.4). The more distal parts of the spinal nerves pass into all other regions of the body, including certain regions of the head.
Fig. 2.4Nervous system.
COMPONENT PARTS
Bones
The major bones of the back are the 33 vertebrae (Fig. 2.5). The number and specific characteristics of the vertebrae vary depending on the body region with which they are associated. There are seven cervical, twelve thoracic, five lumbar, five sacral, and three to four coccygeal vertebrae. The sacral vertebrae fuse into a single bony element, the sacrum. The coccygeal vertebrae are rudimentary in structure, vary in number from three to four, and often fuse into a single coccyx.
Fig. 2.5Vertebrae.
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Page 60
A typical vertebra
A typical vertebra consists of a vertebral body and a vertebral arch (Fig. 2.6).
Fig. 2.6A typical vertebra. A. Superior view. B. Lateral view.
The vertebral body is anterior and is the major weightbearing component of the bone. It increases in size from vertebra CI to vertebra LV. Fibrocartilaginous intervertebral discs separate the vertebral bodies of adjacent vertebrae.
The vertebral arch is firmly anchored to the posterior surface of the vertebral body by two pedicles, which form the lateral pillars of the vertebral arch. The roof of the vertebral arch is formed by right and left laminae, which fuse at the midline.
The vertebral arches of the vertebrae are aligned to form the lateral and posterior walls of the vertebral canal, which extends from the first cervical vertebra (CI) to the last sacral vertebra (sacral vertebra V). This bony canal contains the spinal cord and its protective membranes, together with blood vessels, connective tissue, fat, and proximal parts of spinal nerves.
The vertebral arch of a typical vertebra has a number of characteristic projections, which serve as:
attachments for muscles and ligaments;
levers for the action of muscles; and
sites of articulation with adjacent vertebrae.
A spinous process projects posteriorly and generally inferiorly from the roof of the vertebral arch.
On each side of the vertebral arch, a transverse process extends laterally from the region where a lamina meets a pedicle. From the same region, a superior articular process and an inferior articular process articulate with similar processes on adjacent vertebrae.
Each vertebra also contains rib elements. In the thorax, these elements are large and form ribs, which articulate with the vertebral bodies and transverse processes. In all other regions, these rib elements are small and are incorporated into the transverse processes. Occasionally, they develop into ribs in regions other than the thorax, usually in the lower cervical and upper lumbar regions.
Muscles
Muscles in the back can be classified as extrinsic or intrinsic based on their embryological origin and type of innervation (Fig. 2.7).
Fig. 2.7Back muscles. A. Extrinsic muscles. B. Intrinsic muscles.
The extrinsic muscles are involved with movements of the upper limbs and thoracic wall and, in general, are innervated by anterior rami of spinal nerves. The superficial group of these muscles is related to the upper limbs, while the intermediate layer of muscles is associated with the thoracic wall.
All of the intrinsic muscles of the back are deep in position and are innervated by the posterior rami of spinal nerves. They support and move the vertebral column and participate in moving the head. One group of intrinsic muscles also moves the ribs relative to the vertebral column.
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Vertebral canal
The spinal cord lies within a bony canal formed by adjacent vertebrae and soft tissue elements (the vertebral canal) (Fig. 2.8):
the anterior wall is formed by the vertebral bodies of the vertebrae, intervertebral discs, and associated ligaments;
the lateral walls and roof are formed by the vertebral arches and ligaments.
Fig. 2.8Vertebral canal.
Within the vertebral canal, the spinal cord is surrounded by a series of three connective tissue membranes (the meninges):
the pia mater is the innermost membrane and is intimately associated with the surface of the spinal cord;
the second membrane, the arachnoid mater, is separated from the pia by the subarachnoid space, which contains cerebrospinal fluid;
the thickest and most external of the membranes, the dura mater, lies directly against, but is not attached to, the arachnoid mater.
In the vertebral canal, the dura mater is separated from surrounding bone by an extradural (epidural) space containing loose connective tissue, fat, and a venous plexus.
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Spinal nerves
The 31 pairs of spinal nerves are segmental in distribution and emerge from the vertebral canal between the pedicles of adjacent vertebrae. There are eight pairs of cervical nerves (C1 to C8), twelve thoracic (T1 to T12), five lumbar (L1 to L5), five sacral (S1 to S5), and one coccygeal (Co). Each nerve is attached to the spinal cord by a posterior root and an anterior root (Fig. 2.9).
Fig. 2.9Spinal nerves (transverse section).
After exiting the vertebral canal, each spinal nerve branches into:
a posterior ramus—collectively, the small posterior rami innervate the back; and
an anterior ramus—the much larger anterior rami innervate most other regions of the body except the head, which is innervated predominantly, but not exclusively, by cranial nerves.
The anterior rami form the major somatic plexuses (cervical, brachial, lumbar, and sacral) of the body. Major visceral components of the PNS (sympathetic trunk and prevertebral plexus) of the body are also associated mainly with the anterior rami of spinal nerves.
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RELATIONSHIP TO OTHER REGIONS
Head
Cervical regions of the back constitute the skeletal and much of the muscular framework of the neck, which in turn supports and moves the head (Fig. 2.10).
Fig. 2.10Relationships of the back to other regions.
The brain and cranial meninges are continuous with the spinal cord meninges at the foramen magnum of the skull. The paired vertebral arteries ascend, one on each side, through foramina in the transverse processes of cervical vertebrae and pass through the foramen magnum to participate, with the internal carotid arteries, in supplying blood to the brain.
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Thorax, abdomen, and pelvis
The different regions of the vertebral column contribute to the skeletal framework of the thorax, abdomen, and pelvis (Fig. 2.10). In addition to providing support for each of these parts of the body, the vertebrae provide attachment for muscles and fascia, and articulation sites for other bones. The anterior rami of spinal nerves associated with the thorax, abdomen, and pelvis pass into these parts of the body from the back.
Limbs
The bones of the back provide extensive attachment for muscles associated with anchoring and moving the upper limbs on the trunk. This is less true of the lower limbs, which are firmly anchored to the vertebral column through articulation of the pelvic bones with the sacrum. The upper and lower limbs are innervated by anterior rami of spinal nerves that emerge from cervical and lumbosacral levels, respectively, of the vertebral column.
KEY FEATURES
Long vertebral column and short spinal cord
During development, the vertebral column grows much faster than the spinal cord. As a result, the spinal cord does not extend the entire length of the vertebral canal (Fig. 2.11).
Fig. 2.11Vertebral canal, spinal cord, and spinal nerves.
In the adult, the spinal cord typically ends between vertebrae LI and LII, although it can end as high as vertebra TXII and as low as the disc between vertebrae LII and LIII.
Spinal nerves originate from the spinal cord at increasingly oblique angles from vertebrae CI to Co, and the nerve roots pass in the vertebral canal for increasingly longer distances. Their spinal cord level of origin therefore becomes increasingly dissociated from their vertebral column level of exit. This is particularly evident for lumbar and sacral spinal nerves.
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Intervertebral foramina and spinal nerves
Each spinal nerve exits the vertebral canal laterally through an intervertebral foramen (Fig. 2.12). The foramen is formed between adjacent vertebral arches and is closely related to intervertebral joints:
the superior and inferior margins are formed by notches in adjacent pedicles;
the posterior margin is formed by the articular processes of the vertebral arches and the associated joint;
the anterior border is formed by the intervertebral disc between the vertebral bodies of the adjacent vertebrae.
Fig. 2.12Intervertebral foramina.
Any pathology that occludes or reduces the size of an intervertebral foramen, such as bone loss, herniation of the intervertebral disc, or dislocation of the zygapophysial joint (the joint between the articular processes), can affect the function of the associated spinal nerve.
Innervation of the back
Posterior branches of spinal nerves innervate the intrinsic muscles of the back and adjacent skin. The cutaneous distribution of these posterior rami extends into the gluteal region of the lower limb and the posterior aspect of the head. Parts of dermatomes innervated by the posterior rami of spinal nerves are shown in Fig. 2.13.
Fig. 2.13Dermatomes innervated by posterior rami of spinal nerves.
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Regional anatomy
SKELETAL FRAMEWORK
Skeletal components of the back consist mainly of the vertebrae and associated intervertebral discs. The skull, scapulae, pelvic bones, and ribs also contribute to the bony framework of the back and provide sites for muscle attachment.
Vertebrae
There are approximately 33 vertebrae, which are subdivided into five groups based on morphology and location (Fig. 2.14):
the seven cervical vertebrae between the thorax and skull are characterized mainly by their small size and the presence of a foramen in each transverse process (Figs. 2.14 and 2.15);
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the 12 thoracic vertebrae are characterized by their articulated ribs (Figs. 2.14 and 2.16); although all vertebrae have rib elements, these elements are small and are incorporated into the transverse processes in regions other than the thorax; but in the thorax, the ribs are separate bones and articulate via synovial joints with the vertebral bodies and transverse processes of the associated vertebrae;
inferior to the thoracic vertebrae are five lumbar vertebrae, which form the skeletal support for the posterior abdominal wall and are characterized by their large size (Figs. 2.14 and 2.17);
next are five sacral vertebrae fused into one single bone called the sacrum, which articulates on each side with a pelvic bone and is a component of the pelvic wall;
inferior to the sacrum is a variable number, usually four, of coccygeal vertebrae, which fuse into a single small triangular bone called the coccyx.
Fig. 2.14Vertebrae.
Fig. 2.15Radiograph of cervical region of vertebral column. A. Anterior–posterior view. B. Lateral view.
Fig. 2.16Radiograph of thoracic region of vertebral column. A. Anterior–posterior view. B. Lateral view.
Fig. 2.17Radiograph of lumbar region of vertebral column. A. Anterior–posterior view. B. Lateral view.
In the embryo, the vertebrae are formed intersegmentally from cells called sclerotomes, which originate from adjacent somites (Fig. 2.18). Each vertebra is derived from the cranial parts of the two somites below, one on each side, and the caudal parts of the two somites above. The spinal nerves develop segmentally and pass between the forming vertebrae.
Fig. 2.18Development of the vertebrae.
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Typical vertebra
A typical vertebra consists of a vertebral body and a posterior vertebral arch (Fig. 2.19). Extending from the vertebral arch are a number of processes for muscle attachment and articulation with adjacent bone.
Fig. 2.19Typical vertebra.
The vertebral body is the weightbearing part of the vertebra and is linked to adjacent vertebral bodies by intervertebral discs and ligaments. The size of vertebral bodies increases inferiorly as the amount of weight supported increases.
The vertebral arch forms the lateral and posterior parts of the vertebral foramen.
The vertebral foramina of all the vertebrae together form the vertebral canal, which contains and protects the spinal cord. Superiorly, the vertebral canal is continuous, through the foramen magnum of the skull, with the cranial cavity of the head.
The vertebral arch of each vertebra consists of pedicles and laminae (Fig. 2.19):
the two pedicles are bony pillars that attach the vertebral arch to the vertebral body;
the two laminae are flat sheets of bone that extend from each pedicle to meet in the midline and form the roof of the vertebral arch.
A spinous process projects posteriorly and inferiorly from the junction of the two laminae and is a site for muscle and ligament attachment.
A transverse process extends posterolaterally from the junction of the pedicle and lamina on each side and is a site for articulation with ribs in the thoracic region.
Also projecting from the region where the pedicles join the laminae are superior and inferior articular processes (Fig. 2.19), which articulate with the inferior and superior articular processes, respectively, of adjacent vertebrae.
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Between the vertebral body and the origin of the articular processes, each pedicle is notched on its superior and inferior surfaces. These superior and inferior vertebral notches participate in forming intervertebral foramina.
Cervical vertebrae
The seven cervical vertebrae are characterized by their small size and by the presence of a foramen in each transverse process. A typical cervical vertebra has the following features (Fig. 2.20A):
the vertebral body is short in height and square shaped when viewed from above and has a concave superior surface and a convex inferior surface;
each transverse process is trough shaped and perforated by a round foramen transversarium;
the spinous process is short and bifid;
the vertebral foramen is triangular.
Fig. 2.20Regional vertebrae. A. Typical cervical vertebra.
Regional vertebrae. B. Atlas and axis. C. Typical thoracic vertebra. D. Typical lumbar vertebra.
Regional vertebrae. E. Sacrum. F. Coccyx.
The first and second cervical vertebrae—the atlas and axis—are specialized to accommodate movement of the head.
Atlas and axis
Vertebra CI (the atlas) articulates with the head (Fig. 2.21). Its major distinguishing feature is that it lacks a vertebral body (Fig. 2.20B). In fact, the vertebral body of CI fuses onto the body of CII during development to become the dens of CII. As a result, there is no intervertebral disc between CI and CII. When viewed from above, the atlas is ring-shaped and composed of two lateral masses interconnected by an anterior arch and a posterior arch.
Fig. 2.21Radiograph showing CI (atlas) and CII (axis) vertebrae. Open mouth, anterior-posterior (odontoid peg) view.
Each lateral mass articulates above with an occipital condyle of the skull and below with the superior articular process of vertebra CII (the axis). The superior articular surfaces are bean shaped and concave, whereas the inferior articular surfaces are almost circular and flat.
The atlanto-occipital joint allows the head to nod up and down on the vertebral column.
The posterior surface of the anterior arch has an articular facet for the dens, which projects superiorly from the vertebral body of the axis. The dens is held in position by a strong transverse ligament of atlas posterior to it and spanning the distance between the oval attachment facets on the medial surfaces of the lateral masses of the atlas.
The dens acts as a pivot that allows the atlas and attached head to rotate on the axis, side to side.
The transverse processes of the atlas are large and protrude further laterally than those of the other cervical vertebrae and act as levers for muscle action, particularly for muscles that move the head at the atlanto-axial joints.
The axis is characterized by the large tooth-like dens, which extends superiorly from the vertebral body (Figs. 2.20B and 2.21). The anterior surface of the dens has an oval facet for articulation with the anterior arch of the atlas.
The two superolateral surfaces of the dens possess circular impressions that serve as attachment sites for strong alar ligaments, one on each side, which connect the dens to the medial surfaces of the occipital condyles. These alar ligaments check excessive rotation of the head and atlas relative to the axis.
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Thoracic vertebrae
The twelve thoracic vertebrae are all characterized by their articulation with ribs. A typical thoracic vertebra has two partial facets (superior and inferior costal facets) on each side of the vertebral body for articulation with the head of its own rib and the head of the rib below (Fig. 2.20C). The superior costal facet is much larger than the inferior costal facet.
Each transverse process also has a facet (transverse costal facet) for articulation with the tubercle of its own rib. The vertebral body of the vertebra is somewhat heart-shaped when viewed from above, and the vertebral foramen is circular.
Lumbar vertebrae
The five lumbar vertebrae are distinguished from vertebrae in other regions by their large size (Fig. 2.20D). Also, they lack facets for articulation with ribs. The transverse processes are generally thin and long, with the exception of those on vertebra LV, which are massive and somewhat cone-shaped for the attachment of iliolumbar ligaments to connect the transverse processes to the pelvic bones.
The vertebral body of a typical lumbar vertebra is cylindrical and the vertebral foramen is triangular in shape and larger than in the thoracic vertebrae.
Sacrum
The sacrum is a single bone that represents the five fused sacral vertebrae (Fig. 2.20E). It is triangular in shape with the apex pointed inferiorly, and is curved so that it has a concave anterior surface and a correspondingly convex posterior surface. It articulates above with vertebra LV and below with the coccyx. It has two large L-shaped facets, one on each lateral surface, for articulation with the pelvic bones.
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The posterior surface of the sacrum has four pairs of posterior sacral foramina, and the anterior surface has four pairs of anterior sacral foramina for the passage of the posterior and anterior rami, respectively, of S1 to S4 spinal nerves.
The posterior wall of the vertebral canal may be incomplete near the inferior end of the sacrum.
Coccyx
The coccyx is a small triangular bone that articulates with the inferior end of the sacrum and represents three to four fused coccygeal vertebrae (Fig. 2.20F). It is characterized by its small size and by the absence of vertebral arches and therefore a vertebral canal.
Intervertebral foramina
Intervertebral foramina are formed on each side between adjacent parts of vertebrae and associated intervertebral discs (Fig. 2.22). The foramina allow structures, such as spinal nerves and blood vessels, to pass in and out of the vertebral canal.
Fig. 2.22Intervertebral foramen.
An intervertebral foramen is formed by the inferior vertebral notch on the pedicle of the vertebra above and the superior vertebral notch on the pedicle of the vertebra below. The foramen is bordered:
posteriorly by the zygapophysial joint between the articular processes of the two vertebrae; and
anteriorly by the intervertebral disc and adjacent vertebral bodies.
Each intervertebral foramen is a confined space surrounded by bone and ligament, and by joints.
Pathology in any of these structures, and in the surrounding muscles, can affect structures within the foramen.
Posterior spaces between vertebral arches
In most regions of the vertebral column, the laminae and spinous processes of adjacent vertebrae overlap to form a reasonably complete bony dorsal wall for the vertebral canal. However, in the lumbar region, large gaps exist between the posterior components of adjacent vertebral arches (Fig. 2.23). These gaps between adjacent laminae and spinous processes become increasingly wide from vertebra LI to vertebra LV. The spaces can be widened further by flexion of the vertebral column.
Fig. 2.23Spaces between adjacent vertebral arches in the lumbar region.
These gaps allow relatively easy access to the vertebral canal for clinical procedures.
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In the clinic
Spina bifida
Spina bifida is a disorder in which the two sides of vertebral arches, usually in lower vertebrae, fail to fuse during development, resulting in an “open” vertebral canal (Fig. 2.24). There are two types of spina bifida.
Fig. 2.24T1-weighted MR image in the sagittal plane demonstrating a lumbosacral myelomeningocele. There is an absence of laminae and spinous processes in the lumbosacral region.
The commonest type is spina bifida occulta, in which there is a defect in the vertebral arch of LV or SI. This defect occurs in as many as 10% of individuals and results in failure of the posterior arch to fuse in the midline. In the spina bifida occulta group this is usually an incidental finding, although clinical examination may reveal a tuft of hair over the spinous processes. Clinically, the patient is asymptomatic.
The more severe form of spina bifida involves complete failure of fusion of the posterior arch at the lumbosacral junction with a large outpouching of the meninges. This may contain cerebrospinal fluid (a meningocele) or a portion of the spinal cord (a myelomeningocele). These abnormalities may result in a variety of neurological deficits, including problems with walking and bladder function.
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In the clinic
Vertebroplasty
Vertebroplasty is a new technique in which the body of a vertebra can be filled with bone cement (typically methyl methacrylate). The indications for the technique include vertebral body collapse and pain from the vertebral body, which may be secondary to tumor infiltration. The procedure is most commonly performed for osteoporotic wedge fractures, which are a considerable cause of morbidity and pain in older patients. Early pain free mobilization in patients with osteoporotic wedge fractures has been shown to improve their outcome and ability to return to activities of daily living.
Osteoporotic wedge fractures typically occur at the thoracolumbar region, and the approach to performing vertebroplasty is novel and relatively straightforward. The procedure is performed under sedation or light general anesthetic. Using X-ray guidance the pedicle is identified on the anterior–posterior image. A metal cannula is placed through the pedicle into the vertebral body. It may be necessary to place a cannula in both pedicles of the collapsed vertebra.
Liquid bone cement is injected in the cannula, and this can be demonstrated filling the vertebral body. The function of the bone cement is two-fold. First, it increases the strength of the vertebral body and prevents further loss of height. Furthermore, as the bone cement sets, there is a degree of heat generated that is believed to disrupt pain nerve endings.
In the clinic
Scoliosis
Scoliosis is an abnormal lateral curvature of the vertebral column (Fig. 2.25).
Fig. 2.25Radiograph of thoracic scoliosis.
A true scoliosis involves not only the (right- or left-sided) curvature but also a rotational element of one vertebra upon another.
The commonest types of scoliosis are those for which we have little understanding about how or why they occur and are termed idiopathic scoliosis. These are never present at birth and tend to occur in either the infantile, juvenile, or adolescent age groups. The vertebral bodies and posterior elements (pedicles and laminae) are normal in these patients.
When a scoliosis is present from birth (congenital scoliosis) it is usually associated with other developmental abnormalities. In these patients, there is a strong association with other abnormalities of the chest wall, genitourinary tract, and heart disease. This group of patients needs careful evaluation by many specialists.
Scoliosis may also be the manifestation of central or peripheral nerve abnormalities (neuropathic scoliosis), as in children with cerebral palsy or polio.
A rare but important group of scoliosis is that in which the muscle is abnormal. Muscular dystrophy is the commonest example. The abnormal muscle does not retain the normal alignment of the vertebral column and curvature develops as a result. A muscle biopsy is needed to make the diagnosis.
Other disorders that can produce scoliosis include bone tumors, spinal cord tumors, and localized disc protrusions.
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In the clinic
Kyphosis
Kyphosis is abnormal curvature of the vertebral column in the thoracic region, producing a “hunchback” deformity. This condition occurs in certain disease states, the most dramatic of which is usually secondary to tuberculosis infection of a thoracic vertebral body, where the kyphosis becomes angulated at the site of the lesion. This produces the gibbus deformity, a deformity that was prevalent before the use of antituberculous medication.
Lordosis
Lordosis is abnormal curvature of the vertebral column in the lumbar region, producing a swayback deformity.
In the clinic
Variation in vertebral numbers
There are usually seven cervical vertebrae, although in certain diseases these may be fused. Fusion of cervical vertebrae (Fig. 2.26A) can be associated with other abnormalities, for example Klippel–Feil syndrome, in which there is abnormal fusion of vertebrae CI and CII or CV and CVI, and may be associated with a high-riding scapula (Sprengel’s shoulder) and cardiac abnormalities.
Fig. 2.26Variations in vertebral number. A. Fused vertebral bodies of cervical vertebrae. B. Hemivertebra.
Variations in the number of thoracic vertebrae are well described. One of the commonest abnormalities in the lumbar vertebrae is a partial fusion of vertebra LV with the sacrum (sacralization of the lumbar vertebra). Partial separation of vertebra SI from the sacrum (lumbarization of first sacral vertebra) may also occur (Fig. 2.26B).
No definite correlation exists between the number of vertebrae and back pain, although the two cannot be dissociated. Surgeons, radiologists, and other physicians must be able to accurately define vertebral levels to prevent errors.
A hemivertebra occurs when a vertebra develops only on one side (Fig. 2.26B).
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In the clinic
The vertebrae and cancer
The vertebrae are common sites for metastatic disease (secondary spread of cancer cells). When cancer cells grow within the vertebral bodies and the posterior elements, they destroy the mechanical properties of the bone. A minor injury may therefore lead to vertebral collapse. Importantly, vertebrae that contain extensive metastatic disease may extrude fragments of tumor into the vertebral canal, so compressing nerves and the spinal cord.
In the clinic
Osteoporosis
Osteoporosis is a pathophysiologic condition in which bone quality is normal, but the quantity of bone is deficient. It is a metabolic bone disorder that commonly occurs in women in their 50s and 60s and in men in their 70s.
Many factors influence the development of osteoporosis including genetic predetermination, level of activity and nutritional status, and, in particular, estrogen levels in women.
Typical complications of osteoporosis include “crush” vertebral body fractures, distal radial fractures, and hip fractures.
With increasing age and poor quality bone, patients are more susceptible to fracture. Healing tends to be impaired in these elderly patients, who consequently require long hospital stays and prolonged rehabilitation.
Identifying patients at risk of osteoporosis and instituting appropriate drug therapy and preventive care can prevent injuries. There are no specific clinical signs or symptoms of osteoporosis. Often the diagnosis is made in retrospect when the patient presents with a pathological fracture.
Patients likely to develop osteoporosis can be identified by a dual-photon X-ray absorptiometry (DXA) scanning. Low-dose X-rays are passed through the bone, and by counting the number of photons detected and knowing the dose given, the number of X-rays absorbed by the bone can be calculated. The amount of X-ray absorption can be directly correlated with the bone mass, and this can be used to predict whether a patient is at risk for osteoporotic fractures.
JOINTS
Joints between vertebrae in the back
The two major types of joints between vertebrae are:
symphyses between vertebral bodies (Fig. 2.27); and
synovial joints between articular processes (Fig. 2.28).
Fig. 2.27Intervertebral joints.
Fig. 2.28Zygapophysial joints.
A typical vertebra has a total of six joints with adjacent vertebrae: four synovial joints (two above and two below) and two symphyses (one above and one below). Each symphysis includes an intervertebral disc.
Although the movement between any two vertebrae is limited, the summation of movement among all vertebrae results in a large range of movement by the vertebral column.
Movements by the vertebral column include flexion, extension, lateral flexion, rotation, and circumduction.
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Movements by vertebrae in a specific region (cervical, thoracic, and lumbar) are determined by the shape and orientation of joint surfaces on the articular processes and on the vertebral bodies.
Symphyses between vertebral bodies (intervertebral discs)
The symphysis between adjacent vertebral bodies is formed by a layer of hyaline cartilage on each vertebral body and an intervertebral disc, which lies between the layers.
The intervertebral disc consists of an outer anulus fibrosus, which surrounds a central nucleus pulposus (Fig. 2.27).
The anulus fibrosus consists of an outer ring of collagen surrounding a wider zone of fibrocartilage arranged in a lamellar configuration. This arrangement of fibers limits rotation between vertebrae.
The nucleus pulposus fills the center of the intervertebral disc, is gelatinous, and absorbs compression forces between vertebrae.
Degenerative changes in the anulus fibrosus can lead to herniation of the nucleus pulposus. Posterolateral herniation can impinge on the roots of a spinal nerve in the intervertebral foramen.
Joints between vertebral arches (zygapophysial joints)
The synovial joints between superior and inferior articular processes on adjacent vertebrae are the zygapophysial joints (Fig. 2.28). A thin articular capsule attached to the margins of the articular facets encloses each joint.
In cervical regions, the zygapophysial joints slope inferiorly from anterior to posterior. This orientation facilitates flexion and extension. In thoracic regions, the joints are oriented vertically and limit flexion and extension, but facilitate rotation. In lumbar regions, the joint surfaces are curved and adjacent processes interlock, thereby limiting range of movement, though flexion and extension are still major movements in the lumbar region.
“Uncovertebral” joints
The lateral margins of the upper surfaces of typical cervical vertebrae are elevated into crests or lips termed uncinate processes. These may articulate with the body of the vertebra above to form small “uncovertebral” synovial joints (Fig. 2.29).
Fig. 2.29Uncovertebral joint.
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In the clinic
Back pain
Back pain is an extremely common disorder. It is often difficult to determine whether back pain relates to direct mechanical problems or to a disc protrusion impinging on a nerve. In cases involving discs, it may be necessary to operate and remove the disc that is pressing on the nerve.
Not infrequently, patients complain of pain and no immediate cause is found; the pain is therefore attributed to mechanical discomfort, which may be caused by degenerative disease. One of the treatments is to pass a needle into the facet joint and inject it with local anesthetic and corticosteroid.
In the clinic
Herniation of intervertebral discs
The discs between the vertebrae are made up of a central portion (the nucleus pulposus) and a complex series of fibrous rings (anulus fibrosus). A tear can occur within the anulus fibrosus through which the material of the nucleus pulposus can track. After a period of time, this material may track into the vertebral canal or into the intervertebral foramen to impinge on neural structures (Fig. 2.30). This is a common cause of back pain. A disc may protrude posteriorly to directly impinge on the cord or the roots of the lumbar nerves, depending on the level, or may protrude posterolaterally adjacent to the pedicle and impinge on the descending root.
Fig. 2.30Disc protrusion—T2-weighted magnetic resonance images of the lumbar region of the vertebral column. A. Sagittal plane. B. Axial plane.
In cervical regions of the vertebral column, cervical disc protrusions often become ossified and are termed disc osteophyte bars.
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In the clinic
Joints
Some diseases have a predilection for synovial joints rather than symphyses. A typical example is rheumatoid arthritis, which primarily affects synovial joints and synovial bursae, resulting in destruction of the joint and its lining. Symphyses are usually preserved.
LIGAMENTS
Joints between vertebrae are reinforced and supported by numerous ligaments, which pass between vertebral bodies and interconnect components of the vertebral arches.
Anterior and posterior longitudinal ligaments
The anterior and posterior longitudinal ligaments are on the anterior and posterior surfaces of the vertebral bodies and extend along most of the vertebral column (Fig. 2.31).
Fig. 2.31Anterior and posterior longitudinal ligaments of vertebral column.
The anterior longitudinal ligament is attached superiorly to the base of the skull and extends inferiorly to attach to the anterior surface of the sacrum. Along its length it is attached to the vertebral bodies and intervertebral discs.
The posterior longitudinal ligament is on the posterior surfaces of the vertebral bodies and lines the anterior surface of the vertebral canal. Like the anterior longitudinal ligament, it is attached along its length to the vertebral bodies and intervertebral discs. The upper part of the posterior longitudinal ligament that connects CII to the intracranial aspect of the base of the skull is termed the tectorial membrane.
Ligamenta flava
The ligamenta flava, on each side, pass between the laminae of adjacent vertebrae (Fig. 2.32). These thin, broad ligaments consist predominantly of elastic tissue and form part of the posterior surface of the vertebral canal. Each ligamentum flavum runs between the posterior surface of the lamina on the vertebra below to the anterior surface of the lamina of the vertebra above. The ligamenta flava resist separation of the laminae in flexion and assist in extension back to the anatomical position.
Fig. 2.32Ligamenta flava.
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Supraspinous ligament and ligamentum nuchae
The supraspinous ligament connects and passes along the tips of the vertebral spinous processes from vertebra CVII to the sacrum (Fig. 2.33). From vertebra CVII to the skull, the ligament becomes structurally distinct from more caudal parts of the ligament and is called the ligamentum nuchae.
Fig. 2.33Supraspinous ligament and ligamentum nuchae.
The ligamentum nuchae is a triangular, sheet-like structure in the median sagittal plane:
the base of the triangle is attached to the skull, from the external occipital protuberance to the foramen magnum;
the apex is attached to the tip of the spinous process of vertebra CVII;
the deep side of the triangle is attached to the posterior tubercle of vertebra CI and the spinous processes of the other cervical vertebrae.
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The ligamentum nuchae supports the head. It resists flexion and facilitates returning the head to the anatomical position. The broad lateral surfaces and the posterior edge of the ligament provide attachment for adjacent muscles.
Interspinous ligaments
Interspinous ligaments pass between adjacent vertebral spinous processes (Fig. 2.34). They attach from the base to the apex of each spinous process and blend with the supraspinous ligament posteriorly and the ligamenta flava anteriorly on each side.
Fig. 2.34Interspinous ligaments.
In the clinic
Ligamenta flava
The ligamenta flava are important structures within the vertebral canal. In degenerative conditions of the vertebral column, the ligamenta flava may hypertrophy. This is often associated with hypertrophy and arthritic change of the zygapophysial joints. In combination, zygapophysial joint hypertrophy, ligamenta flava hypertrophy, and a mild disc protrusion can reduce the dimensions of the vertebral canal. The occurrence of all three of these conditions together is not uncommon and produces the syndrome of spinal stenosis.
In the clinic
Vertebral fractures
Vertebral fractures can occur anywhere throughout the vertebral column. In most instances, the fracture will heal under appropriate circumstances. At the time of injury, it is not the fracture itself, but related damage to the contents of the vertebral canal and the surrounding tissues that determines the severity of the patient’s condition.
Vertebral column stability is divided into three arbitrary clinical “columns”: the anterior column consists of the vertebral bodies and the anterior longitudinal ligament; the middle column comprises the vertebral body and the posterior longitudinal ligament; and the posterior column is made up of the ligamenta flava, interspinous ligaments, supraspinous ligaments, and the ligamentum nuchae in the cervical vertebral column.
Destruction of one of the clinical columns is usually a stable injury requiring little more than rest and appropriate analgesia. Disruption of two columns is highly likely to be unstable and requires fixation and immobilization. A three-column spinal injury usually results in a significant neurological event and requires fixation to prevent further extension of the neurological defect and to create vertebral column stability.
At the craniocervical junction, a complex series of ligaments create stability. If the traumatic incident disrupts craniocervical stability, the chances of a significant spinal cord injury are extremely high. The consequences are quadriplegia, although in the short term, respiratory function could be compromised by paralysis of the phrenic nerve (which arises from spinal nerves C3 to C5), and severe hypotension (low blood pressure) may result from central disruption of the sympathetic part of the autonomic division of the PNS.
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Mid and lower cervical vertebral column disruption may produce a range of complex neurological problems involving the upper and lower limbs, although below the level of C5, respiratory function is unlikely to be compromised.
Vertebral injuries may also involve the soft tissues and supporting structures between the vertebrae. Typical examples of this are the unifacetal and bifacetal cervical vertebral dislocations that occur in hyperflexion injuries.
Similarly, lesions in the vertebral column, depending on their level, produce corresponding neurological deficits.
Lumbar vertebral column injuries are rare. When they occur, they usually involve significant force. Knowing that a significant force is required to fracture a vertebra, the abdominal organs and the rest of the axial skeleton need to be assessed for further fractures and visceral rupture.
Pars interarticularis fractures
The pars interarticularis is a clinical term to describe the specific region of a vertebra between the superior and inferior facet (zygapophysial) joints (Fig. 2.35A). This region is susceptible to trauma, especially in athletes.
Fig. 2.35Radiograph of lumbar region of vertebral column, oblique view (“Scottie dog”). A. Normal
radiograph of lumbar region of vertebral column, oblique view. In this view, the transverse process (nose), pedicle (eye), superior articular process (ear), inferior articular process (front leg) and pars interarticularis (neck) resemble a dog. A fracture of the pars interarticularis is visible as a break in the neck of the dog, or the appearance of a collar. B. Fracture of pars interarticularis.
If a fracture occurs around the pars interarticularis, the vertebral body may slip anteriorly and compress the vertebral canal.
The most common sites for pars interarticularis fractures are the LIV and LV levels (Fig. 2.35B). (Clinicians often refer to parts of the back in shorthand terms that are not strictly anatomical; for example, facet joints and apophyseal joints are terms used instead of zygapophysial joints, and spinal column is used instead of vertebral column.)
It is possible for a vertebra to slip anteriorly upon its inferior counterpart without a pars interarticularis fracture. Usually this is related to abnormal anatomy of the facet joints, facet joint degenerative change. This disorder is termed spondylolisthesis.
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In the clinic
Surgical procedures on the back
Discectomy/laminectomy
A prolapsed intervertebral disc may impinge upon the meningeal (thecal) sac, cord, and most commonly the nerve root, producing symptoms attributable to that level. In some instances the disc protrusion will undergo a degree of involution that may allow symptoms to resolve without intervention. In some instances pain, loss of function, and failure to resolve may require surgery to remove the disc protrusion.
There are a number of ways in which the surgeon may approach the disc within the vertebral canal, and there are a number of procedures that can be performed to relieve the patient’s symptoms.
It is of the utmost importance that the level of the disc protrusion is identified before surgery. This may require MRI scanning and on-table fluoroscopy to prevent operating on the wrong level. A midline approach to the right or to the left of the spinous processes will depend upon the most prominent site of the disc bulge. In some instances removal of the lamina will increase the potential space and may relieve symptoms. Some surgeons perform a small fenestration (windowing) within the ligamentum flavum. This provides access to the canal. The meningeal sac and its contents are gently retracted, exposing the nerve root and the offending disc. The disc is dissected free, removing its effect on the nerve root and the canal.
Spinal Fusion
Spinal fusion is performed when it is necessary to fuse one vertebra with the corresponding superior or inferior vertebra, and in some instances multilevel fusion may be necessary. Indications are varied, though they include stabilization after fracture, stabilization related to tumor infiltration, and stabilization when mechanical pain is produced either from the disc or from the posterior elements.
There are a number of surgical methods in which a fusion can be performed, through either a posterior approach and fusing the posterior elements, an anterior approach by removal of the disc and either disc replacement or anterior fusion, or in some instances a 360° fusion where the posterior elements and the vertebral bodies are fused. These procedures are not without risk and require considerable surgical skill and expertise.
BACK MUSCULATURE
Muscles of the back are organized into superficial, intermediate, and deep groups.
Muscles in the superficial and intermediate groups are extrinsic muscles because they originate embryologically from locations other than the back. They are innervated by anterior rami of spinal nerves:
the superficial group consists of muscles related to and involved in movements of the upper limb;
the intermediate group consists of muscles attached to the ribs and may serve as a respiratory function.
Muscles of the deep group are intrinsic muscles because they develop in the back. They are innervated by posterior rami of spinal nerves and are directly related to movements of the vertebral column and head.
Superficial group of back muscles
The muscles in the superficial group are immediately deep to the skin and superficial fascia (Figs. 2.36-2.39). They attach the superior part of the appendicular skeleton (clavicle, scapula, and humerus) to the axial skeleton (skull, ribs, and vertebral column). Because these muscles are primarily involved with movements of this part of the appendicular skeleton, they are sometimes referred to as the appendicular group.
Fig. 2.36Superficial group of back muscles—trapezius and latissimus dorsi.
Fig. 2.37Superficial group of back muscles—trapezius and latissimus dorsi, with rhomboid major, rhomboid minor, and levator scapulae located deep to trapezius in the superior part of the back.
Fig. 2.38Innervation and blood supply of trapezius.
Fig. 2.39Rhomboid muscles and levator scapulae.
Muscles in the superficial group include trapezius, latissimus dorsi, rhomboid major, rhomboid minor, and levator scapulae. Rhomboid major, rhomboid minor, and levator scapulae are located deep to trapezius in the superior part of the back.
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Trapezius
Each trapezius muscle is flat and triangular, with the base of the triangle situated along the vertebral column (the muscle’s origin) and the apex pointing toward the tip of the shoulder (the muscle’s insertion) (Fig. 2.37 and Table 2.1). The muscles on both sides together form a trapezoid.
Table 2.1Superficial (appendicular) group of back muscles
The superior fibers of trapezius, from the skull and upper portion of the vertebral column, descend to attach to the lateral third of the clavicle and to the acromion of the scapula. Contraction of these fibers elevates the scapula. In addition, the superior and inferior fibers work together to rotate the lateral aspect of the scapula upward, which needs to occur when raising the upper limb above the head.
Motor innervation of trapezius is by the accessory nerve [XI], which descends from the neck onto the deep surface of the muscle (Fig. 2.38). Proprioceptive fibers from trapezius pass in the branches of the cervical plexus and enter the spinal cord at spinal cord levels C3 and C4.
The blood supply to trapezius is from the superficial branch of the transverse cervical artery, the acromial branch of the suprascapular artery, and dorsal branches of posterior intercostal arteries.
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Latissimus dorsi
Latissimus dorsi is a large, flat triangular muscle that begins in the lower portion of the back and tapers as it ascends to a narrow tendon that attaches to the humerus anteriorly (Figs. 2.36-2.39 and Table 2.1). As a result, movements associated with this muscle include extension, adduction, and medial rotation of the upper limb. Latissimus dorsi can also depress the shoulder, preventing its upward movement.
The thoracodorsal nerve of the brachial plexus innervates the latissimus dorsi muscle. Associated with this nerve is the thoracodorsal artery, which is the primary blood supply of the muscle. Additional small arteries come from dorsal branches of posterior intercostal and lumbar arteries.
Levator scapulae
Levator scapulae is a slender muscle that descends from the transverse processes of the upper cervical vertebrae to the upper portion of the scapula on its medial border at the superior angle (Fig. 2.37 and 2.39, and Table 2.1). It elevates the scapula and may assist other muscles in rotating the lateral aspect of the scapula inferiorly.
Levator scapulae is innervated by branches from the anterior rami of spinal nerves C3 and C4 and the dorsal scapular nerve, and its arterial supply consists of branches primarily from the transverse and ascending cervical arteries.
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Rhomboid minor and rhomboid major
The two rhomboid muscles are inferior to levator scapulae (Fig. 2.39 and Table 2.1). Rhomboid minor is superior to rhomboid major, and is a small, cylindrical muscle that arises from the ligamentum nuchae of the neck and the spinous processes of vertebrae CVII and TI and attaches to the medial scapular border opposite the root of the spine of the scapula.
The larger rhomboid major originates from the spinous processes of the upper thoracic vertebrae and attaches to the medial scapular border inferior to rhomboid minor.
The two rhomboid muscles work together to retract or pull the scapula toward the vertebral column. With other muscles they may also rotate the lateral aspect of the scapula inferiorly.
The dorsal scapular nerve, a branch of the brachial plexus, innervates both rhomboid muscles (Fig. 2.40).
Fig. 2.40Innervation and blood supply of the rhomboid muscles.
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Intermediate group of back muscles
The muscles in the intermediate group of back muscles consist of two thin muscular sheets in the superior and inferior regions of the back, immediately deep to the muscles in the superficial group (Fig. 2.41 and Table 2.2). Fibers from these two serratus posterior muscles (serratus posterior superior and serratus posterior inferior) pass obliquely outward from the vertebral column to attach to the ribs. This positioning suggests a respiratory function, and at times, these muscles have been referred to as the respiratory group.
Fig. 2.41Intermediate group of back muscles—serratus posterior muscles.
Table 2.2Intermediate (respiratory) group of back muscles
Serratus posterior superior is deep to the rhomboid muscles, whereas serratus posterior inferior is deep to latissimus dorsi. Both serratus posterior muscles are attached to the vertebral column and associated structures medially, and either descend (the fibers of serratus posterior superior) or ascend (the fibers of serratus posterior inferior) to attach to the ribs. These two muscles therefore elevate and depress the ribs.
The serratus posterior muscles are innervated by segmental branches of anterior rami of intercostal nerves. Their vascular supply is provided by a similar segmental pattern through the intercostal arteries.
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Deep group of back muscles
The deep or intrinsic muscles of the back extend from the pelvis to the skull and are innervated by segmental branches of the posterior rami of spinal nerves. They include:
the extensors and rotators of the head and neck—the splenius capitis and cervicis (spinotransversales muscles);
the extensors and rotators of the vertebral column—the erector spinae and transversospinales; and
the short segmental muscles—the interspinales and intertransversarii.
The vascular supply to this deep group of muscles is through branches of the vertebral, deep cervical, occipital, transverse cervical, posterior intercostal, subcostal, lumbar, and lateral sacral arteries.
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Thoracolumbar fascia
The thoracolumbar fascia covers the deep muscles of the back and trunk (Fig. 2.42). This fascial layer is critical to the overall organization and integrity of the region:
superiorly, it passes anteriorly to the serratus posterior muscle and is continuous with deep fascia in the neck;
in the thoracic region, it covers the deep muscles and separates them from the muscles in the superficial and intermediate groups;
medially, it attaches to the spinous processes of the thoracic vertebrae and, laterally, to the angles of the ribs.
Fig. 2.42Thoracolumbar fascia and the deep back muscles (transverse section).
The medial attachments of the latissimus dorsi and serratus posterior inferior muscles blend into the thoracolumbar fascia. In the lumbar region, the thoracolumbar fascia consists of three layers:
the posterior layer is thick and is attached to the spinous processes of the lumbar vertebrae, sacral vertebrae, and the supraspinous ligament—from these attachments, it extends laterally to cover the erector spinae;
the middle layer is attached medially to the tips of the transverse processes of the lumbar vertebrae and intertransverse ligaments—inferiorly, it is attached to the iliac crest and, superiorly, to the lower border of rib XII;
the anterior layer covers the anterior surface of the quadratus lumborum muscle (a muscle of the posterior abdominal wall) and is attached medially to the transverse processes of the lumbar vertebrae—inferiorly, it is attached to the iliac crest and, superiorly, it forms the lateral arcuate ligament for attachment of the diaphragm.
The posterior and middle layers of the thoracolumbar fascia come together at the lateral margin of the erector spinae (Fig. 2.42). At the lateral border of the quadratus lumborum, the anterior layer joins them and forms the aponeurotic origin for the transversus abdominis muscle of the abdominal wall.
Spinotransversales muscles
The two spinotransversales muscles run from the spinous processes and ligamentum nuchae upward and laterally (Fig. 2.43 and Table 2.3):
splenius capitis is a broad muscle attached to the occipital bone and mastoid process of the temporal bone;
splenius cervicis is a narrow muscle attached to the transverse processes of the upper cervical vertebrae.
Fig. 2.43Deep group of back muscles—spinotransversales muscles (splenius capitis and splenius cervicis).
Table 2.3Spinotransversales muscles
Together the spinotransversales muscles draw the head backward, extending the neck. Individually, each muscle rotates the head to one side—the same side as the contracting muscle.
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Erector spinae muscles
The erector spinae is the largest group of intrinsic back muscles. The muscles lie posterolaterally to the vertebral column between the spinous processes medially and the angles of the ribs laterally. They are covered in the thoracic and lumbar regions by thoracolumbar fascia and the serratus posterior inferior, the rhomboid, and splenius muscles. The mass arises from a broad, thick tendon attached to the sacrum, spinous processes of the lumbar and lower thoracic vertebrae, and the iliac crest (Fig. 2.44 and Table 2.4). It divides in the upper lumbar region into three vertical columns of muscle, each of which is further subdivided regionally (lumborum, thoracis, cervicis, and capitis), depending on where the muscles attach superiorly.
Fig. 2.44Deep group of back muscles—erector spinae muscles.
Table 2.4Erector spinae group of back muscles
| Muscle | Origin | Insertion |
:---
| Iliocostalis lumborum | Sacrum, spinous processes of lumbar and lower two thoracic vertebrae and their supraspinous ligaments, and the iliac crest | Angles of the lower six or seven ribs |
| Iliocostalis thoracis | Angles of the lower six ribs | Angles of the upper six ribs and the transverse process of CVII |
| Iliocostalis cervicis | Angles of ribs III to VI | Transverse processes of CIV to CVI |
| Longissimus thoracis | Blends with iliocostalis in lumbar region and is attached to transverse processes of lumbar vertebrae | Transverse processes of all thoracic vertebrae and just lateral to the tubercles of the lower nine or ten ribs |
| Longissimus cervicis | Transverse processes of upper four or five thoracic vertebrae | Transverse processes of CII to CVI |
| Longissimus capitis | Transverse processes of upper four or five thoracic vertebrae and articular processes of lower three or four cervical vertebrae | Posterior margin of the mastoid process |
| Spinalis thoracis | Spinous processes of TX or TXI to LII | Spinous processes of TI to TVIII (varies) |
| Spinalis cervicis | Lower part of ligamentum nuchae and spinous process of CVII (sometimes TI to TII) | Spinous process of CII (axis) |
| Spinalis capitis | Usually blends with semispinalis capitis | With semispinalis capitis |
The outer or most laterally placed column of the erector spinae muscles is the iliocostalis, which is associated with the costal elements and passes from the common tendon of origin to multiple insertions into the angles of the ribs and the transverse processes of the lower cervical vertebrae.
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The middle or intermediate column is the longissimus, which is the largest of the erector spinae subdivision extending from the common tendon of origin to the base of the skull. Throughout this vast expanse, the lateral positioning of the longissimus muscle is in the area of the transverse processes of the various vertebrae.
The most medial muscle column is the spinalis, which is the smallest of the subdivisions and interconnects the spinous processes of adjacent vertebrae. Spinalis is most constant in the thoracic region and is generally absent in the cervical region. It is associated with a deeper muscle (the semispinalis capitis) as the erector spinae group approaches the skull.
The muscles in the erector spinae group are the primary extensors of the vertebral column and head. Acting bilaterally, they straighten the back, returning it to the upright position from a flexed position, and pull the head posteriorly. They also participate in controlling vertebral column flexion by contracting and relaxing in a coordinated fashion. Acting unilaterally, they bend the vertebral column laterally. In addition, unilateral contractions of muscles attached to the head turn the head to the actively contracting side.
Transversospinales muscles
The transversospinales muscles run obliquely upward and medially from transverse processes to spinous processes, filling the groove between these two vertebral projections (Fig. 2.45 and Table 2.5). They are deep to the erector spinae and consist of three major subgroups—the semispinalis, multifidus, and rotatores muscles.
Fig. 2.45Deep group of back muscles—transversospinales and segmental muscles.
Table 2.5Transversospinales group of back muscles
| Muscle | Origin | Insertion |
:---
| Semispinalis thoracis | Transverse processes of TVI to TX | Spinous processes of upper four thoracic and lower two cervical vertebrae |
| Semispinalis cervicis | Transverse processes of upper five or six thoracic vertebrae | Spinous processes of CII (axis) to CV |
| Semispinalis capitis | Transverse processes of TI to TVI (or TVII) and CVII and articular processes of CIV to CVI | Medial area between the superior and inferior nuchal lines of occipital bone |
| Multifidus | Sacrum, origin of erector spinae, posterior superior iliac spine, mammillary processes of lumbar vertebrae, transverse processes of thoracic vertebrae, and articular processes of lower four cervical vertebrae | Base of spinous processes of all vertebrae from LV to CII (axis) |
| Rotatores lumborum | Mammillary processes of lumbar vertebrae | Spinous processes of lumbar vertebrae |
| Rotatores thoracis | Transverse processes of thoracic vertebrae | Spinous processes of thoracic vertebrae |
| Rotatores cervicis | Articular processes of cervical vertebrae | Spinous processes of cervical vertebrae |
The semispinalis muscles are the most superficial collection of muscle fibers in the transversospinales group. These muscles begin in the lower thoracic region and end by attaching to the skull, crossing between four and six vertebrae from their point of origin to point of attachment. Semispinalis muscles are found in the thoracic and cervical regions, and attach to the occipital bone at the base of the skull.
Deep to semispinalis is the second group of muscles, the multifidus. Muscles in this group span the length of the vertebral column, passing from a lateral point of origin upward and medially to attach to spinous processes and spanning between two and four vertebrae. The multifidus muscles are present throughout the length of the vertebral column, but are best developed in the lumbar region.
The small rotatores muscles are the deepest of the transversospinales group. They are present throughout the length of the vertebral column, but are best developed in the thoracic region. Their fibers pass upward and medially from transverse processes to spinous processes crossing two vertebrae (long rotators) or attaching to adjacent vertebra (short rotators).
When muscles in the transversospinales group contract bilaterally, they extend the vertebral column, an action similar to that of the erector spinae group. However, when muscles on only one side contract, they pull the spinous processes toward the transverse processes on that side, causing the trunk to turn or rotate in the opposite direction.
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One muscle in the transversospinales group, the semispinalis capitis, has a unique action because it attaches to the skull. Contracting bilaterally, this muscle pulls the head posteriorly, whereas unilateral contraction pulls the head posteriorly and turns it, causing the chin to move superiorly and turn toward the side of the contracting muscle. These actions are similar to those of the upper erector spinae.
Segmental muscles
The two groups of segmental muscles (Fig. 2.45 and Table 2.6) are deeply placed in the back and innervated by posterior rami of spinal nerves.
Table 2.6Segmental back muscles
The first group of segmental muscles are the levatores costarum muscles, which arise from the transverse processes of vertebra CVII and TI to TXI. They have an oblique lateral and downward direction and insert into the rib below the vertebra of origin in the area of the tubercle. Contraction elevates the ribs.
The second group of segmental muscles are the true segmental muscles of the back—the interspinales, which pass between adjacent spinous processes, and the intertransversarii, which pass between adjacent transverse processes. These postural muscles stabilize adjoining vertebrae during movements of the vertebral column to allow more effective action of the large muscle groups.
Suboccipital muscles
A small group of deep muscles in the upper cervical region at the base of the occipital bone move the head. They connect vertebra CI (the atlas) to vertebra CII (the axis) and connect both vertebrae to the base of the skull. Because of their location they are sometimes referred to as suboccipital muscles (Figs. 2.45 and 2.46 and Table 2.7). They include, on each side:
rectus capitis posterior major;
rectus capitis posterior minor;
obliquus capitis inferior; and
obliquus capitis superior.
Fig. 2.46Deep group of back muscles—suboccipital muscles. This also shows the borders of the suboccipital triangle.
Table 2.7Suboccipital group of back muscles
Contraction of the suboccipital muscles extends the head at the atlanto-axial joint.
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The suboccipital muscles are innervated by the posterior ramus of the first cervical nerve, which enters the area between the vertebral artery and the posterior arch of the atlas (Fig. 2.46). The vascular supply to the muscles in this area is from branches of the vertebral and occipital arteries.
The suboccipital muscles form the boundaries of the suboccipital triangle, an area that contains several important structures (Fig. 2.46):
rectus capitis posterior major forms the medial border of the triangle;
obliquus capitis superior forms the lateral border;
obliquus capitis inferior muscle forms the inferior border.
The contents of the area outlined by these muscles are the posterior ramus of C1, the vertebral artery, and associated veins.
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In the clinic
Nerve injuries affecting superficial back muscles
Weakness in the trapezius, caused by an interruption of the accessory nerve [XI], may appear as drooping of the shoulder, inability to raise the arm above the head because of impaired rotation of the scapula, or weakness in attempting to raise the shoulder (i.e., shrug the shoulder against resistance).
A weakness in, or an inability to use, the latissimus dorsi, resulting from an injury to the thoracodorsal nerve, diminishes the capacity to pull the body upward while climbing or doing a pull-up.
An injury to the dorsal scapular nerve, which innervates the rhomboids, may result in a lateral shift in the position of the scapula on the affected side (i.e., the normal position of the scapula is lost because of the affected muscle’s inability to prevent antagonistic muscles from pulling the scapula laterally).
SPINAL CORD
The spinal cord extends from the foramen magnum to approximately the level of the disc between vertebrae LI and LII in adults, although it can end as high as vertebra TXII or as low as the disc between vertebrae LII and LIII (Fig. 2.47). In neonates, the spinal cord extends approximately to vertebra LIII, but can reach as low as vertebra LIV. The distal end of the cord (the conus medullaris) is cone shaped. A fine filament of connective tissue (the pial part of the filum terminale) continues inferiorly from the apex of the conus medullaris.
Fig. 2.47Spinal cord.
The spinal cord is not uniform in diameter along its length. It has two major swellings or enlargements in regions associated with the origin of spinal nerves that innervate the upper and lower limbs. A cervical enlargement occurs in the region associated with the origins of spinal nerves C5 to T1, which innervate the upper limbs. A lumbosacral enlargement occurs in the region associated with the origins of spinal nerves L1 to S3, which innervate the lower limbs.
The external surface of the spinal cord is marked by a number of fissures and sulci (Fig. 2.48):
the anterior median fissure extends the length of the anterior surface;
the posterior median sulcus extends along the posterior surface;
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the posterolateral sulcus on each side of the posterior surface marks where the posterior rootlets of spinal nerves enter the cord.
Fig. 2.48Features of the spinal cord.
Internally, the cord has a small central canal surrounded by gray and white matter:
the gray matter is rich in nerve cell bodies, which form longitudinal columns along the cord, and in cross-section these columns form a characteristic H-shaped appearance in the central regions of the cord;
the white matter surrounds the gray matter and is rich in nerve cell processes, which form large bundles or tracts that ascend and descend in the cord to other spinal cord levels or carry information to and from the brain.
Vasculature
Arteries
The arterial supply to the spinal cord comes from two sources (Fig. 2.49). It consists of:
longitudinally oriented vessels, arising superior to the cervical portion of the cord, which descend on the surface of the cord; and
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feeder arteries that enter the vertebral canal through the intervertebral foramina at every level; these feeder vessels, or segmental spinal arteries, arise predominantly from the vertebral and deep cervical arteries in the neck, the posterior intercostal arteries in the thorax, and the lumbar arteries in the abdomen.
Fig. 2.49Arteries that supply the spinal cord. A. Anterior view of spinal cord (not all segmental spinal arteries are shown).
Arteries that supply the spinal cord. B. Segmental supply of spinal cord.
After entering an intervertebral foramen, the segmental spinal arteries give rise to anterior and posterior radicular arteries (Fig. 2.49). This occurs at every vertebral level. The radicular arteries follow, and supply, the anterior and posterior roots. At various vertebral levels, the segmental spinal arteries also give off segmental medullary arteries (Fig. 2.49). These vessels pass directly to the longitudinally oriented vessels, reinforcing these.
The longitudinal vessels consist of:
a single anterior spinal artery, which originates within the cranial cavity as the union of two vessels that arise from the vertebral arteries—the resulting single anterior spinal artery passes inferiorly, approximately parallel to the anterior median fissure, along the surface of the spinal cord; and
two posterior spinal arteries, which also originate in the cranial cavity, usually arising directly from a terminal branch of each vertebral artery (the posterior inferior cerebellar artery)—the right and left posterior spinal arteries descend along the spinal cord, each as two branches that bracket the posterolateral sulcus and the connection of posterior roots with the spinal cord.
The anterior and posterior spinal arteries are reinforced along their length by eight to ten segmental medullary arteries (Fig. 2.49). The largest of these is the arteria radicularis magna or the artery of Adamkiewicz (Fig. 2.49). This vessel arises in the lower thoracic or upper lumbar region, usually on the left side, and reinforces the arterial supply to the lower portion of the spinal cord, including the lumbar enlargement.
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Veins
Veins that drain the spinal cord form a number of longitudinal channels (Fig. 2.50):
two pairs of veins on each side bracket the connections of the posterior and anterior roots to the cord;
one midline channel parallels the anterior median fissure;
one midline channel passes along the posterior median sulcus.
Fig. 2.50Veins that drain the spinal cord.
These longitudinal channels drain into an extensive internal vertebral plexus in the extradural (epidural) space of the vertebral canal, which then drains into segmentally arranged vessels that connect with major systemic veins, such as the azygos system in the thorax. The internal vertebral plexus also communicates with intracranial veins.
Meninges
Spinal dura mater
The spinal dura mater is the outermost meningeal membrane and is separated from the bones forming the vertebral canal by an extradural space (Fig. 2.51). Superiorly, it is continuous with the inner meningeal layer of cranial dura mater at the foramen magnum of the skull. Inferiorly, the dural sac dramatically narrows at the level of the lower border of vertebra SII and forms an investing sheath for the pial part of the filum terminale of the spinal cord. This terminal cord-like extension of dura mater (the dural part of the filum terminale) attaches to the posterior surface of the vertebral bodies of the coccyx.
As spinal nerves and their roots pass laterally, they are surrounded by tubular sleeves of dura mater, which merge with and become part of the outer covering (epineurium) of the nerves.
Arachnoid mater
The arachnoid mater is a thin delicate membrane against, but not adherent to, the deep surface of the dura mater (Fig. 2.51). It is separated from the pia mater by the subarachnoid space. The arachnoid mater ends at the level of vertebra SII (see Fig. 2.47).
Fig. 2.51Meninges.
Subarachnoid space
The subarachnoid space between the arachnoid and pia mater contains CSF (Fig. 2.51). The subarachnoid space around the spinal cord is continuous at the foramen magnum with the subarachnoid space surrounding the brain. Inferiorly, the subarachnoid space terminates at approximately the level of the lower border of vertebra SII (see Fig. 2.47).
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Delicate strands of tissue (arachnoid trabeculae) are continuous with the arachnoid mater on one side and the pia mater on the other, span the subarachnoid space and interconnect the two adjacent membranes. Large blood vessels are suspended in the subarachnoid space by similar strands of material, which expand over the vessels to form a continuous external coat.
The subarachnoid space extends further inferiorly than the spinal cord. The spinal cord ends at approximately the disc between vertebrae LI and LII, whereas the subarachnoid space extends to approximately the lower border of vertebra SII (see Fig. 2.47). The subarachnoid space is largest in the region inferior to the terminal end of the spinal cord where it surrounds the cauda equina. As a consequence, CSF can be withdrawn from the subarachnoid space in the lower lumbar region without endangering the spinal cord.
Pia mater
The spinal pia mater is a vascular membrane that firmly adheres to the surface of the spinal cord (Fig. 2.51). It extends into the anterior median fissure and reflects as sleeve-like coatings onto posterior and anterior rootlets and roots as they cross the subarachnoid space. As the roots exit the space, the sleeve-like coatings reflect onto the arachnoid mater.
On each side of the spinal cord, a longitudinally oriented sheet of pia mater (the denticulate ligament) extends laterally from the cord toward the arachnoid and dura mater (Fig. 2.51).
Medially, each denticulate ligament is attached to the spinal cord in a plane that lies between the origins of the posterior and anterior rootlets.
Laterally, each denticulate ligament forms a series of triangular extensions along its free border, with the apex of each extension being anchored through the arachnoid mater to the dura mater.
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The denticulate ligaments generally occur between the exit points of adjacent posterior and anterior rootlets and position the spinal cord in the center of the subarachnoid space.
Arrangement of structures in the vertebral canal
The vertebral canal is bordered:
anteriorly by the bodies of the vertebrae, intervertebral discs, and posterior longitudinal ligament (Fig. 2.52);
laterally, on each side by the pedicles and intervertebral foramina; and
posteriorly by the laminae and ligamenta flava, and in the median plane the roots of the interspinous ligaments and vertebral spinous processes.
Fig. 2.52Arrangement of structures in the vertebral canal and the back.
Between the walls of the vertebral canal and the dural sac is an extradural space containing a vertebral plexus of veins embedded in fatty connective tissue.
The vertebral spinous processes can be palpated through the skin in the midline in thoracic and lumbar regions of the back. Between the skin and spinous processes is a layer of superficial fascia. In lumbar regions, the adjacent spinous processes and the associated laminae on either side of the midline do not overlap, resulting in gaps between adjacent vertebral arches.
When carrying out a lumbar puncture (spinal tap), the needle passes between adjacent vertebral spinous processes, through the supraspinous and interspinous ligaments, and enters the extradural space. The needle continues through the dura and arachnoid mater and enters the subarachnoid space, which contains CSF.
In the clinic
Lumbar cerebrospinal fluid tap
A lumbar tap (puncture) is carried out to obtain a sample of CSF for examination. In addition, passage of a needle or conduit into the subarachnoid space (CSF space) is used to inject antibiotics, chemotherapeutic agents, and anesthetics.
The lumbar region is an ideal site to access the subarachnoid space because the spinal cord terminates around the level of the disc between vertebrae LI and LII in the adult. The subarachnoid space extends to the region of the lower border of the SII vertebra. There is therefore a large CSF-filled space containing lumbar and sacral nerve roots, but no spinal cord.
Depending on the clinician’s preference, the patient is placed in the lateral or prone position. A needle is passed in the midline in between the spinous processes into the extradural space. Further advancement punctures the dura and arachnoid mater to enter the subarachnoid space. Most needles push the roots away from the tip without causing the patient any symptoms. Once the needle is in the subarachnoid space, fluid can be aspirated. In some situations, it is important to measure CSF pressure.
Local anesthetics can be injected into the extradural space or the subarachnoid space to anesthetize the sacral and lumbar nerve roots. Such anesthesia is useful for operations on the pelvis and the legs, which can then be carried out without the need for general anesthesia. When procedures are carried out the patient must be in the erect position and not lying on his or her side or in the head-down position. If a patient lies on his or her side, the anesthesia is likely to be unilateral. If the patient is placed in the head-down position, the anesthetic can pass cranially and potentially depress respiration.
In some instances, anesthesiologists choose to carry out extradural anesthesia. A needle is placed through the skin, supraspinous ligament, interspinous ligament, and ligamenta flava into the areolar tissue and fat around the dura mater. Anesthetic agent is introduced and diffuses around the vertebral canal to anesthetize the exiting nerve roots and diffuse into the subarachnoid space.
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Spinal nerves
Each spinal nerve is connected to the spinal cord by posterior and anterior roots (Fig. 2.53):
the posterior root contains the processes of sensory neurons carrying information to the CNS—the cell bodies of the sensory neurons, which are derived embryologically from neural crest cells, are clustered in a spinal ganglion at the distal end of the posterior root, usually in the intervertebral foramen;
the anterior root contains motor nerve fibers, which carry signals away from the CNS—the cell bodies of the primary motor neurons are in anterior regions of the spinal cord.
Fig. 2.53Basic organization of a spinal nerve.
Medially, the posterior and anterior roots divide into rootlets, which attach to the spinal cord.
A spinal segment is the area of the spinal cord that gives rise to the posterior and anterior rootlets, which will form a single pair of spinal nerves. Laterally, the posterior and anterior roots on each side join to form a spinal nerve.
Each spinal nerve divides, as it emerges from an intervertebral foramen, into two major branches: a small posterior ramus and a much larger anterior ramus (Fig. 2.53):
the posterior rami innervate only intrinsic back muscles (the epaxial muscles) and an associated narrow strip of skin on the back;
the anterior rami innervate most other skeletal muscles (the hypaxial muscles) of the body, including those of the limbs and trunk, and most remaining areas of the skin, except for certain regions of the head.
Near the point of division into anterior and posterior rami, each spinal nerve gives rise to two to four small recurrent meningeal (sinuvertebral) nerves. These nerves re-enter the intervertebral foramen to supply dura, ligaments, intervertebral discs, and blood vessels.
All major somatic plexuses (cervical, brachial, lumbar, and sacral) are formed by anterior rami.
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Because the spinal cord is much shorter than the vertebral column, the roots of spinal nerves become longer and pass more obliquely from the cervical to coccygeal regions of the vertebral canal (Fig. 2.54).
Fig. 2.54Course of spinal nerves in the vertebral canal.
In adults, the spinal cord terminates at a level approximately between vertebrae LI and LII, but this can range between vertebra TXII and the disc between vertebrae LII and LIII. Consequently, posterior and anterior roots forming spinal nerves emerging between vertebrae in the lower regions of the vertebral column are connected to the spinal cord at higher vertebral levels.
Below the end of the spinal cord, the posterior and anterior roots of lumbar, sacral, and coccygeal nerves pass inferiorly to reach their exit points from the vertebral canal. This terminal cluster of roots is the cauda equina.
Nomenclature of spinal nerves
There are approximately 31 pairs of spinal nerves (Fig. 2.54), named according to their position with respect to associated vertebrae:
eight cervical nerves—C1 to C8;
twelve thoracic nerves—T1 to T12;
five lumbar nerves—L1 to L5;
five sacral nerves—S1 to S5;
one coccygeal nerve—Co.
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The first cervical nerve (C1) emerges from the vertebral canal between the skull and vertebra CI (Fig. 2.55). Therefore cervical nerves C2 to C7 also emerge from the vertebral canal above their respective vertebrae. Because there are only seven cervical vertebrae, C8 emerges between vertebrae CVII and TI. As a consequence, all remaining spinal nerves, beginning with T1, emerge from the vertebral canal below their respective vertebrae.
Fig. 2.55Nomenclature of the spinal nerves.
In the clinic
Herpes zoster
Herpes zoster is the virus that produces chickenpox in children. In some patients the virus remains dormant in the cells of the spinal ganglia. Under certain circumstances, the virus becomes activated and travels along the neuronal bundles to the areas supplied by that nerve (the dermatome). A rash ensues, which is characteristically exquisitely painful. Importantly, this typical dermatomal distribution is characteristic of this disorder.
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In the clinic
Back pain—alternative explanations
Back pain is an extremely common condition affecting almost all individuals at some stage during their life. It is of key clinical importance to identify whether the back pain relates to the vertebral column and its attachments or relates to others structures.
The failure to consider other potential structures that may produce back pain can lead to significant mortality and morbidity. Pain may refer to the back from a number of organs situated in the retroperitoneum. Pancreatic pain in particular refers to the back and may be associated with pancreatic cancer and pancreatitis. Renal pain, which may be produced by stones in the renal collecting system or renal tumors, also typically refers to the back. More often than not this is usually unilateral; however, it can produce central posterior back pain. Enlarged lymph nodes in the pre- and para-aortic region may produce central posterior back pain and may be a sign of solid tumor malignancy, infection, or Hodgkin’s lymphoma. An enlarging abdominal aorta (abdominal aortic aneurysm) may cause back pain as it enlarges without rupture. Therefore it is critical to think of this structure as a potential cause of back pain, because treatment will be lifesaving. Moreover, a ruptured abdominal aortic aneurysm may also cause acute back pain in the first instance.
In all patients back pain requires careful assessment not only of the vertebral column but also of the chest and abdomen in order not to miss other important anatomical structures that may produce signs and symptoms radiating to the back.
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Surface anatomy
Back surface anatomy
Surface features of the back are used to locate muscle groups for testing peripheral nerves, to determine regions of the vertebral column, and to estimate the approximate position of the inferior end of the spinal cord. They are also used to locate organs that occur posteriorly in the thorax and abdomen.
Absence of lateral curvatures
When viewed from behind, the normal vertebral column has no lateral curvatures. The vertical skin furrow between muscle masses on either side of the midline is straight (Fig. 2.56).
Fig. 2.56Normal appearance of the back. A. In women. B. In men.
Primary and secondary curvatures in the sagittal plane
When viewed from the side, the normal vertebral column has primary curvatures in the thoracic and sacral/coccygeal regions and secondary curvatures in the cervical and lumbar regions (Fig. 2.57). The primary curvatures are concave anteriorly. The secondary curvatures are concave posteriorly.
Fig. 2.57Normal curvatures of the vertebral column.
Useful nonvertebral skeletal landmarks
A number of readily palpable bony features provide useful landmarks for defining muscles and for locating structures associated with the vertebral column. Among these features are the external occipital protuberance, the scapula, and the iliac crest (Fig. 2.58).
Fig. 2.58Back of a woman with major palpable bony landmarks indicated.
The external occipital protuberance is palpable in the midline at the back of the head just superior to the hairline.
The spine, medial border, and inferior angle of the scapula are often visible and are easily palpable.
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The iliac crest is palpable along its entire length, from the anterior superior iliac spine at the lower lateral margin of the anterior abdominal wall to the posterior superior iliac spine near the base of the back. The position of the posterior superior iliac spine is often visible as a “sacral dimple” just lateral to the midline.
How to identify specific vertebral spinous processes
Identification of vertebral spinous processes (Fig. 2.59A) can be used to differentiate between regions of the vertebral column and facilitate visualizing the position of deeper structures, such as the inferior ends of the spinal cord and subarachnoid space.
Fig. 2.59The back with the positions of vertebral spinous processes and associated structures indicated. A. In a man. B. In a woman with neck flexed. The prominent CVII and TI vertebral spinous processes are labeled. C. In a woman with neck flexed to accentuate the ligamentum nuchae.
The spinous process of vertebra CII can be identified through deep palpation as the most superior bony protuberance in the midline inferior to the skull.
Most of the other spinous processes, except for that of vertebra CVII, are not readily palpable because they are obscured by soft tissue.
The spinous process of CVII is usually visible as a prominent eminence in the midline at the base of the neck (Fig. 2.59B).
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Extending between CVII and the external occipital protuberance of the skull is the ligamentum nuchae, which is readily apparent as a longitudinal ridge when the neck is flexed (Fig. 2.59C).
Inferior to the spinous process of CVII is the spinous process of TI, which is also usually visible as a midline protuberance. Often it is more prominent than the spinous process of CVII.
The root of the spine of the scapula is at the same level as the spinous process of vertebra TIII, and the inferior angle of the scapula is level with the spinous process of vertebra TVII.
The spinous process of vertebra TXII is level with the midpoint of a vertical line between the inferior angle of the scapula and the iliac crest.
A horizontal line between the highest point of the iliac crest on each side crosses through the spinous process of vertebra LIV. The LIII and LV vertebral spinous processes can be palpated above and below the LIV spinous process, respectively.
The sacral dimples that mark the position of the posterior superior iliac spine are level with the SII vertebral spinous process.
The tip of the coccyx is palpable at the base of the vertebral column between the gluteal masses.
The tips of the vertebral spinous processes do not always lie in the same horizontal plane as their corresponding vertebral bodies. In thoracic regions, the spinous processes are long and sharply sloped downward so that their tips lie at the level of the vertebral body below. In other words, the tip of the TIII vertebral spinous process lies at vertebral level TIV.
In lumbar and sacral regions, the spinous processes are generally shorter and less sloped than in thoracic regions, and their palpable tips more closely reflect the position of their corresponding vertebral bodies. As a consequence, the palpable end of the spinous process of vertebra LIV lies at approximately the LIV vertebral level.
Visualizing the inferior ends of the spinal cord and subarachnoid space
The spinal cord does not occupy the entire length of the vertebral canal. Normally in adults, it terminates at the level of the disc between vertebrae LI and LII; however, it may end as high as TXII or as low as the disc between vertebrae LII and LIII. The subarachnoid space ends at approximately the level of vertebra SII (Fig. 2.60A).
Fig. 2.60Back with the ends of the spinal cord and subarachnoid space indicated. A. In a man.
Back with the ends of the spinal cord and subarachnoid space indicated. B. In a woman lying on her side in a fetal position, which accentuates the lumbar vertebral spinous processes and opens the spaces between adjacent vertebral arches. Cerebrospinal fluid can be withdrawn from the subarachnoid space in lower lumbar regions without endangering the spinal cord.
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Because the subarachnoid space can be accessed in the lower lumbar region without endangering the spinal cord, it is important to be able to identify the position of the lumbar vertebral spinous processes. The LIV vertebral spinous process is level with a horizontal line between the highest points on the iliac crests. In the lumbar region, the palpable ends of the vertebral spinous processes lie opposite their corresponding vertebral bodies. The subarachnoid space can be accessed between vertebral levels LIII and LIV and between LIV and LV without endangering the spinal cord (Fig. 2.60B). The subarachnoid space ends at vertebral level SII, which is level with the sacral dimples marking the posterior superior iliac spines.
Identifying major muscles
A number of intrinsic and extrinsic muscles of the back can readily be observed and palpated. The largest of these are the trapezius and latissimus dorsi muscles (Fig. 2.61A and 2.61B). Retracting the scapulae toward the midline can accentuate the rhomboid muscles (Fig. 2.61C), which lie deep to the trapezius muscle. The erector spinae muscles are visible as two longitudinal columns separated by a furrow in the midline (Fig. 2.61A).
Fig. 2.61Back muscles. A. In a man with latissimus dorsi, trapezius, and erector spinae muscles outlined.
Back muscles. B. In a man with arms abducted to accentuate the lateral margins of the latissimus dorsi muscles. C. In a woman with scapulae externally rotated and forcibly retracted to accentuate the rhomboid muscles.
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Clinical cases
Case 1SCIATICA VERSUS LUMBAGO
A 50-year-old woman visited her local family practitioner with severe lower back pain radiating into her right buttock.
Low back pain is a common problem in family practice.
Of the many common causes of low back pain some need to be identified early to commence appropriate treatment. The common causes include an anular disc tear, a disc prolapse that impinges directly on a nerve root, spinal stenosis, and mechanical zygapophysial joint pain. Overall, the main causes can be distilled into three central groups: mechanical back pain, degenerative joint disease, and neuronal compression.
Sciatica and lumbago are not the same. Lumbago is a generic term referring to low back pain. Sciatica is a name given to pain in the area of distribution of the sciatic nerve (L4 to S3), which is commonly felt in the buttock and over the posterolateral aspects of the leg.
Case 2CERVICAL SPINAL CORD INJURY
A 45-year-old man was involved in a serious car accident. On examination he had a severe injury to the cervical region of his vertebral column with damage to the spinal cord. In fact, his breathing became erratic and stopped.
If the cervical spinal cord injury is above the level of C5, breathing is likely to stop. The phrenic nerve takes origin from C3, C4, and C5 and supplies the diaphragm. Breathing may not cease immediately if the lesion is just below C5, but does so as the cord becomes edematous and damage progresses superiorly. In addition, some respiratory and ventilatory exchange may occur by using neck muscles plus the sternocleidomastoid and trapezius muscles, which are innervated by the accessory nerve [XI].
The patient was unable to sense or move his upper and lower limbs.
The patient has paralysis of the upper and lower limbs and is therefore quadriplegic. If breathing is unaffected, the lesion is below the level of C5 or at the level of C5. The nerve supply to the upper limbs is via the brachial plexus, which begins at the C5 level. The site of the spinal cord injury is at or above the C5 level.
It is important to remember that although the cord has been transected in the cervical region, the cord below this level is intact. Reflex activity may therefore occur below the injury, but communication with the brain is lost.
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Case 3PSOAS ABSCESS
A 25-year-old woman complained of increasing lumbar back pain. Over the ensuing weeks she was noted to have an enlarging lump in the right groin, which was mildly tender to touch. On direct questioning, the patient also complained of a productive cough with sputum containing mucus and blood, and she had a mild temperature.
The chest radiograph revealed a cavitating apical lung mass, which explains the pulmonary history.
Given the age of the patient a primary lung cancer is unlikely. The hemoptysis (coughing up blood in the sputum) and the rest of the history suggest the patient has a lung infection. Given the chest radiographic findings of a cavity in the apex of the lung, a diagnosis of tuberculosis (TB) was made. This was confirmed by bronchoscopy and aspiration of pus, which was cultured.
During the patient’s pulmonary infection, the tuberculous bacillus had spread via the blood to vertebra LI. The bone destruction began in the cancellous bone of the vertebral body close to the intervertebral discs. This disease progressed and eroded into the intervertebral disc, which became infected. The disc was destroyed, and the infected disc material extruded around the disc anteriorly and passed into the psoas muscle sheath. This is not an uncommon finding for a tuberculous infection of the lumbar portion of the vertebral column.
As the infection progressed, the pus spread within the psoas muscle sheath beneath the inguinal ligament to produce a hard mass in the groin. This is a typical finding for a psoas abscess.
Fortunately for the patient, there was no evidence of any damage within the vertebral canal.
The patient underwent a radiologically guided drainage of the psoas abscess and was treated for over 6 months with a long-term antibiotic regimen. She made an excellent recovery with no further symptoms, although the cavities within the lungs remain. It healed with sclerosis.
Case 4DISSECTING THORACIC ANEURYSM
A 72-year-old fit and healthy man was brought to the emergency department with severe back pain beginning at the level of the shoulder blades and extending to the mid lumbar region. The pain was of relatively acute onset and was continuous. The patient was able to walk to the gurney as he entered the ambulance; however, at the emergency department the patient complained of inability to use both legs.
The attending physician examined the back thoroughly and found no significant abnormality. He noted that there was reduced sensation in both legs, and there was virtually no power in extensor or flexor groups. The patient was tachycardic, which was believed to be due to pain, and the blood pressure obtained in the ambulance measured 120/80 mm Hg. It was noted that the patient’s current blood pressure was 80/40 mm Hg; however; the patient did not complain of typical clinical symptoms of hypotension.
On first inspection, it is difficult to “add up” these clinical symptoms and signs. In essence we have a progressive paraplegia associated with severe back pain and an anomaly in blood pressure measurements, which are not compatible with the clinical state of the patient.
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It was deduced that the blood pressure measurements were obtained in different arms, and both were reassessed.
The blood pressure measurements were true. In the right arm the blood pressure measured 120/80 mm Hg and in the left arm the blood pressure measured 80/40 mm Hg. This would imply a deficiency of blood to the left arm.
The patient was transferred from the emergency department to the CT scanner, and a scan was performed that included the chest, abdomen, and pelvis.
The CT scan demonstrated a dissecting thoracic aortic aneurysm. Aortic dissection occurs when the tunica intima and part of the tunica media of the wall of the aorta become separated from the remainder of the tunica media and the tunica adventitia of the aorta wall. This produces a false lumen. Blood passes not only in the true aortic lumen but also through a small hole into the wall of the aorta and into the false lumen. It often re-enters the true aortic lumen inferiorly. This produces two channels through which blood may flow. The process of the aortic dissection produces considerable pain for the patient and is usually of rapid onset. Typically the pain is felt between the shoulder blades and radiating into the back, and although the pain is not from the back musculature or the vertebral column, careful consideration of other structures other than the back should always be sought.
The difference in the blood pressure between the two arms indicates the level at which the dissection has begun. The “point of entry” is proximal to the left subclavian artery. At this level a small flap has been created, which limits the blood flow to the left upper limb, giving the low blood pressure recording. The brachiocephalic trunk has not been affected by the aortic dissection, and hence blood flow remains appropriate to the right upper limb.
The paraplegia was caused by ischemia to the spinal cord.
The blood supply to the spinal cord is from a single anterior spinal artery and two posterior spinal arteries. These arteries are fed via segmental spinal arteries at every vertebral level. There are a number of reinforcing arteries (segmental medullary arteries) along the length of the spinal cord—the largest of which is the artery of Adamkiewicz. This artery of Adamkiewicz, a segmental medullary artery, typically arises from the lower thoracic or upper lumbar region, and unfortunately during this patient’s aortic dissection, the origin of this vessel was disrupted. This produces acute spinal cord ischemia and has produced the paraplegia in the patient.
Unfortunately, the dissection extended, the aorta ruptured, and the patient succumbed.
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Case 5SACRAL TUMOR
A 55-year old woman came to her physician with sensory alteration in the right gluteal (buttock) region and in the intergluteal (natal) cleft. Examination also demonstrated low-grade weakness of the muscles of the foot and subtle weakness of the extensor hallucis longus, extensor digitorum longus, and fibularis tertius on the right. The patient also complained of some mild pain symptoms posteriorly in the right gluteal region.
A lesion was postulated in the left sacrum.
Pain in the right sacroiliac region could easily be attributed to the sacroiliac joint, which is often very sensitive to pain. The weakness of the intrinsic muscles of the foot and the extensor hallucis longus, extensor digitorum longus, and the fibularis tertius muscles raises the possibility of an abnormality affecting the nerves exiting the sacrum and possibly the lumbosacral junction. The altered sensation around the gluteal region toward the anus would also support these anatomical localizing features.
An X-ray was obtained of the pelvis.
The X-ray appeared on first inspection unremarkable. However the patient underwent further investigation, including CT and MRI, which demonstrated a large destructive lesion involving the whole of the left sacrum extending into the anterior sacral foramina at the S1, S2 and S3 levels. Interestingly, plain radiographs of the sacrum may often appear normal on first inspection, and further imaging should always be sought in patients with a suspected sacral abnormality.
The lesion was expansile and lytic.
Most bony metastasis are typically nonexpansile. They may well erode the bone producing lytic type of lesions or may become very sclerotic (prostate metastases and breast metastases). From time to time we see a mixed pattern of lytic and sclerotic.
There are a number of uncommon instances in which certain metastases are expansile and lytic. These typically occur in renal metastases and may be seen in multiple myeloma. The anatomical importance of these specific tumors is that they often expand and impinge upon other structures. The expansile nature of this patient’s tumor within the sacrum was the cause for compression of the sacral nerve roots, producing her symptoms.
The patient underwent a course of radiotherapy, had the renal tumor excised, and is currently undergoing a course of chemoimmuno therapy. |
189741 | https://www.youtube.com/watch?v=PXeTs1nf8YQ | Evaluating Limits by variable Substitution | mathematicaATD
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Evaluating Limits by variable Substitution | mathematicaATD
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dear friends our topic is evaluating limits using substitution method please follow two examples in both the examples you find that the numerator and denominator are simultaneously zero for the given limiting conditions see when X is equal to 0 then numerator of this example is 0 plus 8 whole to the power one-third minus 2 means 2 minus 2 that is 0 and for X is equal to 0 the denominator is 0 so in this case this is of 0 by 0 form that means the indeterminate form in the second example also if X is equal to 1 then 1 to the power 1/6 minus 1 means 1 minus 1 this is 0 and here the denominator is 1 minus 1 again 0 so this is also in 0x0 form this is also the indeterminate form ok so in such cases we need to simplify our expressions and one technique is a rationalization but the radical is difficult here X plus 8 whole to the power one-third and the 1 is X to the power 1/6 so we we have to use some substitution method so in the first case we shall substitute X plus 8 whole to the power one-third as Zed or in the second example X to the power one-sixth is equal to Jade so friends eveyone new here please subscribe the channel and please see the complete video for a proper knowledge it's considered that X plus 8 whole to the power one-third is equal to Jade okay so X plus 8 is equal to that cube so when X tends to 0 this is the limiting condition when X tends to 0 the did is standing - 0 + 8 0 + 8 whole to the power one-third that is that is standing - 2 so here limit Z tends to 2x + 8 whole to the power one-third is equal to Jade so Jade -2 divided by X what is X X is equal to Z cube minus 8 so jet cube minus 8 we can write limit there tends to - jet minus 2 divided by this is jet cube minus 2 Q actually this one so can be written Z minus 2 whole into Z square plus 2 Z plus 4 because we have applied here one formula I want to write this formula we have we know that a cube minus B cube is equal to a minus b whole into a square plus a B plus B Square okay this is a formula algebraic formula we have used it here you see that this is Jack cube minus 2 cube so Jade - - holy into Z square plus 2 into Z plus 2 squared 2 square means 4 so now friends in both the numerator and denominator we'll find that J is standing - - since that is standing - 2 we can write judge is naught is equal to 2 therefore Z minus 2 is not is equal to 0 so J minus 2 is a factor in the denominator so we can cancel Z minus 2 okay so here we find that limit Z tends to 2 1 divided by Z square plus 2 Z plus 4 okay so it is equal to 1/2 square plus 2 into 2 plus 4 so getting the result 1/4 plus 4 plus 4 means 1 upon 12 so the limiting value of the given function the function is this one is 1 by 12 are 1 1 upon 12 okay when limit X tends to 0 this is very important when X tends to 0 then the limiting value of this given function is equal to 1 upon 12 we shall now see the second example our second example is a limit X tends to 1 X to the power 1/6 minus 1 divided by X minus 1 so here we shall substitute X to the power 1/6 is equal to did therefore X is equal to Jack to the power 6 when X tends to 1 then jedd is tending to 1 to the point 1 6 means 1 so this is limit Z tends to 1 X to the power one sixth means Jade so this is Jade minus 1 divided by X minus 1 what is X X is equal to Jack to the power 6 so Jack to the power 6 minus 1 we could write limit Z tends to 1 J minus 1 divided by this J to the power 6 can be written jet cube whole square minus 1 square so it is limit Z tends to 1 J minus 1 divided by Jack Q plus 1 whole into jet cube minus 1 we have actually used one formula the formula I want to write this is also algebraic formula the formula is this one a square minus B square is equal to a plus B whole into a minus B we have used here jet cube as a and 1s B so a square minus B Square that means a plus B whole into a minus B now friends we want to break jet cube minus 1 okay we can write limit Z tends to 1 J minus 1 whole divided by Z cube plus 1 whole into this one can be written 1 cube now so this is jet cube minus 1 cube so this can be written as minus 1 whole into jet square plus Z plus 1 okay so now friends since God is standing to 1 since God is tending to 1 so that is naught is equal to 1 since Jolly's naught is equal to 1 so we can write jet minus 1 is not is equal to 0 so friends see that jet minus 1 is a factor in the denominator so sin jet minus 1 is not is equal to 0 so we can cancel jet minus 1 so again I want to write this one limit Z tends to 1 1 upon Z cube plus 1 whole into Z square plus Z plus 1 so for the limiting condition there tends to 1 the limit of the given function will be now 1 upon 1 cube plus 1 whole into 1 square plus 1 plus 1 so this is 1 upon 2 into 3 that is equal to 1 by 6 friends we see that for the function X to the power 1/6 minus 1 divided by X minus 1 when a limit X tends to one we find the limiting value of the given function is 10.6 so in this way you can find many more examples and you can solve these examples using substitution method thank you have a nice day |
189742 | https://pi.math.cornell.edu/~bowman/invfnthm.pdf | INVERSE FUNCTION THEOREM Before we recall the exact statement of the Inverse Function Theorem, let’s think about what we’d like for it to say. We’ve been talking about solving equations. Na¨ ıvely, given a function f : Rn →Rn and a value b in the range, we simply want to solve f(x) = b.
Newton’s method gives us a way to do this. But in the linear case, we have a much stronger situation: when f is invertible, we just have to find f −1 to get solutions to all equations f(x) = b. By examples, we know that it’s generally hopeless to expect this to happen for non-linear functions. But if we know a solution exists for some b0, we might hope that solutions also exist near b0. The Inverse Function Theorem tells us that this hope is (often) justified, and that the solutions depend differentiably on b near b0.
Theorem 1 (Inverse Function Theorem). Let f : U ⊂Rn →Rn be a C1 function on a neighborhood of x0. Suppose that Df(x0) is invertible. Then f has a local C1 inverse on a neighborhood of x0, i.e., f(x) = b has a solution for b in some ball around f(x0).
The concept of “locally invertible” may be difficult. First, you should realize that a property being “local” on a set simply means that every point in that set is contained in a neighborhood on which the property holds. (As opposed to “pointwise”, which only has to hold at each point: continuity is an example of a pointwise property.) Some examples may help explain why local invertibility is such an important concept.
Example 1 (A function that is everywhere locally invertible, but does not have a global inverse). Define f : R →R by f(x) = ( x2 if x ≥0 x2 −1 if x < 0.
At every point of R, f has a local inverse. For x > 0, it is y 7→√y; for x < 0, it is y 7→−√y + 1. There is also an inverse on the interval (−1, 1), given by y 7→ ( −√y + 1 if y ∈(−1, 0) √y if y ∈[0, 1).
However, f has no global inverse, because it is not one-to-one.
Example 2 (A differentiable example). Consider the exponential function exp : C →C.
As you saw in an earlier homework, the derivative of exp as a function R2 →R2 at z0 = x0 + iy0 is D exp x0 y0 = ex0 cos y0 −sin y0 sin y0 cos y0 , This matrix is always invertible—its determinant is e2x0 ̸= 0. Thus the Inverse Function Theorem guarantees a local inverse of exp at each point of C, and the inverse will even be differentiable! (Aside: such a local inverse for exp is called, naturally, a logarithm. But as we’ll see in a moment, logarithms are far from unique.) However, exp is not one-to-one on C: if z1 = x + iy1 and z2 = x + iy2, where y1 and y2 differ by a multiple of 2π, then ez1 = ez2; exp is periodic in the imaginary direction. (Wow!) Any point of C is contained in a ball of radius π on which exp is invertible. (In your spare time, you might think about what the image of this ball would look like.) |
189743 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_11?srsltid=AfmBOor0QRn9nFshToCarc9qGgVXqafOw1t5tVeIAfz034KMmb7b-GgA | Art of Problem Solving
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2018 AIME II Problems/Problem 11
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (Recursion)
5 Solution 4 (PIE)
6 Solution 5 (Recursion)
7 Solution 6 (Complementary)
Problem
Find the number of permutations of such that for each with , at least one of the first terms of the permutation is greater than .
Solution 1
If the first number is , then there are no restrictions. There are , or ways to place the other numbers.
If the first number is , can go in four places, and there are ways to place the other numbers. ways.
If the first number is , ....
4 6 _ _ _ _ 24 ways
4 _ 6 _ _ _ 24 ways
4 _ _ 6 _ _ 24 ways
4 _ _ _ 6 _ 5 must go between and , so there are ways.
ways if 4 is first.
If the first number is , ....
3 6 _ _ _ _ 24 ways
3 _ 6 _ _ _ 24 ways
3 1 _ 6 _ _ 4 ways
3 2 _ 6 _ _ 4 ways
3 4 _ 6 _ _ 6 ways
3 5 _ 6 _ _ 6 ways
3 5 _ _ 6 _ 6 ways
3 _ 5 _ 6 _ 6 ways
3 _ _ 5 6 _ 4 ways
ways
If the first number is , ....
2 6 _ _ _ _ 24 ways
2 _ 6 _ _ _ 18 ways
2 3 _ 6 _ _ 4 ways
2 4 _ 6 _ _ 6 ways
2 5 _ 6 _ _ 6 ways
2 5 _ _ 6 _ 6 ways
2 _ 5 _ 6 _ 4 ways
2 4 _ 5 6 _ 2 ways
2 3 4 5 6 1 1 way
ways
Grand Total:
Solution 2
If is the first number, then there are no restrictions. There are , or ways to place the other numbers.
If is the second number, then the first number can be or , and there are ways to place the other numbers. ways.
If is the third number, then we cannot have the following:
1 _ 6 _ _ _ 24 ways
2 1 6 _ _ _ 6 ways
ways
If is the fourth number, then we cannot have the following:
1 _ _ 6 _ _ 24 ways
2 1 _ 6 _ _ 6 ways
2 3 1 6 _ _ 2 ways
3 1 2 6 _ _ 2 ways
3 2 1 6 _ _ 2 ways
ways
If is the fifth number, then we cannot have the following:
_ _ _ _ 6 5 24 ways
1 5 _ _ 6 _ 6 ways
1 _ 5 _ 6 _ 6 ways
2 1 5 _ 6 _ 2 ways
1 _ _ 5 6 _ 6 ways
2 1 _ 5 6 _ 2 ways
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 3 ways
ways
Grand Total:
Solution 3 (Recursion)
Note the condition in the problem is equivalent to the following condition: for each with , the first terms is not a permutation (since it would mean there must be some integer in the first terms such that ). Then, let denote the number of permutations of satisfying the condition in the problem. We use complementary counting to find . Notice that in order to not satisfy the condition in the problem, there are cases: the first (we don't include since the condition in the problem only holds up to ) terms are a permutation of , and for all , the condition in the problem still holds. Then, for each of these cases, there are ways to arrange the first terms, and then ways to arrange the to terms (assume by contradiction that terms from to is a permutation of . Then, since the first terms are already a permutation of , the first terms form a permutation of . This contradicts our assumption that for all , the condition still holds. Thus, we can only include the permutations of these terms). Now, summing the cases up and subtracting from , we have: . From this recursion, we derive that , , , , , and finally .
~CrazyVideoGamez
~ (Frank FYC)
Solution 4 (PIE)
Let be the set of permutations such that there is no number greater than in the first places. Note that for all and that the set of restricted permutations is .
We will compute the cardinality of this set with PIE. To finish,
Solution 5 (Recursion)
Define the function as the amount of permutations with maximum digit and string length that satisfy the condition within bounds. For example, would be the amount of ways to make a string with length with the highest digit being . We wish to obtain .
To generate recursion, consider how we would get to from for all such that . We could either jump from the old maximum to the new by concatenating the old string and the new digit , or one could retain the maximum, in which case . To retain the maximum, one would have to pick a new available digit not exceeding .
In the first case, there is only one way to pick the new digit, namely picking . For the second case, there are digits left to choose, because there are digits between 1 and total and there are digits already chosen below or equal to . Thus, . Now that we have the recursive function, we can start evaluating the values of until we get to .
Our requested answer is thus ~sigma
Solution 6 (Complementary)
We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first terms is greater than for . We can further simplify this method by approaching it through casework on the first terms.
Case 1: None of the first one terms is greater than 1
The first term must obviously be one. Since there are five spaces left for numbers, there are a total of permutations for this case.
Case 2: None of the first two terms is greater than 2
The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with . Since there are four spaces left, there are a total of permutations for this case.
Case 3: None of the first three terms is greater than 3
The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now ways to order the first three numbers of the permutation, and ways to order the last three numbers, for a total of permutations.
Case 4: None of the first four terms is greater than 4
You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are ), the cases starting with 21 (there are ), and the 3 cases from case 3 (there are ) have been accounted for, so there are a total of permutations for this case.
Case 5: None of the first five terms is greater than 5
This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. start with 1, start with 2, start with 3, and start with 4. Subtracting, we get a total of permutations.
Now, we subtract the total number of permutations from our cases from the total number of permutations, which is : .
~TGSN/curiousmind_888
2018 AIME II (Problems • Answer Key • Resources)
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189744 | https://www.youtube.com/watch?v=eEwZeY51mT0 | 2d curl intuition
Khan Academy
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Posted: 27 May 2016
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A description of how vector fields relate to fluid rotation, laying the intuition for what the operation of curl represents.
21 comments
Transcript:
hello everyone so I'm going to start talking about curl curl is one of those very cool Vector calculus Concepts and you'll be pretty happy that you've learned it once you have if for no other reason because it's kind of artistically pleasing and there's two different versions there's a two-dimensional curl and a three-dimensional curl and naturally enough I'll start talking about the two-dimensional version and kind of build our way up to the 3D one and in this particular video I just want to lay down the intuition for what's what's visually going on and curl has to do with the fluid flow interpretation of vector Fields now this is something that I've talked about in other videos especially the ones on Divergence if you watch that but just as a reminder you kind of imagine that each point in space is a particle like an air molecule or a water molecule and since what a vector field does is associate each point in space with some kind of vector and remember we don't we don't always draw every single Vector we just draw a small subsample but in principle every single point in space has a vector attached to it you can think of each particle each one of these water molecules or air molecules as moving over time in such a way that the the velocity Vector of its movement at any given point in time is the vector that it's attached to so as it moves to a different location in space and that velocity Vector changes it might be turning or it might be accelerating and that velocity might change and you end up getting kind of a a trajectory for your point and since every single point is moving in this way you can start thinking about a flow kind of a global view of the vector field and for this particular example this particular Vector field that I have pictured I'm going to go ahead and put a blue dot at various points in space and each one of these you can think of as representing a water molecule or something and I'm just going to let it play and at any given moment if you look at the movement of one of these blue dots it's moving along the vector that it's attached to at that point or if that Vector is not pictured you know the vector that would be attached to it at that point and as we get kind of a feel for what's going on in this entire flow I want you to notice a couple particular regions um first let's take a look at this region over here on the right kind of around here and just kind of concentrate on what's going on there and I'll go ahead and start playing the animation over here and what's most notable about this region is that there's counterclockwise rotation and this corresponds to an idea that the vector field has a curl here and I I'll I'll go very specifically into what curl means but just right now you should have the idea that in a region where there's counterclockwise rotation we want to say the curl is positive whereas if you look at a region that also has rotation but clockwise going the other way we think of that as being negative curl here I'll start it over here um and in contrast if you look at a place where there's no rotation where like at the center here you have some points coming in from the top right and from the bottom left and then going out from the other corners but there's no net rotation if you were to just put a uh you know if you were to put like a twig somewhere in this water it wouldn't really be rotating these are regions where you think of them as having zero curl so with that as a general idea you know clockwise rotation regions correspond to positive curl counterclockwise rotation regions correspond to negative curl and then no rotation corresponds to zero curl in the next video I'm going to start going through what this means in terms of the underlying function defining the vector field and how we can start looking at the partial differential information of that function to quantify this intuition of fluid rotation and what's neat is that it's not just about fluid rotation right if um if you have Vector fields in other context and you just imagine that they represent a fluid even though they don't this idea of rotation and curling actually has certain importance in ways that you totally wouldn't wouldn't expect the gradient turns out to relate to the curl even though you wouldn't necessarily think the gradient has something to do with fluid rotation in electromagnetism this idea of fluid rotation has a certain importance even though fluids aren't actually involved so it's more General than just the representation that we have here but it's a very strong visual to have in your mind as you as you study Vector fields |
189745 | https://arxiv.org/pdf/2503.21988 | arXiv:2503.21988v1 [math.CO] 27 Mar 2025
A Linear Representation for Constant Term Sequences mod pa with Applications to Uniform Recurrence
Nadav Kohen Indiana University Bloomington, IN, USA nkohen@iu.edu February 14, 2025
Abstract
Many integer sequences including the Catalan numbers, Motzkin numbers, and the Apr´ ey numbers can be expressed in the form ConstantTermOf[ P nQ] for Laurent polynomials P and Q. These are often called “constant term sequences”. In this paper, we characterize the prime powers, pa, for which sequences of this form modulo pa, and others built out of these sequences, are uniformly recurrent. For all other prime powers, we show that the frequency of 0 is 1. This is accomplished by introducing a novel linear representation of constant term sequences modulo pa, which is of independent interest.
1 Introduction
Numerous famous combinatorial sequences admit descriptions of the form sn = ct [ P nQ]where P and Q are integral Laurent polynomials (of possibly many variables) and ct ex-tracts the constant term of a Laurent polynomial. For example, the Catalan numbers can be described by Cn = ct [( x−1 + 2 + x)n(1 − x)], the Motzkin numbers can be described by
Mn = ct [( x−1 + 1 + x)n(1 − x2)], and the Apr´ ey numbers (which were used to prove the irrationality of ζ(3)) can be described by A(n) = ct
[( (1+ x)(1+ y)(1+ z)(1+ y+z+yz +xyz )
xyz
)n]
.This paper is primarily focused on showing when sequences mod prime powers are uni-formly recurrent, which can be thought of as a weaker form of periodicity:
Definition 1. A sequence sn is called uniformly recurrent if for every word in sn (i.e., con-tiguous subsequence), w = sisi+1 · · · si+ℓ−1, there is a constant Cw such that every occurrence of w is followed by another occurrence of w at a distance of at most Cw. I.e., there is a j ≤ Cw
such that w = si+j si+j+1 · · · si+j+ℓ−1. If for all w, Cw is bounded by a constant multiple of the length of w, then we say that sn is linearly recurrent .1For background on uniform recurrence in automatic sequences, see Section 10.9 of and Section 10.8.8 of . In , Rowland and Zeilberger give an algorithm for constructing a Deterministic Finite Automaton with Output (DFAO) that computes such constant term sequences modulo a prime power. In , Rampersad and Shallit use these DFAOs in concert with the Wal-nut library to automatically prove theorems about these sequences using computers. In Theorem 10 of that paper the authors prove that, under certain conditions, the sequence
Tn = ct [( x−1 + 1 + x)n] mod p is uniformly recurrent if and only if it is never 0. They then pose Problem 6 in which they conjecture that the Motzkin numbers mod p are uniformly recurrent if and only if Tn mod p is never 0. In , this author showed that this conjecture is true and that more generally that if P is a symmetric trinomial in one variable, then ct [ P nQ] mod p is uniformly recurrent for every single-variable Q if and only if ct [ P n] mod p
is never 0 (which can be checked by finite means, see Proposition 6). In Theorem 20 and Corollary 21 of Section 5 of this paper, we extend the results of to show that ct [ P nQ] mod pa and every sequence built up of such sequences is linearly recur-rent if and only if ct [ P n] mod p is never 0, and here, no constraints are put on P and both polynomials may have any number of variables. Furthermore, if there exists an n such that ct [ P n] mod p = 0 then the frequency of 0 is 1 in every ct [ P nQ] mod pa and every sequence built up of such sequences. (The frequency of a value, c, in a sequence, ( sn)n∈N, is equal to the asymptotic density of {n ∈ N | sn = c}). This is accomplished by generalizing the proof method of Theorem 10 of by using constructions from Shallit’s book The Logical Approach to Automatic Sequences along with some novel constructions introduced in Sections 3 and 4. In Section 3, we construct a p-linear representation (Definition 4) for sequences of the form ct [ P nQ] mod p. This p-linear representation is in some sense equivalent to the Rowland-Zeilberger Automaton, but it makes many properties of these sequences more apparent and allows us to do away with an assumption made in Theorem 10 of . More generally, this p-linear representation seems to be of independent interest as it generalizes a family of Lucas congruences specified in Proposition 1 of , is amenable to efficient computation associated with constant term sequences mod p, and is largely independent of Q. This construction is generalized to constant term sequences modulo prime powers in Section 5.
1.1 Notation and Conventions
Throughout this paper, P and Q denote Laurent polynomials in r variables with integer coefficients, coef k1,...,k r [Q] denotes the coefficient of xk1
1
· · · xkr
r
, ct [ Q] denotes the constant term of Q, and deg Q denotes the largest absolute value of any exponent on any single vari-able in Q (for example, deg(5 x−3y−2 + 2 xy −1) = 3). For a fixed P , we let ank1,...,k r denote coef −k1,..., −kr [P (x1, . . . , x r)n] = ct [P (x1, . . . , x r)n · xk1
1
· · · xkr
r
] so that the ( r+1)-dimensional array ank1,...,k r forms an ( r + 1)-dimensional pyramid that generalizes Pascal’s triangle (which is achieved when P = x−1 + x). To make notation less verbose, we use the bold xk1,...,k r to 2refer to xk1
1
· · · xkr
r
and when k = ( k1, . . . , k r) ∈ Zr, we simply use xk. Furthermore, for such
k we denote by p | k that p divides every entry of k. Unless otherwise specified, vectors are column vectors and ~utr is the transpose of ~u (which is a row if ~u is a column). If Σ is a set, Σ ∗ denotes the set of words (i.e., strings) of any length whose characters are from Σ (including the empty word), and let |w| denote the length of a word in w ∈ Σ∗.If n is a non-negative integer, and p is a prime, then let np ∈ F∗
p
be the word whose char-acters are the digits of n in base p, with the least significant digit first. That is, if we let
np[i] denote the ith digit in the base-p expansion of n so that n = ∑
i∈Z≥0
np[i]pi, then
np = ( np)( np) · · · (np [|np| − 1]). Please note that in this author’s previous paper , most-significant-digit first (the reverse of this convention) was used, but here we use least-significant-digit first to more closely follow constructions in . Also note that when working with strings, exponents denote repetition. For example, ( p − 1) k ∈ F∗
p
denotes a run of k
characters that are all the character ( p − 1). Lastly, note that every statement made in this paper about np should also hold for np0k for every k.All morphisms in this paper refer to morphisms of strings, which are maps, f : Σ ∗
1
→ Σ∗
2
,such that f (c1c2 · · · ) = f (c1)f (c2) · · · , which can be defined by their action on the ele-ments of Σ 1. If for every c ∈ Σ1, |f (c)| = ℓ is constant then we say that f is ℓ-uniform .If a morphism is 1-uniform we call it a coding . We say that a morphism, f : Σ ∗ → Σ∗,is prolongable on c ∈ Σ if f (c) = cw where w ∈ Σ∗ and f n(w) is non-empty for all
n ≥ 0; in this case, f n(c) = cwf (w)f 2(w) · · · f n−1(w) for every n ≥ 0 and the infinite word f ω(c) = cwf (w)f 2(w) · · · is a fixed point of f (i.e., f (f ω(c)) = f ω(c)). A sequence, an, is p-automatic if it has values in some finite alphabet, and there exists a Deterministic Finite Automaton with Output (DFAO) that, given an index n in its base-p
representation, np, ends on a state whose output is an.A p-linear representation of rank m of a sequence an over Fp is a triple ( v, γ, w ) consisting of row vector from Fmp , a matrix-valued morphism, and column vector from Fmp , respectively. The string morphism γ : F∗
p
→ Mat m×m(Fp) uses matrix multiplication instead of concate-nation in its co-domain, such that v · γ(np) · w ≡ an (mod p) for all n ∈ N. We frequently replace the domain Fp above with Z/p aZ except in the input alphabet of γ, which is always
Fp. If an has such a p-linear representation, we say it is p-regular . Theorem 9.6.1 of says that a sequence with finitely many values is p-automatic if and only if it is p-regular.
2 Preliminaries
The Rowland-Zeilberger automaton is wonderfully described, with examples, in . Its ex-istence shows that constant term sequences modulo prime powers are p-automatic. The general construction is specified in the section titled Teaching the Computer How to Create Automatic p-schemes . We repeat some of that exposition here:
Definition 2. For every positive integer k, define Λ k to be the map from (and to) Laurent 3polynomials over Z (or the induced map on quotient rings) defined by Λk
∑
i∈[−m,m ]r
qixi
= ∑
k|i
qix i
k
.
That is, Λ k deletes terms whose exponents are not all multiples of k and then performs the change of variables xkj 7 → xj for every j.
Lemma 1. For all Laurent polynomials P and Q, all primes p and all integers n and k,
ct [P pn +k · Q] ≡ ct [P n · Λp
(P kQ)] (mod p).
Proof. First note the equivalence often called the generalized Freshman’s dream, P (x1, . . . , x r )p ≡
P (xp
1
, . . . , x pr ) (mod p), and then note that the change of variables xpi 7 → xi does not affect constant terms. From these two observations, it follows that ct [P (x1, . . . , x r )pn +k · Q(x1, . . . , x r)] ≡ ct [P (xp
1
, . . . , x pr )n · P (x1, . . . , x r)kQ(x1, . . . , x r)]
= ct [P (x1, . . . , x r)n · Λp
(P (x1, . . . , x r)kQ(x1, . . . , x r))] .
Using this congruence, we can construct a Deterministic Finite Automaton with Output (DFAO) whose states are labeled with Laurent polynomials by beginning with the state Q,and then having transitions labeled k for k ∈ Fp that lead to (possibly new) states labeled by Q′ = Λ p
(P kQ), and repeating this process until no new states can be reached (see the following lemma to understand why this terminates). The output of a state labeled Q is ct [ Q]. This DFAO is known as the Rowland-Zeilberger automaton, and its construction constitutes a proof that an = ct [ P nQ] mod p is a p-automatic sequence.
Lemma 2. For a fixed P , if we define Λkp(Q) = Λ p(P kQ), then
m := max(deg( P ) − 1, deg( Q)) ≥ deg Λ kt
p
(· · · Λk0
p
(Q) · · · ).
That is, m is a bound on the degree of every polynomial that is the result of iterating Λkp
on input Q for all values of k ∈ Fp. In particular, m bounds the degree of every state, Q,appearing in the Rowland-Zeilberger construction. Proof. If Q0 = ∑
i∈[−m0,m 0]r
qixi is of degree m0, then for all k ≤ p − 1, Λ kp is Fp-linear so that Q1 = Λ kp (Q0) = ∑
i∈[−m0,m 0]r
qiΛp(P kxi) and deg Q1 = max i
(deg(Λ p(P kxi)) ), which depends only on k, m 0, and P . For i ≥ 0, letting |(j1, . . . , j r)| = max k jk, and recalling that
ak
−j
= coef j
[P k],Λp(P kxi) = Λ p
∑
|j|<k deg P
ak
−j
xi+j
= ∑
|j|≤ kdeg P,p |i+j
ak
−j
xi+j
p
.
4From this, since |j| ≤ k deg P ⇒ | i+j
p
| ≤ k deg P +|i|
p
≤ (p−1) deg P +|i|
p
, we can see that deg Q1 ≤ p − 1
p deg P + deg Q0
p ,
independently of k. And in general, a simple induction shows that every Qn reached from
Q0 in n steps has deg Qn ≤ deg P + deg Q0 − deg P
pn .
From this, we can immediately see that if deg Q0 < deg P then deg Qn < deg P as well for all n. Furthermore, the degrees of our Qis are decreasing when deg Q0 > deg P since this implies that deg Q1 ≤ deg P + deg Q0−deg P
p
= (p−1) deg P +deg Q0
p
< deg Q0; in particular, if log p(deg Q0 − deg P ) < n then deg Qn ≤ deg P for every Qn reachable from Q0 in n steps. Lastly, if deg Q0 = deg P , then deg Q1 = deg Λ p(P kQ0) < deg P for all k < p − 1, while deg Λ p(P p−1Q0) = deg P .
For a different analysis yielding Lemma 2, see Theorem 2.4 of . We use m as defined in Lemma 2 for the remainder of the paper when P and Q are understood from context. In general, p-automatic sequences have (at least) three useful descriptions. First, as a sequence determined by a DFAO as above. Second, as the result of coding a fixed point of a prolongable uniform morphism, whose equivalence to the first description is known as Cobham’s Little Theorem (or see Theorem 5.4.1 of ). Third, every p-automatic se-quence is always p-regular and hence has a p-linear representation ( v, γ, w ), and the converse (p-regular ⇒ p-automatic) is true for sequences taking on finitely many values. This equiva-lence is Theorem 9.6.1 of , and the proof is constructive: using the p-linear representation of our sequence, a prolongable uniform morphism is built whose infinite fixed-point has a coding yielding the same sequence. We make use of all three types of descriptions to prove our results, but the “free” construc-tions are not sufficient for our purposes. Hence, rather than using the Rowland-Zeilberger automaton for an = ct [ P nQ] mod p directly, we instead derive a p-linear representation for
an by using Lemma 1 directly. This yields a generalization of the Lucas congruences for cen-tral binomial and trinomial coefficients, which were the primary tools used in to derive similar uniform recurrence results when P is of degree at most 1. We then use this p-linear representation to build our other descriptions of an.
3 A Novel Linear Representation
Our p-linear representation is the result of interpreting the Rowland-Zeilberger machine within the vector space of Laurent polynomials over Fp of degree bounded by m. Note that this construction differs from the standard construction creating a p-linear representation from a DFAO, which uses the states of the DFAO as a basis rather than, for example, the standard basis of powers of x.5Definition 3. Recall the definition of m from Lemma 2. We wish to encode Laurent poly-nomials with coefficients in Fp of degree ≤ m as a product of copies of Fp, and so we must choose a computational basis: Impose an order on T = [ −m, m ]r (e.g. lexicographical), where T indexes the set of coefficients in a polynomial with degree ≤ m. If Q = ∑
i∈T
qixi
then let vec( Q) be the row vector ( qj )j∈T and let code Q : ( FTp )∗ → F∗
p
be the coding defined by code Q(~u) = vec( Q) · ~u.
Proposition 3. For every k ∈ Fp and i ∈ T , Λp(P kxi) = ∑
j∈T
aki−pj xj .Proof. Using the definition of Λ p,coef j
[Λp(P kxi)] = coef pj
[P kxi] = coef pj −i
[P k] = aki−pj .
Definition 4. Notice that the map Q 7 → Λp(P kQ) is Fp-linear and maps FTp (the space of bounded-degree Laurent polynomials) to itself by Lemma 2. Thus, this map corresponds to a matrix with entries in Fp, which we denote γ(k). This map is determined by its action on the standard basis of monomials (written as row vectors to the left of γ(k)). Therefore, we can deduce from Proposition 3 that
γ(k) := ( aki−pj )i,j ∈T .
The matrix-valued string morphism, γ, defined in this way along with the row vector vec( Q)and the column vector, V (0), whose only non-zero entry is at the index corresponding to the constant term constitute a p-linear representation for the sequence an = ct [ P nQ] mod p.This is because the computation vec( Q) · γ(np) · V (0) when multiplied from left to right is exactly the computation executed by the Rowland-Zeilberger machine yielding an.Put another way: in Shallit’s The Logical Approach to Automatic Sequences , section 9.3, he describes how to convert a p-linear representation of a p-automatic sequence into a set of relations of its finite p-kernel, and then Theorem 5.5.1 (which shows constructively that a sequence has finite p-kernel if and only if it is p-automatic) defines a process for converting this finite set of relations into a DFAO describing the sequence. Successively applying these two algorithms to the p-linear representation (vec( Q), γ, V (0)) exactly yields the Rowland-Zeilberger DFAO described in this section. The matrices, γ(k), of Definition 4 can also be interpreted as acting on the left of column vectors that correspond to linear functionals. That is, we begin with V (0), which corresponds to the linear functional Q 7 → ct [ Q], and then read the most significant base-p digit, k, of n
in order to update to the new linear functional Q 7 → ct [P kQ]. This process is then repeated until the final linear functional is Q 7 → ct [ P nQ], as desired. This observation motivates the following definition and lemma.
Definition 5. Let V (n) be the column vector ( ani )i∈T ∈ FTp , which stores the coefficients of
P n whose terms have degree at most m.
Lemma 4. For all non-negative integers n and all k ∈ Fp,
γ(k) · V (n) = V (pn + k).
6Proof. In view of Lemma 1 and Proposition 3, for all i ∈ T ,
V (pn + k)[ i] = apn +ki
= ct [P pn +kxi]
≡ ct [P n · Λp(P kxi)] (mod p)= ct
[
P n · ∑
j∈T
aki−pj xj
]
= ∑
j∈T
aki−pj · ct [P nxj ]
= ∑
j∈T
aki−pj anj
= ((aki−pj )i,j ∈T · V (n)) [i]= ( γ(k) · V (n)) [ i].
The result follows.
Remarkably, this lemma shows that modulo p, the low-degree coefficients of P n can be efficiently computed without reference to higher degree coefficients (except for those of P k
with k < p ). It also details the process executed by our new reverse Rowland-Zeilberger ma-chine whose states are the V (i), which have output code Q(V (i)). This machine is essentially an instantiation of the reversal of a generic DFAO given in Theorem 4.3.3 of , except that we keep the algebraic information inherited from the fact that we are computing constant term sequences. This description also shows that directly computing the reverse Rowland-Zeilberger machine is not, in principle, more expensive than computing the usual machine. Finally, the primary benefit of this description of ct [ P nQ] mod p is that it is independent of Q until the output is computed from the final state. That is, we can study the sequence (V (n)) n∈N whose elements are in FTp and think of all of our constant term sequences as the images of this one sequence under various codings. To this end, we introduce a description of the sequence ( V (n)) n∈N as a fixed point of a prolongable uniform morphism.
Definition 6. When a Laurent polynomial P , a prime p, and an integer m ≥ deg( P ) − 1(and m ≥ deg Q for every Q that we may be considering) are fixed, we define the p-uniform morphism σ : (FTp
)∗ → (FTp
)∗ by
σ(~u) := ( γ(0) · ~u)( γ(1) · ~u) · · · (γ(p − 1) · ~u).σ is prolongable on V (0) since γ(0) · V (0) = V (0) by Lemma 4, and so
α := σω (V (0)) = V (0) V (1) V (2) · · · ∈ (FTp
)∗
whose nth character is V (n), is a fixed point of σ, i.e., σ(α) = α.7To see that α[n] = V (n), note that
σ(V (j)) = ( γ(0) · V (j)) · · · (γ(p − 1) · V (j)) = V (pj ) · · · V (pj + p − 1) and thus σ(V (0) V (1) · · · V (pn − 1)) is equal to (V (0) · · · V (p − 1))( V (p) · · · V (p2 − 1)) · · · (V (pn+1 − p) · · · V (pn+1 − 1)) .
In essence, we can fluidly and interchangeably utilize all four (compatible) representations of ct [ P nQ]; as the Rowland-Zeilberger DFAO (Lemma 1), as the reverse Rowland-Zeilberger DFAO (Lemma 4), using the p-linear representation of Definition 4, and using a coding of the fixed point of Definition 6. The dictionary allowing us to go between these representations is most easily stated by relating each description to the p-linear representation as follows: The states of the Rowland-Zeilberger DFAO are labeled by polynomials Q, corresponding to vec( Q) in the p-linear representation; transitions from Q to Q′ labeled by k in the DFAO are such that vec( Q) · γ(k) = Q′. Likewise, the states of the reverse Rowland-Zeilberger DFAO are labeled by the V (i) with transitions from V (i) to V (j) labeled by k such that
γ(k)·V (i) = V (j). Lastly, the relation γ(k)·V (n) = V (pn +k) describes the progression of the fixed-point, α, which can be useful for reasoning about properties such as uniform recurrence. We end this section with a fundamental lemma about γ(0):
Lemma 5. Let C be the index associated with the polynomial 1 in T ’s order. There is an integer s ∈ N such that for every s′ ≥ s, γ(0) s′
= γ(0 s′
) = eC,C , that is to say it has a 1 in the row and column corresponding to the constant term and 0s everywhere else. Proof. The matrix eC,C is characterized by the fact that for every vector ~u, ~utr · eC,C =(~u[C]) etr C since, in particular, this being true for the basis vectors etr i implies that all rows other than the C row are zero and that the C row is etr C . Every row vector can be written as vec( Q) for some Q, and vec( Q)·γ(0) = vec(Λ p(P 0Q)) = vec(Λ p(Q)). Furthermore, Λ p(Q) has degree at most 0·deg P +deg Q
p
= deg Q
p
as seen in Lemma 2. So letting s′ ≥ s = ⌊log p(deg Q)⌋ + 1 yields vec( Q) · γ(0) s′
= vec((Λ 0
p
)s′
(Q)) = vec(ct [ Q]) = (vec( Q)[ C]) etr C , as desired.
This lemma can be thought of as partially explaining some of the connection between the sequences ct [ P nQ] and ct [ P n · 1] since, beginning with Q, we can always reach a state corresponding to Q′ ∈ Fp ⊂ Fp[x1, . . . , x r, x −11 , . . . , x −1
r
] in some fixed number of steps using
γ(0 s).
4 Classifying Uniformly Recurrent Constant Term Se-quences mod p
We begin by analyzing the case in which ct [ P nQ] mod p is not uniformly recurrent, in which case the frequency of 0 is 1. First, we note that checking whether a 0 appears in ct [ P n] mod p
can be done by only inspecting a prefix of this sequence. 8Proposition 6. Let P be any Laurent polynomial and p be prime. There is a BP,p ∈ N,depending on p and deg( P ), so that if there exists some n ∈ N such that p | ct [ P n], then there exists an n0 ∈ N with n0 < B P,p such that p | ct [ P n0 ].Proof. The Rowland-Zeilberger automaton for ct [ P n] mod p contains at most p|T | = p(2 m+1) r
=
p(2 ·deg( P )−1) r
states. If one of the states has output 0, then there must be a path to it from the starting state of length less than the number of states. Thus, if we let BP,p = pp(2 ·deg( P )−1) r
,then there must be a 0 in our sequence for some n0 < B P,p .
However, this bound of pp(2 ·deg( P )−1) r
seems far too large. In the author’s brief computer experimentation in one variable, no sequence or prime was found violating the following bound:
Conjecture 1. Proposition 6 holds for BP,p = pdeg( P ) when P has one variable ( r = 1 ).
Now, we show one direction of our main result: constant term sequences are not uniformly recurrent when the corresponding Q = 1 constant term sequence has an entry divisible by p.
Theorem 7. If P is any Laurent polynomial and there exists some n ∈ N such that ct [ P n] ≡
0 (mod p), then the frequency of 0 is 1 in every sequence of the form ct [ P nQ] mod p for every Laurent polynomial Q. That is, the set of indices, n, such that ct [ P nQ] ≡ 0 (mod p) has density 1. In particular, ct [ P nQ] mod p contains arbitrarily large runs of 0s and is not uniformly recurrent. Proof. If p | ct [ P n0 ], then there exists s ∈ N by Lemma 5 such that γ(0) sγ(n0)γ(0) s = 0. Because ct [ P nQ] ≡ vec( Q) · V (n) = vec( Q) · γ(n) · V (0), all n such that np contains 0 s(n0)p0s
must have ct [ P nQ] ≡ 0. If β = |(n0)p|, then consider each n ∈ [0, p (β+2 s)·k) as corresponding to an element of (Fβ+2 sp
)k via its base-p representation chunked into blocks of size β + 2 s.If any of the k entries in Fβ+2 sp for n correspond to 0 s(n0)p0s, then we know that V (n) = 0 and ct [ P nQ] ≡ 0, so at least p(β+2 s)·k − (pβ+2 s − 1)k of such n have ct [ P nQ] ≡ 0. Therefore, the frequency of 0 in ct [ P nQ] mod p is at least p(β+2 s)·k−(pβ+2 s−1)k
p(β+2 s)·k
= 1 −
(pβ+2 s−1
pβ+2 s
)k
→ 1 as
k → ∞ . In fact, this shows that the frequency of 0 in α (of Definition 6) is 1.
We wish to prove that if 0 does not appear in ct [ P n] mod p, then ct [ P nQ] mod p is uni-formly recurrent. Our main result follows from the following proposition about the morphism
σ of Definition 6:
Proposition 8. There exists t0 ∈ N such that for all n ∈ N, if p ∤ ct [ P n] then σt0 (V (n)) =
x0V (0) y0 with x0, y 0 ∈ (FTp
)∗. That is, V (0) is a character in σt0 (V (n)) for every such n.Proof. Let s ∈ N such that γ(0) s = eC,C . Next, fix s′ ∈ N such that C-term entries (i.e., indexed by C) of σs′
(V (0)) contain all C-term values of V (i) that occur for any i ∈ N.Recall that the C-term value of V (i) is the constant coefficient of P i. An s′ exists since there are finitely many (namely, p) values that C-term entries of V (i) can take on, each initially occurring at some finite value of i. Let V (n)[ C] ∈ Fp denote the entry of V (n) corresponding to the constant term, which is not 0 by our assumption that p ∤ ct [ P n]. Finally, we have that
γ(0) s · V (n) = ( V (n)[ C]) V (0), σs′+s(V (0)) contains ( V (n)[ C]) V (0), and ( V (n)[ C]) p−1 ≡ 1(mod p) by Fermat’s Little Theorem. Therefore, we can conclude that σ(s′+s)( p−2)+ s(V (n)) contains V (0), as desired.
9Note that t0 above does not depend on n. We now have everything we need to prove our main result and more.
Theorem 9. If P and Q are Laurent polynomials with integer coefficients and p is prime, then the sequence ct [ P nQ] mod p is linearly recurrent if and only if p ∤ ct [ P n] for all n ∈ N,and this can be checked for bounded n < B P,p . When the sequence is not linearly recurrent, it is 0 with frequency 1. This classification is independent of Q.Proof. If there exists an n ∈ N such that p | ct [ P n], then Proposition 6 tells us that there is such an n < B P,p . And Theorem 7 tells us that ct [ P nQ] mod p is not uniformly recurrent. For the converse, we show that α from Definition 6 is linearly recurrent, from which the result follows since our sequence is the image of α under code Q. Theorem 10.8.4 of states that if a p-automatic sequence is uniformly recurrent, it is linearly recurrent. Thus, it is sufficient for us to show that α is uniformly recurrent. If w appears as a word in α, and its first occurrence ends by index i < p t1 , then w
appears in σt1 (V (0)) = V (0) V (1) · · · V (pt1 − 1) since σ is p-uniform. We can conclude from Proposition 8 that for all n ∈ N, and all words w appearing in α,
σt1 +t0 (V (n)) = σt1 (x0V (0) y0) = σt1 (x0)σt1 (V (0)) σt1 (y0) = σt1 (x0)x1wy 1σt1 (y0).
In other words, w appears in every σt1 +t0 (V (n)), and since
α = σt1+t0 (α) = σt1+t0 (V (0)) σt1+t0 (V (1)) σt1+t0 (V (2)) · · ·
and σ is uniform (so that σt1 +t0 is also uniform), this implies that α = V (0) V (1) V (2) · · · is uniformly recurrent.
Corollary 10. Let F : Fℓp → X (which is deterministic) for some ℓ ∈ N and let bn =
F (ct [P n+k1 Q1
] mod p, . . . , ct [P n+kℓ Qℓ
] mod p) where the ki are arbitrary non-negative in-tegers and the Qi are arbitrary Laurent polynomials. Then bn is linearly recurrent if and only if for all n ∈ N, p ∤ ct [ P n]. Furthermore, in the case where there exists n such that
p | ct [ P n], then the frequency of F (0 , . . . , 0) in bn is 1. For example, if F takes a linear combination of its inputs then we get that bn = ∑ℓi=0 βi · (ct [P n+ki Qi
] mod p) for arbitrary
βi is linearly recurrent if and only if p ∤ ct [ P n] for all n.Proof. The proof of Theorem 9 actually shows that α = V (0) V (1) V (2) · · · is linearly recurrent. Since each word beginning with bn0 of length L is determined by the word
V (n0 − min( ki)) · · · V (n0 + max( ki) + L), we can conclude that the sequence bn is linearly recurrent when α is. Furthermore, if α is not linearly recurrent, it contains 0 vectors with frequency 1 (by Theorem 7), and hence bn contains F (0 , . . . , 0) with frequency 1.
We conclude this section by noting that even in the case where p | ct [ P n0 ] for some n0,resulting in α containing 0 vectors with frequency 1, we still have that α is recurrent (see below) so long as there exists an n such that p ∤ ct [ P n]. 10 Definition 7. A sequence sn is called recurrent if for every occurrence of a word, w =
sisi+1 · · · si+ℓ−1, there is another later occurrence. I.e., there is a j > 0 such that w =
si+j si+j+1 · · · si+j+ℓ−1. This is equivalent to saying that every word that appears recurs infinitely often, but there is no bound on gaps between occurrences.
Proposition 11. If P and Q are Laurent polynomials and p is prime, and there exists some
n0 > 0 such that p ∤ ct [ P n0 ], then the sequence bn = ct [ P nQ] mod p is recurrent. Proof. Proposition 8 assures us that σt0 (V (n0)) contains V (0). Therefore, if w is a word in α
and we consider an occurrence of w that ends by index i < p t2 , then w appears in σt2 (V (0)) and thus w appears in σt2+t0 (V (n0)). Finally, since
α = σt2+t0 (α) = σt2+t0 (V (0)) σt2+t0 (V (1)) · · · σt2+t0 (V (n0)) · · ·
we can conclude that w recurs and consequently the corresponding words in bn do as well. Specifically, the occurrence of w we began with is in the first segment, σt2+t0 (V (0)), of the above decomposition of α and there is a subsequent occurrence in the σt2+t0 (V (n0)) segment. Note that the obstruction to uniform recurrence is the dependence of t2 on the index of the occurrence of w since there can be arbitrarily large gaps between constructed occurrences.
Question 1. What can be said about the recurrence (or non-recurrence) of ct [ P n] when
ct [ P n] ≡ 0 for all n > 0? For example, this is true for P (x) = xp−1 + x−1.
5 Extending Results to Prime Powers
Analogous results to Theorem 9 and Corollary 10 hold when we consider constant term se-quences modulo general prime powers, but some extra care is required in this case. To avoid unnecessary confusion in the case of primes (as presented above), the discussion of prime powers has been excised to this section. The construction of the Rowland-Zeilberger automaton in becomes mildly more com-plicated mod pa for a > 1 because Lemma 1 no longer holds. However, after reading fewer than a base-p digits, we have an analogous result.
Definition 8. When the Laurent polynomial P with coefficients in Z/p aZ is fixed, we define ˜P (x) to be such that P (x1, . . . , x r)pa−1
≡ ˜P (xpℓ
1
, . . . , x pℓ
r
) (mod pa) where ℓ is taken to be as large as possible. Note that since P (x1, . . . , x r)p ≡ P (xp
1
, . . . , x pr ) (mod p), it follows that
P (x1, . . . , x r)pa
= ( P (x1, . . . , x r)p)pa−1
≡ P (xp
1
, . . . , x pr )pa−1
(mod pa).
Thus, ˜P (xpℓ
1
, . . . , x pℓ
r
)p ≡ ˜P (xpℓ·p
1
, . . . , x pℓ·pr ) (mod pa), or equivalently, by a change of vari-ables, ˜P (x1, . . . , x r)p ≡ ˜P (xp
1
, . . . , x pr ) (mod pa).
Consequently, ˜P is stable under Rowland-Zeilberger transitions, i.e., ˜˜P = ˜P . Also note that when a = 1, this section coincides with the previous one if we take ˜P = P .11 Lemma 12. For all Laurent polynomials P and primes p, there exists an n0 ∈ N such that
p | ct [ P n0 ] if and only if there exists an n1 ∈ N such that p | ct
[ ˜P n1
]
. Furthermore, if n0
exists one can take n1 to be n0.Proof. Because ct
[ ˜P (x1, . . . , x r )n
]
= ct
[ ˜P (xpℓ
1
, . . . , x pℓ
r
)n
]
= ct
[
P (x1, . . . , x r)pa−1n
]
, the se-quence (ct
[ ˜P n
]
)n∈N is a subsequence of (ct [ P n]) n∈N. Thus, it is sufficient to show that when ct [ P n0 ] 6 ≡ 0 (mod p), then ct
[ ˜P n0
]
6≡ 0 (mod p). By Lemma 5 and Definition 5 we have that γ(0) s ·V (n0) 6 = 0. Furthermore, using Lemmas 5 and 4,
γ(0) s · V (n0) = γ(0) s+a−1 · V (n0) = γ(0) s · γ(0) a−1 · V (n0) = γ(0) s · V (pa−1n0) 6 = 0 ,
from which we can conclude that ct
[ ˜P n0
]
≡ ct
[
P pa−1n0
]
6≡ 0 (mod p).
We now apply some simple tricks to describe ct [ P nQ] (mod pa) in terms of sequences of the form ct
[ ˜P nQ
]
(mod pa). And for ˜P , since ct
[ ˜P (x1, . . . , x r)p
]
≡ ct
[ ˜P (xp
1
, . . . , x pr )
]
(mod pa), all of our completed analysis applies so long as we substitute Fp coefficients with coefficients from Z/p aZ and various symbols with their (mod pa)-counterparts, which are labeled with the subscript 2 in this section. These translations are so direct that we omit proofs for most of the results and instead reference the analogous results above.
Proposition 13. For all Laurent polynomials P and Q with coefficients in Z/p aZ, all n ∈ N,and all k ∈ Z/p a−1Z,
ct
[
P pa−1n+kQ
]
≡ ct
[ ˜P n · Λpℓ (P kQ)
]
(mod pa).
Proof.
ct
[
P (x1, . . . , x r)pa−1n+kQ(x1, . . . , x r)
]
= ct
[
(P (x1, . . . , x r)pa−1
)nP (x1, . . . , x r)kQ(x1, . . . , x r)
]
≡ ct
[ ˜P (xpℓ
1
, . . . , x pℓ
r
)nP (x1, . . . , x r)kQ(x1, . . . , x r)
]
≡ ct
[ ˜P (x1, . . . , x r)nΛpℓ (P (x1, . . . , x r)kQ(x1, . . . , x r))
]
.
Now we can compute ct [P n′
Q] mod pa by reading the a − 1 least significant base-p
digits of n′, call this number k, and then use the digits remaining, call this number n, and then compute ct
[ ˜P n · Λpℓ (P kQ)
]
mod pa where ˜P satisfies ˜P (x1, . . . , x r)p ≡ ˜P (xp
1
, . . . , x pr )(mod pa). Using this congruence, we obtain lemmas analogous to Lemma 1 and Lemma 2 for ˜P :12 Lemma 14. For all k ∈ Fp, n ∈ N and all Laurent polynomials, ˜P (x), with coefficients in
Z/p aZ satisfying ˜P (x1, . . . , x r )p ≡ ˜P (xp
1
, . . . , x pr ) (mod pa),
ct
[ ˜P pn +kQ
]
≡ ct
[ ˜P n · Λp( ˜P kQ)
]
(mod pa).
Lemma 15. For a fixed ˜P ,
˜m := max(deg( ˜P ) − 1, deg( Q0)) ≥ deg Λ kt
p
(· · · Λk0
p
(Q0) · · · ).
That is, ˜m is a bound on the degree of every polynomial that is the result of iterating Λkp on input Q0 for all values of k ∈ Fp. In particular, ˜m bounds the degree of every Q0 appearing in the Rowland-Zeilberger construction.
Note that by Proposition 13, the maximum that defines ˜ m must be taken over deg( ˜P )−1and Q0 = deg(Λ pℓ (P kQ)) for all k ∈ Z/p a−1Z for all values of Q we are considering.
Definition 9. We wish to encode Laurent polynomials with coefficients in Z/p aZ of degree
≤ m as a product of copies of Z/p aZ, and so we must choose a computational basis: Impose an order on ˜T = [ − ˜m, ˜m]r (e.g. lexicographical), where ˜T indexes the set of coefficients in a polynomial with degree ≤ ˜m. If Q = ∑
i∈˜T
qixi then let vec( Q) be the row vector ( qi)i∈ ˜T
and let code Q : (( Z/p aZ) ˜T )∗ → (Z/p aZ)∗ be the coding defined by code Q(~u) = vec( Q) · ~u.Just as in Definition 4, the map Q 7 → Λp( ˜P kQ) is ( Z/p aZ)-linear and has the following matrix description:
Definition 10. If ˜ γ(k) := (coef pj −i
[ ˜P k
]
)i,j ∈ ˜T , then vec( Q) · ˜γ(k) = vec(Λ p( ˜P kQ)). Defin-ing ˜ γ as a matrix-valued string morphism in this way yields the p-linear representation (vec( Q), ˜γ, ˜V (0)) for the sequence
(
ct
[ ˜P nQ
]
mod pa
)
n∈N
, where the only non-zero entry of ˜V (0) is at the index corresponding to the constant term, where it is 1. More generally, define ˜V (n) to be the column vector (ct
[ ˜P nxi
]
mod pa)i∈ ˜T ∈ (Z/p aZ) ˜T .Just as with Lemma 4:
Lemma 16. For all non-negative integers n and all k ∈ Fp,
˜γ(k) · ˜V (n) = ˜V (pn + k).
Definition 11. Just as in Definition 6, we use ˜ γ to define the p-uniform string morphism ˜σ :
(
(Z/p aZ) ˜T
)∗
→
(
(Z/p aZ) ˜T
)∗
by ˜σ(~u) := (˜ γ(0) · ~u)(˜ γ(1) · ~u) · · · (˜ γ(p − 1) · ~u).
Then define ˜ α := ˜ σω( ˜V (0)) = ˜V (0) ˜V (1) ˜V (2) · · · ∈
(
(Z/p aZ) ˜T
)∗
.Just as with Lemma 5: 13 Lemma 17. Let C be the index of the polynomial 1 in ˜T ’s order. There is an integer s ∈ N
such that for every s′ ≥ s ∈ N, ˜γ(0) s = ˜ γ(0 s) = eC,C , that is to say, it has a 1 in the row and column corresponding to the constant term and 0s everywhere else.
In analogy with Theorem 7:
Lemma 18. If ˜P is a Laurent polynomial satisfying ˜P (x1, x 2, . . . , x r)p ≡ ˜P (xp
1
, x p
2
, . . . , x pr )(mod pa) and there exists some n0 ∈ N such that ct
[ ˜P n0
]
≡ 0 (mod p), then the frequency of 0 is 1 in every sequence of the form ct
[ ˜P nQ
]
mod pa for every Laurent polynomial Q.Proof. If p | ct
[ ˜P n0
]
, then ˜ γ(0) s ˜γ(n0)˜ γ(0) s is a zero-divisor multiple of eC,C , so that there exists some t such that (˜ γ(0) s ˜γ(n0)˜ γ(0) s)t = 0. Since ct
[ ˜P nQ
]
≡ vec( Q) · ˜V (n) = vec( Q) ·
˜γ(n) · ˜V (0) all n such that np contains (0 s(n0)p0s)t must have ct
[ ˜P nQ
]
≡ 0 (mod pa). If β = |(n0)p|, then consider each n ∈ [0, p t(β+2 s)·k) as corresponding to an element of
(
Ft(β+2 s)
p
)k
via its base-p representation chunked into blocks of size t(β + 2 s). If any of the
k entries for n correspond to (0 s(n0)p0s)t, then we know that ˜V (n) = 0 and ct
[ ˜P nQ
]
≡ 0, so at least pt(β+2 s)·k − (pt(β+2 s) − 1)k of such n have ct
[ ˜P nQ
]
≡ 0. Therefore, the frequency of 0 in ct
[ ˜P nQ
]
mod pa is at least pt(β+2 s)·k−(pt(β+2 s)−1)k
pt(β+2 s)·k
= 1 −
( pt(β+2 s)−1
pt(β+2 s)
)k
→ 1 as k → ∞ .In fact, this shows that the frequency of 0 in ˜ α is 1.
The last thing we need before we can prove our main results is the following proposition analogous to Proposition 8:
Proposition 19. There exists t0 ∈ N such that for all n ∈ N, if p ∤ ct
[ ˜P n
]
then ˜σt0 ( ˜V (n)) =
x0 ˜V (0) y0 with x0, y 0 ∈
(
(Z/p aZ) ˜T
)∗
. That is, ˜V (0) is a character in ˜σt0 ( ˜V (n)) for every such n.Proof. Let s ∈ N such that ˜ γ(0) s = eC,C . Next, fix s′ ∈ N such that C-term entries (i.e., indexed by C) of ˜ σs′
( ˜V (0)) contain all C-term values of ˜V (i) that ever occur for any
i ∈ N. Recall that the C-term value of ˜V (i) is the constant coefficient of ˜P i. Note that ˜V (n)[ C] 6 ≡ 0 (mod pa) by our assumption that p ∤ ct
[ ˜P n
]
. Finally, we have that ˜ γ(0) s ·
˜V (n) =
( ˜V (n)[ C]
) ˜V (0), ˜ σs′+s( ˜V (0)) contains
( ˜V (n)[ C]
) ˜V (0), and
( ˜V (n)[ C]
)ϕ(pa)
= 1 by Euler’s theorem (where ϕ is Euler’s totient function). Therefore, we can conclude that ˜σ(s′+s)( ϕ(pa )−1)+ s( ˜V (n)) contains ˜V (0), as desired.
Note that t0 does not depend on n. Finally, our main result:
Theorem 20. If P and Q are Laurent polynomials with integer coefficients and pa is a prime power, then the sequence ct [ P nQ] mod pa is linearly recurrent if and only if p ∤ ct [ P n] for
14 all n ∈ N, and this can be checked for bounded n < B P,p . When the sequence is not linearly recurrent, it is 0 with frequency 1. This classification is independent of Q, as well as the exponent a.Proof. For every k ∈ Z/p a−1Z and Laurent polynomial Q, define code Q,k (~u) = vec(Λ pℓ (P kQ)) · ~u
where ~u is from ( Z/p aZ) ˜T . Then for Laurent polynomials P and Q, define the coding code P,Q (~u) = (code Q, 0(~u), code Q, 1(~u), . . . , code Q,p a−1−1(~u)) ∈ (Z/p aZ)pa−1
.
Let ι :
(
(Z/p aZ)pa−1
)∗
→ (Z/p aZ)∗ denote the morphism determined by the canonical injection on letters,
ι(j1, j 2, . . . , j pa−1) = j1j2 · · · jpa−1.
Then, by Proposition 13, the sequence
β := ι(code P,Q (˜ α)) has ct [ P nQ] mod pa as its nth term. If there is some n0 ∈ N such that p | ct [ P n0 ] then Lemma 12 gives us an n1 ∈ N such that p | ct
[ ˜P n1
]
so that Lemma 18 tells us that the frequency of 0 is 1 in ˜ α and thus 0 also has frequency 1 in β so that, in particular, β is not uniformly recurrent. If there is no such n0, and if w appears as a word in β, then we can extend w to a word
w′ in β corresponding to a word in ˜ α. That is, w′ contains w but starts on an index that is a multiple of pa−1 and ends on an index one less than a multiple of pa−1. Thus, it is sufficient to show that ˜ α is uniformly recurrent as this forces uniform recurrence for w′ and consequently for w.Given a word w in ˜ α, choose t1 such that w has its first occurence ending before the index
pt1 . Thus, w appears in ˜ σt1 ( ˜V (0)). We can conclude from Proposition 19 that for all n ∈ N,˜σt1+t0 ( ˜V (n)) = ˜ σt1 (x0 ˜V (0) y0) = ˜ σt1 (x0)x1wy 1 ˜σt1 (y0).
In other words, w appears in every ˜ σt1 +t0 ( ˜V (n)), and since ˜α = ˜ σt1+t0 (˜ α) = ˜ σt1+t0 ( ˜V (0))˜ σt1+t0 ( ˜V (1))˜ σt1+t0 ( ˜V (2)) · · ·
and ˜ σ is uniform (so that ˜ σt1+t0 is also uniform), this implies that ˜ α is uniformly recurrent, and in turn β is uniformly recurrent. Finally, Theorem 10.8.4 of tells us that p-automatic uniformly recurrent sequences are linearly recurrent.
Counterintuitively, this theorem shows that ct [ P nQ] mod pa is either linearly recurrent or else 0 has frequency 1 for all values a ≥ 1 simultaneously. Furthermore, just as in Corollary 10, the same generalization applies to prime powers: 15 Corollary 21. Let F : ( Z/p aZ)ℓ → X (which is deterministic) for some ℓ ∈ N and let bn =
F (ct [P n+k1 Q1
] mod pa, . . . , ct [P n+kℓ Qℓ
] mod pa) where the ki are arbitrary non-negative integers and the Qi are arbitrary Laurent polynomials. Then bn is linearly recurrent if and only if for all n ∈ N, p ∤ ct [ P n]. Furthermore, in the case where there exists n such that
p | ct [ P n], then the frequency of F (0 , . . . , 0) in bn is 1.
Lastly, using the same construction of β as in the proof of Theorem 20, we have the analogous result to Proposition 11 where we can refer to P and not ˜P because we have Lemma 12:
Proposition 22. If P and Q are Laurent polynomials and pa is a prime power, and there exists some n0 > 0 such that ct [ P n0 ] mod p 6 = 0 , then the sequence bn = ct [ P nQ] mod pa is recurrent.
6 Application to Representation Theory
In this section, we briefly interpret Theorem 20 (or more precisely, Corollary 21) in the con-text of sequential tensor powers of semisimple Lie algebra representations. This context is a great source of constant term sequences. Let g be a semisimple Lie algebra, and let V be a representation of g whose formal character is P ∈ Z [x, x−1]. If W is an irreducible representation of g, then let aW,n denote the multiplicity of W in the direct sum decomposition of V ⊗n, which has character P n. It is a consequence of the Weyl Integral Formula (see, for example, Section IV.1 of ) that a multiple of aW,n is a constant term sequence; in particular, if Q is the product of the formal character of W with the square of the Weyl denominator, then C · aW,n = ct [ P nQ], where
C is the order of the Weyl group. The following theorem follows as a result of Corollary 21.
Theorem 23. Let g be a semisimple Lie algebra whose irreducible representations are {Wi}
and let V be a representation of g whose formal character is P ∈ Z [x, x−1]. Let ai,n denote the multiplicity of Wi in the direct sum decomposition of V ⊗n. Then for each prime p,either all of the sequences (ai,n mod pa)n∈N are simultaneously linearly recurrent for every
a ≥ 1 and every i or else all of those sequences simultaneously have 0 with frequency 1. In particular, the sequences are linearly recurrent if and only if there exists an n0 < B P,p such that p | ct [ P n0 ].
16 References
Jean-Paul Allouche and Jeffrey Shallit. Automatic Sequences: Theory, Applications, Generalizations . Cambridge University Press, 2003. Theodor Br¨ ocker and Tammo Dieck. Representations of Compact Lie Groups . Graduate Texts in Mathematics. Springer Berlin, Heidelberg, 1985. Alan Cobham. Uniform tag sequences. Mathematical systems theory , 6:164–192, 1972. Nadav Kohen. Uniform recurrence in the Motzkin numbers and related sequences mod
p. 2024. Preprint arXiv:2403.00149. Hamoon Mousavi. Automatic theorem proving in Walnut. 2021. Preprint arXiv:1603.06017. Narad Rampersad and Jeffrey Shallit. Congruence properties of combinatorial sequences via Walnut and the Rowland-Yassawi-Zeilberger automaton. Electronic Journal of Com-binatorics , 29(3):Paper No. 3.36, 13, 2022. Eric S. Rowland and Doron Zeilberger. A case study in meta-automation: automatic generation of congruence automata for combinatorial sequences. Journal of Difference Equations and Applications , 20:973–988, 2013. Jeffrey Shallit. The Logical Approach to Automatic Sequences: Exploring Combinatorics on Words with Walnut . London Mathematical Society Lecture Note Series. Cambridge University Press, 2022. Armin Straub. On congruence schemes for constant terms and their applications. 2022. Preprint arXiv:2205.09902. 17 |
189746 | https://www.hunter.cuny.edu/dolciani/pdf_files/brushup-materials/solving-mixture-and-solution-verbal-problems.pdf | SOLVING SOLUTION AND MIXTURE VERBAL PROBLEMS This type of problem involves mixing two different solutions of a certain ingredient to get a desired concentration of the ingredient. Before we can solve problems that involve concentrations, we must review certain concepts about percents. If you need to do this, go to the brush-up materials for solving percent problems on the Dolciani website. 1. Solution Problems Basic Equation: amount of solution concentration of substance = amount of substance Example: 40 ounces (amount of solution) of a 25% solution of acid (concentration) contains 25(40) = 10 ounces of acid Usual equation to solve for the variable: Amount of substance in solution 1 + Amount of substance in solution 2 = Amount of substance in solution 2. Mixture Problems Basic Equation: unit price # units = cost (or value) Example: 5 pounds of apples (# units) that sell for $1.20 per pound (unit price) costs 5(1.20) = $6 Using equation to solve for the variable: Cost of ingredient 1 + Cost of ingredient 2 = Cost of mixture Now let’s look at an actual mixture problem. It is easiest when solving a mixture problem to make a table to get the information organized. No matter the story line of the problem, the table can be used and labelled as necessary. Let’s look at a few examples. Example 1: Fatima’s chemistry lab stocks an 8% acid solution and a 20% acid solution. How many ounces of each must she combine to produce 60 ounces of a mixture that is 10% acid? Solution: We want to find how much of each solution must be in the mixture. We assign variables to these two unknowns: Let x = amount of 8% solution needed and let y = amount of 20% solution needed Now we can make a table showing what we know about the two solutions and the mixture Amount of Solution Amount of Acid 8% solution x 0.08x 20% solution y 0.20y 10% mixture 60 (0.10)(60) The right hand column tells us how much acid is in the ingredients and the mixture. We obtain the right hand column by remembering that “and 8% acid solution” means that 8% of the solution is acid. Thus 8% of x ounces of solution is 0.08x ounces of acid. Similarly, 20% of y ounces is 0.20y ounces of acid; and 10% of 60 is (0.10)(60) ounces of acid. Now we need to write two equations. We do so by noting that the amounts put into any mixture must add up to the total amount that is in the mixture. Therefore, for the solutions (middle column), x + y = 60 and for the acid only (right column), 0.08x + 0.20y = (0.10)(60) We thus have a systems of two equations in two unknowns that we can solve by substitution. We rewrite the first equation and substitute in the second. y = 60 – x First equation 0.08x + 0.20(60 – x) = 6 Substituting for y 0.08 + 12 – 0.20x = 6 Multiplying out –0.12x + 12 = 6 Combining like terms –0.12x = –6 Adding –12 to both sides x = 50 Multiplying by −1 0.12 Now, by substituting 50 for x in the first equation, we have 50 + y = 60 y = 10 We then need to check the values for x and y to see that they total 60 ounces. 50 + 10 = 60 And we need to check that the concentration of the mixture is correct. Amount of acid Amount of solution × 100% = 0.08𝑥+ 0.20𝑦 60 × 100% = 0.08(50) + 0.20(10) 60 × 100% = 4 + 2 60 × 100% = 6 60 × 100% = 1 10 × 100% = 10% Correct So the 10% mixture contains 50 ounces of 8% acid solution and 10 ounces of 20% acid solution. Example 2: A piggy bank has 40 nickels and dimes. The total amount of money in the bank $2.50. Find the amount of each coin. Solution: We need to find how many of each coin there are in the bank. We assign the variables that way. Let x = number of nickels Let y = number of dimes Then we make a table showing what we know about the two coins, their monetary value, and the total amount in the bank. Because each nickel is worth 5 cents, the monetary value of all the nickels is (0.05)x, or 0.05x. Similarly, because each dime is worth 10 cents, the monetary value of all the dimes in (0.10)y, or 0.10y. Number Of Coins Monetary Value Nickels x 0.05x Dimes y 0.10y Bank 40 2.50 Next we need to write the two equations. We can write one for the number of coins and one for their value. x + y = 40 0.05x + 0.10y = 2.50 We solve the first equation for y: y = 40 – x Solving the system by substituting into the second equation, we get 0.05x + 0.10(40 – x) = 2.50 Substituting for y 0.05x + 4 – 0.10x = 2.50 Multiplying out –0.05x + 4 = 2.50 Combining like terms –0.05x = –1.50 Adding –4 to both sides x = 30 Multiplying by −1 0.05 Substituting 30 for x in the first equation yields 30 + y = 40 y = 10 Checking the value, we have 30 nickels: 30(0.05) = $1.50 10 dimes: 10(0.10) = $1.00 Total: $2.50 Correct So there are 30 nickels and 10 dimes in the piggy bank. Example 3: How many pounds of a 15% aluminum alloy must be added to 500 pounds of a 22% alloy to make a 20% alloy? Solution: Let x = amount of 15% alloy that must be added to the 500 pounds of the 22% alloy. Then the total number of pounds of the new solution is x + 500. A table or a diagram is very useful for these problems: amount of alloy Concentration amount of silver 15% alloy x .15 .15x 22% alloy 500 .22 .22(500) solution x + 500 .20 .20(x + 500) The equation is ) ( .
) ( .
.
500 20 500 22 15 x x Solving for x, you find that 200 pounds of a 15% alloy of aluminum must be added to a 500 pounds of a 22% alloy to obtain a 20% alloy. Example 4: The owner of a fruit stand combined cranberry juice that cost $4.60 per gallon with 50 gallons of apple juice that cost $2.24 per gallon. How much cranberry juice was used to make the cranapple juice if the mixture costs $3.00 per gallon. Round to the nearest tenth. Let x = # of gallons of cranberry juice that must be added. Then the total # of gallons of the mixture is x + 50. # gallons price per gallon total cost cranberry juice x 4.6 4.6x apple juice 50 2.24 2.24(50) cranapple juice x + 500 3.0 3.0(x + 50) The equation is: ) ( .
) ( .
.
500 0 3 50 24 2 6 4 x x Solving for x, you find that 23.8 gallons of cranberry juice costing $4.60 per pound is needed to combine with 50 gallons of apple juice costing $2.24 per gallon to make cranapple juice costing $3.00 per gallon. Practice: 1. How many liters of a 25% solution of a drug must be mixed with a 55% solution of the drug to produce 50 liters of a 46% solution of this drug? 2. A pocketful of coins includes dimes and quarters. The total value of the 36 coins is $5.10. Find the number of each coin. 3. A hospital laboratory needs a 28% hydrogen chloride solution, but only 15% and 35% HCl solutions are in stock. How much of the 35% solution should be mixed with 100 fl oz of the 15% solution to get a 28% solution? 4. How many liters of a 10% dextrose solution should be mixed with 20 L of a 15% dextrose solution to obtain a 12% dextrose solution? 5. How many liters of a 30% NaCl solution should be mixed with 1 L of a solution that contains no NaCl to obtain a 24% NaCl solution? 6. 80 pounds of a dry powder drug with 30% concentration needs to be mixed with how many pounds of the dry powder drug with a 10% concentration to get a 20% dry powder drug? 7. An 8% solution of local anesthesia is needed. However, 40 fl oz of a 10% solution is the only type of this anesthesia in stock. How much neutral solution containing no anesthesia should be added to the 40 fl oz container of the 10% solution? 8. A pharmacist needs to fill an order for a 4% lidocaine solution topical. However, only 2% and 9% concentrations are in stock. How much of the 9% concentration should be mixed with 150g of the 2% concentration to get a 4% concentration? 9. How many ounces of a certain chemical that is 40% alcohol must be mixed with a second chemical that is 60% alcohol to get 40 ounces of a mixture that is 55% alcohol? 10. A sample of 200 ml of 15% HCl is combined with 200 ml of an unmarked concentration of HCl. What is the concentration if the resulting solution has a 14% concentration? 11. Mixed Nuts has 10% pecans. Nestor Nuts has 18% pecans. How many ounces of each should be mixed to get 20 ounces of a nut mixture that is 15% pecans? 12. Chocolate that sell for $1.50 per pound is added to mints that sell for $2.75 per pound to make a 25 pound mix that sells for $2.50 per pound. How many pounds of each are there in the mix? 13. Sean has $3.70 with quarters, nickels and dimes. The number of dimes is one less than the number of nickels, and the number of quarters is 3 less than the number of dimes. How much of each does he have? Answers: 1. 15 liters of 25% and 35 liters of 55% 2. 26 dimes and 10 quarters 3. 1300 7 or 185.71 oz of 35% solution 4. 30 liters of 10% solution 5. 4 liters of 30% solution 6. 80 pounds 7. 10oz is needed 8. 60g of 9% solution 9. 30 ox of 60%, 10 oz of 40% 10. 13% 11. Mixed: 7.5 ounces; Nestor Nuts: 12.5 ounces 12. 20 pounds of mints; 5 pounds of chocolate 13. 12 nickels, 11 dimes, 8 quarters |
189747 | https://www.khanacademy.org/math/algebra-basics/basic-alg-foundations/alg-basics-roots/v/simplifying-square-roots-comment-response | Simplifying square roots (variables) | Algebra (video) | Khan Academy
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Skip to lesson content
Algebra basics
Course: Algebra basics>Unit 1
Lesson 4: Square roots
Intro to square roots
Square roots of perfect squares
Square roots
Simplifying square roots
Simplifying square roots of fractions
Simplify square roots
Simplifying square-root expressions: no variables
Simplifying square roots (variables)
Simplify square roots (variables)
Math>
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Square roots
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Simplifying square roots (variables)
CCSS.Math: HSN.RN.A.2
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About About this video Transcript
A worked example of simplifying radical with a variable in it. In this example, we simplify 3√(500x³).Created by Sal Khan and Monterey Institute for Technology and Education.
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Les Howles 8 years ago Posted 8 years ago. Direct link to Les Howles's post “This is hard stuff and he...” more This is hard stuff and he seems to be blowing through it rapidly like it's just review. What is the previous video he keeps referring to in his commentary? The video I see before this one is entitled "Simplifying Square Roots" and he doesn't seem to cover a lot of this. Am I missing something?
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Erik 5 years ago Posted 5 years ago. Direct link to Erik's post “yes, i don't understand w...” more yes, i don't understand why this is so rushed... and how is he getting to "30 and the absolute value of x" this site is losing me.
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Javares 2 years ago Posted 2 years ago. Direct link to Javares's post “This video was useless fo...” more This video was useless for the practice problems.
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tonytakeru a year ago Posted a year ago. Direct link to tonytakeru's post “Highly recommend looking ...” more Highly recommend looking up the video "Square Roots with Variables (Simplifying Math)" on YouTube by Eric Buffington if this didn't make any sense to you. Made much more sense to me.
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learneverydayayay 10 months ago Posted 10 months ago. Direct link to learneverydayayay's post “Thanks! Big help.” more Thanks! Big help.
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ecerv002 2 years ago Posted 2 years ago. Direct link to ecerv002's post “i have no idea what you j...” more i have no idea what you just said and i have rewatched this multiple times.
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Ishaan 2 years ago Posted 2 years ago. Direct link to Ishaan's post “This video is confusing” more This video is confusing
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Pi is the best 11 years ago Posted 11 years ago. Direct link to Pi is the best's post “Around 2:18 of the video ...” more Around 2:18 of the video Sal mentions the absolute value of | x | which he got from the x^2 under the radical. if he were to use the absolute value for the x^2 why would he not use the absolute value of the lone | x | under the radical?
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Kanmani 11 years ago Posted 11 years ago. Direct link to Kanmani's post “Because the lone x was no...” more Because the lone x was not a perfect square, he could not simplify the radical. Only if you are taking the principle root and you are SIMPLIFYING the radical can you put in the absolute value. If he would have put the absolute value sign for the x under the radical it would've become:
sqrt(|x|) = +/- (x^1/2)
Which still gives you the negative root which is extraneous for principle roots. Therefore putting the absolute value under the radical is never done when taking the principle root as it yields unwanted answers.
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tbaugh87 2 years ago Posted 2 years ago. Direct link to tbaugh87's post “For those that are lost, ...” more For those that are lost, look up "the organic chemistry tutor" on youtube.
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Anand Iyer 5 years ago Posted 5 years ago. Direct link to Anand Iyer's post “Can you please type up a ...” more Can you please type up a lesson page for this topic subject? I'm not able to grasp it. Thanks!
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Kate 2 years ago Posted 2 years ago. Direct link to Kate's post “I'm so confused, why is t...” more I'm so confused, why is the principal root of x^2 equal to the absolute value of x ?
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Square roots have two pos...” more Square roots have two possible answers. The default is always the principal root (which is the positive value). If a problem wants you to use the negative root, there will be a "-" front of the radical. For example:
sqrt(9) = +3 because this is asking for the principal root.
-sqrt(9) = -3 because this is asking for the negative root.
When working with variables, unless someone has told you that the variable will not be a negative value, you need to assume it could be positive or negative.
sqrt(x^2) needs to create the principal root (a positive value). Using the absolute value of x ensures that the result is positive (just in case x is a negative number0.
Hope this helps.
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What I want to do in this video is resimplify this expression, 3 times the principal root of 500 times x to the third, and take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you could simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500-- we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5. Or even better, we could rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it, you're going to get another negative number. And then at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this-- if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open it up to complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x, because it's not going to be a negative number. And so if we're assuming that the domain of x is-- or if this expression is going to be evaluatable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers-- and if you don't know what a complex number is, or an imaginary number, don't worry too much about it. But if you were dealing with those, then you would have to keep the absolute value of x there. Because then this would be defined for numbers that are less than 0.
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189748 | http://www.drshirley.org/latin/grote/grote19.pdf | CHAPTER 19 Perfect Passive System of All Verbs; Interrogative Pronouns and Adjectives PERFECT PASSIVE SYSTEM We divide the Latin tense system into two categories. (1) The present system, active and passive, uses the first principal part of the verb. It includes the present, future and imperfect tenses. Notice, these tenses use the first principal part for both the active and passive voices. The only difference between the active and passive voices in the present system is the personal endings. You learned all about this in Chapter 18. (2) The perfect system active uses the third principal part of the verb and attaches different personal endings to get the three different tenses of the perfect system. Write out the endings : Perfect Pluperfect Future Perfect 1 _ _ _ 2 _ __ _ 3 _ _ _ 1 _ __ _ 2 _ __ _ 3 _ __ _ Notice, now, that the third principal part is the stem for the perfect system active only. To form the perfect system tense in the passive voice, Latin uses the fourth principal part of the verb. Since it uses a different principal part, the Perfect System Passive is considered to be a different category of tenses. So there are three tense systems in Latin : (1) the Present System Active and Passive; (2) the Perfect System Active; (3) the Perfect System Passive. The tenses of systems (2) and (3) are the same - Perfect, Pluperfect and Future Perfect; the only difference is in the voice, and the principal part of the verb on which they're built. THE PASSIVE VOICE IN ENGLISH : THE PARTICIPLE DEFINED First, let's look at how English forms its passive voice again. As we saw in Chapter 18, English uses the third principal part of the verb and uses an inflected form of the verb "to be" as the auxiliary or helping verb. That is to say, the verb "to be" will indicate the tense, the number, and the mood of the verb, while the third principal part of the verb will define the specific action involved. For example, for the verb "to see, saw, seen": Betty is seen by George. is being seen will be seen would be seen should be seen was seen was being seen has been seen had been seen should have been seen would have been seen You can clearly see that the constant in all these modifications is the verbal form "seen". The verb "to be" is doing all the work. So let's look at little more closely at the verbal form "seen". The third principal part of the English verb is called a "participle". Now listen closely; this is going to be an important definition : A participle is a "verbal adjective". That is, an adjective which is derived from a verb. In fact, that's why we call it a participle, because it "participates" in the essence of both a verb and of an adjective. So in the constructions of the English passive voice, the participle "seen" is actually "modifying" the subject of the verb "to be". I can say "Betty is tall" and "Betty is seen", and these two sentences are analogous. In the predicate of both these sentences the subject is further modified, since it is linked to an adjective by the verb "to be". It may seem bizarre to be thinking of a verbal construction as being essentially adjectival, but watch how we can use participles where their adjectival force is quite obvious: "the written text", "the spoken word", "the destroyed city", "the bewildered students", "the beleaguered professor", etc. THE LATIN PERFECT PASSIVE PARTICIPLE So where are we? English forms the passive voice of all its tenses by using the participle of the verb which it links to the subject with a conjugated form of the verb "to be". Now you already know that Latin forms the passive voice of some of its tenses - those of the present system - simply by using special passive endings. The formation of the passive voice of the perfect system, however, doesn't work that way. The Latin perfect passive system is perfectly analogous to the formation of the English passive voice. The perfect passive system in Latin uses the fourth principal part of the verb, which is then linked to the subject with an inflected form of the verb "sum". The fourth principal part of a Latin verb is called the "Perfect Passive Participle". Let's zero in on all the parts of this description. (1) We call it "Perfect" because the action is considered to have been completed. This is an important difference with the English participle. In English, we might say "Betty is being seen", and the participle doesn't force us to understand that the action is finished. In this example, the action is clearly still going on. (2) We say "Passive" since whatever the participle is going with had something done to it, rather than being the agent of some action. Again, the English participle can be used in conditions where the passive force is not so obvious. In the sentence "I have seen Betty", the participle "seen" doesn't strike us as passive in force, but rather as a part of an active construction. (3) We say "Participle" because it is a "verbal adjective", and for Latin, this is going to have monumental implications. The participle is an adjective, so it must agree in number, gender, and case with the noun is modifying. And if it must agree with nouns, then the participle must be able to decline to get the different numbers, genders, and cases it needs. (This is the feature of the Perfect System Passive which causes students the most trouble. It's difficult for them to realize that the passive voice in the perfect system is essentially adjectival : the verb "sum" linking the subject of the verb with a predicate adjective.) Now let's look at the fourth principal part of a verb. As you know, the dictionary must give you all the principal parts of the verb you're considering. (1) The first entry is the first person singular of the present tense. (2) The second is the present infinitive, from which you drop the "-re" to get the present system stem. (3) The third is the first person singular of the perfect tense, from which you get the perfect active stem by dropping the "-i". (4) The fourth entry is the perfect passive participle, which is used with the auxiliary verb "sum" in the formation of the perfect system passive. We've said that the perfect passive participle is a verbal adjective, so it must be able to decline, just like adjectives, in order to agree with the nouns they're modifying. The perfect passive participles of all verbs declines just like the first adjectives you learned : just like "magnus, -a, -um". That is, it uses endings of the first declension to modify feminine nouns, endings of the second declension "-us" type to modify masculine nouns, and endings of the second declension "-um" type to modify neuter nouns. The dictionaries tell you this in a number of different ways; but they're all telling you the same thing. Some write out the whole "-us, -a, -um"; others abbreviate it by using only the neuter "-um" or the masculine "-us". So you may see the entry for the fourth principal part of "laudo", for example, given in these three different ways : (1) laudatus, -a, -um (2) laudatum (3) laudatus PERFECT TENSE PASSIVE So let's put this participle to work. How would you translate this in Latin : "I was praised". Well, the tense is obviously perfect - that is, the action was completed before it was reported - so we must use the perfect passive participle : "laudatus, -a, -um". The person is first and the number is singular. Let's assume that the "I" is male. What case is "I"? Obviously nominative - it's the subject of the verb - so the form of the participle will be "laudatus" - masculine, nominative singular. Got that? The participle is going to agree with the subject of the verb. The subject of the verb is nominative, so the participle must be nominative, too. Now what form of the verb "sum" should we use. Of course, we'll use the first person singular, but what tense? Did you guess "eram" - "I was"? If you did, that's one demerit. Look, the fourth principal part is the "perfect passive participle" and the "perfect" tells you that the action is considered to have been already completed. That is, in the participle itself is the notion of a past event, so "laudatus" could be translated as "having been praised". Therefore you needn't repeat the idea of past completion in the auxiliary verb "to be". So the correct form of the auxiliary is the present tense: "sum". Think of it this way, and I admit this may seem clumsy: "Laudatus sum" means "I am now in the condition of having been praised". We can bring this over into English as either "I was praised" or "I have been praised". So to form the perfect tense passive in Latin, you use the perfect passive participle + the verb "sum" as the auxiliary in the present tense. Now let's suppose that the subject "I" is feminine. What changes would this necessitate? Well, the participle is a verbal adjective, so it must agree in number, gender and case with whatever it's modifying. If the subject of the verb is feminine, then the participle has to be feminine, nominative, singular to agree with it. So the participle will have be "laudata". Therefore, if a woman is speaking, she would say "Laudata sum" for "I was praised". PLUPERFECT TENSE PASSIVE So how do you imagine Latin forms the passive of the pluperfect tense? Think. You're still going to use the perfect passive participle linked to the subject with a conjugated form of the verb "sum". All perfect system passive tenses do that. But what tense will the verb "sum" be in? Right! Now you use the auxiliary verb "sum" in the imperfect tense. What you're doing is adding an additional past idea in auxiliary to the past idea already implicit in the participle. Therefore "Laudatus eram" means "I was in the condition of having been praised" or "I had been praised". And if the subject were feminine: "Laudata eram". FUTURE PERFECT TENSE PASSIVE And the future perfect tense? Yes. You use the future of the verb "sum", thus attaching a future idea to the past idea in the participle, and that's the definition of the future perfect tense. "Laudatus ero" therefore means "I will be in the condition of having been praised", which comes out "I will have been praised". And if the subject were feminine "Laudata ero". PERFECT SYSTEM PASSIVE SUMMARIZED So let's look at all this. Conjugate in full the three tenses of the perfect system passive, using the verb "laudo". (Carry all the possible genders and check your work against the lists on page 88.) PARTICIPLE PERFECT PLUPERFECT FUTURE PERFECT laudatus, -a, -um sum eram ero __ _ _ _ __ _ _ _ laudati, -ae, -a _ _ _ __ _ _ _ __ _ _ _ THE FOURTH PRINCIPAL PART OF VERBS In Chapter 12, you realized that you were going to have to memorize the third principal part of all your verbs if you wanted to be able to work with them in all their tense systems. Similarly, now you must go back and memorize the fourth principal parts of your verbs if you want to work with them in the perfect system passive. As with the third principal parts, the formation of the fourth will follow some regular patterns, so the task of memorization will not be as tedious as it at first might seem. FIRST CONJUGATION VERBS The vast majority of first conjugation verbs, as you know, are regular. This means that its principal parts are formed regularly using the first principal part as the stem. The third principal part, as you recall, is just the first principal part + "vi". The fourth principal part also is a regular derivation from the first principal part : it's the first principal part + "t" plus the adjectival endings "-us, -a, -um". So for "laudo", the fourth principal part is "laudatus, -a, -um" (lauda + t + us, -a, -um) which is often abbreviated just as "laudatus" or "laudatum". Here are all the first conjugation verbs you've had up to this chapter. Fill out the principal parts, and double check your work. You can use these lists to review from. II III IV amo __ ___ ___ cogito ___ ___ ___ conservo ___ ___ ___ do ___ ___ datus erro ___ __ ___ exspecto ___ ___ ___ iuvo ___ ___ ___ laudo ___ ___ ___ libero ___ ___ ___ muto ___ ___ ___ paro ___ ___ ___ servo ___ ___ ___ supero ___ ___ ___ tolero ___ ___ ___ voco ___ ___ ___ (The two exceptions to this regularity of the first conjugation verbs is "do, dare, dedi, datus", and "[ad]iuvo, -iuvare, -iuvi, -iutus". If you look closely, however, you'll see that "do" isn't really a first conjugation verb, since the stem vowel "-a-" is not long.) SECOND CONJUGATION VERBS Although second conjugation verbs are slightly less regular than first conjugation verbs, they do tend to follow a pattern in their formation of the second, third, and fourth principal parts. But because there are occasional irregularities in third and second conjugation verbs, the dictionary will list all four principal parts of a second conjugation verb. Often the third principal part of a third conjugation verb is the first principal part + vi", which then becomes simplified from "-evi" to just "-ui". The fourth principal part very often ends "-itus, -a, -um". So for the paradigm verb "moneo", the principal parts are "moneo, monere, monui, monitus". Again, here is the complete list of the second conjugation verbs you've had till now. I've left the principal parts of the regular verbs blank for you to fill in on your own. When a verb lacks one of the principal parts, I've left no blank. Some verbs have unusual principal parts, which would involve some explanation. Where verbs have principal parts which are outside our interest here, I've inserted dashes. For now, pretend they don't exist and just memorize the principal parts the verbs do have. audeo ___ --------------- --------------- debeo ___ ___ ___ deleo ___ delevi deletus doceo ___ ___ doctus habeo ___ ___ ___ moneo ___ ___ ___ moveo ___ movi motus remaneo ___ remansi remansus teneo ___ ___ tentus terreo ___ ___ ___ timeo ___ ___ --------------- valeo ___ ___ --------------- video ___ vidi visus THIRD CONJUGATION VERBS The third conjugation (-i- stem and non -i- stem) displays several different ways of forming third and fourth principal parts. Each verb is best treated individually as if they were irregular, but certain patters are obvious. Additionally, a great many of our English derivations come from the fourth principal part of the original Latin verb. If you keep this in mind as you try to memorize these forms, you'll find they'll stick more readily. ago ___ ___ actus capio ___ ___ captus coepi coeptus committo ___ ___ commissus curro ___ ___ cursus dico ___ ___ dictus duco ___ ___ ductus diligo ___ ___ dilectus eicio ___ ___ eiectus facio ___ ___ factus fugio ___ ___ ------- gero ___ ___ gestus iacio ___ ___ iactus incipio ___ ___ inceptus intellego ___ ___ -tellectus iungo ___ ___ iunctus lego ___ ___ lectus mitto ___ ___ missus neglego ___ ___ neglectus scribo ___ ___ scriptus traho ___ ___ tractus vinco ___ ___ victus vivo ___ ___ victus FOURTH CONJUGATION VERBS The fourth conjugation sometimes forms third and fourth principal parts regularly by adding "-vi" to the present stem for the third and by adding "-tus, -a, -um" for the fourth. But there are so many irregularities that fourth conjugation verbs are listed with all four principal parts. Here's your list of all the four conjugation verbs you've had up to Chapter 19. audio ___ ___ auditus invenio ___ ___ inventus sentio ___ ___ sensus venio ___ ___ ventus THE INTERROGATIVE PRONOUN Do you remember how Latin asks a question? You've learned that enclitic "-ne" is attached to the end of the first word of the sentence to indicate a question. Latin must do this because the word order is so flexible that no rearrangement of the words will indicate necessarily that a question is coming up. In English, we ask a simple question by inverting the subject of the verb with an auxiliary. The statement "You are walking the dog" becomes a question like this: "Are you walking the dog?" But Latin doesn't have all these handy auxiliary verbs, and besides, since Latin doesn't rely on word order much to tell you the syntax of the words in the sentence, inverting words won't help. So Latin uses the enclitic, and the word the enclitic is attached to is the focus of the question. For example, in the question "Laudatisne filios huius viri?" the point of inquiry is whether you are performing the action of praising. But if we begin the sentence with "the sons" - "Filiosne huius viri laudatis?" then the focus of the question changes: "Are you praising this man's sons? We can accomplish this effect in English by inflecting our voice when we reach the word that is the point of the question. Now look more closely at each of these questions. Even though each has a different emphasis, all the questions are essentially asking one thing : "If I should turn this question into a statement, would it be true?" That is, the question is about the validity of the predication. The question "Are you praising this man's sons" is asking whether it is true to say "You are praising this man's sons". We call this kind of question a simple question; it asks for no information that is not contained in its structure. Now look at these questions : (1) "Why are you praising this man's sons"? (2) "When are you praising this man's sons"? (3) "How are you praising this man's sons"? Here it is taken for granted that the predication is true - you are praising this man's sons - and the questions being asked are not whether you're praising the sons, but why, when, or how? These questions are calling for information that is not contained within the syntax of the question; they are asking for specific kinds of additional information. And the kind of information they're asking for is indicated in the words "why, when, and how". We call words which ask for specific kinds of information "interrogatives". Some more questions with another kind of interrogative : (1) "Who's there?" (2) "What's that?" (3) "Whose mess is this?" (4) "Whom are you accusing?" (5) "What are you trying to say?" In these questions, the predication is taken as true : (1) someone is there; (2) that is something; (3) the mess does belong to someone; (4) you are accusing someone; (5) you are trying to say something. The information the questions are asking for, however, is temporarily replaced with another word, and the hope is that soon the information will be plugged into the spot where its replacement now stands. What do we call a word which takes the place of another word or idea? Right! We call them pronouns, so these words are interrogative (because they're asking questions) and pronouns (because they're replacing other nouns or ideas ) : "interrogative pronouns". The English interrogative pronouns, as you can see in the examples above, have different cases and even genders. The gender is determined by what is be filled in for, but the case is determined by the way the pronoun is being used in the question. MASCULINE AND FEMININE INANIMATE Nom. who what Gen. whose whose Acc. whom what Do you see any similarity between the interrogative pronouns and the relative pronouns? Of course you do. "Who, whose, and whom" are all forms that can also be used as relative pronouns. Only the interrogative pronoun "what" has no use as an relative pronoun. The Latin interrogative pronoun also resembles the Latin relative pronoun. In the plural, the forms of the interrogative pronoun are identical to those of the relative pronoun. In the singular many of the forms of the interrogative pronouns overlap with those of the relative pronouns, but there are some differences : (1) For one, the forms for the masculine and feminine are the same. Consequently, there are only two forms for the nominative singular : one for the masculine and feminine genders, and one for the neuter. Similarly, there are only two forms for the genitive singular - one masculine and feminine, and one neuter. And so on or all the cases in the singular. Only two forms. (2) Next, two of the forms are just plain different from those of the relative pronoun. (a) For the masculine and feminine nominative singular, the form is "quis", not "qui" or "quae" as you might expect. (b) You might expect "quod" for neuter nominative and accusative singular, but the form is "quid". (c) For the remaining cases of the masculine/feminine forms, the interrogative pronoun uses the masculine forms of the relative pronoun. Look this description closely over and try to write out the Latin interrogative pronoun (see Wheelock, page 89). MASCULINE AND FEMININE NEUTER Nom. ___ ___ Gen. ___ ___ Dat. ___ ___ Acc. ___ ___ Abl. ___ ___ MASCULINE FEMININE NEUTER Nom. ___ ___ ___ Gen. ___ ___ ___ Dat. ___ ___ ___ Acc. ___ ___ ___ Abl. ___ ___ ___ Let's look at some examples of how the interrogative pronoun works in Latin. You'll that it has some surprising properties, which the English interrogative pronoun "who, what", etc. doesn't have. "Quis librum tibi dedit?" ("Who gave you the book?") You can tell this sentence is a question, obviously, because it is introduced with the interrogative pronoun and because it ends with a question mark. But the English translation isn't as precise as the Latin. Why not? Look at "quis". It's nominative because it is used as the subject of the verb. But what about its number and gender? It's masculine/feminine in gender and singular in number. That means that the question was formed in such a way as to imply that there was only one person who gave you the book. Now look at the English "who". Can you tell whether the person asking the question expects there to be only one person who gave you the book? No, you can't. So, in Latin, the questioner reveals more about the kind of answer expected because the pronoun reveals more about the possible antecedent. How would we translate these into English : (a) "Qui librum tibi dederunt?" (b) "Quae librum tibi dederunt?" We'd have to translate them both as "Who gave you the book?", but look more closely at the Latin. In (a), the question implies that more than one person gave you the book and that they are either all male or mixed male and female. In (b), those who gave you the book are implied to be plural and all feminine. Look at another example. All of these Latin question can be translated into English as "Whose book did Cicero give you?": "Cuius librum Cicero tibi dedit?" "Quorum librum Cicero tibi dedit?" "Quarum librum Cicero tibi dedit?" The interrogative pronoun in each of these question is in the genitive case because the point of the question is to learn more about the owner(s) of the book. But each question suggests an different kind of answer. Can you spot the different expectations? INTERROGATIVE ADJECTIVE Okay, you know that the interrogative pronoun is a word which takes the place of another noun or idea about which certain information is being sought. Because it asks a question we call it "interrogative"; because it stands in for something else, we call it a "pronoun": "interrogative pronoun". So what is an "interrogative adjective". Start from the beginning. "Interrogative" means that it will be asking a question. "Adjective" means that it will be modifying a noun in the sentence and to modify a noun an adjective must agree with it in number, gender, and case. Putting these two parts together, we come up with this: an "interrogative adjective" is a word which modifies an noun in a way that asks more information about it. How does this work? Look at these English questions : (a) "What child is this?" (b) "Which way did he go?" (c) "For what reason are we doing this?" In each of these questions, more information is being requested about something which is already expressed in the question. Like this. What's the difference between "What is this"? and "What child is this"? In (a), the answer sought is not restricted to anything specified in the sentence itself. But in the second, the potential responder is directed to limit his reply to something in particular; namely, "the child". The same is true with (b) and (c). (b) is not asking whether he's gone, but which way he went; (c) is not asking what we're doing, but for what reason. So English uses the adjective "which or what" to ask for information specific to something already expressed in the sentence. Latin also has interrogative adjectives for this purpose, but because Latin is a fully inflected language, the interrogative adjective has many more forms than its English analogue. After all, the Latin interrogative adjective is going to have to agree with masculine, feminine, or neuter nouns in any one of the ten cases and numbers. You'll be pleased to know, however, that you're not going to have to learn anything new, because the Latin interrogative adjective uses the forms of its relative pronoun. Go ahead and write out the forms of the interrogative adjective to refresh your memory. (Remember, it's exactly the same as the relative pronoun). INTERROGATIVE ADJECTIVE MASCULINE FEMININE NEUTER Nom. ___ ___ ___ Gen. ___ ___ ___ Dat. ___ ___ ___ Acc. ___ ___ ___ Abl. ___ ___ ___ Nom. ___ ___ ___ Gen. ___ ___ ___ Dat. ___ ___ ___ Acc. ___ ___ ___ Abl. ___ ___ ___ Because the interrogative adjective is an adjective, its form is determined entirely by the noun with which it is agreeing in the sentence. Like this: "Quem librum legebatis?" (What (or which) book were you reading?) The interrogative adjective "quem" is singular, accusative, masculine because the noun about which the question is seeking more information is singular, accusative, and masculine. Study these examples : (a) "Quibus feminis libros illos dedistis?" (To which women did you give those books?) (b) "A quo viro admoniti sunt?" (By which (or what) man were they warned?) (c) "A quibus viris admoniti sunt?" (By which (or what) men were they warned?) (d) "A qua femina admoniti sunt?" (By which woman were they warned?) DRILLS Translate the following short sentences. 1. Cui libros dederunt? ________ 2. Qui ei libros dederunt? ________ 3. A quo libri dati sunt? ________ 4. A quibus hi libri lecti erant? ________ 5. A quibus discipulis hi libri lecti sunt? ________ 6. Quis ab omnibus civibus amatus est? ________ 7. Cuius civitatis ille homo erat? ________ 8. E qua urbe iste tyrannus venit? ________ 9. E quorum urbe iste tyrannus venit? ________ 10. Qui vir ab omnibus civibus amatus est? ________ 11. Who came from that city? ________ 12. Which books did you read? ________ 13. To whom were these books given? ________ 14. Which students read these books? ________ 15. Which citizens loved this man? ________ 16. Whose city was loved by that tyrant? ________ 17. By whom were those books given to the students? ________ 18. By whom was this city loved? ________ 19. To which women was the book given? ________ 20. To which woman was the book given? ____________ VOCABULARY PUZZLES senex, senis This word is much more bizarre than Wheelock lets on. You'll see it mainly as a noun, meaning "old man" or "old woman". Don't expect to see it modifying a neuter noun. It'll always be masculine or feminine. Because it's really a third declension adjective, it'll decline like : senex senes senis senium seni senibus senem senes seni senibus novus, -a, -um Like most ancient civilizations, ancient Rome didn't care much for change. So a way of asking "What's wrong"? was "Quid novum est"? 01/08/20 |
189749 | https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/1.12/primary/lesson/order-of-operations-with-negative-real-numbers-alg-i-hnrs/ | Skip to content
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1.12
Order of Operations with Negative Real Numbers
Written by:Brenda Meery | Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Ginny walked into Math class and saw the following question on the board:
Find the value of the following expression when @$\begin{align}a = -2, b = 3, c = -1, d = 1\end{align}@$:
@$$\begin{align}\boxed{(4a+c)\div b-(bd)\div (ac)}\end{align}@$$
Can you answer this question?
Order of Operations with Negative Numbers
The standard order of operations involves specific steps for performing the mathematical calculations presented in a mathematical statement. These steps are represented by the letters PEMDAS.
P – Parentheses – Do all the calculations within parentheses.
E – Exponents – Do all calculations that involve exponents.
M/D – Multiplication/Division – Do all multiplication and division, in the order it occurs, working from left to right.
A/S – Addition/Subtraction – Do all addition and subtraction, in the order it occurs, working from left to right.
These steps do not change whether they are being applied to positive real numbers or to negative real numbers. The rules for adding, subtracting, multiplying and dividing real negative numbers must be applied when evaluating expressions that require PEMDAS to be used. A good way to learn the order of operations is to observe examples where it is correctly applied. The majority of this lesson is a number of problems that you should follow step-by-step in order to help cement the order in your memory.
Let's complete the following problems using PEMDAS:
Simplify the expression: @$\begin{align}32 \div 4^2 \times 2-21\end{align}@$
There are no calculations inside parentheses. The first step is to evaluate the number with the exponent.
@$$\begin{align}=32\div {\color{blue}16}\times 2-21\end{align}@$$
The next step is to perform any division or multiplication, working from left to right.
@$$\begin{align}&={\color{blue}2} \times 2-21 \
&={\color{blue}4}-21\end{align}@$$
The final step is to rewrite the expression as an addition problem and to change the sign of the original number being subtracted.
@$$\begin{align}=4{\color{blue}+}(\text{-}21 )\end{align}@$$
When adding two numbers with unlike signs, subtract the smaller number from the larger, then the sign of the number with the greater magnitude will be the sign of the answer.
@$$\begin{align}= {\color{blue}\text{-}17}\end{align}@$$
Find the value of the following expression when @$\begin{align}a = \text{-}2, b = 3, c = \text{-}1, \text{ and }d = 1.\end{align}@$
@$$\begin{align}(4a^2c^2)-(3ac^3)\end{align}@$$
Begin by substituting the variables with the given values. Use brackets to group the operations with parentheses.
@$$\begin{align}=[4(\text{-}2)^2(\text{-}1)^2] - [3(\text{-}2)(\text{-}1)^3]\end{align}@$$
In the first set of brackets, do the calculations with exponents.
@$$\begin{align}=[4({\color{blue}4})({\color{blue}1})] - [3(\text{-}2)(\text{-}1)^3] \end{align}@$$
In the second set of brackets, do the calculations with exponents.
@$$\begin{align}=[4(4)(1)]-[3(\text{-}2)({\color{blue}\text{-}1})] \end{align}@$$
In the first set of brackets, do the multiplication. The brackets can now be removed.
@$$\begin{align}={\color{blue}16}-[3(\text{-}2)({\color{blue}\text{-}1})] \end{align}@$$
In the second set of brackets, do the multiplication. The brackets can now be removed.
@$$\begin{align}=16-{\color{blue}6} \end{align}@$$
Subtract the numbers.
@$$\begin{align}={\color{blue}10}\end{align}@$$
What is the value of @$\begin{align}3-2 \left[\frac{8(\text{-}1)-7}{\text{-}3(2)-4}\right]\end{align}@$?
Simplify within the brackets first:
@$$\begin{align}&=3-2\left[\frac{{\color{blue}\text{-}8}-7}{\text{-}3(2)-4}\right] \
&=3-2\left[\frac{\text{-}8-7}{{\color{blue}\text{-}6}-4}\right]\end{align}@$$
Rewrite subtraction as addition of the opposite:
@$$\begin{align}&=3-2 \left[\frac{\text{-}8{\color{blue}+}(\text{-}7)}{\text{-}6{\color{blue}+}(\text{-}4)}\right]\
&=3-2\left[\frac{\text{-}8-7}{{\color{blue}\text{-}6}-4}\right]\
&=3-2\left[{\color{blue}\frac{\text{-}15}{\text{-}10}}\right] \
&=3- {\color{blue}\frac{\text{-}30}{\text{-}10}}\
&=3-{\color{blue}3}\
&={\color{blue}0}\end{align}@$$
Examples
Example 1
Earlier, you read about the math question that Ginny saw written on the board in her math class:
Find the value of the following expression when @$\begin{align}a = -2, b = 3, c = -1, d = 1\end{align}@$:
@$$\begin{align}\boxed{(4a+c)\div b-(bd)\div (ac)}\end{align}@$$
Ginny felt good about answering the problem because she remembered the steps involved in the standard order of operations. When she noticed that the values for two of the variables were negative numbers, she realized that she would have to be careful doing the calculations because she would also have to apply the rules that she had learned for adding, subtracting, multiplying and dividing negative real numbers.
@$$\begin{align}(4a+c)\div b-(bd) \div(ac)\end{align}@$$
Ginny began by substituting the variables with the given values.
@$$\begin{align}(4(-2)+(-1))\div 3-((3)(1))\div((-2)(-1))\end{align}@$$
To reduce errors in her calculations, Ginny wrote all of the values in parentheses. The statement now has parentheses within parentheses. This may seem confusing and the order in which to perform the operations may become skewed. Ginny asked her teacher about writing the expression another way. Her teacher advised her to replace the outer parentheses with brackets [ ]. Brackets are another type of grouping symbol. When evaluating an expression that has grouping symbols (parentheses) within grouping symbols (brackets), perform the operations within the innermost set of symbols first.
Ginny rewrote the expression using both brackets and parentheses.
@$$\begin{align}[4(-2)+(-1)]\div 3-[(3)(1)] \div [(-2)(-1)]\end{align}@$$
Then it was easier to keep track of the steps and the proper order to simplify the expression:
@$$\begin{align}&=[{\color{blue}-8}+(-1)]\div 3 - [(3)(1)] \div [(-2)(-1)]\end{align}@$$
@$$\begin{align}&=[\text{-}8+(\text{-}1)] \div 3-\div\
&=[{\color{blue}\text{-}9}] \div 3- \div \
&=\text{-}9\div 3-3\div 2\
&={\color{blue}\text{-}3}-3 \div 2\
&=\text{-}3- \color{blue}1.5 \color{black}\rightarrow \text{-}3+ \text{-}1.5\
&={\color{blue}\text{-}4.5}\end{align}@$$
Example 2
Perform the following operations: @$\begin{align}8 \times -9 +19 \div (-30+11)-14 \times (-1)^2\end{align}@$
@$$\begin{align} & 8 \times -9+19 \div (-30+11)-14 \times (-1)^2\
& =8 \times 9 +19 \div {\color{blue}-19}-14 \times (-1)^2 \
& =8 \times9+19 \div {\color{blue}-19}-14 \times (-1)^2 \
& ={\color{blue}72}+19 \div -19-14\times 1\
& =72+{\color{blue}-1}-14 \times 1 \
& =72+-1-{\color{blue}14} \
& ={\color{blue}71}-14 \
& ={\color{blue}57}\end{align}@$$
Example 3
Simplify: @$\begin{align}\left(\frac{-12-6}{6+3}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\end{align}@$
@$$\begin{align}& \left(\frac{-12-6}{6+3}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\
& =\left({\color{blue}\frac{-18}{9}}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\
& ={\color{blue}-2}+\left(\frac{-36}{-4}\right)+(-8 \times 2)\
& =-2+{\color{blue}9}+(-8 \times 2)\
& =-2+9+{\color{blue}-16}\
& ={\color{blue}-9}\end{align}@$$
Example 4
A formula from geometry is @$\begin{align}V=\frac{h}{6}(B+4M+b)\end{align}@$. Find @$\begin{align}V\end{align}@$ when @$\begin{align}h = -15, B = 12,M = 8, b = 4\end{align}@$.
@$$\begin{align}& V=\frac{h}{6}(B+4M+b).\
& V=\frac{-15}{6}(12+4(8)+4)\
& V=\frac{-15}{6}(12+{\color{blue}32}+4)\
& V=\frac{-15}{6}(\color{blue}48)\
& V={\color{blue}-2.5}(48)\
& V={\color{blue}-120}\end{align}@$$
Review
If @$\begin{align}a = -3, b = 1\end{align}@$ and @$\begin{align}c = 2\end{align}@$, what is the value of @$\begin{align}2a^3-3b+2c^2\end{align}@$? _______
221
59
–49
43
Evaluate @$\begin{align}a-bc\end{align}@$ when @$\begin{align}a=\frac{1}{2}, b=\frac{1}{3}\end{align}@$ and @$\begin{align}c=\frac{5}{4}\end{align}@$: _______
1
@$\begin{align}\frac{1}{12}\end{align}@$
@$\begin{align}\frac{2}{5}\end{align}@$
@$\begin{align}\frac{5}{2}\end{align}@$
Evaluate: @$\begin{align}a(-b^2-a^2)\end{align}@$ when @$\begin{align}a=-3\end{align}@$ and @$\begin{align}b=4\end{align}@$: _______
75
2
3
4
Simplify the following: @$\begin{align}[3-[5-(6-8)]+4]-2\end{align}@$ = _______
–6
–2
2
–19
Perform the following operations and evaluate: @$\begin{align}(5\times 3-7)^2 \div 4+9\end{align}@$ = _______
109
5
11
25
Perform the indicated calculations.
6. @$\begin{align}\frac{6-(24-14)}{-10-[2-(-4)^2]}\end{align}@$
7. @$\begin{align}\frac{[12-(-15+6)]\times 4 -16}{-4}\end{align}@$
8. @$\begin{align}3\left(-2 \times \frac{20}{-4 \times -1}-5-7\right)\end{align}@$
9. @$\begin{align}-3 \times 5 -[3(3+9)-9\times -3]\end{align}@$
10. @$\begin{align}4^2-(4+8-2-4-2\times 7)\end{align}@$
11. @$\begin{align}2\times -18+9\div -3-5\times -2\end{align}@$
12. @$\begin{align}-4\times (-12-8)\div -2\end{align}@$
13. @$\begin{align}-15-6\div 3\end{align}@$
14. @$\begin{align}\frac{-2-(15-5)}{-4\times -2-12+-6\div 3}\end{align}@$
15. Which expression has the greatest value if @$\begin{align}a=-2\end{align}@$ and @$\begin{align}b=3\end{align}@$?
1. @$\begin{align}3[a^2+b^2-2ab-2(a^2-b^2)]\end{align}@$
2. @$\begin{align} 3(a-b)-2(b-a)+3(a-2b)\end{align}@$
3. @$\begin{align} b(a-b)-a(b-a)-ab\end{align}@$
Review (Answers)
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189750 | https://www.chegg.com/homework-help/questions-and-answers/bainbridge-mass-spectrometer-ions-pass-electromagnetic-field-b1-e-velocity-selector-deflec-q97008651 | Solved In a Bainbridge mass spectrometer, ions pass through | Chegg.com
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In a Bainbridge mass spectrometer, ions pass through the electromagnetic field B1, E of a velocity selector and are then deflected by a purely magnetic field B2. Knowing that E = 3.6×105 V/m and that B1 = B2 = 0.42 T, calculate the difference of the positions on the photographic plate for the ions of charge q = e of the carbon isotopes of masses 12 u and 14
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Question: In a Bainbridge mass spectrometer, ions pass through the electromagnetic field B1, E of a velocity selector and are then deflected by a purely magnetic field B2. Knowing that E = 3.6×105 V/m and that B1 = B2 = 0.42 T, calculate the difference of the positions on the photographic plate for the ions of charge q = e of the carbon isotopes of masses 12 u and 14
In a Bainbridge mass spectrometer, ions pass through the electromagnetic field B1, E of a velocity selector and are then deflected by a purely magnetic field B2. Knowing that E = 3.6×105 V/m and that B1 = B2 = 0.42
T, calculate the difference of the positions on the photographic plate for the ions of charge q = e of the carbon isotopes of masses 12 u and 14 u.
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Équations Tables m₁ m₂ m3 Consignes E Énoncé (II) Dans un spectromètre de masse de Bainbridge, lesions traversent le champ électromagnétique B₁. E) d'un sélecteur de vitesse et sont ensuite déviés par un champ purement magnétique (B₂) Sachant que E = 3,6x 10 V/m et que B, =B₂ = 0,42 T, calculez la différence des positions sur la plaque photographique pour les ions de charge q = e des isotopes de carbone de masses 12 u et 14 u. C Question Trouvez la distance qui sépare les isotopes des deux types sur la plaque photographique. d=cm Effacer tout Vérifier la réponse
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189751 | https://www.youtube.com/watch?v=Bnp1-Xd-Eo4 | The Weird Graph of y = x^x
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Intro in this video we're going to talk about the weird graph of y equals x raised to the x so this is an online graphing calculator at desmos.com and you can graph any function with this website you can graph the linear function y equals x y equals x squared trigonometric functions like sine x or tangent x but what we want to focus on is this one x to the x and we get something that looks like that now one thing i want to mention for those of you who may want to graph this function yourself if you're going to use this website you may want to use a very large monitor because when i try to minimize the screen for some reason this website will only show the upper portion of the graph and not the lower portion so just wanted to let you know about that but now here's some questions for you why are some of the values positive on the left side of the y-axis and some of the values negative why do we see some values below the x-axis while others are above the x-axis and also on the left side of the graph why is it discontinuous but it's continuous on the right side we're going to answer those questions shortly Rough sketch so here's a rough sketch of that same graph now it's continuous on the right because we can plug in any positive value for x for instance if we plug in one one to the first power is one if we plug in two two squared is four if we plug in three three to the third power is not nine but twenty-seven so we can plug in anything any positive number and we're gonna get an output now the situation is not the same when plugging in a negative number for instance if you have your calculator with you try plugging in negative 0.1 raised to the negative 0.1 power negative one raised to the negative point one will give you an error the question is why why do we get an error when we plug this in negative point one raised to the negative point one is 1 over negative 0.1 raised to the positive 0.1 to get rid of the negative exponent you need to move this number in the denominator of the fraction now 0.1 as a decimal is one over ten as a fraction now converting that fractional exponent into a radical we get one over the tenth root of negative point one because this index number is even we're going to get an imaginary number not a real number so we can get an imaginary number like i or something and that's why we see the discontinuity on the left because some negative values will not exist it won't give us a real number output so that's why it's discontinuous on the left side of the graph however we do get some values not all values on the left are imaginary numbers for instance if we plug in negative 0.2 raised to the negative 0.2 into our calculator this is the value that we're going to get this is going to be negative 1.3797 and then there's some more numbers so that's going to be somewhere in this region so let's say that's this point here so why do we get an actual number in this case well first let's move this number to the bottom to change the negative exponent into a positive one this is equivalent to one over negative point two raised to the positive point two now point two as a fraction is one over five and so we can write this as the fifth root of negative point two now it's important to understand that when you take the square root of a negative number like the square root of negative four you're going to get an imaginary number not a real number and that's the case anytime you have an even index number but if the index number is odd you'll get a real number the cube root of negative eight is negative two so the cube root of negative i mean the fifth root of negative 0.2 will give us a real negative value and so that's why we have some values below the x-axis because we're going to get a negative output for some of these numbers now what about the positive numbers what are some situations where we can get a positive result Positive values let's plug in this number negative 0.4 raised to the negative 0.4 power and you should get positive 1.44267 if you round it so this is a number that would be somewhere in this region it's above the x-axis on the left now let's see why we get a positive answer for this particular number so negative point four raised to the negative point four power that's one over negative 0.4 raised to the positive 0.4 power now what is point 4 as a fraction to convert a decimal to a fraction you can multiply it by ten over ten point four times ten is four so we get four over ten four is two times two ten is two times five so if we cancel a two this is two over five so we can write this as negative point four raised to the two over five now let's convert the fractional exponent into a radical so this is going to be the fifth root of negative point four squared now the square is going to make the negative point four value positive negative point four squared is well we can write this as a fraction it's four over twenty five which is point sixteen and then point sixteen raised to the one over five or the fifth root of point sixteen gives us this number point six nine three one four four eight and so one over that will give us one point four four two seven so that's why we get a positive number is because we can turn this decimal into a fraction where the numerator is even and the denominator is odd when the numerator is even it will cause the negative value to become positive and so that's why we're going to get a number above the x-axis and so that's it for this particular type of graph so just to recap the reason why the curve of y equals x to the x is continuous on the right is we can is because we can plug in any positive value for x we can plug in three to the third four to the fourth fifth to the fifth point five to the point five and we're always going to get a positive result now the reason why it's discontinuous on the left side is because for certain values we're going to get an imaginary number we're not going to get a real number now the reason why it's negative for some values is because we can convert this decimal into a fractional exponent like one over five or maybe one over seven or one over nine and for those values you're going to get a negative result and the reason why some values are positive is because we can get a fraction that is even in a numerator like two over five for instance or let's say three over seven if you use i mean not three of a seven but like four of a seven and so that's it for this video hopefully you found it to be educational and informative and for those of you who want more videos on interest in mathematical topics make sure to check out the links in the description section below i'm going to be posting you know a few other videos related to math and some that are not really related to math but are still interesting as well so that's it for this video thanks again for watching and don't forget to subscribe to this channel if you haven't done so already and also click on that notification bell as well you |
189752 | https://www.ehn.org/great-lakes-pollution-birds | Long-banned toxics are still accumulating in Great Lakes birds—as new chemical threats emerge - EHN
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Long-banned toxics are still accumulating in Great Lakes birds—as new chemical threats emerge
We're "at the mercy of a shifting chemical landscape."
Hannah Seo
Sep 22, 2020
5 min read
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Decades ago several bird species in the Great Lakes—including the iconic bald eagle—faced an uncertain future because toxic chemicals were threatening their populations.
While several bans and policies have offered some protection, the same chemicals threatening these birds 60 years ago continue to accumulate in their bodies—and new chemical threats are adding to their toxic burdens, according to two new studies.
Lokta paper provides low-cost arsenic test
The two studies add to evidence that pollutants not only persist in the Great Lakes, but continue to travel up food chains to reach and endanger apex predators; and suggest that birds in the Great Lakes continue to impart toxic loads to their offspring—results that do not bode well for long-term bird populations. In addition, as birds are sentinel species, the studies show how previous and existing regulations have been inadequate at protecting all biota of the Great Lakes, including humans, from the harms of these industrial chemical pollutants.
In the first study, published online in Environment International last month, researchers examined how legacy pollutants around Lake Erie such as polychlorinated biphenyls (PCBs), polybrominated diphenyl ethers (PBDEs), and dichlorodiphenyltrichloroethane (DDT) continue to bioaccumulate in common terns. These migratory seabirds spend winters in Central and South America but stay near the Great Lakes from late April to mid-October, relying on the area as breeding grounds.
"Every common tern we looked at had some level of PCBs and PBDEs," Diana Aga, an analytical chemist at the University of Buffalo and co-author of the study, told EHN.
The U.S. phased out PBDEs in 2013, PCBs in 1979, and DDT way back in 1972. PBDEs were most commonly used as flame retardants; PCBs were used as coolants and electrical insulation; and DDT was a widely used insecticide. The chemicals were all banned because they were showing up in wildlife and humans and are linked to a variety of health impacts.
All three are persistent organic pollutants, chemicals that remain stable in the environment, lingering in sediments and plants rather than breaking down. Over the years, these chemicals move up food chains as predators eat contaminated prey, bioaccumulating in apex predators like the common tern. That is why birds of the Great Lakes region continue to feel the aftereffects of decades-old pollution.
To mitigate the accumulation of toxics in their bodies, female common terns, like other birds and animals exposed to chemical pollutants, will offload chemicals to their babies, said Aga—"out of all the birds at different life stages, tern eggs birds had the highest accumulation of these pollutants."
Persistent organic pollutants are fat soluble, and egg yolks are full of fat, so the eggs they lay provide mother terns with an opportunity to rid themselves of those harmful compounds. This results in chicks that are born with already very high loads of toxics, if they manage to hatch at all.
While there is very little research that explicitly looks at the toxicology of PBDEs and PCBs in terns, a few older studies suggest a link between these contaminants and physical abnormalities like twisted beaks or deformities in eyes and feet in several other bird species like chickens, kestrels and cormorants. DDT, on the other hand, has long been linked to egg shell thinning and reduced hatching success.
Peregrine falcons in Detroit. (Credit: pverdonk/flickr)
“It's a little bit like whack-a-mole”
The second study, published in Environmental Research this month, showed how peregrine falcons accumulate perfluorinated chemical pollutants like perfluoroalkyl substances (PFAS), which have made headlines in recent years for being "forever chemicals"—since they take exceedingly long times to break down. Human PFAS exposure is linked to cancer, endocrine disruption, and elevated cholesterol, among other health detriments.
The study's results suggest that peregrines—birds of prey and one of the most widely found bird species—get exposed to these chemicals through maternal transfer to eggs, their diets via contaminated smaller birds and mammals, and also simply by living in areas with higher levels of these substances.
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"We're also at the mercy of a shifting chemical landscape," Robert Letcher, an environmental toxicologist at Environment and Climate Change Canada and co-author of the study, told EHN.
Despite being largely phased out in North America, perfluorooctanesulfonic acid (PFOS), a kind of PFAS that was officially added to the Stockholm Convention's list of persistent organic pollutants in 2010, is "still singularly the most bioaccumulative, highest level perfluorinated compound that's being reported in the environment and in the biota," said Letcher. He said the chemical industry is consistently developing new chemicals to replace those that are phased out, chemicals that stay largely secret until they are investigated and identified by scientists like himself.
"We have to do a little bit of chemical sleuthing to figure out our targets," Letcher said. Through that sleuthing, he says, scientists have been able to trace PFOS replacements like F53B, GenX, ADONA, and a number of others that are all different minor variations of PFOS, but it's hard to keep up with the sheer number of replacements that just keep coming.
This perpetual conveyor belt of toxics, on top of the already varied arsenal of legacy pollutants, makes zeroing in on what each chemical is doing to the falcons' health extremely difficult, Kim Fernie, another environmental toxicologist at Environment and Climate Change Canada and one of Letcher's co-authors, told EHN. Not to mention that there are a multitude of other environmental factors like climate change that could be acting in tandem, she added.
Despite the unknown, the results are concerning, as PFAS chemicals have been previously shown to be overtly toxic. Research in other animals reveals that PFAS decreases reproductive success, can cause developmental defects, endanger kidney and liver health, and even cause cancer—effects that researchers suspect may translate to the health of these falcons and other Great Lakes birds.
"As we continue to have these new chemicals crop up, it's a little bit like whack-a-mole," Laura Rubin, the director of the Healing Our Waters—Great Lakes Coalition, told EHN. The Coalition is a group of about 165 NGOs and nonprofits throughout the Great Lakes region that pushes for clean water and the protection of landscapes and their wildlife.
"What we need now more than ever, is a precautionary principle, the idea that we should not be introducing new pollutants or chemicals until we know the effects," said Rubin. Currently, the onus to prove that a chemical is safe does not lie with the company manufacturing it. Without a precautionary principle, industries are free to release chemicals under presumed ignorance of the harms their products cause to the environment. Scientists and activists then have to identify and raise alarms about each individual pollutant. It becomes a repetitive story with different faces.
The destructive chemical pollutants of decades past yielded a few really seminal pieces of legislation, Rubin said—The Clean Water Act, The Clean Air Act, and the National Environmental Policy Act, to name a few. "But those are starting to show their age and they need some updating in terms of responding to these newer classes of chemicals." Getting a precautionary principle in place would be a monumental step forward for protecting our landscapes, she said.
A warning for us all
Pollution in the Great Lakes has significant implications for human health as well. According to the Great Lakes regional Water Use Database, 30 million people depend on the 40.4 billion gallons of water withdrawn from the Great Lakes-St. Lawrence River Basin each day. The land around that area is also home to roughly 10 percent of the U.S. population and more than 30 percent of the Canadian population, according to the U.S. Environmental Protection Agency.
A 2016 study also found that public water supplies with PFAS concentrations at or above the EPA's recommendations reached as many as 6 million U.S. residents. PFAS in food is also of concern, especially via manufactured food, food packaging, and fish consumption. One report of PFOS concentrations in fish found the highest levels in lake trout collected from Lake Ontario.
These birds being contaminated by all these chemicals are really acting as canaries in the coal mine for us humans, Alicia Perez-Fuentetaja, an aquatic ecologist at SUNY Buffalo State University and one of Diana Aga's co-authors, told EHN.
To Perez-Fuentetaja, the birds should be a warning to us all.
"These birds are not even smoking, they're not drinking, they're not doing any of the things we do," she said "They're just eating their food in the wild, and look at the exposure they have."
Banner photo: Common terns in Hueston Woods State Park, College Corner, Ohio. (Credit: Andrew Cannizzaro/flickr)
About the author(s):
Hannah Seo
Hannah Seo is a science journalist, essayist, podcast writer, and poet based in Brooklyn.
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189753 | https://www.quora.com/How-do-you-find-the-equation-of-the-tangent-of-the-curve-y-x-4-2x-1-at-the-point-where-the-curve-crosses-the-y-axis-calculus-derivatives-math | How to find the equation of the tangent of the curve y = x-4/ 2x+1 at the point where the curve crosses the y- axis (calculus, derivatives, math) - Quora
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How do you find the equation of the tangent of the curve y = x-4/ 2x+1 at the point where the curve crosses the y- axis (calculus, derivatives, math)?
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1y
To find the equation of the tangent line to the curve y=x−4 2 x+1 y=x−4 2 x+1 at the point where it crosses the y-axis, we follow these steps:
Step 1: Find the Point of Intersection with the Y-Axis
The curve crosses the y-axis when x=0 x=0. We can find the corresponding y y-coordinate by substituting x=0 x=0 into the equation:
y=0−4 2(0)+1=−4 1=−4 y=0−4 2(0)+1=−4 1=−4
Thus, the point where the curve crosses the y-axis is (0,−4)(0,−4).
Step 2: Find the Derivative of the Curve
To find the slope of the tangent line at this point, we need to compute the derivative of y y. We will use the quotient rule, which s
Continue Reading
To find the equation of the tangent line to the curve y=x−4 2 x+1 y=x−4 2 x+1 at the point where it crosses the y-axis, we follow these steps:
Step 1: Find the Point of Intersection with the Y-Axis
The curve crosses the y-axis when x=0 x=0. We can find the corresponding y y-coordinate by substituting x=0 x=0 into the equation:
y=0−4 2(0)+1=−4 1=−4 y=0−4 2(0)+1=−4 1=−4
Thus, the point where the curve crosses the y-axis is (0,−4)(0,−4).
Step 2: Find the Derivative of the Curve
To find the slope of the tangent line at this point, we need to compute the derivative of y y. We will use the quotient rule, which states that if y=u v y=u v, then
d y d x=u′v−u v′v 2 d y d x=u′v−u v′v 2
where u=x−4 u=x−4 and v=2 x+1 v=2 x+1.
Calculating u′u′ and v′v′:
u′=1 u′=1
v′=2 v′=2
Now applying the quotient rule:
d y d x=(1)(2 x+1)−(x−4)(2)(2 x+1)2 d y d x=(1)(2 x+1)−(x−4)(2)(2 x+1)2
Simplifying this expression:
d y d x=(2 x+1)−(2 x−8)(2 x+1)2=2 x+1−2 x+8(2 x+1)2=9(2 x+1)2 d y d x=(2 x+1)−(2 x−8)(2 x+1)2=2 x+1−2 x+8(2 x+1)2=9(2 x+1)2
Step 3: Evaluate the Derivative at x=0 x=0
Next, we evaluate the derivative at the point where x=0 x=0:
d y d x∣∣∣x=0=9(2(0)+1)2=9 1 2=9 d y d x|x=0=9(2(0)+1)2=9 1 2=9
So, the slope of the tangent line at the point (0,−4)(0,−4) is 9 9.
Step 4: Write the Equation of the Tangent Line
The equation of a line in point-slope form is given by:
y−y 1=m(x−x 1)y−y 1=m(x−x 1)
where m m is the slope and (x 1,y 1)(x 1,y 1) is the point of tangency. Plugging in our values:
m=9 m=9
(x 1,y 1)=(0,−4)(x 1,y 1)=(0,−4)
The equation becomes:
y−(−4)=9(x−0)y−(−4)=9(x−0)
Simplifying this, we get:
y+4=9 x y+4=9 x
Thus, the equation of the tangent line is:
y=9 x−4 y=9 x−4
Summary
The equation of the tangent line to the curve y=x−4 2 x+1 y=x−4 2 x+1 at the point where it crosses the y-axis is:
y=9 x−4 y=9 x−4
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What is the equation of the tangent to the curve [y = x^3 + 6(x) ^2 - 34x + 44] at the point where the curve crosses the y-axis? What are the coordinates of the point where the tangent meets the curve again?
What are the intercepts of the equation y=x^2 -4?
Which is the point where the graph intersects x-axis and y-axis of the equation y=2x +4?
At what point does the curve y=-x^2 -3x +2 cut the x-axis?
Given y=2x^2-8x+6, what are the coordinates of the y intercept, coordinate of the x intercept, line of symmetry or axis of symmetry, minimum or maximum value and coordinate of a vertex or turning point?
Robert Peasley
Former Code Monkey, Electronics Tech & Reactor Tech (USN) · Author has 245 answers and 770.9K answer views
·4y
The overall process for this is to:
Find the point the tangent line needs to go through.
Determine the slope of that tangent line at that point.
Using the point from 1, the slope from 2 and the “point slope equation” for a line, determine the equation for the tangent line.
For step 1, we need to recognize that the point where “the curve crosses the y-axis” is equivalent to saying that x is zero. So evaluate the function for the curve with x set to zero. I get -4. So our point for step 1 is (0,-4).
Step 2 is to find the slope of the tangent line. The derivative is the slope of the tangent line (Note
Continue Reading
The overall process for this is to:
Find the point the tangent line needs to go through.
Determine the slope of that tangent line at that point.
Using the point from 1, the slope from 2 and the “point slope equation” for a line, determine the equation for the tangent line.
For step 1, we need to recognize that the point where “the curve crosses the y-axis” is equivalent to saying that x is zero. So evaluate the function for the curve with x set to zero. I get -4. So our point for step 1 is (0,-4).
Step 2 is to find the slope of the tangent line. The derivative is the slope of the tangent line (Note: it’s only the slope, not the tangent line itself.) So use the quotient rule to find the derivative of the function. Nope, not going to do that for you but it’s not hard. You can do it.
Of course the derivative is some function of x. That is to say, the derivative tells you the slope of any and all x values. We want the slope at a specific x value so we’ll need to plug in our specific x value. That is the x value for our specific point, zero. The value of the derivative when x is zero is the slope of the tangent line for our curve when it passes through (0,-4).
Now we’ve got all we need for step 3. The point slope equation is:
y-y0=m(x-x0) where m is the slope of the line. (Note: if you weren’t already familiar with this equation, it’s a good one to remember.)
For us, x0 is zero, y0 is minus four and m is the value you found for the slope in step 2.
Plugging in values for x0 and y0 we get
y+4=mx or stated more properly y=mx-4.
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Mark Regan
Currently Enrolled in Computer Science&Mathematics, Currently Attending Www.codacademy.com (Graduated 2022) · Author has 95 answers and 74K answer views
·4y
y = x-4/ 2x+1
y = (x-4) (2 x+1)−1(2 x+1)−1
Let u = x - 4 …… du/dx = 1
Let v = 1/(2x+1) = (2 x+1)−1(2 x+1)−1 ……… dv/dx = (-1) (2 x+1)−2(2 x+1)−2 (2) = -2/(2 x+1)2(2 x+1)2
dy/dx = u dv/dx + v du/dx
= (x - 4) (-2/(2 x+1)2(2 x+1)2) + 1/(2x+1) 1
= 2(4 - x)/(2 x+1)2(2 x+1)2) + 1/(2x+1)
= (8 - 2x + 2x + 1)/(2 x+1)2(2 x+1)2
= 9/(2 x+1)2(2 x+1)2
the slope is equal to dy/dx at any point x. The curve crosses the y axis where x is equal to zero, at point (0, y).
y = (0 - 4) / (2 0 + 1) = -4/1 = -4
P(0, -4)
The slope at (0, -4) is
m = 9/(2 x+1)2(2 x+1)2 = 9/(2∗0+1)2(2∗0+1)2 = 9/1 = 9
(y - 9)/(x - 0) = 9
(y - 9)/x = 9
((y - 9)/x) x = 9 x
y - 9 = 9x
y = 9x + 9
Upvote ·
Lowie Tambis
Watch some awesome Math videos. · Author has 372 answers and 186.4K answer views
·4y
Take derivative first
y=x−4 2 x+1=(x−4)(2 x+1)−1,y=x−4 2 x+1=(x−4)(2 x+1)−1,use chain rule
y′=−(x−4)(2 x+1)−2+(2 x+1)−1=−x−4(2 x−1)2+1(2 x−1)y′=−(x−4)(2 x+1)−2+(2 x+1)−1=−x−4(2 x−1)2+1(2 x−1)
=−x+4+2 x−1(2 x−1)2=−x+4+2 x−1(2 x−1)2
y′=x+3(2 x−1)2 y′=x+3(2 x−1)2 since we find slope when pass y -axis, x=0 x=0
m=3(−1)2=3 m=3(−1)2=3
substitute x =0 in the original equation to find y intercept
b=−4 1=−4 b=−4 1=−4
y=m x+b y=m x+b
hence our tangent line at x =0 is
y=3 x−4 y=3 x−4
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Related questions
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What angle does the tangent to the curve y=x^2-5x+4 at the point (2,4)? At what point does normal cut the y-axis?
How can we find the equation of a tangent to a curve at some point by using its gradient or by using points on the curve (calculus, math)?
At what angle does the curve y (1 + x) = x cut the x-axis?
How do you find points at which tangent intersects curve for variable x and y values (calculus, derivatives, tangent line, math)?
What is the meaning of "curve crosses the y-axis?"
Francesco Amato
Studied at University of Bari (Graduated 1999) · Author has 4.5K answers and 1M answer views
·Feb 15
Related
What is the equation of the tangent line to f(x) = sqrt (2x + 1) x = 4?
f(x)=√2 x+1 f(x)=2 x+1
f(4)=√8+1=3 f(4)=8+1=3
D[f(x)]=2 2√2 x+1 D[f(x)]=2 2 2 x+1
D[f(x)]|x=4=1 3 D[f(x)]|x=4=1 3
Tangent line
y−3=1 3(x−4)y−3=1 3(x−4)
Continue Reading
f(x)=√2 x+1 f(x)=2 x+1
f(4)=√8+1=3 f(4)=8+1=3
D[f(x)]=2 2√2 x+1 D[f(x)]=2 2 2 x+1
D[f(x)]|x=4=1 3 D[f(x)]|x=4=1 3
Tangent line
y−3=1 3(x−4)y−3=1 3(x−4)
Upvote ·
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·1y
Related
What is the equation of the tangent of a curve y 2x +1 at point (1,1)?
What is the equation of the tangent of a curve y 2x +1 at point (1, 1)?
If you meant y = 2x + 1, then the equation is a straight line and the point (1, 1) is not on the curve.
If you meant y² = x + 1, then (1, 1) is still not on the curve and since it is inside the curve, no tangent to the parabola can be drawn through it.
Continue Reading
What is the equation of the tangent of a curve y 2x +1 at point (1, 1)?
If you meant y = 2x + 1, then the equation is a straight line and the point (1, 1) is not on the curve.
If you meant y² = x + 1, then (1, 1) is still not on the curve and since it is inside the curve, no tangent to the parabola can be drawn through it.
Upvote ·
99 13
9 2
Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·Mar 11
Related
What is the equation of the tangent to the curve y=x^(3)-4x^ (2) +2x-1 at the point where x=2?
Given curve
y=x 3−4 x 2+2 x−1…(1)y=x 3−4 x 2+2 x−1…(1)
Let is see if we can get this without calculus.
We need a tangent at a point where x=2 x=2
x=2⟹y=2 3−4(2)2+2(2)−1=−5 x=2⟹y=2 3−4(2)2+2(2)−1=−5
We have the point of tangency T(2,−5)T(2,−5). Let the equation of tangent be,
y=m x+c y=m x+c
For passage of tangent through T T,
−5=m(2)+c⟹c=−5−2 m−5=m(2)+c⟹c=−5−2 m
y=m x−5−2 m…(2)y=m x−5−2 m…(2)
From eqn. (1)
m x−5−2 m=x 3−4 x 2+2 x−1 m x−5−2 m=x 3−4 x 2+2 x−1
x 3−4 x 2+(2−m)x+2 m+4=0 x 3−4 x 2+(2−m)x+2 m+4=0
For the tangent to touch a single point on curve, x=2 x=2 must be repeated root. Let α α be the third root
By Vieta’s rule,
2+2+α=−(−4)1⟹α=0 2+2+α=−(−4)1⟹α=0
2⋅2⋅α=−(2 m+4)1 2⋅2⋅α=−(2 m+4)1
From above,
2⋅2⋅
Continue Reading
Given curve
y=x 3−4 x 2+2 x−1…(1)y=x 3−4 x 2+2 x−1…(1)
Let is see if we can get this without calculus.
We need a tangent at a point where x=2 x=2
x=2⟹y=2 3−4(2)2+2(2)−1=−5 x=2⟹y=2 3−4(2)2+2(2)−1=−5
We have the point of tangency T(2,−5)T(2,−5). Let the equation of tangent be,
y=m x+c y=m x+c
For passage of tangent through T T,
−5=m(2)+c⟹c=−5−2 m−5=m(2)+c⟹c=−5−2 m
y=m x−5−2 m…(2)y=m x−5−2 m…(2)
From eqn. (1)
m x−5−2 m=x 3−4 x 2+2 x−1 m x−5−2 m=x 3−4 x 2+2 x−1
x 3−4 x 2+(2−m)x+2 m+4=0 x 3−4 x 2+(2−m)x+2 m+4=0
For the tangent to touch a single point on curve, x=2 x=2 must be repeated root. Let α α be the third root
By Vieta’s rule,
2+2+α=−(−4)1⟹α=0 2+2+α=−(−4)1⟹α=0
2⋅2⋅α=−(2 m+4)1 2⋅2⋅α=−(2 m+4)1
From above,
2⋅2⋅0=−(2 m+4)1⟹m=−2 2⋅2⋅0=−(2 m+4)1⟹m=−2
From eqn. (2),
y=(−2)x−5−2(−2)y=(−2)x−5−2(−2)
y=−2 x−1 y=−2 x−1
Upvote ·
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·Mar 11
Related
How do you find the value of when the line y = mx + 1 is a tangent to the curve y2 = 4x?
Let us assume a parabola with equation,
y 2=4 a x y 2=4 a x
A line having equation y=m x+c y=m x+c is a tangent to this parabola if,
c=a m c=a m
We are given y 2=4 a x y 2=4 a x and y=m x+1 y=m x+1
We have a=1 a=1, c=1 c=1
Therefore,
1=1 m 1=1 m
m=1 m=1
Continue Reading
Let us assume a parabola with equation,
y 2=4 a x y 2=4 a x
A line having equation y=m x+c y=m x+c is a tangent to this parabola if,
c=a m c=a m
We are given y 2=4 a x y 2=4 a x and y=m x+1 y=m x+1
We have a=1 a=1, c=1 c=1
Therefore,
1=1 m 1=1 m
m=1 m=1
Upvote ·
9 8
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·3y
Related
What is the equation of the line perpendicular to y=4x-2 and the tangent of y=1/x?
Draw a picture! It clears your mind so that you can focus on what is needed!
Now it is clear we need to find the line with a gradient of –¼ which is a tangent to the hyperbola.
Of course we have TWO tangents y = –¼ x – 1 and y = –¼ x + 1
Continue Reading
Draw a picture! It clears your mind so that you can focus on what is needed!
Now it is clear we need to find the line with a gradient of –¼ which is a tangent to the hyperbola.
Of course we have TWO tangents y = –¼ x – 1 and y = –¼ x + 1
Upvote ·
9 6
9 1
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·2y
Related
How can I find the slope and equation of the tangent to the curve y = 3x² + 2x + 1 at point (2,17) using m = ∆y/∆x?
If equation of a curve is …
f(x)=A x n+B x m+C x p f(x)=A x n+B x m+C x p
its first derivative is…
f′(x)=n A x n−1+m B x m−1+p C x p−1 f′(x)=n A x n−1+m B x m−1+p C x p−1
The given equation is…
f(x)=3 x 2+2 x+1 f(x)=3 x 2+2 x+1
In this equation, the last term is a constant but we know that anything(not zero) raised to zero is one so it can also be written as…
f(x)=3 x 2+2 x 1+1 x 0 f(x)=3 x 2+2 x 1+1 x 0
Thus…
Δ y Δ x=(2)(3)x 2−1+(1)(2)x 1−1+(0)(1)x 0−1 Δ y Δ x=(2)(3)x 2−1+(1)(2)x 1−1+(0)(1)x 0−1
=6 x+2+0=6 x+2+0
The point where tangent touches the curve must be substituted into first derivative to get numerical value of slope of tangent.
m=6(2)+2=14 m=6(2)+2=14
Equation of tangent…
14 x−y=14(2)−17 14 x−y=14(2)−17
14 x−y=11 14 x−y=11
Continue Reading
If equation of a curve is …
f(x)=A x n+B x m+C x p f(x)=A x n+B x m+C x p
its first derivative is…
f′(x)=n A x n−1+m B x m−1+p C x p−1 f′(x)=n A x n−1+m B x m−1+p C x p−1
The given equation is…
f(x)=3 x 2+2 x+1 f(x)=3 x 2+2 x+1
In this equation, the last term is a constant but we know that anything(not zero) raised to zero is one so it can also be written as…
f(x)=3 x 2+2 x 1+1 x 0 f(x)=3 x 2+2 x 1+1 x 0
Thus…
Δ y Δ x=(2)(3)x 2−1+(1)(2)x 1−1+(0)(1)x 0−1 Δ y Δ x=(2)(3)x 2−1+(1)(2)x 1−1+(0)(1)x 0−1
=6 x+2+0=6 x+2+0
The point where tangent touches the curve must be substituted into first derivative to get numerical value of slope of tangent.
m=6(2)+2=14 m=6(2)+2=14
Equation of tangent…
14 x−y=14(2)−17 14 x−y=14(2)−17
14 x−y=11 14 x−y=11
Upvote ·
9 3
Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·4y
Related
The curve y= (9-4x) ^1/2 cuts the y-axis at point P. What is the equation of the tangent to the curve at the point P?
The curve y= (9-4x) ^1/2 cuts the y-axis at point P. What is the equation of the tangent to the curve at the point P?
At the y-axis where x = 0, y = (9 - 4(0))^(1/2) = 3 for the y-intercept.
The first derivative of y= (9-4x) ^(1/2) uses the power rule and the chain rule.
y’ = (1/2)(9 - 4x)^(-1/2) (-4);
y’ = -2 / sqrt(9 - 4(0)) = -2/3 = slope at x = 0
y = (-2/3)x +3 or 2x + 3y = 9 is the equation for the tangent line to the curve at P(0,3)
Continue Reading
The curve y= (9-4x) ^1/2 cuts the y-axis at point P. What is the equation of the tangent to the curve at the point P?
At the y-axis where x = 0, y = (9 - 4(0))^(1/2) = 3 for the y-intercept.
The first derivative of y= (9-4x) ^(1/2) uses the power rule and the chain rule.
y’ = (1/2)(9 - 4x)^(-1/2) (-4);
y’ = -2 / sqrt(9 - 4(0)) = -2/3 = slope at x = 0
y = (-2/3)x +3 or 2x + 3y = 9 is the equation for the tangent line to the curve at P(0,3)
Upvote ·
9 4
9 1
9 3
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·1y
Related
How do I find the equation of the common tangent to the curves y = x^2 + 4 and y = (x – 2) ^2?
Let the common tangent be y = mx + c
Now we must solve EQU A and EQU B simultaneously…
The common tangent is y = – 2x +3
Graphical...
Continue Reading
Let the common tangent be y = mx + c
Now we must solve EQU A and EQU B simultaneously…
The common tangent is y = – 2x +3
Graphical...
Upvote ·
9 9
9 2
Bhalchandra Khare
Former Self Employed Chemical Engineer (1975–2016) · Author has 4.3K answers and 1.4M answer views
·1y
Related
What is the equation of tangent and normal to the curve, given the curve y^2=5x-1 at point (1,-2)?
Slope of tangent = dy/dx at (1,-2)
2ydy/dx=5
dy/dx=5/2y=5/-4=-1.25
Slope of normal to curve =0.8
Equation of tangent y+2=-5(x-1)/4
4y+8+5x-5=0
5x+4y+3=0
Equation of normal
y+2=4(x-1)/5
4x-4-5y-10=0
4x-5y-14=0
Continue Reading
Slope of tangent = dy/dx at (1,-2)
2ydy/dx=5
dy/dx=5/2y=5/-4=-1.25
Slope of normal to curve =0.8
Equation of tangent y+2=-5(x-1)/4
4y+8+5x-5=0
5x+4y+3=0
Equation of normal
y+2=4(x-1)/5
4x-4-5y-10=0
4x-5y-14=0
Upvote ·
9 2
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Finding the maximum squared distance between a pair of coordinates from a set of coordinates
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I am trying to write an algorithm which finds the maximum squared distance between a pair of coordinates from a set of given coordinates.
Note : The coordinates are integers
Let's suppose I have a set of coordinates as a:(1,2),b:(2,3),c:(3,4)a:(1,2),b:(2,3),c:(3,4), I naturally took the process of finding the squared distance between each pair;
D a b=2 D a b=2
D b c=2 D b c=2
D a c=8 D a c=8
Here we can easily pick D a c D a c as our maximum squared distance.
Now while trying to write the code for this, I am a little confused as to how to generalize this approach, here's what I tried,
Say we enter in n-coordinates, (x 1,y 1),(x 2,y 2),⋯(x n,y n)(x 1,y 1),(x 2,y 2),⋯(x n,y n).
Our squared distance is defined by;
D=(x i−x j)2+(y i−y j)2 D=(x i−x j)2+(y i−y j)2
Now I went for the obvious way which is finding the right conditions which maximizes the above expression using the known inputs from (x 1,y 1),(x 2,y 2),⋯(x n,y n)(x 1,y 1),(x 2,y 2),⋯(x n,y n).
I naturally thought of picking the value of x i x i as the largest of the x-coordinates and x j x j as the most smallest of the x-coordinates to maximize (x i−x j)2(x i−x j)2.
So I applied the above approach by taking in n-coordinates and then storing all x and y coordinates in two different arrays, and then sorting the x-array using the best sort algorithm; quick sort with time complexity O(n l o g(n))O(n l o g(n)), So now I have the advantage of picking the largest and smallest of the x-coordinates, which are present in the last and first position of the array. But this method did not guarantee the maximized value of D D when i stored the previous position of the largest and smallest of the x-array and mapped it to the unsorted y-array to keep a track of the y-coordinate associated with the largest and smallest x-coordinate. This method kept failing as i varied the number of coordinates i entered, primarily because of the constraint on picking the y-coordinates this method drastically is flawed, and I am unable to move further.
I would like to mention I am trying my best to keep the time complexity as low as possible, unlike my previous tries where I ran multiple loops and found my maximum distance easily but this is causing a time complexity of O(n 2)O(n 2) or more. I am extremely interested in finding the most optimized algorithm that can solve this problem.
calculus
multivariable-calculus
optimization
contest-math
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asked Aug 16, 2024 at 11:34
Amrut AyanAmrut Ayan
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1 One can generate the convex hull first, which has order n log n n logn complexity according to en.wikipedia.org/wiki/Convex_hull_algorithms . The final minimization over all pairs then needs to be applied only on the vertices of the convex hull.R. J. Mathar –R. J. Mathar 2024-08-16 11:48:48 +00:00 Commented Aug 16, 2024 at 11:48
How large is n n?user1353692 –user1353692 2024-08-16 11:50:08 +00:00 Commented Aug 16, 2024 at 11:50
@YvesDaoust, in my assignment it is 3≤n≤100 3≤n≤100, does a large upper bound bring difficulties?Amrut Ayan –Amrut Ayan 2024-08-16 11:52:32 +00:00 Commented Aug 16, 2024 at 11:52
1 100 2/2=5000 100 2/2=5000, while 100 log 2(100)=664 100 log 2(100)=664, the ratio is about 7.5 7.5, not counting that the sorting step in Andrew's algorithm performs no computation. So the convex hull solution is worth the try.user1353692 –user1353692 2024-08-16 11:58:20 +00:00 Commented Aug 16, 2024 at 11:58
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If the number of points is low, don't worry about quadratic complexity.
Otherwise
build the convex hull of the point set for instance by the Monotone Chain (Andrew's) algorithm. This takes time O(n log n)O(n logn) and leaves h h extreme points.
if h h is low, you can use brute force (O(h 2)O(h 2)). Otherwise, the Rotating Calipers method, which will run in time O(h)O(h), is fine.
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answered Aug 16, 2024 at 11:48
user1353692 user1353692
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189755 | https://www.mashupmath.com/blog/how-to-subtract-fractions-with-different-denominators | How to Subtract Fractions with Different Denominators — Mashup Math
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How to Subtract Fractions with Different Denominators
Math Skills: How to Subtract Fractions with Unlike Denominators in 3 Easy Steps
Learning how to add and subtract fractions with different denominators is an important math skill that requires a strong understanding of fractions overall. After you learn how to subtract fractions with the same denominator, the next step is to learn how to subtract fractions with different denominators (i.e. how to subtract fractions with unlike denominators).
Subtracting fractions with the same denominator is a pretty straightforward task, but problems involving subtracting fractions with unlike denominators can be more challenging. However, the overall process for subtracting fractions with unlike denominators is a skill that can be mastered by learning a few simple steps and by working through practice problems (and this is exactly what this step-by-step guide will entail).
This free guide on How to Add Subtract with Different Denominators will teach you everything you need to know about subtracting fractions. Together, we will learn a simple 3-step strategy for subtracting fractions that you will be able to use to solve any subtracting fractions with unlike denominators problem.
While we highly recommend that you follow each section of the guide in order, you can click on any of the quick-links below to jump to a specific section.
Review: Subtracting Fractions with the Same Denominator
How to Subtract Fractions with Different Denominators: 3-Step Strategy
How to Subtract Fractions with Different Denominators Example #1
How to Subtract Fractions with Different Denominators Example #2
How to Subtract Fractions with Different Denominators Example #3
Before we get started with learning the 3-step strategy or working on the practice problems, we will do a quick recap of fractions and what it means to subtract fractions with unlike denominators.
Recap: Subtracting Fractions with the Same Denominator
Let’s start by recapping two important vocabulary terms related to fractions:
Definition: The numerator of a fraction is the top number. For example, 2/3 has a numerator of 2.
Definition: The denominator of a fraction is the bottom number. For example, 2/3 has a denominator of 3.
Throughout this tutorial, we will make several references to the numerator and denominator of a fraction, so be sure that you have a firm understanding of these two vocabulary terms before going any further.
Subtracting fractions with the same denominator is a relatively simple task.
For example, consider the following problem where you have to subtract one fraction from another:
3/4 - 1/4 = ?
Both of the fractions, 3/4 and 1/4, have the same denominator (i.e. both fractions have 4 as the denominator).
For examples such as this, if both fractions have a common denominator, all you have to do to solve the problem is to subtract the second numerator from the first numerator and keep the denominator the same. This can be done as follows:
3/4 - 1/4 = (3-1)/4 = 2/4
Notice that our result, 2/4, can be simplified to 1/2 (since both the numerator and the denominator share a greatest common factor of 2, we can reduce by dividing both numbers by 2).
So, we can conclude that:
Final Answer: 3/4 - 1/4 = 1/2
Figure 02 below illustrates how we solved this problem. For a deeper dive into subtracting fractions with the same denominator, we suggest reviewing our free How to Subtract Fractions Student Guide before going any further.
The key takeaway from this review of how to subtract fractions with the same denominator is that whenever you are subtracting fractions with the same denominator, find the difference of the numerators and leave the denominator the same.
Now we are ready to move onto the next section, which will focus on how to subtract fractions with unlike denominators using a 3-step strategy.
How to Subtract Fractions with Unlike Denominators Step-by-Step
First, let’s recall the sample problem from the previous section when we reviewed how to subtract fractions with a common denominator:
3/4 - 1/4 = ?
For this problem, we concluded that the answer is 1/2. Now, let’s look at another practice problem where the fractions have unlike denominators:
3/4 - 2/8 = ?
Notice that, in this problem, one fraction has a denominator of 4 and the other has a denominator of 8 (i.e. the denominators are different).
However, if we compare both problems, we should notice that they are equivalent to each other. Namely, the second fraction in both problems represent “one-quarter” since 1/4 and 2/8 are equivalent fractions.
What does this mean? Both problems represent the difference of three-quarters and one quarter. However, the first problem has common denominators and the section problem does not.
With this difference in mind, we will go ahead and learn how to use a 3-step strategy for subtracting fractions with unlike denominators to solve this problem to see if we get the same result (i.e. the final answer is 1/2).
How to Subtract Fractions with Unlike Denominators:
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
Step Three: Reduce the result if possible.
Now that you know the 3-step strategy for subtracting fractions with unlike denominators, let’s apply them to solving the problem 3/4 - 2/8 = ?.
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
For step one, we have to find a common denominator.
Definition: In math, a common denominator is a common value that the denominators of both fractions can be divided into evenly.
While there are a few ways to find a common denominator between two fractions, the simplest way is to multiply the denominator of the first fraction by the second fraction and vice versa (i.e. multiply the denominator of the second fraction by the first fraction).
For the problem 3/4 - 2/8 = ?, we can use this approach to finding a common denominator as follows:
3/4 - 2/8 = (3x8)/(4x8) - (2x4)/(8x4) = 24/32 - 8/32
Now we have a new equivalent expression, 24/32 - 8/32, where the denominators are the same. And, since the denominators of this new equivalent expression are the same, we can say that we have found a common denominator and we can move onto step two.
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
For the second step, we will focus on our new expression where the fractions share a common denominator of 32:
3/4 - 2/8 → 24/32 - 8/32
From here, we can solve 24/32 - 8/32 by finding the difference of the numerators and leaving the denominator the same:
24/32 - 8/32 = (24-8)/32 = 16/32
Step Three: Reduce the result if possible.
Lastly, we have to check and see whether or not our result, 13/15, can be simplified. Since both 16 and 32 share a greatest common factor of 16, we can divide both the numerator, 16, and the denominator, 32, by 16 to get an answer in reduced form:
(16÷16)/(32÷16) = 1/2
So, we can conclude that:
Final Answer: 3/4 - 2/8 = 1/2
Our final answer should look familiar. Remember that we already knew that our answer would be 1/2 since we know that three-quarters minus one-quarter equals one-half. The key takeaway here is understanding that our 3-step strategy will allow us to correctly solve any problem where you are tasked with subtracting fractions with the same denominator.
Now, let’s go ahead and get some more practice using the 3-step strategy for subtracting fractions with unlike denominators.
How to Subtract Fractions with Different Denominators Example #1
Example #1: 4/5 - 2/3
For this first example, we want to find the difference between four-fifths (4.5) and two-thirds (2/3). To solve this problem, we can use of 3-step strategy as follows:
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
First, we have to find a common denominator by multiplying the denominator of the first fraction (5) by the second fraction (2/3) and then by multiplying the denominator of the second fraction (3) by the first fraction (4/5):
4/5 - 2/3 = (3x4)/(3x5) - (5x2)/(5x3) = 12/15 - 10/15
Notice that we now have a new equivalent expression with common denominators (i.e. both fractions have the same denominator, which, in this case, is 15).
4/5 - 2/3 → 12/15 - 10/15
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
For the second step, we can solve 12/15 - 10/15 = ? as follows:
12/15 - 10/15 = (12-10)/15 = 2/15
Step Three: Reduce the result if possible.
And now, for the final step, we just have to see if the result, 2/15, can be simplified. Since 2 and 15 do not have any factors in common (other than 1), this fraction is already in reduced form and can not be simplified further. Therefore…
Final Answer: 4/5 - 2/3 = 2/15
The entire 3-step process for solving this first example is illustrated in Figure 06 below:
How to Subtract Fractions with Unlike Denominators Example #2
Example #2: 3/7 - 2/9
Just like the last example, we can use our 3-step strategy to solve this problem.
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
To find a common denominator, we can multiply the first fraction, 3/7, by the denominator of the second fraction, 9, and then multiply the second fraction, 2/9, by the denominator of the first fraction, 7, as follows:
3/7 - 2/9 = (9x3)/(9x7)-(7x2)/(7x9) = 27/63 - 14/63
Now we have found an expression that is equivalent to the original problem:
3/7 - 2/9 → 27/63 - 14/63
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
Next, we can go ahead and solve 27/63 - 14/63 as follows:
27/63 - 14/63 = (27-14)/63 = 13/63
Step Three: Reduce the result if possible.
For the very last step, we just have to see if our result, 13/63, can be simplified any further. Notice that 13 and 63 do not share any common factors besides 1, so we can conclude that 13/63 can not be simplified and:
Final Answer: 3/7 - 2/9 = 13/63
All of the steps for solving this second example are detailed in Figure 07 below.
Are you feeling more confident with using the 3-step strategy to subtract fractions with unlike denominators? If not, let’s go ahead and gain some more experience by working through one more practice problem.
How to Subtract Fractions with Different Denominators Example #3
Example #3: 7/8 - 4/24
For this last example, we can again use the 3-step strategy to solve it:
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
First, we can find a common denominator by taking each fraction and multiplying it by the denominator of the opposite fraction:
7/8 - 4/24 = (24x7)/(24x8) - (8x4)/(8x24) = 168/192 - 32/192
Now we have a new expression where both fractions share a common denominator:
7/8 - 4/24 → 168/192 - 32/192
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
Next, we have to solve:
168/192 - 32/192 = (168-32)/192 = 136/192
Step Three: Reduce the result if possible.
One more step! To wrap this problem up, we have to try and simplify 136/192. We know that this fraction can be reduced since both 136 and 192 share 2 as a common factor. However, they actually share a greater common factor, which is 8. So, after we divide the numerator, 136, and the denominator, 192, by 8, we are left with 17/24, and we can conclude that:
Final Answer: 7/8 - 4/24 = 17/24
The complete step-by-step process of solving Example #3 is illustrated in Figure 08 below.
Conclusion: How to Subtract Fractions with Different Denominators
Subtracting fractions with unlike denominators is a foundational math skill that every student must master when learning how to perform operations with fractions.
This step-by-step tutorial on subtracting fractions focused on teaching you how to find the difference between two fractions that do not have the same denominator.
To solve problems where you have to subtract fractions with unlike denominators, we used the following 3-step strategy:
Step One: Find a common denominator by multiplying each fraction by the opposite fraction’s denominator.
Step Two: Subtract the second numerator from the first numerator and keep the denominator.
Step Three: Simplify the result if possible.
These three steps were used to solve the three example problems in this guide. They can also be used to solve any subtracting fractions problems that you may come across, so be sure to keep them in your notes and practice using them as often as you can!
Keep Learning:
How to Add Fractions with Different Denominators in 3 Easy Steps
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189756 | https://chem.libretexts.org/Courses/Southwestern_College/Atoms_First_-_Introductory_Chemistry_for_Science_and_Engineering/09%3A_Stoichiometry_-_Quantities_in_Chemical_Reactions/9.01%3A_Stoichiometry_-_Moles_are_Key_to_Understanding/9.1.01%3A_Stoichiometry_-_Mass_to_Mass_Conversions | 9.1.1: Stoichiometry - Mass to Mass Conversions - Chemistry LibreTexts
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9.1: Stoichiometry - Moles are Key to Understanding
9: Stoichiometry - Quantities in Chemical Reactions
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Contents
1. Home
2. Campus Bookshelves
3. Southwestern College
4. Atoms First - Introductory Chemistry for Science and Engineering
5. 9: Stoichiometry - Quantities in Chemical Reactions
6. 9.1: Stoichiometry - Moles are Key to Understanding
7. 9.1.1: Stoichiometry - Mass to Mass Conversions
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Atoms First - Introductory Chemistry for Science and Engineering
Front Matter
1: Chemistry, Measurements, and Calculations
2: Atomic Structure
3: Light, Electrons, and the Periodic Table
4: Matter, Compounds, and Nomenclature
5: The Mole, Chemical Formulas, and Composition
6: Lewis Structures, Molecular Shape, and Intermolecular Forces
7: Aqueous Solutions
8: Chemical Reactions
9: Stoichiometry - Quantities in Chemical Reactions
10: Solids, Liquids, and Gases
11: Acids and Bases
12: Chemical Equilibrium
13: Radioactivity and Nuclear Chemistry
Back Matter
9.1.1: Stoichiometry - Mass to Mass Conversions
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9.1.2: Stoichiometry - Volume to Volume Conversions
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Table of contents
1. Learning Objectives
2. Mass to Mass Stoichiometry Conversions
1. Example 9.1.1.2: Decomposition of Ammonium Nitrate
2. Exercise 9.1.1.2: Carbon Tetrachloride
Summary
Learning Objectives
Convert from mass of one substance to mass of another substance in a chemical reaction.
Mass to Mass Stoichiometry Conversions
The first pathway we will look at is starting with grams of one chemical in an equation and ending with grams of another. See the highlighted portion below.
Figure 9.1.1.1: Mass stoichiometry pathway
Here's an example of how it will work.
Example 9.1.1.2: Decomposition of Ammonium Nitrate
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.
NHA 4NOA 3(s)→NA 2O(g)+2HA 2O(l)
In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Solutions to Example 8.5.2| Steps for Problem Solving | Example 9.1.1.2 |
--- |
| Identify the "given" information and what the problem is asking you to "find." | Given: 45.7 gNHA 4NOA 3 Find: Mass NA 2O=?g Mass HA 2O=?g |
| List other known quantities. | 1 mol NHA 4NOA 3=80.06 g/mol 1 mol NA 2O=44.02 g/mol 1 mol HA 2O=18.02 g/mol 1 mol NH 4 NO 3 to 1 mol N 2 O to 2 mol H 2 O |
| Prepare two concept maps and use the proper conversion factor. | Flowchart of conversion factors: 1 mole NH4NO3 to 80.06 grams NH4NO3, 1 mole N2O to 1 mole NH4NO3, 44.02 grams N2O to 1 mole N2O Flowchart of conversion factors: 1 mole NH4NO3 to 80.06 grams NH4NO3, 2 moles H2O to 1 mole NH4NO3, 18.02 grams H2O to 1 mole H2O |
| Cancel units and calculate. | 45.7 gNHA 4NOA 3×1 mol NHA 4NOA 3 80.06 gNHA 4NOA 3×1 mol NA 2O 1 mol NHA 4NOA 3×44.02 gNA 2O 1 mol NA 2O=25.1 gNA 2O 45.7 gNHA 4NOA 3×1 mol NHA 4NOA 3 80.06 gNHA 4NOA 3×2HA 2O 1 mol NHA 4NOA 3×18.02 gHA 2O 1 mol HA 2O=20.6 gHA 2O |
| Think about your result. | The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures. |
Exercise 9.1.1.2: Carbon Tetrachloride
Methane can react with elemental chlorine to make carbon tetrachloride (CClA 4). The balanced chemical equation is as follows:
CHA 4(g)+4 ClA 2(g)CClA 2(l)+4HCl(l)
How many grams of HCl are produced by the reaction of 100.0 g of CHA 4?
Answer 908.7g HCl
Summary
A balanced chemical reaction can be used to determine mass relationships between substances.
9.1.1: Stoichiometry - Mass to Mass Conversions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.
8.4: Making Molecules- Mass-to-Mass Conversions by Henry Agnew, Marisa Alviar-Agnew is licensed CC BY-NC-SA 3.0.
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189757 | https://home.ttic.edu/~madhurt/courses/infotheory2021/l4.pdf | Information and Coding Theory Winter 2021 Lecture 4: January 21, 2021 Lecturer: Madhur Tulsiani 1 Fano’s inequality We first prove an important inequality that lets us understand how well can some “ground truth” random variable X be predicted based on some observed data Y. We state the inequality in the language of Markov chains, which we saw before in the context of data processing inequality. We will denote the Markov chain as X →Y →b X. We can think of X as the choice of an unknown parameter from some finite set X . We think of Y as the “data” generated from this, say a sequence independent samples. Finally, we think of b X as a “guess” for X, which depends only on the data. Fano’s inequality is concerned with the probability of error in the guess, defined as pe = P h b X ̸= X i . We have the following statement Lemma 1.1 (Fano’s inequaity). Let X →Y →b X be a Markov chain, and let pe = P ˆ X ̸= X .
Let H2(pe) denote the binary entropy function computed at pe. Then, H2(pe) + pe · log (|X | −1) ≥H(X| b X) ≥H(X|Y) .
Proof: We define a binary random variable, which indicates an error i.e E := 1 if b X ̸= X 0 if b X = X The bound in the ineuality then follows from considering the undertainty that still remains after our prediction, i.e., the entroy H(X, E| b X).
H(X, E| b X) = H(X| b X) + H(E|X, b X) = H(X| b X) , since H(E|X, b X) = 0 (why?) Another way of computing this entropy is H(X, E| b X) = H(E| b X) + H(X|E, b X) = H(E| b X) + pe · H(X|E = 1, b X) + (1 −pe) · H(X|E = 0, b X) ≤H(E) + pe · H(X|E = 1, b X) ≤H2(pe) + pe · log (|X | −1) .
Comparing the two expressions then proves the claim.
1 Fano’s inequality provides a useful way of lower bounding the error of a predictor, partic-ularly in the case when |X | > 2. As we will see later, in the case when |X | = 2, we will be able to obtain better bounds using the concept of KL-divergence considered later.
2 Graph Entropy We now consider an application of mutual information, using the concept of Graph En-tropy defined by Körner [Kör73], and later used by Newman and Wigderson [NW95] for certain circuit (formula) lower bound problems. This also provides an example of the sce-nario we discussed in the previous lecture, when the mutual information I(X; Y is being optimized over our choice of random variables X, Y, rather than being computed for given random variables.
Given a graph G = (V, E), we define the graph entropy H(G) as min X,Y I(X; Y) s. t. X is uniformly distributed over V Y is an independent set in G containing X Note that while the concept is called “entropy”, we are defining it as a mutual informa-tion. The name entropy comes from the original definition related to the best (asymptotic) transmission rate for a random variable distributed over the vertices of the graph, when we are required to use different symbols for vertices connected by edges (but not neces-sarily otherwise). It can be proved that this asymptotic limit comes out to be equal to the mutual information above, and we will use this version of the definition. Also, while the graph entropy can be defined with respect to any distribution P on the vertex set V, we will restric our discussion to the uniform distribution. Let us check a couple of examples.
Example 2.1 (Complete graph). Let Kn denote the complete graph on n vertices. Then H(Kn) = log n. This follows from the fact that any independent set is of size at most one, and thus, we must have Y = X. This gives I(X; Y) = H(X) −H(X|Y) = log n −0 = log n .
Also note that log n is the maximum possible value for a graph with |V| = n.
Example 2.2 (Bipartite graph). Let G be a bipartite graph, with n1 vertices on one side and n2 vertices on the other. Then, for any vertex v, all the vertices on the side of v form an indepdent set containing v. If X is a uniformly random vertex, and Y equals all the vertices on the side of X, then I(X; Y) ≤H(Y) = n1 n1 + n2 · log n1 + n2 n1 + n2 n1 + n2 · log n1 + n2 n2 ≤1 .
Since H(G) is the minimum of I(X; Y) over all (X, Y), we get that H(G) ≤1.
2 Exercise 2.3. Let α(G) denote the size of the maximum independent set in a graph G. Prove that H(G) ≥log n α(G) .
An important property of graph entropy that we need, is that it is sub-additive under union of edges.
Proposition 2.4 (Sub-additivity of graph entropy). Let G1 = (V, E1) and G2 = (V, E2) be two graphs, and let G = (V, E1 ∪E2), which we denote by G = G1 ∪G2. Then, H(G) = H(G1 ∪G2) ≤H(G1) + H(G2) .
Proof: Let (X, Y1) and (X, Y2) be pairs of random variables achieving H(G1) and H(G2) (note that in both cases X is a uniform vertex from V). We can define (why?) a joint distribution on the tuple (X, Y1, Y2) such that Y1 and Y2 are independent conditioned on any value of X. Take this to be the joint distribution of the tuple (X, Y1, Y2) and let Y = Y1 ∩Y2. Note that if Y1, Y2 are independent sets containing X in G1 and G2 respectively, then Y1 ∩Y2 is an indepdent set in G, containing X. This gives, H(G1 ∪G2) ≤I(X; Y) ≤I(X; (Y1, Y2)) (data processing inequality) = H(Y1, Y2) −H(Y1, Y2 | X) = H(Y1, Y2) −H(Y1 | X) −H(Y2 | X) (conditional independence) ≤H(Y1) + H(Y2) −H(Y1 | X) −H(Y2 | X) (sub-additivity of entropy) = H(G1) + H(G2) , which proves the claim.
2.1 Covering the complete graph with bipartite graphs The properties of graph entropy considered so far can be used to provide a very simple answer to the following combnatorial question: what is the minimum number of bipartite graphs G1, . . . , Gr such that their edges cover all the edges of the complete graph i.e., Kn = G1 ∪· · · ∪Gr .
Note that just counting edges does not give a very strong bound since Kn has n(n −1)/2 edges, while even a single bipartite graph can have n2/4 edges. On the other hand, graph entropy will yield a (tight!) boound of log n. This also proves a special case of the formula-size lower bounds considered by Newman and Wigderson [NW95], when considering W V W formulas (three alternating layers of OR, AND, and OR gates, with AND gates hav-ing fan-in 2) for the threshold function checking ∑n i=1 xi ≥2. Take a look at the paper for more details.
3 Back to the case of graphs, when Kn = G1 ∪· · · ∪Gr, we have log n = H(Kn) ≤H(G1) + · · · + H(Gr) ≤r , where we used the bounds on the graph entropy of complete and bipartite graphs, as computed earlier.
Exercise 2.5. Prove that the above bound is tight. In particular, when n is a power of 2, find a covering of Kn with log n bipartite graphs (Hint: Think of each vertex as a (log n)-bit string).
3 Kullback Leibler divergence The Kullback-Leibler divergence (KL-divergence), also known as relative entropy, is a measure of how different two distributions are. Note that here we will talk in terms of distributions instead of random variables, since this is how KL-divergence is most com-monly expressed. It is of course easy to think of a random variable corresponding to a given distribution and vice-versa. We will use capital letters like P(X) to denote a distri-bution for the random variable X and lowercase letters like p(x) to denote the probability for a specific element x.
Let P and Q be two distributions on a universe X , then the KL-divergence between P and Q is defined as: D(P||Q) := ∑ x∈U p(x) log p(x) q(x) Let us consider a simple example.
Example 3.1. Suppose X = {a, b, c}, and p(a) = 1 3, p(b) = 1 3, p(c) = 1 3 and q(a) = 1 2, q(b) = 1 2, q(c) = 0. Then D(P||Q) = 2 3 log 2 3 + ∞= ∞.
D(Q||P) = log 3 2 + 0 = log 3 2 .
The above example illustrates two important facts: D(P||Q) and D(Q||P) are not necessar-ily equal, and D(P||Q) may be infinite. Even though the KL-divergence is not symmetric, it is often used as a measure of “dissimilarity” between two distribution. Towards this, we first prove that it is non-negative and is 0 if and only if P = Q.
Lemma 3.2. Let P and Q be distributions on a finite universe X . Then D(P||Q) ≥0 with equality if and only if P = Q.
4 Proof: Let Supp(P) = {x | p(x) > 0}. Then, we must have Supp(P) ⊆Supp(Q) if D(P, Q) < ∞. We can then assume without loss of generality that Supp(Q) = X . Using the fact the log is a (strictly) concave function, with Jensen inequality, we have: D(P||Q) = ∑ x∈X p(x) log p(x) q(x) = ∑ x∈Supp(P) p(x) log p(x) q(x) = − ∑ x∈Supp(P) p(x) log q(x) p(x) ≥−log ∑ x∈Supp(P) p(x) · q(x) p(x) = −log ∑ x∈Supp(P) q(x) ≥−log 1 = 0 .
For the case when D(P||Q) = 0, we note that this implies p(x) = p(x) ∀x ∈Supp(P), which in turn gives that p(x) = q(x) ∀x ∈X .
Like entropy and mutual information, we can also derive a chain rule for KL-divergence.
Let P(X, Y) and Q(X, Y) be two distributions for a pair of variables X and Y. We then have the following expression for D(P(X, Y)||Q(X, Y)).
Proposition 3.3 (Chain rule for KL-divergence). Let P(X, Y) and Q(X, Y) be two distributions for a pair of variables X and Y. Then, D(P(X, Y) ∥Q(X, Y)) = D(P(X) ∥Q(X)) + E x∼P [D(P(Y|X = x) ∥Q(Y|X = x))] = D(P(X) ∥Q(X)) + D(P(Y|X) ∥Q(Y|X)) Here P(X) and Q(X) denote the marginal distributions for the first variable, and P(Y|X = x) denotes the conditional distribution of Y.
Proof: The proof follows from (by now) familiar manipulations of the terms inside the 5 log function.
D(P(X, Y) ∥Q(X, Y)) = ∑ x,y p(x, y) log p(x, y) q(x, y) = ∑ x,y p(x)p(y|x) log p(x) q(x) · p(y|x) q(y|x) = ∑ x p(x) log p(x) q(x) ∑ y p(y|x) + ∑ x p(x)∑ y p(y|x) log p(y|x) q(y|x) = D(P(X) ∥Q(X)) + ∑ x p(x) · D(P(Y|X = x) ∥Q(Y|X = x)) = D(P(X) ∥Q(X)) + D(P(Y|X) ∥Q(Y|X)) Note that if P(X, Y) = P1(X)P2(Y) and Q(X, Y) = Q1(X)Q2(Y), then D(P||Q) = D(P1||Q1) + D(P2||Q2).
We note that KL-divergence also has an interesting interpretation in terms of source cod-ing. Writing D(P||Q) = ∑p(x) log p(x) q(x) = ∑p(x) log 1 q(x) −∑p(x) log 1 p(x) , we can view this as the number of extra bits we use (on average) if we designed a code according to the distribution P, but used it to communicate outcomes of a random variable X distributed according to Q. The first term in the RHS, which corresponds to the average number of bits used by the “wrong” encoding, is also referred to as cross entropy.
3.1 Total variation distance and Pinsker’s inequality We can now relate KL-divergence to some other notions of distance between two proba-bility distributions.
Definition 3.4. Let P and Q be two distributions on a finite universe X . Then the total-variation distance or statistical distance between P and Q is defined as δTV(P, Q) = 1 2 · ∥P −Q∥1 = 1 2 · ∑ x∈X |p(x) −q(x)| .
The quantity ∥P −Q∥1 is referred to as the ℓ1-distance between P and Q.
6 The total variation distance of P and Q represents the maximum probability with which any test can distinguish between the two distributions given one random sample. It may seem that the restriction to one sample severely limits the class of tests, but we can always think of an n-sample test for P and Q as getting one sample from one of the product distributions Pn or Qn.
Let f : X →{0, 1} be any classifier, which given one sample x ∈X , outputs 1 if the guess is that the sample came from P, and 0 if the guess is that it came from Q. The difference in its behavior over the two distributions can be measured by the quantity (which can be thought of as the rate of true positive minus the rate of false positive) |EP [ f (x)] −EQ [ f (x)]|. The following lemma bounds this in terms of the total variation distance.
Lemma 3.5. Let P, Q be any distributions on X . Let f : X →[0, B]. Then E P [ f (x)] −E Q [ f (x)] ≤B 2 · ∥P −Q∥1 = B · δTV(P, Q) .
Proof: E P [ f (x)] −E Q [ f (x)] = ∑ x∈X p(x) · f (x) −∑ x∈X q(x) · f (x) = ∑ x∈X (p(x) −q(x)) · f (x) = ∑ x∈X (p(x) −q(x)) · f (x) −B 2 + B 2 · ∑ x∈X p(x) −q(x) ! ≤∑ x∈X |p(x) −q(x)| · f (x) −B 2 ≤B 2 · ∥P −Q∥1 Exercise 3.6. Prove that the above inequality is tight. What is the optimal classifier f?
In many applications, we want to actually bound the ℓ1-distance between P and Q but it’s easier to analyze the KL-divergence. The following inequality helps relate the two.
Lemma 3.7 (Pinsker’s inequality). Let P and Q be two distributions defined on a universe X .
Then D(P ∥Q) ≥ 1 2 ln 2 · ∥P −Q∥2 1 .
7 We will prove the inequality in two steps. Let us first consider a special case when X = {0, 1} and P, Q are distributions as below P = ( 1 w.p. p 0 w.p. 1 −p and Q = ( 1 w.p. q 0 w.p. 1 −q In this case, we have D(P∥Q) = p · log p q + (1 −p) · log 1 −p 1 −q and ∥P −Q∥1 = 2 · |p −q| .
We will first prove Pinsker’s inequality for this special case.
Proposition 3.8 (Pinsker’s inequality for X = {0, 1}). Let P and Q be distributions as above.
Then, p · log p q + (1 −p) · log 1 −p 1 −q ≥ 2 ln 2 · (p −q)2 .
Proof: Let f (p, q) := p · log p q + (1 −p) · log 1 −p 1 −q − 2 ln 2 · (p −q)2 .
We have, ∂f ∂q = −(p −q) ln 2 1 q(1 −q) −4 .
Since 1 q(1−q) −4 ≥0 for all q, we have that ∂f ∂q ≤0 when q ≤p and ∂f ∂q ≥0 when q ≥p.
Moreover, f (p, q) = ∞when q = 0 and f (p, q) = 0 when q = p. Thus, the function achieves its minimum value at q = p and is always non-negative, which proves the desired inequality.
We can now reduce the general case of Pinsker’s inequality, to the case of X = {0, 1} considered above.
Proposition 3.9. Let P and Q be distributions on a finite set X . Then, there exist distributions P′, Q′ on {0, 1} such that P′ −Q′ 1 = ∥P −Q∥1 and D(P∥Q) ≥D(P′∥Q′) Proof: Let A ⊂X be A = {x | p(x) ≥q(x)} .
8 and P′ and Q′ be P′ := 1 w.p.
∑ x∈A p(x) 0 w.p.
∑ x/ ∈A p(x) and Q′ := 1 w.p.
∑ x∈A q(x) 0 w.p.
∑ x/ ∈A q(x) Then, ∥P −Q∥1 = ∑ x∈X |p(x) −q(x)| = ∑ x∈A (p(x) −q(x)) + ∑ x/ ∈A (q(x) −p(x)) = ∑ x∈A p(x) −∑ x∈A q(x) + 1 −∑ x∈A p(x) !
− 1 −∑ x∈A q(x) ! = P′ −Q′ 1 To calculate the KL-divergence, we define a random variable Z (which is a function of X) as Z = ( 1 if x ∈A 0 if x / ∈A .
Since Z is a function of X, we can also think of the two distributions P and Q as joint distributions for the random variables (X, Z). Also, note that the marginal distributions of Z are P′ and Q′. Applying the chain rule for KL-divergence gives D(P∥Q) = D(P(X, Z) ∥Q(X.Z)) = D(P(Z) ∥Q(Z)) + D(P(X|Z) ∥Q(X|Z) ≥D(P(Z) ∥Q(Z)) = D P′∥Q′ which completes the proof.
Finally, we can complete the proof of Pinkser’s inequality for the general case, by noting that D(P∥Q) ≥D(P′∥Q′) ≥ 1 2 ln 2 · P′ −Q′ 2 1 = 1 2 ln 2 · ∥P −Q∥2 1 .
References [Kör73] János Körner, Coding of an information source having ambiguous alphabet and the en-tropy of graphs, 6th Prague conference on information theory, 1973, pp. 411–425.
2 9 [NW95] Ilan Newman and Avi Wigderson, Lower bounds on formula size of Boolean functions using hypergraph entropy, SIAM Journal on Discrete Mathematics 8 (1995), no. 4, 536–542. 2, 3 10 |
189758 | https://www.youtube.com/watch?v=GOAbVl5wtlk | On a farm, the ratio to cows to goats is 1:3, pigs to goats 3:4. What is the ratio of cows to pigs?
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How to solve a ratio math word problem - 3 ratios involved.
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Transcript:
okay so I think a lot of people are going to be very confused on how to solve this math problem but uh maybe you will be the exception let's go and take a look at our problem so the main question here is which is the correct ratio and obviously this is a multiple choice math question but let me go ahead and read the problem it says on a farm the ratio of cows to goats is 1 to three and the ratio of pigs to goats is 3 to4 what is the ratio of cows to pigs all right now if you can figure this out we go ahead and put your answer into the comment section and feel free to use a calculator but I'm going to show you the exact answer in just one second then of course we're going to walk through exactly how to solve this interesting problem but before we get started let me quickly introduce myself my name is John and I have been teaching middle and high school math for a decades and if you need help learning math well check out my math help program at TCM academy.com you can find a link to that in the description below and if this video helps you out or if you just enjoy this content make sure to like And subscribe as that definitely helps me out all right so one more time we have a uh multiple choice math question so our first uh answer option here a is one to4 B is 4 to9 C is 5 to 4 d is 9 to4 and E is is 4 to1 so once again on a farm the ratio of cows to goats is 1 to3 and the ratio of goats uh of pigs to goats is 3 to four what is the ratio of cows to pigs all right so let's go and take a look at the answer the correct answer is b429 all right now if you got this right well you're definitely going to get a happy face and a plusy 100% and a certificate of excellence for being a certified professional in the area of ratios and proportions and uh this problem is not that easy a lot of uh even really strong people that are strong in math are going to be very confused on how to get the anwers so if you're like all right Mr you to math man just hurry up and get to the solution well yeah let's go ahead and start but once again don't feel bad if you are confused and maybe you're in doubt of this answer well I'll go ahead and show you at the end of this video we'll kind of like verify the in fact this is the correct solution all right so let's go ahead and get into the solution right now so the first thing is we are dealing with a ratio problem now what is a ratio well in simple terms a ratio is a fraction so right here for example I'm saying 1 2 3 so we can express a ratio as 1 2 3 or we could write it this way 1 2 3 or we could even write it this way 1 2 3 now what makes a ratio a ratio well basically the numerator and denominator of these respective fractions um have the same units of measure so here you might be saying well we're we're talking about cows and goats well we're really counting animals okay so this is a big broader Topic in mathematics called uh ratios rates and proportions but effectively that is what a ratio is all right now uh getting back to the problem now one thing that you always want to be aware of when it comes to a math multiple choice uh problem is to see if you can use the answers to figure this problem out okay so you can go through a process of elimination but in this case I think a lot of people will be confused on you know how they could possibly use the answer choices to get to the right answer so uh probably the best option here is just to go ahead and solve it directly all right now let me go ahead and show you a very simple ratio and proportion problem for uh problem for those of you that are like yes indeed I understand ratios Mr YouTube Math Man well let's review a basic problem then of course we'll get into this solution right now okay so what if we had this situation right here so the ratio of uh cows to goats is 1 to three all right so let's say this is given information and let's say the question was all right so the ratio of cows goats is 1 to three how many cows do we have if we have 15 goats all right so the way you solve this is to set up a proportion now a proportion is two equal fractions or two equal uh rates or ratios so what we're going to what we're going to do is uh write C okay this is our variable we're looking for how many cows when we have 15 goats and the ratio is 1 to three one cow for every uh three goats so what we're going to do here is simply use the cross product or cross multiply so 3 C is 3 c and 1 15 is 15 and then we solve for C by dividing both sides of the equation by three so we get C is equal to five all right so that means that there is five cows when we have 15 goats and that makes sense because this fraction here 5 over 15 we can simplify to 1 2 3 which of course is the ratio of cows to goats now hopefully most of you out there are familiar with uh solving these types of uh ratio and proportion problems and if you are not well this is a huge deal in mathematics I'll give you some suggestions on how you can improve in this material but anyways this is a simple example of how to solve a ratio problem but this ratio is uh or this problem is more interesting because we have two different ratios going on and we're asked to find another ratio so what do we do well let's go ahead and uh Define the problem here of course you always want to use the rule of three when you're solving any math word problem read the thing at least three times uh obviously if you read this thing one time it's probably going to be confusing so reread it take your time and now let's go ahead and start uh defining uh this uh the various aspects of this problem all right so we have one ratio cows to goats and then we have another ratio here of pigs to goats and we're given uh these uh specific specific ratios and we're um asked to find this other ratio of of uh cows to pigs even even I have to reread this here not to be confused all right so let's use Simple variables like a c for cows and a g for goats and a for pigs just to kind of uh you know Define everything that's going on in this problem all right so here we go so the ratio of cows to goats is 1 to three so this is one ratio and then we have another ratio of for pigs uh to goats that's 3 to4 and the question is what is the ratio uh for cows to Pig so this is what we're looking for right here all right so what do you think uh how can we approach this problem well let's uh go ahead and discuss that right now now by the way there is multiple ways different paths that you can take to solve this problem now uh the way I'm going to solve it may not be the way that you solved it if you got this right but uh again here's the thing whatever you do in mathematics in terms of solving a problem it's really important that you're able to kind of justify your conclusion so write these things down write these steps down and act as if you were going to turn this into let's say a math teacher on a test you need to justify your results all right so what can we do here well we have this given information and we're asked to find this ratio so does anything stand out here well some of you might be saying hey Mr YouTube math man it looks like uh the goats are common between these two ratios so maybe we need to focus in on the goats uh the number of goats that we have in this situation on this farm right so we have cows goats and pigs but we do know the ratios these two ratios and we have goats uh goats in this ratio and goats in this ratio but uh over here we have three goats and over here we have four goats so how can we kind of look at this where we can look at both ratios that you know basically have them both be in common well this is a little bit of a hint common so what we want to do here is find the common denominator so in other words we're going to rewrite these fractions such that the denominators here are the same okay when we uh do that we're going to have basically equivalent fractions but we're going to be able to uh basic basically have ratios two ratios that we can comp compare the same number of goats and then we can look at the number of cows and pigs that we have all right so how can we change uh these fractions now here um obviously kind of wrote this out the LCD the lowest common denominator is 12 so hopefully you know if you for example were're trying to add 1/3 plus 3/4s that the LCD would in fact be 12 all right so what do we need to do here to fix these fractions up such that the LCD is 12 but here we're going to have to multiply this by four and if we're going to multiply this denominator by four we're going to multiply this numerator by four and here we have to multiply this denominator by three and we're going to multiply this numerator by three okay so let's go and do that right now and here is what we have all right so the results of doing that is going to be 4 over2 and 9 over 12 for our ratios okay so I did the work so keep that in mind because I'm going to show you the answer right now okay so here we go so so cows to goats okay our ratio for cows to goats is 1 to three okay remember it's 1 to three but an equivalent uh fraction for that is 4 to2 because we can reduce 4 to 12 down to 1: 3 okay so if you have four cows and 12 goats that is a ratio of 1 to3 how about pigs to goats well it's 9 to2 and that is equivalent to the ratio 3 to 4 all right but the advantage of doing this is now we have the same number of goats in this situation okay so now we can kind of compare how many cows and pigs we have because the question is asking what is the ratio of cows to pigs well how many cows and how many pigs do we have now because we have the same number of goats in common well we have four cows and nine pigs so we need to compare those two numbers and this is becomes very easy so our ratio for cows to pigs our cows we have four cows we're going to put that here and for pigs we have nine pigs so here is our final answer for tonight all right so hopefully this makes sense but let's suppose you were saying I don't know Mr YouTube math man I don't think this is right you know uh let's kind of you maybe verify this well let's go ahead and do that right now and uh before we do that though I'm hoping that you can go ahead and smash that subscribe button now I'm definitely not um shy to ask for help matter of fact can't even be shy be on YouTube just got to put your content out and uh if any of you are interested in doing videos for YouTube well I strongly uh would encourage you just to you know all of us are experts in something so if you're an expert in a particular you know uh subject or a skill get on YouTube it's a great platform to help others but I definitely need your help uh to grow my channel okay now why would I want to grow my channel well as a math teacher I want to teach as many people as possible and the only way I can grow this channel uh well on YouTube is to have people subscrib so this is a big deal and it's a great way to show support for my work so hit that subscribe button and hit that notification Bell as well so you can get my latest videos all right now before we get into the rest of this prom we're going to look at verifying is if you are confused with uh ratios uh rates proportions basically we're talking about is a basic algebra now this is taught in basic math as well so I'm going to leave links to all my full main math courses in the description of this video uh I would suggest for this particular level of math uh check out like my pre-algebra or algebra one course now if you are not a math student and you just want to kind of relearn math well then check out my math skills rebuild a course I start with basic math we start from the very very beginning in this course and then we work our way up through algebra geometry even some trigonometry and some probability and statistics all right so let's go and finish this up now we already have the right answer uh 4 to9 but let's just double check this right so this is a good way of doing this okay so here uh we have uh the same number of goats so we have 12 goats right so let's just kind of write out 12 goats so we have one two uh three four five six so these are 12 goats right here now we have four cows so here's our four cows one 2 three four and here is our nine pigs right so uh we have nine p pigs we have 12 cows and we have four four cows excuse me nine pigs and 12 goats all right so let's just kind of put all these animals on the farm and now let's go ahead and compare these ratios right so again we have four uh 12 and 9 all right so remember the given information uh was uh the following okay so for cows to goats the ratio is 1 to three so what would we you know how do we you know kind of get to this well we would say okay we have four cows and we're going to compare that to our goats which is what well we have uh 12 goats so that's 4 over 12 which is 1 2 3 okay so this ratio checks out because we have four cows to 12 goats we reduce this fraction down it is one two three and we could do the same thing with pigs to goats we know the ratio 3 to four but we could just count these up so here we have nine pigs and 12 goats so when we compare 9 over 12 well this fraction could be simplified to 3 over 4 and then of course if we are you know pretty sure that we have the correct number of animals here if we want to define the ratio of uh let me go ah and show you down here um cows to pigs well we just count up well we have four cows and nine pigs so that is going to be 4 over 9 okay so hopefully this makes sense and this is a type of problem that uh those of you that have to still take test in life uh this is definitely something that you will encounter this type of problem would uh likely be like on a SAT or act or some sort of college admissions exam well outside of the SAT or act or maybe graduate school so again little bit trickier problem than a a regular run-of-the mail ratio problem but don't feel bad if you got this wrong the main idea here is that you learn something and if that's the case don't forget to like And subscribe and with that being said I definitely wish you all the best in your math Adventures thank you for your time and have a great day |
189759 | https://brainly.in/question/52256265 | convert 220 K to ° C and ° F - Brainly.in
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220 Kelvin is equal to -53.15 Celsius. Formula to convert 220 K to °C is 220 - 273.15
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JEE Main & Advanced
Mathematics
Conic Sections
Equation of the Chord Joining any Two Points on the Parabola
JEE Main & Advanced Mathematics Conic Sections Equation of the Chord Joining any Two Points on the Parabola
Equation of the Chord Joining any Two Points on the Parabola
Category :JEE Main & Advanced
Let P(at_{1}^{2},2a{{t}{1}}),Q(at{2,}^{2},2a{{t}_{2}})
P(a t 2 1,2 a t 1),Q(a t 2 2,,2 a t 2)
be any two points on the parabola {{y}^{2}}=4ax
y 2=4 a x
. Then, the equation of the chord joining these points is, y-2a{{t}{1}}=\frac{2}{{{t}{1}}+{{t}{2}}}\left( x-at{1}^{2} \right)
y−2 a t 1=2 t 1+t 2(x−a t 2 1)
or y({{t}{1}}+{{t}{2}})=2x+2a{{t}{1}}{{t}{2}}
y(t 1+t 2)=2 x+2 a t 1 t 2
.
(1) Condition for the chord joining points having parameters {{\mathbf{t}}{\mathbf{1}}} and {{\mathbf{t}}{\mathbf{2}}} to be a focal chord:If the chord joining points (at_{1}^{2},2a{{t}_{1}})
and (at_{2}^{2},2a{{t}{2}}) on the parabola passes through its focus, then (a,0) satisfies the equation y({{t}{1}}+{{t}{2}})=2x+2a{{t}{1}}{{t}_{2}}
\Rightarrow
0=2a+2a{{t}{1}}{{t}{2}}\Rightarrow {{t}{1}}{{t}{2}}=-1 or {{t}{2}}=-\frac{1}{{{t}{1}}}.
(2) Length of the focal chord:The length of a focal chord having parameters {{t}_{1}}
and {{t}{2}}for its end points is a{{({{t}{2}}-{{t}_{1}})}^{2}}.
keyboard_arrow_left Equation of the Chord of the Parabola Which is Bisected at a Given Point
Diameter of a Parabola keyboard_arrow_right
Other Topics
play_arrowDefinition
play_arrowDefinitions of Various Important Terms
play_arrowGeneral Equation of a Conic Section When Its Focus, Directrix and Eccentricity are Given
play_arrowRecognisation of Conics
play_arrowDefinition
play_arrowStandard Equation of The Parabola
play_arrowSpecial form of Parabola {{\mathbf{(yk)}}^{\mathbf{2~}}}\mathbf{= 4a}\,\,\mathbf{(xh)= a}
play_arrowParametric Equations of a Parabola
play_arrowPosition of a Point and a Line With Respect to a Parabola
play_arrowEquations of Tangent in Different Forms
play_arrowPoint of Intersection of Tangents at Any Two Points on The Parabola
play_arrowEquation of Pair of Tangents From a Point to a Parabola
play_arrowEquations of Normal in Different Forms
play_arrowPoint of Intersection of Normals at Any Two Points on The Parabola
play_arrowRelation Between \mathbf{'}{{\mathbf{t}}{\mathbf{1}}}\mathbf{'} and \mathbf{'}{{\mathbf{t}}{\mathbf{2}}}\mathbf{'} if Normal at \mathbf{'}{{\mathbf{t}}{\mathbf{1}}}\mathbf{'} Meets the Parabola Again at \mathbf{'}{{\mathbf{t}}{\mathbf{2}}}\mathbf{'}
play_arrowCo-normal Points
play_arrowEquation of The Chord of Contact of Tangents to a Parabola
play_arrowEquation of the Chord of the Parabola Which is Bisected at a Given Point
play_arrowEquation of the Chord Joining any Two Points on the Parabola
play_arrowDiameter of a Parabola
play_arrowLength of Tangent, Subtangent, Normal, Subnormal
play_arrowLength of Tangent, Subtangent, Normal and Subnormal to {{\mathbf{y}}^{\mathbf{2}}}\mathbf{= 4ax}\,\,\mathbf{at }\,\mathbf{(a}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{,2at)}
play_arrowPole and Polar
play_arrowDefinition
play_arrowStandard Equation of the Ellipse
play_arrowParametric Form of The Ellipse
play_arrowSpecial Forms of an Ellipse
play_arrowPosition of a Point with Respect to an Ellipse
play_arrowIntersection of a Line and an Ellipse
play_arrowEquations of Tangent in Different Forms
play_arrowEquation of Pair of Tangents \mathbf{S}{{\mathbf{S}}_{\mathbf{1}}}\mathbf{=}{{\mathbf{T}}^{\mathbf{2}}}
play_arrowEquations of Normal in Different Forms
play_arrowAuxiliary Circle
play_arrowChord of Contact
play_arrowEquation of Chord With Mid Point \mathbf{(}{{\mathbf{x}}{\mathbf{1}}}\mathbf{,}{{\mathbf{y}}{\mathbf{1}}}\mathbf{)}
play_arrowEquation of The Chord Joining Two Points on an Ellipse
play_arrowPole and Polar
play_arrowDiameter of The Ellipse
play_arrowSubtangent and Subnormal
play_arrowDefinition
play_arrowStandard Equation of the Hyperbola
play_arrowConjugate Hyperbola
play_arrowSpecial form of Hyperbola
play_arrowAuxiliary Circle of Hyperbola
play_arrowParametric Equations of Hyperbola
play_arrowPosition of a Point With Respect to a Hyperbola
play_arrowIntersection of a Line and a Hyperbola
play_arrowEquations of Tangent in Different Forms
play_arrowEquation of Pair of Tangents
play_arrowEquations of Normal in Different Forms
play_arrowEquation of Chord of Contact of Tangents Drawn from a Point to a Hyperbola
play_arrowEquation of the Chord of the Hyperbola Whose mid point \mathbf{(}{{\mathbf{x}}{\mathbf{1}}}\mathbf{,}\,\,{{\mathbf{y}}{\mathbf{1}}}\mathbf{)} is given
play_arrowEquation of The Chord Joining Two Points on The Hyperbola
play_arrowPole and Polar
play_arrowDiameter of The Hyperbola
play_arrowSubtangent and Subnormal of the Hyperbola
play_arrowAsymptotes of a Hyperbola
play_arrowRectangular or Equilateral Hyperbola
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189761 | https://www.mcgill.ca/oss/article/medical-student-contributors-did-you-know/how-have-twins-different-fathers | McGill University
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How to Have Twins with Different Fathers
Mating, mixing, and matching.
Haleh Cohn | 16 Jun 2023
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Yes, it is possible to have twins with different biological fathers. The scientific term for this anomaly is “heteropaternal superfecundation,” and it’s super cool. “Heteropaternal” signifies different fathers and “superfecundation” means the fertilization of two ova during the same menstrual cycle by separate mating actions. In other words, it is a phenomenon where a second egg is released, and two acts of sexual intercourse can lead to the fertilization of these eggs by two different sources of sperm. The short story is they’re just fraternal twins with an extra twist.
The key ingredient is hyperovulation – which is a fancy way of saying two (or more) eggs are released by the ovaries in the same month, as opposed to the usual one egg. At the beginning of a cycle, the ovaries build up structures called follicles. Typically, at around day 8 of the cycle, all but one of these follicles degenerate. The last one standing develops the ovum (egg), and will then rupture near day 14, ejecting the egg – that’s called ovulation. This process is regulated by hormones, particularly follicle-stimulating hormone (FSH). High FSH levels are thought to be responsible for multiple ovulation, causing more than one follicle to become dominant, or multiple instances of follicle development to occur in the same period. This means the two eggs can get ejected by the same ovary, or one egg can get ejected from each. You can think of the ovaries like a busy restaurant kitchen, receiving a high number of orders. With all that overwhelming input, you may have an accidental duplicate order, which is harmless, it just means someone’s belly might get a little extra full. That’s how you get fraternal twins – two eggs, two sperm, two babies. Note that identical twins arise from one egg splitting into two after fertilization. Identical twins share 100% of their genes, classic fraternals share around 50%, and fraternal twins from different fathers have an average of 25% of the same genes (just like half siblings).
The capacity to have fraternal twins, hence, to double ovulate, is believed to be a complex trait, influenced by environment and genetics. That’s why twins are observed to run in families. But research on candidate genes is still premature. Age is also a factor – higher FSH is more likely in younger menstruators and those approaching menopause.
It’s also important to note that two sperm can’t fertilize the same egg. Double fertilization means there would be too much DNA in the cell, hence it wouldn’t be viable. There is also the case of “superfetation,” when someone with child becomes pregnant again during separate cycles. But this is also very unlikely in humans because the pregnant body creates an environment that suppresses ovulation. While few proposed cases have been reported, the topic remains controversial due to the difficulty in collecting evidence.
The occurrence of heteropaternal superfecundation is extremely rare, though difficult to estimate because it is only discovered if paternity is contested. To reiterate, it relies on many events. When multiple ovulation occurs, it then requires the fertilization of both ova, from different sources of sperm. For many, it’s already difficult enough to conceive one child, let alone twice in one cycle. Couples trying to have children have an approximate conception rate of 30% in the first month, though this varies according to many factors. Ultrasound scans are able to detect multiple ovulation, though its incidence can generally be estimated by the frequency of fraternal twin births. In 2021, around 3% of total live births in Canada were multiple births, of which the vast majority were twins; fraternal twins make up 70% of twin births. The frequency of twin births has also increased overtime with the rise of infertility treatments. That said, in 2020, only 19 cases of heteropaternal superfecundation were reported worldwide.
This phenomenon is more common in animals like cows that ovulate more often and aren’t as monogamous as most humans. Who knew bovine biology could be so bountiful?
@HalehCohn
Haleh Cohn is studying Anatomy and Cell Biology at McGill University with a minor in Economics.
Keywords:
twins
heteropaternal superfecundation
genetics
fraternal twins
identical twins
fertilization
ovulation
siblings
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189762 | https://dqydj.com/median-calculator/ | Median Calculator: What's the Middle Value in a Set?
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Written by:
PK
Below is a median calculator, which will calculate the median from a set of numbers. Enter any numbers (including decimals or negatives) and the tool will return the median.
Need a different tool? Try another:
Mode Calculator
Average Calculator
Range Calculator
Standard Deviation Calculator
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Median Calculator
Table of Contents show ▼
1 Median Calculator
2 What is the Median of a Set of Numbers?
3 Finding a Median in a Set
3.1 Examples of Finding Medians
4 Using the Median Calculator
What is the Median of a Set of Numbers?
The median of a set of numbers is the number that splits the higher numbers from the lower numbers in an ordered set. When a set of numbers has an even length, the median is the average of the middle two numbers.
Along with arithmetic average (the sum of the numbers divided by the number of numbers) and mode (number(s) that appear most in the set) it's a colloquial way to describe the central tendency of a set of numbers.
Finding a Median in a Set
To find the median of a set of numbers, first sort the numbers from low to high, then pick out the "middle" number. In the case your list is an even length, pick the middle two numbers and average them.
This is better shown than told; let's walk through some examples.
Examples of Finding Medians
Let's say you have the numbers: -3, -5, 6, 2, 2. Here's how you find the median:
−3,−5,6,2,2=−5,−3,2◯,2,6=median=2
That works well for odd length lists. However, what if you have an even length list?
Let's add 1 to the above list: 1, -3, -5, 6, 2, 2
1,−3,−5,6,2,2=−5,−3,1◯,2◯,2,6=median=21+2=1.5
Using the Median Calculator
To use the median calculator, enter your list of numbers in the box. The input is very forgiving – as long as you separate numbers with non-numbers it should work. Try commas, spaces, new lines, tabs, etc.
When you're done, hit the Compute Median button and we'll show you the median of the list. To check we understood your input, look at the list length in the Number of Values Input box.
Like this? Visit our other calculators and tools.
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PK started DQYDJ in 2009 to research and discuss finance and investing and help answer financial questions. He's expanded DQYDJ to build visualizations, calculators, and interactive tools.
PK lives in New Hampshire with his wife, kids, and dog.
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189763 | https://logicmadness.com/what-is-dc-gain/ | What is DC Gain?
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DC gain refers to the gain of a system when the input is a direct current (DC), meaning a constant signal with zero frequency. In simple terms, it’s how much the system amplifies a steady input over time. This concept is vital in electronics, where amplifiers boost signals, and in control theory, where it helps analyze how systems respond to constant inputs.
Table of Contents
Toggle
Why is it Called DC Gain?
Unexpected Detail: Infinite Gain in Ideal Systems
Detailed Exploration of DC Gain and Its Naming
Defining DC Gain in Electronics and Control Theory
Origins of the Term “DC Gain”
Significance in System Analysis
Comparison with AC Gain: A Detailed Table
Historical Context and Naming Convention
Practical Implications and Modern Usage
Unexpected Detail: Integration and Infinite Gain
Conclusion
Why is it Called DC Gain?
The term “DC gain” comes from “direct current,” which has a frequency of zero. In electronics, DC signals are constant, and the gain at this frequency is what we call DC gain. In control systems, it’s the steady-state output ratio for a step input, reflecting the system’s long-term behavior. Because it deals with the system’s response to these zero-frequency, constant signals.
Unexpected Detail: Infinite Gain in Ideal Systems
An interesting aspect is that ideal integrators have infinite DC gain, as their transfer function ( G(s) = \frac{1}{s} ) gives ( G(0) = \infty ). This doesn’t occur in physical systems due to practical limitations, but it highlights the theoretical underpinnings of DC gain in system design, as discussed on Electrical Engineering Stack Exchange.
Detailed Exploration of DC Gain and Its Naming
In the fields of electronics and control theory, the term “DC gain” is a fundamental concept that describes a system’s behavior under specific conditions. This section delves into its definition, the reasoning behind its name, calculation methods, significance, and comparisons with related concepts, providing a comprehensive understanding for both novices and experts.
Defining DC Gain in Electronics and Control Theory
Gain, in electronics, measures how much a two-port circuit, often an amplifier, increases the amplitude or power of a signal from input to output. It can be expressed as voltage gain, current gain, or power gain, often in decibels (dB) for convenience. DC gain specifically refers to this amplification when the input is a direct current (DC) signal, which is constant and has a frequency of zero.
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Field-programmable gate array
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In control theory, DC gain is defined as the ratio of the steady-state output to the steady-state input for a step input, essentially the gain at zero frequency. This is calculated using the transfer function ( G(s) ) evaluated at ( s = 0 ), provided the system is stable.
Origins of the Term “DC Gain”
The name “DC gain” derives from “direct current,” a term rooted in electrical engineering for signals that do not vary over time, corresponding to zero frequency. This naming convention likely emerged as engineers analyzed systems’ responses to constant inputs, particularly in amplifiers and control systems. For example, in audio power amplifiers, DC gain might be set to 1, while AC gain could be around 30, highlighting the distinction based on signal type.
Research suggests that the term became standardized as systems were modeled using frequency domain analysis, where DC corresponds to the lowest frequency point, zero Hz. This is evident in discussions on platforms like Electrical Engineering Stack Exchange, where users clarify that DC gain is the gain at zero frequency, contrasting with AC gain at higher frequencies.
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Significance in System Analysis
DC gain is crucial for understanding steady-state performance, especially in control systems. It affects how well a system tracks constant inputs, impacting stability and accuracy. For instance, in amplifier design, a high DC gain ensures low-level signals are amplified sufficiently, but excessive gain can introduce noise or instability, as noted in Fiveable.
In control theory, DC gain is linked to steady-state error coefficients, particularly for open-loop systems. This is important for applications like temperature control, where maintaining a constant output is critical. The concept also ties to frequency response, where DC gain is the magnitude at zero frequency, aiding in low-frequency signal analysis.
Comparison with AC Gain: A Detailed Table
To clarify the distinction, let’s compare DC gain and AC gain using a table:
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| Aspect | DC Gain | AC Gain |
---
| Definition | Gain at zero frequency, for DC signals | Gain at non-zero frequencies, for AC signals |
| Frequency Range | ( f = 0 ) Hz | ( f > 0 ) Hz |
| Calculation | ( G (0) ) for transfer function | Varies with frequency, often mid-band gain |
| Significance | Steady-state response, stability | Dynamic response, signal amplification at frequencies |
| Example Application | Audio amplifier DC offset, control systems | Radio frequency amplifiers, audio signal processing |
This comparison highlights that while DC gain is static, AC gain is dynamic, varying with frequency, which is crucial for applications like radio transmission where signals oscillate.
Historical Context and Naming Convention
The term “DC gain” likely solidified in the mid-20th century as frequency domain analysis became prevalent in engineering. Discussions on forums like Edaboard suggest it might have been a “sloppy expression” initially, but it stuck due to its clarity in referring to gain at zero frequency. The connection to DC (direct current) is intuitive, as it aligns with the physical meaning of constant signals in electrical systems.
Practical Implications and Modern Usage
Today, DC gain is analyzed using tools like MATLAB for control systems or Cadence for circuit design, where AC analysis can extrapolate to DC conditions. For instance, in operational amplifiers, DC gain is critical for ensuring the output doesn’t drift, affecting applications from audio to power systems. The concept remains relevant, with ongoing discussions on platforms like ResearchGate exploring its relation to system poles and transient response.
Unexpected Detail: Integration and Infinite Gain
An interesting aspect is that ideal integrators have infinite DC gain. This doesn’t occur in physical systems due to practical limitations, but it tells us about ideal behavior, highlighting the theoretical underpinnings of DC gain in system design, as discussed on Electrical Engineering Stack Exchange.
Conclusion
In conclusion, DC gain is called that because it measures the system’s gain at zero frequency, corresponding to direct current signals, reflecting steady-state behavior in both electronics and control theory. Its calculation is straightforward for stable systems, and it plays a vital role in design and analysis, contrasting with AC gain for dynamic responses. This understanding aids engineers in creating robust systems, ensuring performance under constant inputs.
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189764 | https://www.frontiersin.org/journals/medicine/articles/10.3389/fmed.2024.1328014/full | Your new experience awaits. Try the new design now and help us make it even better
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Volume 11 - 2024 |
This article is part of the Research TopicT Cells in Pregnancy Adverse OutcomesView all 4 articles
Effects of vaginal progesterone and placebo on preterm birth and antenatal outcomes in women with singleton pregnancies and short cervix on ultrasound: a meta-analysis
Limin Peng1,2Yan Gao1Chengkun Yuan1Hongying Kuang2,3
1Department of Obstetrics, The First Affiliated Hospital, Heilongjiang University of Chinese Medicine, Harbin, Heilongjiang, China
2Medical Department of First Clinical Medical College, Heilongjiang University of Chinese Medicine, Harbin, China
3Department of Gynecology, The First Affiliated Hospital, Heilongjiang University of Chinese Medicine, Harbin, China
Background: Vaginal progesterone in preterm birth and adverse outcomes caused by cervical insufficiency remains controversial. To address it, the effect of vaginal progesterone on preterm delivery and perinatal outcome of single pregnancy women with short cervix (less than 25 mm) was systematically evaluated by meta-analysis.
Methods: “Vaginal progesterone,” “placebo,” “ultrasound,” “cervix,” “singleton pregnancy,” “preterm birth,” and “antenatal outcomes” were entered to screen clinical studies PubMed, Embase, and the Chinese Biomedical Literature Database (CBM). The study population consisted of women with singleton pregnancies and a short cervix on ultrasound, and were assigned into the progesterone group (n = 1,368) and the placebo group (n = 1,373). Treatment began after the patient was diagnosed with short cervix until delivery. Neonatal survival rate, Neonatal Intensive Care Unit (NICU) admission rate, respiratory distress syndrome (RDS), intraventricular hemorrhage (IVH), neonatal mortality, and birth weight <1,500 g were analyzed.
Results: A total of 8 articles, totaling 2,741 study subjects, were enrolled. The progesterone group exhibited an obvious reduced rate of preterm birth at <34 weeks (OR = 0.67, 95% CI: 0.53∼0.84; Z = 3.53, P = 0.004), preterm birth at <32 weeks (OR = 0.46, 95% CI: 0.28∼0.77; Z = 2.99, P = 0.003), NICU admission rate (OR = 0.45, 95% CI: 0.30∼0.66; Z = 0.15, P < 0.0001), RDS rate (OR = 0.42, 95% CI: 0.28∼0.63; Z = 4.25, P < 0.0001), IVH incidence rate (OR = 0.40, 95% CI: 0.17∼0.95; Z = 2.08, P = 0.04), neonatal mortality (OR = 0.25, 95% CI: 0.13∼0.46; Z = 4.39, P < 0.0001), and proportion of neonates with birth weight < 1,500 g (OR = 0.45, 95% CI: 0.32∼0.64; Z = 4.50, P < 0.0001).
Conclusion: Vaginal progesterone lowered the incidences of preterm birth and adverse pregnancy outcomes in women with singleton pregnancies and a short cervix.
Background
Preterm birth is a significant pregnancy complication, and according to statistics from the World Health Organization (WHO), nearly 10 million preterm births occur each year, accounting for approximately 10% of all global newborns (1). Research indicates that factors such as male gender, advanced maternal age (35 years and older), ethnic origin, socioeconomic factors, and multiple pregnancies are associated with the occurrence of preterm birth (2). The maternal complications during pregnancy, such as intrauterine infection, bleeding, and hypertension, not only increase the risk of preterm birth (3), but it also presents issues like low birth weight, respiratory distress syndrome, neurological problems, and infection risk for the newborn (4). Cervical insufficiency is considered a major factor contributing to preterm birth (5). Under normal circumstances, the cervix remains closed to keep the fetus secure within the uterus. However, in the case of a short cervix, the cervix may fail to stay effectively closed, making the fetus vulnerable to external pressure and infection, potentially leading to preterm birth or miscarriage (6). A short cervix is also known as cervical incompetence and typically refers to a cervix length of less than 25 mm, although some researchers use 3.0 cm as the critical threshold for a short cervix (7, 8). Due to the cervix opening prematurely, pregnant women are more susceptible to intrauterine infections, which can result in preterm birth or endanger the life of the fetus (6). Some studies suggest that a short cervix may also lead to other antenatal complications, such as premature rupture of membranes, early membrane rupture, or malposition of the fetus and the placenta (9).
Progesterone primary pregnancy promoting hormone, is secreted by the corpus luteum in the ovaries and is crucial in maintaining the integrity of uterine lining during early pregnancy and throughout the course of pregnancy (10). Insufficient progesterone levels can lead to cervical insufficiency, early miscarriage, early pregnancy bleeding, and other pregnancy complications, and can also have an impact on the emotional wellbeing of the mother and the development of the fetus (11). In recent years, vaginal progesterone has been widely used as a therapeutic medication during pregnancy for the treatment or prevention of conditions related to cervical insufficiency and preterm birth. Several studies have indicated that vaginal progesterone can significantly reduce the risk of preterm birth caused by cervical insufficiency, especially in cases where pregnant women have a history of preterm birth (12, 13). However, there is still ongoing debate about the effectiveness of progesterone therapy in preventing and managing preterm birth. Some studies suggest that the effectiveness of progesterone therapy may be limited (14).
To gain a better comprehension of progesterone in prevention and management of preterm birth, more comprehensive research and comprehensive analysis are needed. Therefore, this work employed a meta-analysis approach to comprehensively investigate whether progesterone treatment had a remarkable impact on preventing preterm birth and antenatal outcomes. The aim was to evaluate the effectiveness of progesterone treatment in improving cervical function and lowering the possibility of preterm birth, providing additional evidence for medical decision-making, offering better guidance for the health and antenatal management of pregnant women, and further advancing research in this field.
Materials and methods
Research design
This work focused on women with singleton pregnancies who have been diagnosed with a short cervix through ultrasound examination. It employed a meta-analysis approach as the research strategy. Based on the differences in the measures taken for the included study subjects, the participants were assigned into a vaginal progesterone treatment group and a placebo treatment group. The work aimed to compare the differences in preterm birth rates at different gestational weeks and the related antenatal outcomes between these two groups.
Methods for literature screening
Before conducting a literature search, it is essential to clarify the research question and objectives. During the literature search process, keywords related to the research topic, such as “vaginal progesterone,” “placebo,” “ultrasound,” “cervix,” “singleton pregnancy,” “preterm birth,” and “antenatal outcomes,” should be used. These keywords were adopted to search in various medical literature resources and databases, encompassing PubMed, Embase, Cochrane Library, Springer, Science Direct, China National Knowledge Infrastructure (CNKI), Wanfang Data, VIP, and China Biomedical Literature Database (CBM). In each database, it would search for the following string: “Vaginal Progesterone” AND “Placebo” AND “Ultrasound” AND “Cervix” “Singleton Pregnancy” AND “Preterm Birth” AND “Antenatal Outcomes.” The search timeframe should cover from the inception of the database to July 2023. Once the search results were retrieved, the next steps included preliminary screening to exclude literature that did not meet the inclusion criteria. This preliminary screening helped narrow down the selection to potentially relevant studies. After this, a full-text review of the selected literature was conducted to judge whether they met the inclusion criteria. This step involved a more in-depth examination of each paper, including the sections “Materials and methods, Results, and Discussion.” Additionally, manual searches for the references and related literature of the selected papers should be performed to ensure that no key studies were missed. This comprehensive approach would help gather a robust set of relevant studies for your meta-analysis.
Methods for enrolling and excluding the articles
The studies enrolled had to satisfy all the following conditions. (1) Study type: clinical studies, such as randomized controlled trials (RCTs), cohort studies, and case-control studies, were included. (2) Participants must be women with a single pregnancy who had undergone ultrasound testing during pregnancy to determine cervical shortening, with a cervix smaller than 25 mm. (3) When the patient was diagnosed with short cervix, the gestational age should be less than 24 weeks, and at least 34 weeks at delivery. (4) Treatment comparison: enrolled studies must have compared the treatment effects of vaginal progesterone with a placebo. (5) Primary outcomes: the enrolled studies must have reported at least the preterm birth rate and key outcome indicators related to antenatal outcomes, such as neonatal survival rate, neonatal admission to the neonatal intensive care unit (NICU), respiratory distress syndrome (RDS), intraventricular hemorrhage (IVH), neonatal mortality, and birth weight <1,500 g. (6) Complete data: the articles must provide sufficient data. (7) Language of literature: the language of the research literature will be restricted to common languages like English and Chinese to ensure understanding and analysis of the studies.
Articles with any of following conditions had to be excluded. (1) Non-clinical studies, such as reviews, editorial opinion articles, case reports, and animal studies, were excluded. (2) Non-singleton pregnancies: participants with multiple pregnancies (e.g., twins and triplets) were excluded. (3) No cervical length measurement: studies that did not perform ultrasound examinations to determine cervical length in participants were excluded. (4) No control group: studies that did not compare the treatment with vaginal progesterone to a control group were excluded. (5) No primary outcome data: studies that did not provide data on preterm birth rates and antenatal outcomes were excluded. (6) Duplicate publications: those that had already been published in other journals as duplicate research were excluded. (7) Language of literature: literature in languages other than English or Chinese was excluded, unless suitable translations were available for use.
Extraction of required data
Following the screening of literature building upon the above criteria, the following data were collected and extracted:
(1) Record of each study’s title, author(s), publication year, and journal or source.
(2) Study design: specifying the study design type for each study, such as RCTs, cohort studies, or case-control studies.
(3) Sample size and characteristics: this included the number of participants in both the treatment and control groups, their ages, and other relevant characteristics.
(4) Cervical length measurement: details about the range and threshold values for cervical length measurement.
(5) Outcome measures: neonatal survival rate, NICU, RDS, IVH, neonatal mortality, and birth weight <1,500 g, among others.
Quality assessment of studies is a crucial step in meta-analysis, primarily evaluating aspects such as randomization, blinding, and sample size. Tools provided by the Cochrane Collaboration are commonly utilized for a quality assessment on included literature and potential biases, including selection bias, recall bias, and detection bias, among others. For studies that do not meet the criteria, sensitivity analyses can be conducted to examine their impact on the analysis results. For instance, a meta-analysis can be re-conducted after excluding lower-quality studies. Methods like funnel plots are used to check for the potential of publication bias.
Methods for statistical analysis
Data extracted are subjected to statistical analysis using Excel software to calculate the combined effect size (typically using the risk ratio) and its 95% confidence interval (CI). Heterogeneity analysis was employed to assess differences among various studies. An initial assessment for heterogeneity was conducted using the chi-squared test to determine the significance of heterogeneity (α = 0.05). P < 0.05 indicated a remarkable heterogeneity. The I2 statistic, available in Rev Man 5.3, was adopted for quantitative assessment of heterogeneity. I2 < 50% suggested low heterogeneity among studies, and a fixed-effects model (FEM) was typically used for the meta-analysis. I2 > 50% reflected substantial heterogeneity among studies, a random-effects model (REM) was more appropriate. To assess the potential for publication bias, a funnel plot can be created. The analysis results were often presented in the form of a forest plot, where the effect size and confidence intervals for each study are displayed. Z-values and P-values from the results are extracted to determine whether inter-group differences are statistically significant. Typically, P < 0.05 indicated statistically significant inter-group differences.
Results
Retrieved results
The keyword search in online databases yielded a total of 244 relevant articles. After the initial screening, 127 articles were obtained. Further refinement by reviewing article titles and abstracts led to the exclusion of 71 articles that did not align with the research topic, leaving 56 articles. A more in-depth review of the content resulted in the exclusion of 47 articles. After a thorough examination of the remaining 9 articles, one article without access to raw data was excluded. Ultimately, 8 articles were included for analysis. The literature search and selection process were illustrated in Figure 1.
FIGURE 1
Figure 1. The literature searches and selection process.
Brief introduction of enrolled literatures
A total of 8 articles (15–22) were enrolled, comprising a combined population of 2,741 women with ultrasound-diagnosed short cervix in singleton pregnancies. Among them, 1,368 participants were in the progesterone group, and 1,373 were in the placebo group. The basic information was summarized in Table 1.
TABLE 1
Table 1. Brief introduction of these studies.
Basic characteristics of involved patients
The statistical results for the age, cervical range, cervical length, and treatment methods of the women with ultrasound-diagnosed short cervix in singleton pregnancies from the included studies were presented in Table 2.
TABLE 2
Table 2. Basic characteristics of involved patients.
Quality of these studies
Using the Cochrane Reviewer’s Handbook, the quality of the 8 enrolled studies was assessed. A summary chart and bar chart for assessing the risk of bias in the literature were demonstrated in Figure 2. From the figure, it can be observed that all these literatures had “low risk” for random sequence generation (selection bias), allocation concealment (selection bias), blinding of outcome assessment (detection bias), and blinding of participants and personnel (performance bias). One article each had “unclear risk” for incomplete outcome data (attrition bias) and selective reporting (reporting bias), while one article had “high risk” for other bias. Therefore, all the included literature was considered of high quality.
FIGURE 2
Figure 2. Summary of biased risk assessment in included studies.
Impacts of vaginal progesterone on preterm birth of women with ultrasound short cervix and singleton pregnancy
Statistical analysis was performed on the preterm birth rates in ultrasound-diagnosed short cervix singleton pregnancy women under both <34 weeks and <32 weeks of gestational age in the progesterone group and the placebo group. No obvious heterogeneity was observed among these studies for both groups (I2 = 12%, 0%; P = 0.34, 0.52). Therefore, a FEM was adopted for the analysis. The results reflected that the preterm birth rate for ultrasound-diagnosed short cervix singleton pregnancy women under <34 weeks of gestational age in the progesterone group was sharply lower in contrast to that in the placebo group (OR = 0.67, 95% CI: 0.53∼0.84; Z = 3.53, P = 0.004). Similarly, the preterm birth rate for ultrasound-diagnosed short cervix singleton pregnancy women under <32 weeks of gestational age in the progesterone group was lower than that in the placebo group (OR = 0.46, 95% CI: 0.28∼0.77; Z = 2.99, P = 0.003) (Figure 3).
FIGURE 3
Figure 3. Forest plot of the impact of vaginal progesterone on preterm birth rates in ultrasound-confirmed short cervix singleton pregnancies at <34 weeks and <32 weeks.
The impact of vaginal progesterone on pregnancy outcomes in ultrasound-confirmed short cervix singleton pregnancies
Statistical analysis was conducted on the proportion of neonatal admissions to the NICU for ultrasound-diagnosed short cervix singleton pregnancy infants in the progesterone group and the placebo group, as demonstrated in Figure 4. No visible heterogeneity was presented (I2 = 41%; P = 0.15). The FEM analysis results implied that the proportion of neonates born to ultrasound-diagnosed short cervix singleton pregnancies and admitted to the NICU was lower in the progesterone group (OR = 0.45, 95% CI: 0.30∼0.66; Z = 0.15, P < 0.0001).
FIGURE 4
Figure 4. Forest plot of the impact of vaginal progesterone on NICU admission in ultrasound-confirmed short cervix singleton pregnancy after birth.
Figure 5 illustrated the incidence of RDS and IVH in neonates born to ultrasound-diagnosed short cervix singleton pregnancies in the progesterone group and the placebo group. Heterogeneity was not obvious for both RDS and IVH incidence (I2 = 0%, 0%; P = 0.84, 1.00). Consequently, a FEM was employed here, and its findings revealed that the incidence of RDS in neonates born to ultrasound-diagnosed short cervix singleton pregnancies in the progesterone group was lower (OR = 0.42, 95% CI: 0.28∼0.63; Z = 4.25, P < 0.0001). Similarly, the incidence of IVH in neonates was also lower in the progesterone group (OR = 0.40, 95% CI: 0.17∼0.95; Z = 2.08, P = 0.04).
FIGURE 5
Figure 5. Forest plot of the effect of vaginal progesterone on the incidence of RDS and IVH in ultrasound-confirmed short cervix singleton pregnancy.
Figure 6 compared the neonatal mortality rate and the proportion of neonates with a birth weight of <1,500 g among neonates born to ultrasound-diagnosed short cervix singleton pregnancies in the progesterone group and the placebo group. Heterogeneity was not remarkable in these two indicators (I2 = 13%, 0%; P = 0.33, 0.55). The FEM results indicated that the neonatal mortality rate in neonates born to ultrasound-diagnosed short cervix singleton pregnancies in the progesterone group was significantly lower (OR = 0.25, 95% CI: 0.13∼0.46; Z = 4.39, P < 0.0001). Similarly, the proportion of neonates with a birth weight of <1,500 g was also lower in the progesterone group (OR = 0.45, 95% CI: 0.32∼0.64; Z = 4.50, P < 0.0001).
FIGURE 6
Figure 6. Forest plot for effect of vaginal progesterone on neonatal mortality and the proportion of newborns with birth weight <1,500 g in ultrasound-confirmed short cervix singleton pregnancy.
Publication bias
A funnel plot was employed to analyze publication bias in these studies enrolled in this work. The results unveiled that the scatter distribution of the funnel plot for them was symmetrical, indicating no remarkable publication bias.
Discussion
Preterm birth can lead to a range of health issues, depending on the fetal maturity and the degree of prematurity. Newborns born between 28 and 32 weeks of gestation are at risk of RDS, IVH, and other complications (23, 24). These babies may require care in the NICU to monitor and address potential health problems. For infants born mildly premature between 32 and 34 weeks, their risk is relatively lower, and they typically can breathe independently, with a very high survival rate (25, 26). The results in this work demonstrated that in contrast to the placebo group, patients in the progesterone group experienced a reduced preterm birth rate at both <34 weeks of gestation (OR = 0.67, 95% CI: 0.53∼0.84; Z = 3.53, P = 0.004) and <32 weeks of gestation (OR = 0.46, 95% CI: 0.28∼0.77; Z = 2.99, P = 0.003), highlighting the significant reduction in the risk of preterm birth, particularly for births before 34 weeks. This is crucial for extending gestation and improving the maturity of the newborn. The significant reduction in preterm birth before 32 weeks of gestation underscores the effectiveness of progesterone. RDS and IVH are common and critical complications in preterm births. RDS is one of the most common respiratory problems in preterm infants and can impact the normal development of the lungs, especially the maturity of the alveoli and the respiratory tract (27). High IVH is associated with bleeding or hemorrhagic events within the ventricles of the brain, which can lead to damage to brain tissue and have adverse effects on infant neurological development, especially in sensitive brain regions. It may result in cognitive and motor function issues. The study results explicated that progesterone treatment reduces the preterm birth rate, the incidence of RDS and IVH, decreases neonatal mortality, and improves the birth weight of newborns. These are all highly beneficial outcomes for maternal and infant health. Conde-Agudelo et al. (28) evaluated the efficacy of vaginal progesterone in preventing premature delivery and adverse perinatal outcome of twin pregnancy. The results showed that vaginal progesterone could not prevent premature delivery or improve the perinatal outcome of unselected twin pregnancy, but it seemed to reduce the risk of premature delivery in early gestational age and the morbidity and mortality of twins with short cervical ultrasound examination. Romero et al. (29) explored whether vaginal progesterone can prevent premature delivery and improve the perinatal outcome of asymptomatic singleton pregnancy and short-term cervical ultrasound examination in the second trimester. The results showed that vaginal progesterone can reduce the risk of premature delivery and improve the pregnancy outcome of singleton with short-term cervical ultrasound examination in the second trimester, without any obvious harmful effects on children’s neurological development. The results of these meta-analyses are consistent with our research results.
Conclusion
The results herein revealed that prophylactic vaginal progesterone treatment can reduce preterm birth rates before 34 and 32 weeks of gestation, the incidence of RDS, the incidence of IVH, the proportion of newborns requiring admission to the NICU, neonatal mortality, and the proportion of newborns with a birth weight below 1,500 grams. These findings suggest that progesterone treatment had a greatly positive impact on preterm birth and antenatal outcomes in singleton pregnancies. However, this work was subjected to certain limitations, such as sample size restrictions, differences in study designs, or potential publication bias. Future work should aim to further validate the results of this study. In summary, the results offered valuable insights into intervention strategies for preterm birth and adverse pregnancy outcomes in ultrasound-assessed short cervix singleton pregnancies.
Author contributions
LP: Writing – original draft, Writing – review and editing. YG: Conceptualization, Software, Writing – original draft, Writing – review and editing. CY: Conceptualization, Investigation, Software, Writing – original draft, Writing – review and editing. HK: Writing – original draft, Writing – review and editing.
Funding
The author(s) declare that no financial support was received for the research, authorship, and/or publication of this article.
Acknowledgments
We express our sincere gratitude to Heilongjiang University of Chinese Medicine for the diligent guidance and valuable advice provided throughout the entire research process. Additionally, we would like to extend our thanks to Gynecology ward, who collaborated with us and offered crucial support in data analysis and manuscript writing. Finally, we acknowledge the PS and TV and GR of Frontiers in Medicine, whose suggestions and comments have significantly contributed to the continuous improvement of this research.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Keywords: vaginal progesterone, placebo, short cervix, singleton pregnancy, preterm birth, adverse pregnancy outcomes
Citation: Peng L, Gao Y, Yuan C and Kuang H (2024) Effects of vaginal progesterone and placebo on preterm birth and antenatal outcomes in women with singleton pregnancies and short cervix on ultrasound: a meta-analysis. Front. Med. 11:1328014. doi: 10.3389/fmed.2024.1328014
Received: 25 October 2023; Accepted: 28 February 2024;
Published: 05 April 2024.
Edited by:
Panicos Shangaris, King’s College London, United Kingdom
Reviewed by:
Tomislav Vladic, Independent Researcher, Stockholm, Sweden
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189765 | https://www.nm.org/conditions-and-care-areas/neurosciences/movement-disorders/wilson-disease/causes-and-diagnoses | Wilson Disease Causes and Diagnoses | Northwestern Medicine
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189766 | https://pnp.mathematik.uni-stuttgart.de/iaz/iaz1/Pressland/ana1a-win11files/wk8.pdf | Week 8: Cauchy’s Multiplication Theorem This week I’ll attempt to explain the proof of Cauchy’s Multiplication Theorem given in the lectures. The theorem states that if P∞ n=0 an and P∞ n=0 bn are abso-lutely convergent, then the series P∞ n=0 cn, where cn = Pn m=0 ambn−m, is absolutely convergent, and: ∞ X n=0 cn = ∞ X n=0 an ! ∞ X n=0 bn !
Before beginning the proof, it may be worth trying some examples of multiplying finite sums, and see that a similar result holds (multiplying a polynomial in x by a polynomial in y would be a good example). The main idea in the proof is to take a limit of this process.
So to begin the proof, we take N, K ∈N0, and choose K ≥2N, for reasons that will become apparent later. We want a neat way of writing some of the infinite series that will appear later on, so define: A = ∞ X n=0 an A = ∞ X n=0 |an| B = ∞ X n=0 bn B = ∞ X n=0 |bn| If we want to get the result in the theorem, we need to find that PK k=0 ck gets close to the product PN n=0 an PN m=0 bn when N and K are large. So we bound: K X k=0 ck − N X n=0 an ! N X m=0 bm ! ≤ X 0≤n≤K 0≤m≤K n>N or m>N |an||bm| This is perhaps not so easy to see, but if you stare at the definition of cn, you should be able to see that PK k=0 ck is going to be a sum of things that look like ambk−m for k < K, and see that these terms also occur in PN n=0 an PN m=0 bn when m < N and k −m < N. So the only terms left after cancelling look like aibj with either i > N or j > N, which (up to a change of notation) is what we have on the right-hand side, which now follows by applying the triangle inequality. This is also where we use K ≥2N, so we don’t end up subtracting any terms in the product PN n=0 an PN m=0 bn that don’t actually appear in PK k=0 ck. We then notice that: X 0≤n≤K 0≤m≤K n>N or m>N |an||bm| ≤ K X n=0 |an| ! K X m=N |bm| !
+ K X m=0 |bm| ! K X n=N |an| !
1 as all the terms are positive, and all the terms on the left-hand side certainly appear on the right-hand side. (I think this is actually an equality, but as we only need the inequality anyway I’m not going to worry about it too much).
Now we have some idea what’s going on with the partial sums, we can start taking limits. We’d like to take K and N to infinity, but we don’t really know how to do this for a number of reasons. Does the order matter? Does the requirement that K ≥2N affect how we can take the limit? These problems are somewhat compli-cated, so we’ll take a more delicate approach. Taking K →∞on the right-hand side only makes things bigger, so using our notation from above, we get the inequality: K X k=0 ck − N X n=0 an ! N X m=0 bn ! ≤A ∞ X m=N |bm| + B ∞ X n=N |an| =: SN Note that as tails of convergent series go to zero, the limit of the right-hand side as N →∞is zero. Note that we don’t take this limit on the left-hand side because we don’t quite know what to do with the K, which is bigger than 2N so has to go to infinity as well.
Now if we fix ε > 0, there is N1 ∈N such that SN < 1 2ε for all N ≥N1. By algebra of limits, we also know that PN n=0 an PN m=0 bm →AB as N →∞, so we can find N2 ∈N such that: N X n=0 an ! N X m=0 bm !
−AB < 1 2ε for all N ≥N2. Let N3 = max{N1, N2}. Then if K ≥2N3 (recall that we needed K ≥2N for all of the above calculations), we get from the triangle inequality that: K X k=0 ck −AB ≤ K X k=0 ck − N3 X n=0 an ! N3 X m=0 bm ! + N3 X n=0 an ! N3 X m=0 bm !
−AB ≤1 2ε + 1 2ε = ε and hence PK k=0 ck →AB as K →∞. (“Splitting the difference” like this is a standard trick in analysis, so you should remember it).
As mentioned in the notes, the proof of absolute convergence is very similar, and follows by defining ck = Pk n=0 |an||bk−n|, and noting that |ck| ≤ck for all k, so if we can show that P∞ k=0 ck converges, the comparison test will tell us that P∞ k=0 |ck| converges. This can be shown by repeating the above proof with ck, A and B re-placing ck, A and B — it is probably a good idea to actually do this, making sure you understand all the steps.
2 |
189767 | https://math.stackexchange.com/questions/2043690/is-12mod-13-the-same-as-1mod-13 | elementary number theory - Is 12(mod 13) the same as -1(mod 13)? - Mathematics Stack Exchange
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Is 12(mod 13) the same as -1(mod 13)?
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I'm in the process of studying for an exam and this just popped into my head. Sorry if it's a dumb question
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edited Dec 4, 2016 at 19:06
Martin Sleziak
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asked Dec 4, 2016 at 18:35
kaz2y5kaz2y5
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2 I am not sure you understand the "concept" of "a≡b(mod n)a≡b(mod n)" or what this means: "a mod n a mod n". To answer your question, yes because 12≡−1(mod 13).12≡−1(mod 13).CIJ –CIJ 2016-12-04 18:36:45 +00:00 Commented Dec 4, 2016 at 18:36
a≡a±n(mod n)a≡a±n(mod n)kingW3 –kingW3 2016-12-04 18:37:35 +00:00 Commented Dec 4, 2016 at 18:37
Yes, because -1 + 1 = 0 = 0 mod 13; 12 + 1 = 13 = 0 mod 13 Saketh Malyala –Saketh Malyala 2016-12-04 19:12:01 +00:00 Commented Dec 4, 2016 at 19:12
Boom... Here is an upvote too.Vidyanshu Mishra –Vidyanshu Mishra 2016-12-04 19:20:43 +00:00 Commented Dec 4, 2016 at 19:20
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Yes, they are the same. Note that (−1)+13=12(−1)+13=12, so their difference is exactly the thing that is ignored by mod 13 13!
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answered Dec 4, 2016 at 18:37
Noah SchweberNoah Schweber
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Thank you for your explanation. It seems so clear now, I'm not sure why that confused me before kaz2y5 –kaz2y5 2016-12-04 18:40:08 +00:00 Commented Dec 4, 2016 at 18:40
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Indeed they are the same, although I'd write them either without brackets or with brackets around the whole expression.
I really wanted to add that that using the (−1 mod 13)(−1 mod 13) representation can be very useful, especially when try to find the result of a multiplication. For a simple example, 12×7 mod 13 12×7 mod 13 is easily converted to 12×7≡−1×7≡−7≡4 mod 13 12×7≡−1×7≡−7≡4 mod 13
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answered Dec 4, 2016 at 20:38
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189768 | https://www.tiger-algebra.com/drill/m2-m-1=0/ | Copyright Ⓒ 2013-2025
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "m2" was replaced by "m^2".
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring m2-m-1
The first term is, m2 its coefficient is 1 .
The middle term is, -m its coefficient is -1 .
The last term, "the constant", is -1
Step-1 : Multiply the coefficient of the first term by the constant 1 • -1 = -1
Step-2 : Find two factors of -1 whose sum equals the coefficient of the middle term, which is -1 .
| | | | | | | |
--- --- ---
| | -1 | + | 1 | = | 0 | |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 1 :
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = m2-m-1
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Am2+Bm+C,the m -coordinate of the vertex is given by -B/(2A) . In our case the m coordinate is 0.5000
Plugging into the parabola formula 0.5000 for m we can calculate the y -coordinate :
y = 1.0 0.50 0.50 - 1.0 0.50 - 1.0
or y = -1.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = m2-m-1
Axis of Symmetry (dashed) {m}={ 0.50}
Vertex at {m,y} = { 0.50,-1.25}
m -Intercepts (Roots) :
Root 1 at {m,y} = {-0.62, 0.00}
Root 2 at {m,y} = { 1.62, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving m2-m-1 = 0 by Completing The Square .
Add 1 to both side of the equation :
m2-m = 1
Now the clever bit: Take the coefficient of m , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4
Add 1/4 to both sides of the equation :
On the right hand side we have :
1 + 1/4 or, (1/1)+(1/4)
The common denominator of the two fractions is 4 Adding (4/4)+(1/4) gives 5/4
So adding to both sides we finally get :
m2-m+(1/4) = 5/4
Adding 1/4 has completed the left hand side into a perfect square :
m2-m+(1/4) =
(m-(1/2)) • (m-(1/2)) =
(m-(1/2))2
Things which are equal to the same thing are also equal to one another. Since
m2-m+(1/4) = 5/4 and
m2-m+(1/4) = (m-(1/2))2
then, according to the law of transitivity,
(m-(1/2))2 = 5/4
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(m-(1/2))2 is
(m-(1/2))2/2 =
(m-(1/2))1 =
m-(1/2)
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
m-(1/2) = √ 5/4
Add 1/2 to both sides to obtain:
m = 1/2 + √ 5/4
Since a square root has two values, one positive and the other negative
m2 - m - 1 = 0
has two solutions:
m = 1/2 + √ 5/4
or
m = 1/2 - √ 5/4
Note that √ 5/4 can be written as
√ 5 / √ 4 which is √ 5 / 2
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving m2-m-1 = 0 by the Quadratic Formula .
According to the Quadratic Formula, m , the solution for Am2+Bm+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
m = ————————
2A
In our case, A = 1
B = -1
C = -1
Accordingly, B2 - 4AC =
1 - (-4) =
5
Applying the quadratic formula :
1 ± √ 5
m = ————
2
√ 5 , rounded to 4 decimal digits, is 2.2361
So now we are looking at:
m = ( 1 ± 2.236 ) / 2
Two real solutions:
m =(1+√5)/2= 1.618
or:
m =(1-√5)/2=-0.618
Two solutions were found :
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189769 | https://chemcollective.org/activities/autograded/127 | Designing a buffer solution with a specific pH
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What is the interplanar spacing between (200), (220), (111) planes in an FCC crystal of atomic radius 1.246 Å
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d(200) = 1.762 Å, d(220) = 1,24 Å, and d(111) = 2.034 Å
d(200) = 1.24 Å, d(220) =1.762 Å and d(111) = 2.034 Å
d(200) = 2.034 Å, d(220) =1.24 Å and d(111) = 1.762 Å
d(200) = 2.5 Å, d(220) = 4.2 Å and d(111) = 2.6 Å
Answer (Detailed Solution Below)
Option 1 : d(200) = 1.762 Å, d(220) = 1,24 Å, and d(111) = 2.034 Å
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Concept:
The magnitude of the distance between two adjacent and parallel planes of atoms i.e. interpalanar spacing dhkl is a function of Miller indices (h, k, and l) as well as the lattice parameters. For cubic crystal
({d_{\left( {h,\;k,\;l} \right)}} = \frac{a}{{\sqrt {{h^2} + {k^2} + {l^2}} }})
In which a is the lattice parameter (unit cell edge length).
For a FCC crystal the relationship between atomic radius r and lattice parameter a is given by,
(a = 2\sqrt 2 \;r)
Calculation:
Given for FCC crystal, r = 1.246 Å
(\therefore a = 2\sqrt 2 \;r = \;2\sqrt 2 \times 1.246) Å = 3.524 Å
({d_{\left( {2,\;0,\;0} \right)}} = \frac{{3.524\;{Å}}}{{\sqrt {{2^2} + {0^2} + {0^2}} }} = 1.762\;{Å})
({d_{\left( {2,\;2,\;0} \right)}} = \frac{{3.524\;{Å}}}{{\sqrt {{2^2} + {2^2} + {0^2}} }} = 1.245\;{Å})
({d_{\left( {1,\;1,\;1} \right)}} = \frac{{3.524\;{Å}}}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = 2.034\;{Å})
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189771 | https://www.reddit.com/r/math/comments/u0hat/continuity_vs_strong_continuityquotient_map/ | Continuity vs Strong continuity(quotient map) : r/math
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Continuity vs Strong continuity(quotient map)
Can anybody give me an example of a function that is continous, but not a quotient map? I can't tell from the definitions whether a quotient map is just a surjective continous map, but I can't construct any counterexamples to tell one way or the other.
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189772 | https://medium.com/@shihyu-wu/algorithms-studynote-2-divide-and-conquer-counting-inversions-9e0899bc9e26 | Sitemap
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Algorithms StudyNote — 2: Divide and Conquer — Counting Inversions
Shih-Yu (Fiona) Wu
4 min readJul 20, 2021
The Divide and Conquer
Divide into smaller subproblems
Conquer via recursive calls
Combine solutions of subproblems into one for the original problem
Take previous merging sort as an example, we split the list into two sub list recursively, sort each of them in turn, and interleave both results appropriately to obtain the sorted version of the given list (see the picture).
Counting Inversion
Here is an another problem using divide and conquer. First of all, inversions are pairs of numbers, in a disordered list, where the larger of the two is to the left of the smaller. What we have to do is to return the number of the inversions in the input list:
Input: array A containing the numbers 1,2,3,…,n in some arbitrary order
Output: number of inversions = number of pairs (i, j) of array indices with i A[j]
In the following list: [1, 3, 5, 2, 4, 6], there are 3 inversions: (3,2), (5,2), and (5,4). Since larger number come with smaller index, and smaller number come with bigger index. We can think of it this way:
Every intersection we created represents a pair of inversion.
High-Level Approach
Brute-forceusing two “for” loop, making it θ(n²) time complexity
Divide and Conquer
We can certainly do better than the brute-force approach by divide and conquer. Here, we divide the inversion (i, j) into 3 different groups:
Left inversion: if i, j<n/2
Right inversion: if i, j>n/2
Split inversion: if i ≤ n/2 <j
The first two can be compute recursively. However, the split inversion needs separate subroutine. After combine the number of inversions of the three groups, we can get the total number. The steps can be represented as follow:
Count(array A, length n) if n = 1, return 0 else X = Count(1st half of A, n/2) Y = Count(2nd half of A, n/2) Z = CountSplitInv(A,n) return X+Y+Z
Recall that when doing merge sort. First we recursively split the input list in half log₂(n) times until it hits the base case. Then we work our way back up and merge the lists into one. We would like to piggy back our inversion-counting operation on the merge sort so as to implement CountSplitInv() in linear O(n) time. And Count() will run in O(nlog(n)) time, just like merge sort.
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Consider merging a=[1, 3, 5] and b=[2, 4, 6]. Two pointers i, j are initialized pointing at the first element in the two arrays respectively.
i = 0j = 0a = [ 1, 3, 5 ]b = [ 2, 4, 6 ]1. a[i] is smaller than b[j] so a[i] is appended to ‘c’ and ‘i’ is incremented. c = [ 1 ]i = 1, j = 02. a[i] is greater than b[j] so b[j] is appended to ‘c’ and ‘j’ is incremented.c = [ 1, 2 ]i = 1, j = 13. a[i] is smaller than b[j] so a[i] is appended to ‘c’ and ‘i’ is incremented. c = [ 1, 2, 3]1 = 2, j = 1 and so on…
When 2 is copied to output, we discover the split inversions (3,2) and (5,2), and when 4 copied, discover (5,4). To count the number of inversions, all we have to do is subtract ‘i’ from the length of ‘a’ whenever the element in b is copied and sum them up.
Notice that this is essentially the exact same algorithm as earlier on. Merge Sort was already counting inversions for us, we just needed to track it. Merge Sort with inversion counting, just like regular Merge Sort, is O(n log(n)) time.
Application
Inversions show the degree to which our list is out of order. If we have a good way to count inversions then we can map values to numbers and make comparisons between lists of values.
For example, if I wanted to find who in a group I most share the same hobbies with I could have everyone rank hobbies, map them to values based on my hobbies to establish an order, and run them through an inversion counting algorithm to see who’s hobbies most match mine. We would probably have to do a little extra cleanup work to get matching sets of data but after that it could look like this:
I have 4 inversions with person A and 2 inversions with person B. So it looks like I should hangout with Person B.
Of course, such a recommendation engine will have a number of problems in the real world. However, it is an interesting use of what is essentially a side effect of a common sorting algorithm.
Reference
## Counting Inversions with Merge Sort
In my last blog post I ended with Merge Sort and briefly mentioned inversion counting and that it can be useful for a…
medium.com
## Divide-and-conquer algorithm - Wikipedia
In computer science, divide and conquer is an algorithm design paradigm. A divide-and-conquer algorithm recursively…
en.wikipedia.org
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Counting Inversions
## Written by Shih-Yu (Fiona) Wu
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Kris Stern
Jul 26, 2022
```
I have tried out the algo but it seems to give a count of inversions that is off.
```
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189773 | https://www.youtube.com/watch?v=KK2NxTiS9R0 | MCAT General Chemistry: Chapter 11 - Oxidation Reduction Reactions Problems
Professor Eman
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Hello Future Doctors! This video is part of a series for a course based on Kaplan MCAT resources. For each lecture video, you will be able to download the blank and completed notes.
To find the notes, click on the following link:
If you have any questions or have specific problems that you would like to see in a problem set video, feel free to leave a comment below or email me at:
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Good luck. You got this. Happy studying!
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Practice Problem 1 hello everyone my name is Iman welcome back to my YouTube channel today we're going to be tackling a practice problem set that relates to our lecture on Redux reactions let's go ahead and get started with problem number one problem number one says consider the following equation which species acts as an oxidizing agent now to answer this question let's review some terms first remember that oxidation is the process where an element loses electrons and thereby increases its oxidation state and reduction is the process where an element gains electrons and thereby decreases its oxidation state now an oxidizing agent is the species that accepts electrons and gets reduced and a reducing agent is the species that donates electrons and gets oxidized to determine the oxidizing agent the first thing that we're going to do is assign oxidation states to each element in the reactants and in the products so some rules to keep in mind when we do that are the following rule number one the oxidation state of an element in its Elemental form is equal to zero rule number two for monatomic ions the oxidation state is equal to the charge of the ion rule number three hydrogen generally has an oxidation state of + one except when it's bonded to Metals in binary compounds there the oxidation state for that hydrogen is minus1 rule number four is that oxygen generally has an oxidation state of min-2 except in peroxides where it is minus1 or in compounds with Florine rule number five is that the sum of oxidation States in a neutral compound must be zero while in a polyatomic ion it has to equal the ion's charge now let's go ahead and look at our equation and start assigning oxidation States let's start off with sodium in its Elemental form sodium is going to have an oxidation state of zero next let's look at ammonia hydrogen has an oxidation state of + one now to balance the three hydrogens which gives us a total of + three nitrogen must have an oxidation state of Min -3 then we can look at sodium nitride sodium as an ion has an oxidation state of + one and there are three sodium atoms now to balance this plus three charge nitrogen must have an oxidation state of Min -3 then last but not least we can look at hydrogen gas in in its Elemental form this is going to have an oxidation state of zero okay now let's look at the changes that are occurring let's look at sodium sodium goes from an oxidation state of zero to + one so that means that sodium is oxidized if it is oxidized from 0 to plus one this makes it the reducing agent next let's look at nitrogen nitrogen goes from an oxidation state of -3 to Min -3 there is no change in oxidation state for nitrogen what about for hydrogen well here hydrogen has an oxidation state of + one and then it changes to zero because hydrogen here as hydrogen gas has an oxidation state of zero so we're changing from + one to zero that means that the hydrogen is reduced from + one to 0er making it the oxidizing agent so the oxidizing agent here is the hydrogen in ammonia and that makes the correct answer for problem number one answer Choice C beautiful let's move on to problem Practice Problem 2 two problem two asks how many electrons are in involved in the following half reaction after it is balanced we're looking at the reaction between di chromate and chromium 2 ion now to start this off we're not going to concern ourselves with the electrons we're going to leave that at the end when our final step is to balance the charge so we're going to look at this reaction without the electrons and we're going to start by balancing atoms starting off with chromium now on the react side we have our di chromate here where we have two chromium atoms to balance that on the produ side we're going to add a coefficient of two in front of the chromium 2 ion now our chromium atom is balanced the next atom we want to balance is our oxygen atoms there are seven oxygen atoms in the DI chromate ion so we're going to put a coefficient of seven in front of water so that we have seven oxygen atoms on the product side as well now the number of oxygen atoms are equal to each other on both the reactant and product side now our last and final step is to balance the hydrogen atoms on the product side we have 14 hydrogens so to balance that we're going to add a coefficient of 14 in front of our hydrogen ions now all of the atoms are balanced let's go ahead and calculate the charge on the reactant side and the product side so that we can determine how many electrons we need and on which side to balance the charge now on the left side on the reactant side we have a minus 2 charge from our di chromate ion and we have a 14 plus charge from our hydrogen ions 14 plus 2 gives us a total of +2 as our charge for the reactant side what about our product side in our product side we only have a charge of four plus plus two from our two chromium 2 ions gives us a total of plus4 charge now the charges are not balanced to ensure that the charge is balanced we're going to have to add eight electrons to the reactant side so that both sides have a charge that is equal to plus4 so that means we need eight electrons on the reactant side which makes the correct answer B for problem two awesome let's tackle problem three problem three says lithium Practice Problem 3 aluminum hydde is often used in laborator because of its tendency to donate a hydride ion which of the following roles would lithium aluminum hydde likely play in a reaction a says strong reducing agent only B says strong oxidizing agent only C says both a strong reducing agent and a strong oxidizing agent and D says neither now to determine the role that lithium aluminum hydde likely plays in a reaction we need to understand the chemical properties and the common uses of lithium aluminum hydde in a laboratory setting this is a powerful reducing agent that's commonly used in organic chemistry remember that a reducing agent is a substance that donates electrons to another substance in a reduxx reaction thereby reducing the oxidation state of that substance and the reducing power of lithium aluminum hydride it stems from its ability to donate a hydride ion which is a hydrogen atom with an extra electron now as a general hint strong reducing agents they tend to have metals or hydrides and strong oxidizing agents they tend to have oxygen or similarly electronegative elements so in short lithium aluminum hydde acts as a strong reducing agent because it donates electrons in the form of hydride ions to other substances facilitating their reduction it is not known to be a oxidizing agent or to act as an oxidizing agent as it does not accept electrons therefore the correct answer is a it is a strong reducing agent only let's move into problem Practice Problem 4 four problem 4 asks what is the oxidation number of chlorine in sodium hypochloride now to start this let's once again review our rules for assigning oxidation numbers rule number one is that the oxidation state of an element in its Elemental form is zero rule number two for monatomic ions the oxidation state is equal to the charge of the ion rule number three hydrogen generally has an oxidation state of + one except when it is bonded to Metals in binary compounds there it will have an oxidation state of minus one rule number four is that oxygen generally has an oxidation state of min-2 except in peroxides where it is minus1 or in compounds with Florine and then the last rule is that the sum of oxidation States in a neutal compound must be zero while in a polyatomic ion it must equal the ion's charge so now let's take a look at our compound here sodium is an alal metal it's going to have an oxidation number of + one and oxygen typically has an oxidation number of minus 2 now this is a neutral compound which means that the chlorine atom must carry a + one oxidation state in order to balance the overall charge of 0 because + 1 + 1 - 2 gives us a sum of 0er which is what we're looking for since sodium hypochlorite is a neutral compound the correct answer for problem four is C let's move on to problem five problem Practice Problem 5 five says the following electronic configurations represent elements in their neutral form which element is the strongest oxidizing agent to answer this question we should keep in mind that an oxidizing agent is a substance that gains electrons in a chemical reaction and gets reduced strong oxidizing agents they typically have high electro negativity and a strong tendency to accept electrons now considering we're given the electronic configuration for elements in their neutral form what we can do is identify the atoms and then discuss them one by one so for answer Choice a this is the electronic configuration for calcium calcium is an Alkali Earth metal it tends to lose electrons rather than gain them making it a poor oxidizing agent also to keep in mind it has a full 4S orbital meaning that it can only gain an electron if it gains an entire cell subshell so answer Choice a is not what we're looking for answer Choice B gives us the electronic configuration for manganese now manganese can exhibit various oxidation States but in its neutral state it is not a strong oxidizing agent as the other options also it is stable with a half full 3d orbital so it's unlikely to pick up electrons unless it can gain a about five more so this is not the answer we're looking for answer Choice C gives us the electronic configuration for gallium this is a metal and it tends to lose electrons to form cations I making it a poor oxidizing agent also it has a single electron in the outer shell which is more likely lost upon ionization so again this is not what we're looking for answer Choice D gives us the electronic configuration for Bromine bromine as a hallogen has a high electro negativity and a strong tendency to gain electrons halogens are actually well known for being strong oxidizing agents in other words it would fill up its 4p orbital by gaining one electron so it is easily reduced and it is a strong oxidizing agent this is what we're looking for and so the correct answer for problem five is answer Choice D let's move on to problem Practice Problem 6 six problem six asks which of the following is the correct net ionic reaction for the reaction of copper with silver one nitrate to determine the correct net ionic reaction we need to follow our three main steps first we write the molecular equation with all the reactants and all of the products in their undissociated form next we can write the total ionic equation noting that all of the soluble compounds should be written in their dissociated form then last by removing The Spectator ions we get the net ionic equation so let's do that for this reaction first we're going to start with the balanced molecular equation for the reaction we have copper reacts with silver nitrate to form copper 2 nitrate and silver so here we have the balanced molecular equation next we want to write the total ionic equation by dissociating all of the strong electrolytes all of the soluble salts into their constituent ions this gives us the following equation in this total ionic equation what we notice is that the nitrate ions appear on both sides of the equation they do not participate in the actual chemical reaction these are known as spectator ions we can remove these spectator ions to simplify the equation really just focusing on the species that undergo the chemical change and after removing The Spectator ions we are left with the net ionic equation this is our net ionic equation and it matches answer Choice D so the correct answer for problem six is D let's go ahead and move Practice Problem 7 into problem seven Now problem s says one way to test for the presence of iron in solution is by adding potassium thiocyanate to the solution the product when this reagent reacts with iron is iron thiocyanate which creates a dark red color in solution via the following net ionic equation how many grams of iron sulfate would be needed to produce 2 moles of iron thiocyanate now what we're shown is a net ionic equation if 2 moles of iron thoy are created then we're going to need two moles of iron because there is a 1:1 mole ratio between these two species now iron sulfate has the formula fe2 so4 three because the sulfate has a charge of min-2 and iron has a charge of plus three we get this based off of the net ionic equation we're given therefore one mole of iron sulfate is needed to make two moles of iron for the reaction and if we have 2 moles of iron then we can make 2 moles of iron thiocyanate so what we need to figure out is the molar mass of iron sulfate that is the GRS of iron sulfate that we're going to need to prod produce 2 moles of iron thiocyanate to figure out the molar mass of iron sulfate we just need to figure out how many of each atoms we have and what the molar mass of each atom is so we have two iron atoms iron has a molar mass of 55 G per mole we have three sulfate atoms sulfate has a molar mass of 32 G per mole and we have 12 oxygen atoms oxygen has a molar mass of 16 G per mole 2 55 is 110 3 32 is 96 and 12 16 is 192 if we go ahead and add this up we should get approximately 400 G per mole now we know that we just need one mole of iron sulfate to get two moles of iron with 2 moles of iron we can make 2 moles of iron bio cyanate so the answer here is going to be answer Choice c 7 is C let's move on to problem 8 problem 8 says during the assigning of Practice Problem 8 oxidation numbers which of the following elements would most likely be determined last so when we're assigning oxidation numbers we usually start with elements of known oxidation state first and then we determine the oxidation state of other elements by deduction let's work through each of these answer choices a says argon argon is a noble gas it typically does not form compounds so when it's present in a reaction it's usually going to have an oxidation state of zero because it's going to exist in its Elemental Form B says Florine Florine can have an oxidation state of zero when it's on its own or negative one in a compound this can be usually determined first so so it's not what we're looking for C says strontium strontium is an alkaline earth metal with a fixed oxidation state of plus two in compounds this makes it straightforward to assign its oxidation number early in the process and again this is not what we're looking for D says aridium this can exhibit multiple oxidation States from -3 to + 8 and so due to its variable oxidation States the assignment of it oxidation number often depends on the oxidation states of the other elements in the compound therefore it's usually determined last and that makes the correct answer for problem 8 answer Choice D let's move on to problem N9 problem 9 Practice Problem 9 says as methanol is converted to methanol and then methanoic acid the oxidation number of the carbon blank does it increase does it decrease does it increase then decrease or does it decrease then increase in order to answer this problem we need to figure out the oxidation state of carbon in each of these compounds let's go ahead and start with methanol now methanol is a neutral compound so the sum of the oxidation states must add up to zero keeping that in mind methanol has a molecular formula of ch3o hydrogen here has an oxidation state of + one and there's a total of four hydrogen atoms so plus four oxygen here has an oxidation state of minus 2 there's only one oxygen atom in order for the oxidation states to sum up to zero then carbon must have an oxidation state of min-2 so in methanol carbon has an oxidation state of min-2 what about methanol this is going to have a molecular formula of hch again here hydrogen has an oxidation state of + one and there's a total of two hydrogen atoms so plus two oxygen has an oxidation state of min-2 and considering that methanol is a neutral compound that means that the sum of the oxidation states must add up to zero we have plus two from the hydrogen atoms minus 2 from the oxygen atom this already sums up to zero to keep it that way carbon must have an oxidation state of zero then the last compound we're going to look at is methanolic acid again this is a neutral compound so the sum of the oxidation states must add up to zero the molecular formula here is H Co the hydrogen has a oxidation state of + one and there's two hydrogen atoms here so Plus + 2 total oxygen has an oxidation state of min-2 and there are two oxygen atoms so we have a total of -4 + 2 - 4 so far gives us -2 in order to ensure that the sum of the oxidation numbers in methanolic acid is zero carbon must have an oxidation number of + two that way we have + 2 + 2 that's a total of 4 minus 4 that gives us a total of zero so then as we move from methanol to methanol to methanolic acid we are going from an oxidation number of min-2 to 0 to + 2 and so the oxidation number of carbon is increasing from min-2 to 0 and increasing again from 0 to + 2 and that makes the correct answer answer for problem 9 a the oxidation number of the carbon increases going from methanol to methanol to methanolic acid let's move on to problem 10 problem 10 asks in the compound Practice Problem 10 pottassium dihydrogen phosphate which element has the highest oxidation number now something that's important to note is that this is a neutral compound which means that the sum of all the the oxidation numbers must sum up to zero keeping that in mind let's get started pottassium here is an Alkali metal and in compounds it always has an oxidation number of + one hydrogen also typically has an oxidation number of + one when it's bonded to nonmetals and there are two hydrogen atoms total oxygen has an oxidation number of -2 and we have a total of four oxygen atoms so this gives us -8 now the oxidation number of phosphorus will be determined by ensuring that the sum of the oxidation numbers in the compound equals the overall charge of the compound which is neutral that means zero right now we have 3 - 8 that gives us a sum of minus5 to ensure that we get a sum of zero the o oxidation number of phosphorus must be + 5 this makes the correct answer for problem 10 answer Choice C phosphorus has the highest oxidation number now moving on to problem Practice Problem 11 11 problem 11 says if a certain metal has multiple oxidation states its acidity as an oxide generally increases as the oxidation State increases therefore which of the following tungsten compounds is likely to be the strongest acid we're looking at which tungsten compound has a Tungsten atom with the highest oxidation number so let's look at answer Choice a here we have tungsten 4 oxide we have an oxygen atom which has an oxidation state of minus 2 and we have a total of two oxygen atoms so this gives us minus 4 since this is a neutral compound to ensure that the sum of the oxidation States equal zero tungsten must have an oxidation state of +4 then for answer Choice B we have tungsten 6 oxide here we have oxygen again with a min-2 oxidation number we have three total oxygen atoms this gives us minus 6 again this is a neutral compound to ensure that the sum of the oxidation numbers equals zero tungsten must have an oxidation number of Plus plus 6 here for answer Choice C again oxygen with a min-2 oxidation number and there is a total of three oxygen atoms we get min -6 what is the oxidation number of tungsten to ensure that the sum equals zero considering we have two tungsten atoms in this case it will have A+ three oxidation state then for our last and final answer Choice again oxygen minus 2 oxidation number and then we have five oxygen atoms this gives us minus1 this is a neutral compound to ensure that the sum of the oxidation numbers equal zero tungsten has to have an oxidation number of + 5 because we have two tungsten atoms this gives us - 10 + 10 that equals 0er so now looking at these answer choices the compound that has a Tungsten atom with the highest oxidation number is going to be answer Choice B this is tungsten 6 oxide so 11 is B beautiful let's move on to problem 12 problem 12 says consider the Practice Problem 12 following steps in the reaction between oxyc acid and chlorine which of these steps occurring in aquous solution is an example of a disproportionation reaction now a disproportionation reaction reaction is a type of Redux reaction in which a single substance is both oxidized and reduced forming at least two different products let's go ahead and look at step one first here we have chlorine and chlorine is converted to both hypochlorous acid and chloride ions chlorine here has an oxidation state of zero in hypochlorous acid it has an oxidation state of + one and in the chloride ion it has an oxidation state of minus1 so here chlorine is both oxidized going from 0 to + one and reduce going from 0 to minus1 and this makes this reaction a disproportionation reaction what about step two here we see an acid dissociation reaction where oxyc acid dissociates into hydrogen ion and hydrogen oxalate ions now here there is no change in oxidation state so it is not even a Redux reaction let let alone a disproportionation reaction so step two is not what we're looking for what about step three in this reaction we have hypochlorous acid and hydrogen oxalate ions reacting together they give us water chloride ions and carbon dioxide now this reaction is going to involve the reduction of hyp chlorus acid to Chloride because here we're going from + one to minus1 so it is reduced and then we see the oxidation of oxalate ions to carbon dioxide so since the same species is not both oxidized and reduced this is not a disproportionation reaction that means that the only step that represents a disproportionation reaction is step one because chlorine under goes both oxidation and reduction therefore the correct answer to problem 12 is a let's move on to problem Practice Problem 13 13 problem 13 says potentiometry in an oxidation reduction titration is analogous to performing an acid based titration with a blank acidic indicator basic indicator pH meter or an oxidizing agent now in an acid based titration the pH of the solution is monitored to determine the end point of the titration this is typically done using a pH meter or an indicator that changes color at a specific pH value similarly in an oxidation reduction titration the potential or the voltage of the solution is monitored to determine the end point this is where potentiometry comes into play here this involves measuring the electrical potential or voltage of a solution to monitor the progress of a redo reaction it is used in redo titrations to determine the equivalence point which is analogous to how a pH meter is used in acid based titrations so a pH meter all right is used to detect the end point and the equivalence point of a titration by indicating when the pH changes significantly so in the context of this question potentiometry in a Redux titration serves a similar role to a pH meter in an acid base titration both methods involve monitoring a specific property to determine when the reaction has reached its equivalence point and its end point so the correct answer here is going to be C let's move on to problem Practice Problem 14 14 problem 14 says after balancing the following oxidation reduction reaction what is the sum of the stochiometric coeffic eents of all of the reactants and products in order to get the sum of these stochiometric coefficients of all the reactants and products we need to balance this Redux reaction and to do that we're going to use our half reaction method to be able to get our oxidation half reaction and our reduction half reaction we need to look at this reaction and decide what is being oxidized and what is being reduced so our first step here is going to be to assign oxidation numbers to all of our atoms starting off here each sulfur atom in Elemental sulfur has an oxidation number of zero then we can look at our nitrate ion it has a charge of minus1 so that means the sum of the oxidation numbers needs to be equal to minus1 here we have oxygen oxygen has an oxidation number of minus1 and there are a total of three oxygen atoms this gives us min-6 in in order for our sum to be equal to minus1 the oxidation state of nitrogen needs to be + 5 - 6 +5 gives us minus one that is exactly matching our charge for this polyatomic ion next we're going to look at our sulfite ion here this has a charge of 2 minus so the sum of the oxidation States need to be equal to min-2 again here we have oxygen with a minus 2 oxidation number and there are three oxygen atoms so we have -6 in order for the sum to be equal to -2 sulfur needs to have an oxidation state of plus4 then last but not least we can look at nitrogen monoxide oxygen here is going to have a oxidation state of minus 2 so nitrogen will have an oxidation number of plus two now we can look a little bit more in depth and decide what is being oxidized and what is being reduced sulfur here goes from zero to + 4 sulfur is oxidized then we can look at nitrogen nitrogen is going from + 5 to + 2 nitrogen is being reduced so with that information we can go ahead and write our oxidation half reaction and our reduction half reaction then the goal is to balance each half reaction individually and then combine them together to get the overall reaction let's start off with balancing our oxidation half reaction here on the reactant side we have eight sulfur atoms to balance this with the product side we're going to add a coefficient of eight right here now on the product side we have 24 oxygen atoms in order to balance that with the reactant side we're going to go ahead and add 24 water molecules now the sulfur atoms are balanced and the oxygen atoms are balanced but the hydrogen atoms are not on the reactant side we have a total of 48 hydrogen atoms so to balance that on the product side we're going to add 48 hydrogen ions now all the atoms are balanced but our final step is to go ahead and balance the charge on the reactant side we have a charge of zero and on the product side we have a charge of -6 + 48 this gives us a charge of POS 32 to get this positive 32 charge to be equal to Z like it is on the reactant side we're going to add 32 electrons so now we have a balanced oxidation half reaction our next step is to do the same for our reduction half reaction here on both the reactant and the product side we have one nitrogen atom now let's do the oxygen on the reactant side we have three oxygen atoms on the product side we only have one oxygen atom to balance it we're going to add two water molecules now there's three oxygen atoms on both sides the next thing to balance is the hydrogen atoms we have four on the product side so we're going to go ahead and add four hydrogen ions to the reactant side to balance that now all of the atoms are balanced we need to make sure that the charge is balanced on the reactant side we have minus1 + 4 we have a total charge of plus three on the reactant side and zero on the product side to get the reactant side to be equal to the product side we just need to add three electrons to the reactant side now we have a balanced reduction half reaction both of our half reactions are now balanced the next thing we want to do is add them together but in order for us to add them together we need to have an equal number of electrons in both of these reactions now here on the oxidation half reaction we have 32 electrons and here in our reduction half reaction we have three the common number here is going to be 96 which means that we need a multiply our oxidation half reaction by three and we need to multiply our reduction half reaction by 32 so when we do this we can go ahead and begin to modify our value so I'm going to erase all of our coefficients and modify them appropriately now that we're multiplying by a different coefficient so let's go ahead and do that starting off with our oxidation half reaction now we have 72 water molecules we have three of these S8 molecules we have 24 sulfite molecules 144 hydrogen ions and 96 electrons that is our oxidation half reaction what about our reduction half reaction now we have 32 nitrate ions 128 hydrogen ions 96 electrons 32 of these nitrogen monoxide uh molecules and we have 64 water molecules now we can go ahead and add together these two reactions noting that the electrons are going to cancel out what we're going to get now is the following equation we now have 3 S8 + 32 nitrate ions plus eight water all right and that gives us in our product side 24 sulfite all right sulfite ion ions 2 minus plus 32 nitrogen monoxides plus 16 hydrogen ions all right and note that our number of hydrogen ions and for water they have been modified because we made sure to go ahead and cancel out similar numbers of water molecules from both sides so this gets us the following overall reaction now all we have to do is sum up the coefficients so here we have 3 + 32 + 8 + 24 + 32 + 16 this is going to give us a total of 115 which makes the correct answer for problem 14 answer Choice D beautiful now we can move into our last and final problem for this video Problem 15 says an essay is Practice Problem 15 performed to determine the gold content in a supply of crushed ore one method for pulling gold out of ore is to react it in a concentrated cyanide solution the equation is provided below now an indicator is used during this reaction and approximately 100 ml of a 2 molar sodium cyanide solution is used to reach the end point how many moles of gold are present in the crushed ore the first step that we want to do is go ahead and balance the chemical equation I went ahead and did this for us since we've had plenty of practice balancing Chemical Equations but go ahead and check for yourself now that we have the balanced chemical equation the second thing we want to do is determine the number of moles of sodium cyanide that's used in this reaction we can determine this because we know the volume and the marity so we can easily solve for the moles what we're going to have to do is multiply the volume which we're going to convert to liters 0.1 L and we're going to multiply this by the marity which is 2 moles per liter notice that the units of liter cancel out and what we're going to be left with in terms of units is mole 0.1 2 is equal to 0.2 moles of sodium cyanide now we're going to use this amount to do a conversion to number of moles of gold and the way we're going to do that is by taking advantage of our balanced chemical equation and by using mol to Mo ratios so if we have 0.2 moles of sodium cyanide we can use a mul tole ratio between sodium cyanide and gold for 8 moles of sodium cyanide we have 4 moles of gold and what we're going to get here is 0.8 / 8 and this is going to be equal to 0.1 moles of gold that is in our crushed or and that makes the correct answer answer Choice C and that's how we solve problem 15 and with that we've completed our practice problem set for our redo lecture please let me know if you have any questions comments concerns down below I really hope that this was helpful other than that good luck happy studying and have a beautiful beautiful day future doctors |
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18.218 | Spring 2021 | Graduate
Topics in Combinatorics: Analysis of Boolean Functions
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Lecture 1: Course Overview and the Basic Set-up
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Lecture 2: The BLR Lineairty Test and Random Restrictions
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Lecture 3: PAC Learning and Learning Sparse Functions
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Lecture 4: Influences of Boolean Functions
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Lecture 5: Hypercontractive Inequality
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Lecture 6: The FKN Theorem and the KKL Theorem
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Lecture 7: Talagrand’s Version of the KKL Theorem, the Friedgut Junta Theorem, and Isoperimetric Inequalities
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Lectures 8–10: Noise Stability and Arrow’s Impossibility Theorem
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Lectures 11–12: Erdos-Ko-Rado Type Theorem, the p-biased Cube, and Properties of Quasi-random Families
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Lectures 13–14: Invariance Principle, the Berry-Essen Theorem, and Hypercontractivity in Gaussian Space
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Lecture 15: Introduction to Complexity Theory, Approximation Problems, and the PCP Theorem
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Lecture 16: The Goemans-Williamson Algorithm and A Hardness Result for Max-Cut
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Lecture 17: Preliminaries, the Reduction, and Analysis of the Reduction
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189775 | https://www.prepscholar.com/solutions/probability-statistics-for-engineering-science-devore-9th-edition-chapter-8-problem-48E/ | Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 48E from Chapter 8 from Devore's Probability and Statistics for Engineering and Science.
Problem 48E
Step-by-Step Solution
Step 1 of 2
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Step 2 of 2
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Chapter 8, Problem 48E is solved.
Sample size of wine consumers, n = 51.
Number of homes had problems, x = 41.
a)
Significance level, $\alpha = 0.01$.
Parameter of interest:
Consider, p: Population proportion of homes with Chinese drywall have electrical and environmental problems.
Null hypothesis:
Alternative hypothesis:
Test statistic:
Formula to find Z test statistic is given below.
Use below formula to find sample proportion.
Put values in the formula of Z test statistic.
P value:
For left tailed test, the formula of P value is given below.
P value = P(Z > Z test statistic)
The value 4.34 is beyond the range of Z table. The value of probability corresponding to 4.34 will approach to 1.
Decision rule:
If P value $ \le \alpha $ , then reject Ho.
If P value $ > \alpha $ , then do not reject Ho.
Here, 0 < 0.01, hence reject the null hypothesis.
Interpretation:
There is sufficient evidence to support that the true proportion of households with Chinese drywall that have electrical/environmental problem is more than 0.5.
b)
Confidence level, C = 0.99.
The one tailed critical Z value corresponding to left area 0.99 is 2.33.
Formula to find lower limit of confidence interval for population proportion is given below.
Solve it.
c)
It is asked to find type II error.
Left area is found below.
In Z table the values corresponding to area 0.995 is 2.58.
Formula of Z score is given below.
Solve further.
So, find power using below equation.
Power = P( Reject Ho | Ho is false)
In Z table, the value corresponding to row headed -2.1 and column headed 0.04. The value is 0.0162.
So, $P\left( {Z < - 2.14} \right) = 0.0162$.
Formula of type II error is given below.
Type II error = 1 – Power |
189776 | https://www.compadre.org/physlets/mechanics/intro2.cfm | Physlet Physics: Chapter 2: One-Dimensional Kinematics
I.Mechanics
1: Introduction to Physlets
2: One-Dimensional Kinematics
3: Two-Dimensional Kinematics
4: Newton's Laws
5: Newton's Laws 2
6: Work
7: Energy
8: Momentum
9: Reference Frames
10: Rotations about a Fixed Axis
11: General Rotations
12: Gravitation
13: Statics
II.Fluids
14: Static Fluids
15: Fluids in Motion
III.Waves
16: Periodic Motion
17: Waves and Oscillations
18: Sound
IV.Thermodynamics
19: Heat and Temperature
20: Kinetic Theory and Ideal Gas Law
21: Engines and Entropy
V.Electromagnetism
22: Electrostatics
23: Electric Fields
24: Gauss's Law
25: Electric Potential
26: Capacitance and Dielectrics
27: Magnetic Fields and Forces
28: Ampere's Law
29: Faraday's Law
VI.Circuits
30: DC Circuits
31: AC Circuits
VII.Optics
32: EM Waves
33: Mirrors
34: Refraction
35: Lenses
36: Optical Applications
37: Interference
38: Diffraction
39: Polarization
Appx: Optics Appendix
Home » Mechanics » One-Dimensional Kinematics
Chapter 2: One-dimensional Kinematics
;)Illustrations;)
2.1:Position and Displacement
2.2:Average Velocity
2.3:Average and Instantaneous Velocity
2.4:Acceleration and Measurement
2.5:Motion on a Hill or Ramp
2.6:Free Fall
;)Explorations;)
2.1:Compare Position vs. Time and Velocity vs. Time Graphs
2.2:Determine the Correct Graph
2.3:A Curtain Blocks Your View of a Golf Ball
2.4:Set the x(t) of a Monster Truck
2.5:Determine x(t) and v(t) of the Lamborghini
2.6:Toss the Ball to Barely Touch the Ceiling
2.7:Drop Two Balls; One with a Delayed Drop
2.8:Determine the Area Under a(t) and v(t)
;)Problems;)
2.1:Position vs. time graph for the T-bird
2.2:A hockey puck sliding on ice collides and rebounds from a wall
2.3:Which helicopter flies according to the velocity vs. time graph shown?
2.4:Two balls are putted with the same initial velocity on separate greens
2.5:Sketch velocity vs. time graph
2.6:Playing constant velocity putt-putt golf
2.7:Calculate the acceleration of 6 carts depending on the data given
2.8:The purple truck is catching up to the yellow truck
2.9:The animation simulates the motion of a helium balloon
2.10:The rope holding cargo on a hot-air balloon is cut
2.11:A golf ball is putted on a level, but wet, green
2.12:Two springs are attached to the ends of a cart that is on a cart track
2.13:A golf ball is putted up a steep hill on a green
2.14:A putted golf ball rolls on a green
2.15:Playing constant acceleration putt-putt golf
2.16:Playing constant acceleration putt-putt golf
2.17:Drag the black rectangle into position
2.18:A tennis ball launcher shoots a red tennis ball into the air
2.19:A tennis ball launcher shoots a tennis ball into the air
;)Supplements;)
Worksheet for Exploration 2.1
Worksheet for Exploration 2.2
Worksheet for Exploration 2.3
Worksheet for Exploration 2.4
Worksheet for Exploration 2.5
Worksheet for Exploration 2.6
Worksheet for Exploration 2.7
Worksheet for Exploration 2.8
Chapter 2: One-Dimensional Kinematics
Motion along a straight line, also called one-dimensional motion, can be represented in a number of different ways: as a formula, as a graph, as data in a table, or as an animation. All four representations are useful for problem solving.
The study of motion in one, two, or three dimensions is called kinematics. What distinguishes kinematics from the techniques which we will consider later is that, at the moment, we do not care whyan object is moving the way it is. We just care that it is moving the way described. Do not think that this degrades the study of kinematics. The exact opposite is true. Kinematics is powerful precisely because it is independent of the cause of the motion. We will learn to speak using the common language for describing motion irrespective of the cause.
Table of Contents
Illustrations
Illustration 2.1: Position and Displacement.
Illustration 2.2: Average Velocity.
Illustration 2.3: Average and Instantaneous Velocity.
Illustration 2.4: Acceleration and Measurement.
Illustration 2.5: Motion on a Hill or Ramp.
Illustration 2.6: Free Fall.
Explorations
Exploration 2.1: Compare Position vs. Time and Velocity vs. Time Graphs.
Exploration 2.2: Determine the Correct Graph.
Exploration 2.3: A Curtain Blocks Your View of a Golf Ball.
Exploration 2.4: Set the x(t) of a Monster Truck.
Exploration 2.5: Determine x(t) and v(t) of the Lamborghini.
Exploration 2.6: Toss the Ball to Barely Touch the Ceiling.
Exploration 2.7: Drop Two Balls; One with a Delayed Drop.
Exploration 2.8: Determine the Area Under a(t) and v(t).
Problems
Problem 2.1: Position vs. time graph for the T-bird.
Problem 2.2: A hockey puck sliding on ice collides and rebounds from a wall.
Problem 2.3: Which helicopter flies according to the velocity vs. time graph shown?
Problem 2.4: Two balls are putted with the same initial velocity on separate greens.
Problem 2.5: Sketch velocity vs. time graph.
Problem 2.6: Playing constant velocity putt-putt golf.
Problem 2.7: Calculate the acceleration of 6 carts depending on the data given.
Problem 2.8: The purple truck is catching up to the yellow truck.
Problem 2.9: The animation simulates the motion of a helium balloon.
Problem 2.10: The rope holding cargo on a hot-air balloon is cut.
Problem 2.11: A golf ball is putted on a level, but wet, green.
Problem 2.12: Two springs are attached to the ends of a cart that is on a cart track.
Problem 2.13: A golf ball is putted up a steep hill on a green.
Problem 2.14: A putted golf ball rolls on a green.
Problem 2.15: Playing constant acceleration putt-putt golf.
Problem 2.16: Playing constant acceleration putt-putt golf.
Problem 2.17: Drag the black rectangle into position.
Problem 2.18: A tennis ball launcher shoots a red tennis ball into the air.
Problem 2.19: A tennis ball launcher shoots a tennis ball into the air.
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189777 | https://cdn.prod.web.uta.edu/-/media/project/website/science/mathematics/documents/preprint/2014/rep2014_04.pdf | Technical Report 2014-04 Rayleigh Quotient Based Optimization Methods For Eigenvalue Problems Ren-Cang Li Rayleigh Quotient Based Optimization Methods For Eigenvalue Problems Ren-Cang Li∗ January 16, 2014 Abstract Four classes of eigenvalue problems that admit similar min-max principles and the Cauchy interlacing inequalities as the symmetric eigenvalue problem famously does are investigated. These min-max principles pave ways for efficient numerical solutions for extreme eigenpairs by optimizing the so-called Rayleigh quotient functions. In fact, scientists and engineers have already been doing that for computing the eigenvalues and eigenvectors of Hermitian matrix pencils A −λB with B positive definite, the first class of our eigenvalue problems.
But little attention has gone to the other three classes: positive semidefinite pencils, linear response eigenvalue problems, and hyperbolic eigenvalue problems, in part because most min-max principles for the latter were discovered only very recently and some more are being discovered. It is expected that they will drive the effort to design better optimization based numerical methods for years to come.
1 Introduction Eigenvalue problems are ubiquitous. Eigenvalues explain many physical phenomena well such as vibrations and frequencies, (in)stabilities of dynamical systems, and energy levels in molecules or atoms. This article focuses on classes of eigenvalue problems that ad-mit various min-max principles and the Cauchy interlacing inequalities as the symmetric eigenvalue problem famously does [4, 38, 47]. These results make it possible to efficiently calculate extreme eigenpairs of the eigenvalue problems by optimizing associated Rayleigh quotients.
Consider the generalized eigenvalue problem Ax = λBx, (1) where both A and B are Hermitian. The first class of eigenvalue problems are those for which B is also positive definite. Such an eigenvalue problem is equivalent to a symmetric eigenvalue problem B−1/2AB−1/2y = λx and thus, not surprisingly, all min-max principles (Courant-Fischer, Ky Fan trace min/max, Wielandt-Lidskii) and the Cauchy interlacing inequalities have their counterparts in this eigenvalue problem. The associated Rayleigh quotient is ρ(x) = xHAx xHBx.
(2) ∗Department of Mathematics, University of Texas at Arlington, P.O. Box 19408, Arlington, TX 76019.
E-mail: rcli@uta.edu. Supported in part by NSF grants DMS-1115834 and DMS-1317330, and a research gift grant from Intel Corporation.
1 When B is indefinite and even singular, (1) is no longer equivalent to a symmetric eigenvalue problem in general and it may even have complex eigenvalues which clearly admit no min-max representations. But if there is a real scalar λ0 such that A −λ0B is positive semidefinite, then the eigenvalue problem (1) has only real eigenvalues and they admit similar min-max principles and the Cauchy interlacing inequalities [25, 27, 29]. This is the second class of eigenvalue problems and it share the same Rayleigh quotient (2) as the first class. We call a matrix pencil in this class a positive semidefinite pencil. Opposite to the concept of a positive semidefinite matrix pencil, naturally, is that of a negative semidefinite matrix pencil A −λB by which we mean that A and B are Hermitian and there is a real λ0 such that A −λ0B is negative semidefinite. Evidently, if A −λB is a negative semidefinite matrix pencil, then −(A −λB) = (−A) −λ(−B) is a positive semidefinite matrix pencil because (−A) −λ0(−B) = −(A −λ0B). Therefore it suffices to only study either positive or negative semidefinite pencils.
The third class of eigenvalue problems is the so-called linear response eigenvalue prob-lem or random phase approximation eigenvalue problem [ 0 K M 0 ] [y x ] = λ [y x ] , where K and M are Hermitian and positive semidefinite matrices and one of them is definite. Any eigenvalue problem in this class can be turned into one in the second class by permuting the first and second block rows to get [M 0 0 K ] [y x ] = λ [0 I I 0 ] [y x ] , where I is the identity matrix of apt size. In this sense, the third class is a subclass of the second class, but with block substructures. The associated Rayleigh quotient is ρ(x, y) = xHKx + yHMy 2|xHy| .
The first minimization principle for such eigenvalue problems was essentially published by Thouless [50, 1961], but more were obtained only very recently [2, 3].
The fourth class of eigenvalue problems is the hyperbolic quadratic eigenvalue problem (λ2A + λB + C)x = 0 arising from dynamical systems with friction, where A, B, and C are Hermitian and A is positive definite and (xHBx)2 −4(xHAx)(xHCx) > 0 for any nonzero vector x.
The associated Rayleigh quotients are ρ±(x) = −xHBx ± [ (xHBx)2 −4(xHAx)(xHCx) ]1/2 2(xHAx) .
Courant-Fischer type min-max principles were known to Duffin [10, 1955] and the Cauchy type interlacing inequalities to Veseli´ c [52, 2010]. Other min-max principles (Wielandt-Lidskii type, Ky Fan trace min/max type) are being discovered .
2 In the rest of this article, we will explain the steepest descent/ascent methods and nonlinear conjugate gradient methods for the first class of eigenvalue problems, including the incorporation of preconditioning techniques, extending search spaces, and block im-plementations, in detail but only state min-max principles – old and new – for the other three classes. The interested reader can consult relevant references for the corresponding steepest descent/ascent methods and nonlinear conjugate gradient methods or design his own based on the min-max principles stated.
Notation.
Throughout this paper, Cn×m is the set of all n × m complex matrices, Cn = Cn×1, and C = C1, and similarly Rn×m, Rn, and R are for their real counterparts.
In (or simply I if its dimension is clear from the context) is the n × n identity matrix, and ej is its jth column.
The superscript “·T” and “·H” take transpose and complex conjugate transpose of a matrix/vector, respectively. For a matrix X, R(X) and N(X) are the column space and null space of X, respectively.
We shall also adopt MATLAB-like convention to access the entries of vectors and matrices. Let i : j be the set of integers from i to j inclusive. For a vector u and an matrix X, u(j) is u’s jth entry, X(i,j) is X’s (i, j)th entry; X’s submatrices X(k:ℓ,i:j), X(k:ℓ,:), and X(:,i:j) consist of intersections of row k to row ℓand column i to column j, row k to row ℓ, and column i to column j, respectively.
For A ∈Cn×n, A ≻0 (A ≽0) means that A is Hermitian and positive (semi-)definite, and A ≺0 (A ≼0) means −A ≻0 (−A ≽0).
2 Hermitian Pencil A −λB with Definite B In this section, we consider the generalized eigenvalue problem Ax = λBx, (3) where A, B ∈Cn×n are Hermitian with B ≻0. When the equation (3) for a scalar λ ∈C and 0 ̸= x ∈Cn holds, λ is called an eigenvalue and x a corresponding eigenvector.
Theoretically, it is equivalent to the standard Hermitian eigenvalue problem B−1/2AB−1/2y = λy.
(4) Both have the same eigenvalues with eigenvectors related by y = B1/2x, where B−1/2 = (B−1)1/2 is the positive definite square root of B−1 (also B−1/2 = (B1/2)−1) [5, 19].
Numerically, if it has to be done (usually for modest n, up to a few thousands), the conversion of (3) to a standard Hermitian eigenvalue problem is usually accomplished through B’s Cholesky decomposition: B = RHR, where R is upper triangular, rather than B’s square root which is much more expensive to compute but often advantageous for theoretical investigations. The converted eigenvalue problem then is R−HAR−1y = λy (5) with eigenvectors related by y = Rx, and can be solved as a dense eigenvalue problem by LAPACK for modest n.
But calculating the Cholesky decomposition can be very expensive, too, for large n, not to mention possible fill-ins for unstructured sparse B. In this section, we are concerned with Rayleigh Quotient based optimization methods to calculate a few extreme eigenvalues of (3).
3 By the theoretical equivalence of (3) to the standard Hermitian eigenvalue problem (4) or (5), we know that (3) has n real eigenvalues and B-orthonormal eigenvectors.
Throughout the rest of this section, A−λB will be assumed a Hermitian matrix pencil of order n with B ≻0, and its eigenvalues, eigenvectors, and eigen-decomposition are given by (6).
eigenvalues: λ1 ≤λ2 ≤· · · ≤λn, and Λ = diag(λ1, λ2, . . . , λn), B-orthonormal eigenvectors: u1, u2, . . . , un, and U = [u1, u2, . . . , un], eigen-decomposition: U HAU = Λ and U HBU = In.
(6) In what follows, our focus is on computing the first few smallest eigenvalues and their associated eigenvectors. The case for the largest few eigenvalues can be dealt with in the same way by replacing A by −A, i.e., considering (−A) −λB instead.
2.1 Basic Theory Given x ∈Cn, the Rayleigh Quotient for the generalized eigenvalue problem Ax = λBx is defined by ρ(x) = xHAx xHBx.
(7) Similarly for X ∈Cn×k with rank(X) = k, the Rayleigh Quotient Pencil is XHAX −λXHBX.
(8) Theorem 2.1 collects important min-max results and the Cauchy interlacing inequalities for the eigenvalue problem (3). They can be derived via the corresponding results for (4) or (5), the theoretical equivalence of (3) [4, 38, 47].
Theorem 2.1. Let A −λB be a Hermitian matrix pencil of order n with B ≻0.
1 (Courant-Fischer min-max principles) For j = 1, 2, . . . , n, λj = min dim X=j max x∈X ρ(x), (9a) λj = max codim X=j−1 min x∈X ρ(x).
(9b) In particular, λ1 = min x ρ(x), λn = max x ρ(x).
(10) 2 (Ky Fan trace min/max principles) For 1 ≤k ≤n, k ∑ i=1 λi = min XHBX=Ik trace(XHAX), (11a) n ∑ i=n−k+1 λi = max XHBX=Ik trace(XHAX).
(11b) Furthermore if λk < λk+1, then R(X) = R(U(:,1:k)) for any minimizing X ∈Cn×k for (11a); if λn−k < λn−k+1, then R(X) = R(U(:,n−k+1:n)) for any maximizing X ∈Cn×k for (11b).
4 3 (Cauchy interlacing inequalities) Let X ∈Cn×k with rank(X) = k, and denote by µ1 ≤µ2 ≤· · · ≤µk the eigenvalues of the Rayleigh quotient pencil (8). Then λj ≤µj ≤λn−k+j for 1 ≤j ≤k.
(12) Furthermore if λj = µj for 1 ≤j ≤k and λk < λk+1, then R(X) = R(U(:,1:k)); if µj = λn−k+j for 1 ≤j ≤k and λn−k < λn−k+1, then R(X) = R(U(:,n−k+1:n)).
The computational implications of these results are as follows. The equations in (10) or (11) naturally leads to applications of optimization approaches to computing the first/last or first/last few eigenvalues and their associated eigenvectors, while the inequalities in (12) suggest that judicious choices of X can push µj either down to λj or up to λn−k+j for the purpose of computing them.
In pertinent to deflation, i.e., avoiding computing known or already computed eigen-pairs, we have the following results.
Theorem 2.2. Let integer 1 ≤k < n and ξ ∈R.
1. We have λk+1 = min x⊥B ui, 1≤i≤k ρ(x), λk+1 = max x⊥B ui, k+2≤i≤n ρ(x), where ⊥B stands for B-orthogonality, i.e., x⊥B y means ⟨x, y⟩B ≡xHBy = 0.
2. let V = [u1, u2, . . . , uk]. The eigenvalues of matrix pencil [A + ξ(BV )(BV )H] −λB are λj + ξ for 1 ≤j ≤k and λj for k + 1 ≤j ≤n with the corresponding eigenvectors uj for 1 ≤j ≤n. In particular, U H[A + ξ(BV )(BV )H]U = [Λ1 + ξIk 0 0 Λ2 ] , U HBU = In, where Λ1 = Λ(1:k,1:k) and Λ2 = Λ(k+1:n,k+1:n).
The concept of invariant subspace is very important in the standard eigenvalue problem and, more generally, operator theory. In a loose sense, computing a few eigenvalues of a large scale matrix H ∈Cn×n is equivalent to calculating a relevant invariant subspace X of H, i.e., a subspace X ⊆Cn such that HX ⊆X. This concept naturally extends to the generalized eigenvalue problem for A −λB that we are interested in, i.e., A and B are Hermitian and B ≻0.
Definition 2.1. A X ⊆Cn is called a generalized invariant subspaces of A −λB if AX ⊆BX.
Sometimes, it is simply called an invariant subspace.
Some important properties of an invariant subspaces are summarized into the following theorem whose proof is left as an exercise.
Theorem 2.3. Let X ⊆Cn and dim X = k, and let X ∈Cn×k be a basis matrix of X.
1. X is an invariant subspace of A −λB if and only if there is A1 ∈Ck×k such that AX = BXA1.
(13) 5 2. Suppose X is an invariant subspace of A −λB and (13) holds. Then the following statements are true.
(a) A1 = (XHBX)−1(XHAX) and thus it has the same eigenvalues as XHAX − λXHBX. If X has B-orthonormal columns, i.e., XHBX = Ik (one can always pick a basis matrix like this), then A1 = XHAX which is also Hermitian.
(b) For any eigenpair (ˆ λ, ˆ x) of A1: A1ˆ x = ˆ λˆ x, (ˆ λ, Xˆ x) is an eigenpair of A −λB.
(c) Let X⊥∈Cn×(n−k) such that Z := [X, X⊥] is nonsingular and XHBX⊥= 0.
We have ZHAZ = [XHAX XH ⊥AX⊥ ] , ZHBZ = [XHBX XH ⊥BX⊥ ] .
2.2 Rayleigh-Ritz Procedure Theorem 2.3 says that partial spectral information can be extracted from an invariant subspace if known.
But an exact invariant subspace is hard to come by in practice.
Through computations we often end up with subspaces X that 1. are accurate approximate invariant subspaces themselves, or 2. have a nearby lower dimensional invariance subspace.
For the former, it means that ∥AX −BXA1∥is tiny for some matrix A1, where X is a basis matrix X and ∥·∥is some matrix norm. For the latter, it means there is an invariant subspace U of a lower dimension than X such that the canonical angles from U to X are all tiny.
The Rayleigh-Ritz procedure is a way to extract approximate spectral information on A −λB for a given subspace that satisfies either one of the two requirements.
Algorithm 2.1 Rayleigh-Ritz procedure Given a computed subspace X of dimension ℓ in the form of a basis ma-trix X ∈ Cn×ℓ, this algorithm computes approximate eigenpairs of A − λB.
1: compute the projection matrix pencil XHAX −λXHBX which is ℓ× ℓ; 2: solve the eigenvalue problem for XHAX −λXHBX to obtain its eigenpairs (ˆ λi, ˆ xi) which yield approximate eigenpairs (ˆ λi, Xˆ xi), called Rayleigh-Ritz pairs, for the orig-inal pencil A −λB. These ˆ λi are called Ritz values and Xˆ xi Ritz vectors.
If X is a true invariant subspace, the Ritz values and Ritz vectors as rendered by this Rayleigh-Ritz procedure are exact eigenvalues and eigenvectors of A −λB in the absence of roundofferrors, as guaranteed by Theorem 2.3. So in this sense, this Rayleigh-Ritz procedure is a natural thing to do. On the other hand, as for the standard symmetric eigenvalue problem, the procedure retains several optimality properties as we shall now explain.
By Theorem 2.1, λi = min Y⊆Cn dim Y=i max y∈Y ρ(y), (14) 6 where the minimization is taken over all Y ⊂Cn with dim Y = i. So given X ⊂Cn, the natural definition of the best approximation αi to λi is to replace Cn by X to get αi = min Y⊆X dim Y=i max y∈Y ρ(y).
(15) Any Y ⊆X with dim Y = i can be represented by its basis matrix Y ∈Cn×i which in turn can be uniquely represented by b Y ∈Cℓ×i with rank(b Y ) = i such that Y = X b Y . So y ∈Y is equivalent to y = Y ˆ y = X b Y ˆ y =: Xz for some unique z ∈b Y := R(b Y ) ⊆Cℓ. We have by (15) αi = min Y⊆X dim Y=i max y∈Y yHAy yHBy = min b Y⊆Cℓ dim b Y=i max z∈b Y zHXHAXz zHXHBXz = ˆ λi, the ith eigenvalues of XHAX −λXHBX. This is the first optimality of the Rayleigh-Ritz procedure.
Suppose we are seeking λi for 1 ≤i ≤k. By Theorem 11, we have k ∑ i=1 λi = min R(Y )⊆Cn Y HBY =Ik trace(Y HAY ) (16) where the minimization is taken over all Y ∈Cn×k satisfying Y HBY = Ik. So given X ⊂Cn, the natural definition for the best approximation is to replace Cn by X to achieve min R(Y )⊆X Y HBY =Ik trace(Y HAY ).
(17) Any R(Y ) ⊆X with Y HBY = Ik can be represented uniquely by Y = X b Y for some b Y ∈Cℓ×k such that b Y H(XHBX)b Y = Ik. So (16) becomes min R(Y )⊆X Y HBY =Ik trace(Y HAY ) = min b Y H(XHBX)b Y =Ik trace(b Y H(XHBX)b Y ) = k ∑ i=1 ˆ λi.
This gives the second optimality of the Rayleigh-Ritz procedure.
The third optimality is concerned with the residual matrix R(A1) := AX −BXA1.
If R(A1) = 0, then X is an exact invariant subspace. So it would make sense to make ∥R(A1)∥as small as possible for certain matrix norm ∥· ∥. The next theorem says the optimal A1 is XHAX when X is a B-orthonormal basis matrix of X.
Theorem 2.4. Suppose X has B-orthonormal columns, i.e., XHBX = Ik, and let H = XHAX. Then for any unitarily invariant norm1 ∥· ∥ui ∥B−1/2 R(H)∥ui ≤∥B−1/2 R(A1)∥ui for all k-by-k A1.
(18) 1Two common used unitarily invariant norms are the spectral norm ∥· ∥2 and the Frobenius norm ∥· ∥F. It is natural to think ∥B−1/2( · )∥ui as a B−1-unitarily invariant norm induced by a given unitarily invariant norm ∥· ∥ui. For example, the usual Frobenius norm can be defined by ∥C∥F := √ trace(CHC).
Correspondingly, we may define the B−1-Frobenius norm by ∥C∥B−1;F = √ trace(CHB−1C).
7 2.3 Steepest Descent Methods The basic idea of the steepest descent (SD) method to minimize a function value is to perform a line-search along the (opposite) direction of the gradient of the function at each iteration step. Our function is ρ(x) defined by (7) whose gradient is given by ∇ρ(x) = 2 xHBx r(x), (19) where r(x) := Ax −ρ(x) Bx is the residual of (ρ(x), x) as an approximate eigenpair of A −λB. Notice that ∇ρ(x) points to the same direction as r(x). Therefore, given an approximation x x x to u1 and ∥x x x∥B = 1, one step of the steepest descent method for computing (λ1, u1) is simply to perform a line-search: inf t∈C ρ(x x x + tr r r), (20) where r r r = r(x). Since x x xHr r r = 0, x x x and r r r are linearly independent unless r r r = 0 which implies (ρ(x x x),x x x) is already an exact eigenpair. An easy to use stopping criteria is to check if ∥r(x x x)∥2 ∥Ax x x∥2 + |ρ(x x x)| ∥Bx x x∥2 ≤rtol, (21) where rtol is a given relative tolerance. When it is satisfied, (ρ(x x x),x x x) will be accepted as a computed eigenpair.
We have to solve the line-search (20). Since such a problem arises often in the conjugate gradient methods for A −λB, we consider the following more general line-search: inf t∈C ρ(x + tp), (22) where the search direction p is the residual r(x) in the SD method but will be differ-ent in the conjugate gradient method, for example. Suppose that x and p are linearly independent; otherwise ρ(x + tp) ≡ρ(x) = ρ(p). It is not difficult to show that inf t∈C ρ(x + tp) = min |ξ|2+|ζ|2>0 ρ(ξx + ζp).
(23) Therefore the infimum in (22) is the smaller eigenvalue µ of the 2 × 2 matrix pencil XHAX −λXHBX, where X = [x, p]. Let v = [ν1, ν2]T be the corresponding eigenvector.
Then ρ(Xv) = µ. Note Xv = ν1x + ν2p. We conclude arginf t∈C ρ(x + tp) =: topt = { ν2/ν1, if ν1 ̸= 0, ∞, if ν1 = 0.
(24) Here topt = ∞should be interpreted in the sense of taking t →∞: lim t→∞ρ(x + tp) = ρ(p).
Accordingly, we have ρ(y) = inf t∈C ρ(x + tp), y = { x + toptp if topt is finite, p otherwise.
(25) 8 Now the simple SD method can be readily stated. We leave it to the reader.
This simple SD method can be slowly convergent in practice. This happens when the contours of ρ(x) on the sphere {x : xHx = 1} near the eigenvector u1 is very flat: very long stretched in one or a few directions but very short compressed in other directions. So rarely, this plain version is used in practice, but rather as a starting point for designing faster variations of the method. In what follows, we will present three ideas some or all of which can be combined to improve the method in practice. The three ideas are • extending the search space, • preconditioning the search direction, • introducing block implementation.
We now explain the three ideas in detail.
Extending the search space.
The key step of the SD method is the line-search (20) which can be interpreted as seeking the best possible approximation ρ ρ ρnew: ρ ρ ρnew = min z∈span{x x x,r r r} ρ(z) (26) to λ1 through projecting A −λB to the 2-dimensional subspace spanned by x x x, r r r = Ax x x −ρ ρ ρBx x x = (A −ρ ρ ρB)x x x, where ρ ρ ρ = ρ(x x x). This subspace is in fact the 2nd order Krylov subspace K2(A −ρ ρ ρB,x x x) of A −ρ ρ ρB on x x x. Naturally, a way to accelerate the simple SD method is to use a larger Krylov subspace, i.e., the mth order Krylov subspace Km(A −ρ ρ ρB,x x x) which is spanned by x x x, (A −ρ ρ ρB)x x x, . . . , (A −ρ ρ ρB)m−1x x x.
A better approximation to λ1 is then obtained for m ≥3 since now they are achieved by minimizing ρ(x) over a larger subspace that contains span{x x x,r r r} = K2(A −ρ ρ ρB,x x x): ρ ρ ρnew = min z∈Km(A−ρ ρ ρB,x x x) ρ(z) (27) This leads to the inverse free Krylov subspace method of Golub and Ye but we will call it the extended steepest descent method (ESD).
Theorem 2.5 (). Suppose λ1 is simple, i.e., λ1 < λ2, and λ1 < ρ ρ ρ < λ2. Let ω1 < ω2 ≤· · · ≤ωn be the eigenvalues of A −ρ ρ ρB and v1 be an eigenvector corresponding to ω1, and let ρ ρ ρnew be defined by (27). Then ρ ρ ρnew −λ1 ≤(ρ ρ ρ −λ1)ϵ2 m + 2(ρ ρ ρ −λ1)3/2ϵm (∥B∥2 ω2 )1/2 + O(|ρ ρ ρ −λ1|2), (28) where ϵm := min f∈Pm−1,f(ω1)=1 max j>1 |f(ωj)| ≤2 [ ∆m−1 η + ∆−(m−1) η ]−1 , (29) η = ω2−ω1 ωn−ω1 and ∆η = 1+√η 1−√η.
9 There are a few other existing results for m = 2 and B = I. Kantorovich and Akilov [21, p.617,1964] established (ρ ρ ρnew −λ1)/(ρ ρ ρ −λ1) ⪅ϵ2 m for completely continuous operators. Knyazev and Skorokhodov [22, 1991] obtained some-thing that is stronger in the sense that it is a strict inequality (i.e., without the need of ignoring high order terms). Samokish presented an estimate on convergence rate for the preconditioned steepest descent method. Although his technique was for the case B = I, but can be made to work for the case B ̸= I after minor changes (see also [24, 37]).
We omit stating them to limit the length of this paper.
Preconditioning the search direction.
The idea of preconditioning a linear system Ax = b to KAx = Kb such that KA is “almost” the identity matrix before it is iteratively solved is quite natural. After all if KA = I, we would have x = Kb immediately. Here that KA is “almost” the identity matrix is understood either ∥KA −I∥is relatively small or KA −I is near a low rank matrix.
But there is no such an obvious and straightforward way to precondition the eigenvalue problem Ax = λBx. How could any direction be more favorable than the steepest descent one when it comes to minimize ρ(x)? After all, we are attempting to minimize the objective function ρ(x).
In what follows, we shall offer two view points as to understand preconditioning an eigenvalue problem and how an effective preconditioner should be approximately con-structed.
The first view point is more intuitive. The rationale lies as follows. It is well-known that when the contours of the objective function near its optimum are extremely elongated, at each step of the conventional steepest descent method, following the search direction which is the opposite of the gradient gets closer to the optimum on the line for a very short while and then starts to get away because the direction doesn’t point “towards the optimum”, resulting in a long zigzag path of a large number of steps. The ideal search direction p is therefore the one such that with its starting point at x x x, p points to the optimum, i.e., the optimum is on the line {x x x + tp : t ∈C}. Specifically, expand x x x as a linear combination of eigenvectors uj x x x = n ∑ j=1 αjuj =: α1u1 + v v v, v v v = n ∑ j=2 αjuj.
(30) Then the ideal search direction is p = αu1 + βv v v for some scalar α and β ̸= 0 such that α1β −α ̸= 0 (otherwise p = βx x x). Of course, this is impractical because we don’t know u1 and v v v. But we can construct one that is close to it.
One such p is p = (A −σB)−1r r r = (A −σB)−1(A −ρ ρ ρB)x x x, where2 σ is some shift near λ1 but not equal to ρ ρ ρ. Let us analyze this p. By (6), we find p = n ∑ j=1 µjαjuj, µj := λj −ρ ρ ρ λj −σ.
(31) 2We reasonably assume also σ ̸= λj for all other j, too.
10 Now if λ1 ≤ρ ρ ρ < λ2 and σ is also near λ1 but not equal to ρ ρ ρ and if the gap λ2 −λ1 is reasonably modest, then µj ≈1 for j > 1 to give a p ≈αu1 + v v v, resulting in fast convergence. This rough but intuitive analysis suggests that (A −σB)−1 with a suitably chosen shift σ can be used to serve as a good preconditioner. Qualitatively, we have Theorem 2.6. Let x x x be given by (30), and suppose α1 ̸= 0. If σ ̸= ρ ρ ρ such that either µ1 < µj for 2 ≤j ≤n or µ1 > µj for 2 ≤j ≤n, (32) where µj are defined in (31), then tan θB(u1, Km) ≤2 [ ∆m−1 η + ∆−(m−1) η ]−1 tan θB(u1,x x x), (33) 0 ≤ρ ρ ρnew −λ1 ≤4 [ ∆m−1 η + ∆−(m−1) η ]−2 tan θB(u1,x x x), (34) where Km := Km([A −σB]−1(A −ρ ρ ρB),x x x), and η = { λn−σ λn−λ1 · λ2−λ1 λ2−σ , if µ1 < µj for 2 ≤j ≤n, λ2−σ λ2−λ1 · λn−λ1 λn−σ , if µ1 > µj for 2 ≤j ≤n, ∆η = 1 + √η 1 −√η.
Proof. The proof is similar to the one in Saad for the symmetric Lanczos method.
The assumption (32) is one of the two criteria for selecting a shift σ, and the other is to make η close to 1. Three interesting cases are • σ < λ1 ≤ρ < λ2 under which µ1 is smallest • λ1 < σ < ρ < λ2 under which µ1 is biggest • λ1 < ρ < σ < λ2 under which µ1 is smallest.
Often σ is selected as a lower bound of λ1 as in the first case above, but it does not have to be. As for η, it is 1 for σ = λ1, but since λ1 is unknown, the best one can hope is to make σ ≈λ1 through some kind of estimation.
In practice, because of high cost associated with (A −σB)−1, some forms of approxi-mations to (A −σB)−1, such as those by incomplete decompositions LDLH of A −σB or by iterative methods [9, 13, 14] CG, MINRES, or GMRES, are widely used.
The second view point is proposed by Golub and Ye , based on Theorem 2.5 which reveals that the rate of convergence depend on the distribution of the eigenvalues ωj of A −ρ ρ ρB, not those of the pencil A −λB as in the Lanczos algorithm. In particular, if all ω2 = · · · = ωn, then ϵm = 0 for m ≥2 and thus ρ ρ ρnew −λ1 = O(|ρ ρ ρ −λ1|2), suggesting quadratic convergence. Such an extreme case, though highly welcome, is un-likely to happen in practice, but it gives us an idea that if somehow we could transform an eigenvalue problem towards such an extreme case, the transformed problem would be 11 easier to solve. Specifically we should seek equivalent transformations that change the eigenvalues of A −ρ ρ ρB as much as possible to, one smallest isolated eigenvalue ω1, and the rest ωj (2 ≤j ≤n) tightly clustered, (35) but leave those of A −λB unchanged. This goal is much as the one for preconditioning a linear system Ax = b to KAx = Kb for which a similar eigenvalue distribution for KA like (35) will result in swift convergence by most iterative methods.
We would like to equivalently transform the eigenvalue problem for A−λB to L−H(A− λB)L−1 by some nonsingular L (whose inverse or any linear system with L is easy to solve) so that the eigenvalues of L−1(A −ρ ρ ρB)L−H distribute more or less like (35). Then apply one step of ESD to the pencil L−1(A−λB)L−H to find the next approximation ρ ρ ρnew. The process repeats.
Borrowed from the incomplete decomposition idea for preconditioning a linear system, such an L can be constructed using the LDLH decomposition of A −ρ ρ ρB [13, p.139] if the decomposition exists: A −ρ ρ ρB = LDLH, where L is lower triangular and D = diag(±1).
Then L−1(A −ρ ρ ρB)L−H = D has the ideal eigenvalue distribution that gives ϵm = 0 for any m ≥2. Unfortunately, this simple solution is impractical in practice for the following reasons: 1. The decomposition may not exist at all. In theory, the decomposition exists if all of the leading principle submatrices of A −ρ ρ ρB are nonsingular.
2. If the decomposition does exist, it may not be numerically stable to compute, espe-cially when ρ ρ ρ comes closer and closer to λ1.
3. The sparsity in A and B is most likely destroyed, leaving L significantly denser than A and B combined. This makes all ensuing computations much more expensive.
A more practical solution is, however, through an incomplete LU factorization (see [44, Chapter 10]), to get A −ρ ρ ρB ≈LDLH, where “≈” includes not only the usual “approximately equal”, but also the case when (A −ρ ρ ρB) −LDLH is approximately a low rank matrix, and D = diag(±1). Such an L changes from one step of the algorithm to another. In practice, often we may use one fixed preconditioner for all or several iterative steps. Using a constant preconditioner is certainly not optimal: it likely won’t give the best rate of convergence per step and thus increases the number of total iterative steps but it can reduce overall cost because it saves work in preconditioner constructions and thus reduces cost per step. The basic idea of using a step-independent preconditioner is to find a σ that is close to λ1, and perform an incomplete LDLH decomposition of A −σB ≈LDLH and transform A−λB accordingly before applying SD or ESD. Now the rate of convergence is determined by the eigenvalues of b C = L−1(A −σB)L−H + (σ −ρ ρ ρ)L−1BL−H ≈D 12 which would have a better spectral distribution so long as (σ −ρ ρ ρ)L−1BL−H is small relative to b C. When σ < λ1, A−σB ≻0 and the incomplete LDLH factorization becomes incomplete Cholesky factorization.
We have insisted so far about applying SD or ESD straightforwardly to the transformed problem. There is another way, perhaps, better: only symbolically applying SD or ESD to the transformed problem as a derivation stage for a preconditioned method that always projects the original pencil A −λB directly every step. The only difference is now the projecting subspaces are preconditioned.
Suppose A −λB is transformed to b A −λ b B := L−1(A −λB)L−H. Consider a typical step of ESD applied to b A −λ b B. For the purpose of distinguishing notational symbols, we will put hats on all those for b A −λ b B. The typical step of ESD is compute the smallest eigenvalue µ and corresponding eigen-vector v of b ZH( b A−λ b B) b Z, where b Z ∈Cn×m is a basis matrix of Krylov subspace Km( b A −ˆ ρ ρ ρ b B, ˆ x x x).
(36) Notice [ b A −ˆ ρ ρ ρ b B ]j ˆ x x x = LH [ (LLH)−1(A −ˆ ρ ρ ρB) ]j (L−Hˆ x x x) to see L−H · Km( b A −ˆ ρ ρ ρ b B, ˆ x x x) = Km( K(A −ˆ ρ ρ ρB),x x x), where x x x = L−Hˆ x x x and K = (LLH)−1. So Z = L−H b Z is a basis matrix of Krylov subspace Km( K(A −ˆ ρ ρ ρB),x x x). Since also b ZH( b A −λ b B) b Z = (L−H b Z)H(A −λB)(L−H b Z), ˆ ρ ρ ρ = ˆ x x xH b A ˆ x x x ˆ x x xH b B ˆ x x x = x x xHAx x x x x xHB x x x = ρ ρ ρ, the typical step (36) can be reformulated equivalently to compute the smallest eigenvalue µ and corresponding eigen-vector v of ZH(A −λB)Z, where Z ∈Cn×m is a ba-sis matrix of Krylov subspace Km( K(A −ρ ρ ρB),x x x), where K = (LLH)−1.
(37) Introducing block implementation.
The convergence rate of ESD with a precondi-tioner K ≻0 is determined by the eigenvalues ω1 < ω2 ≤· · · ≤ωn of K1/2(A −ρ ρ ρB)K1/2 can still be very slow if λ2 is very close to λ1 relative to λn in which case ω1 ≈ω2.
Often in practice, there are needs to compute the first few eigenpairs, not just the first one. For that purpose, block variations of the methods become particularly attractive for at least the following reasons: 1. they can simultaneously compute the first k eigenpairs (λj, uj); 2. they run more efficiently on modern computer architecture because more computa-tions can be organized into matrix-matrix multiplication type; 3. they have better rates of convergence to the desired eigenpairs and save overall cost by using a block size that is slightly bigger than the number of asked eigenpairs.
13 In summary, the benefits of using a block variation are similar to those of using the simultaneous subspace iteration vs. the power method .
A block variation starts with X X X ∈Cn×nb with rank(X X X) = nb, instead of just one vector x x x ∈Cn previously for the single-vector steepest descent methods.
Here either the jth column of X X X is already an approximation to uj or the subspace R(X X X) is a good approximation to the generalized invariant subspace spanned by uj for 1 ≤j ≤nb or the canonical angles from R([u1, . . . , uk]) to R(X X X) are nontrivial, where k ≤nb is the number of desired eigenpairs. In the latter two cases, a preprocessing is needed to turn the case into the first case: 1. solve the eigenvalue problem X X XH(A −λB)X X X to get (X X XHAX X X) W = (X X XHBX X X) WΩ, where Ω= diag(ρ ρ ρ1,ρ ρ ρ2, . . . ,ρ ρ ρnb) is the diagonal matrix of eigenvalues in ascending order, and W is the eigenvector matrix; 2. reset X X X := X X XW.
So we will assume henceforth the jth column of the given X X X is an approximation to uj.
Now consider generalizing the steepest descent method to a block one. Its typical iterative step may well look like the following. Let X X X = [x x x1,x x x2, . . . ,x x xnb] ∈Cn×nb whose jth column x x xj approximates uj and Ω= diag(ρ ρ ρ1,ρ ρ ρ2, . . . ,ρ ρ ρnb) whose jth diagonal entry ρ ρ ρj = ρ(x x xj) approximates λj. We may well assume X X X has B-orthonormal columns, i.e., X X XHBX X X = I. Define the residual matrix R R R = [r(x x x1), r(x x x2), . . . , r(x x xnb)] = AX X X −BX X XΩ.
The key iterative step of the block steepest descent (BSD) method for computing the next set of approximations is as follows: 1. compute a basis matrix Z of R([X X X,R R R]) by, e.g., MGS in the B-inner product, keeping in mind that X X X has B-orthonormal columns already; 2. find the first nb eigenpairs of ZHAZ−λZHBZ by, e.g., one of LAPACK’s subroutines [1, p.25] because of its small scale, to get (ZHAZ)W = (ZHBZ)WΩnew, where Ωnew = diag(ρ ρ ρnew;1,ρ ρ ρnew;2, . . . ,ρ ρ ρnew;nb); 3. set X X Xnew = ZW.
This is in fact the stronger version of Simultaneous Rayleigh Quotient Minimization Method, called SIRQIT-G2, in Longsine and McCormick . To introduce the block extended steepest descent (BESD) method, we notice that r(x x xj) = (A −ρ ρ ρjB)x x xj and thus R([X X X,R R R]) = nb ∑ j=1 R([x x xj, (A −ρ ρ ρjB)x x xj]) = nb ∑ j=1 K2(A −ρ ρ ρjB,x x xj).
14 BESD is simply to extend each Krylov subspace K2(A −ρ ρ ρjB,x x xj) to a high order one, and of course different Krylov subspaces can be expanded to different orders. For simplicity, we will expand each to the mth order. The new extended search subspace now is nb ∑ j=1 Km(A −ρ ρ ρjB,x x xj).
(38) Define the linear operator R : X ∈Cn×nb →R(X) = AX −BXΩ∈Cn×nb.
Then the subspace in (38) can be compactly written as Km(R, X) = span{X X X, R(X X X), . . . , Rm−1(X X X)}, (39) where Ri( · ) is understood as successively applying the operator R i times, e.g., R2(X) = R(R(X)).
As to incorporate suitable preconditioners, in light of our extensive discussions before, the search subspace should be modified to nb ∑ j=1 Km(Kj(A −ρ ρ ρjB),x x xj), (40) where Kj are the preconditioners, one for each approximate eigenpair (ρ ρ ρj,x x xj) for 1 ≤j ≤ nb. As before, Kj can be constructed in one of the following two ways: • Kj is an approximate inverse of A−˜ ρ ρ ρjB for some ˜ ρ ρ ρj different from ρ ρ ρj, ideally closer to λj than to any other eigenvalue of A−λB. But this requirement on ˜ ρ ρ ρj is impractical because the eigenvalues of A −λB are unknown. A compromise would be to make ˜ ρ ρ ρj close but not equal to ρ ρ ρj than to any other ρ ρ ρj.
• Perform an incomplete LDLH factorization (see [44, Chapter 10]) A−ρjB ≈LjDjLH j , where “≈” includes not only the usual “approximately equal”, but also the case when (A−ρ ρ ρjB)−LjDjLH j is approximately a low rank matrix, and Dj = diag(±1). Finally set Kj = LjLH j .
Algorithm 2.2 is the general framework of a Block Preconditioned Extended Steepest Descent method (BPESD) which embeds many steepest descent methods into one. In particular, 1. With nb = 1, it gives various single-vector steepest descent methods: • Steepest Descent method (SD): m = 2 and all preconditioners Kℓ;j = I; • Preconditioned Steepest Descent method (PSD): m = 2; • Extended Steepest Descent method (ESD): all preconditioners Kℓ;j = I; • Preconditioned Extended Steepest Descent method (PESD).
2. With nb > 1, various block steepest descent methods are born: • Block Steepest Descent method (BSD): m = 2 and all preconditioners Kℓ;j = I; • Block Preconditioned Steepest Descent method (BPSD): m = 2; 15 Algorithm 2.2 Extended Block Preconditioned Steepest Descent method Given an initial approximation X0 ∈Cn×nb with rank(X0) = nb, and an integer m ≥2, the algorithm attempts to compute approximate eigenpairs to (λj, uj) for 1 ≤j ≤nb.
1: compute the eigen-decomposition: (XH 0 AX0)W = (XH 0 BX0)WΩ0, where W H(XH 0 BX0)W = I, Ω0 = diag(ρ0;1, ρ0;2, . . . , ρ0;nb); 2: X0 ≡[x0;1, x0;2, . . . , x0;nb] = X0W; 3: for ℓ= 0, 1, . . . do 4: test convergence and lock up the converged (detail to come later); 5: construct preconditioners Kℓ;j for 1 ≤j ≤nb; 6: compute a basis matrix Z of the subspace (40) with ρ ρ ρj = ρℓ;j and x x xj = xℓ+1;j; 7: compute the nb smallest eigenvalues and corresponding eigenvectors of ZH(A − λB)Z to get (ZHAZ)W = (ZHBZ)WΩℓ, where W H(ZHBZ)W = I, Ωℓ+1 = diag(ρℓ+1;1, ρℓ+1;2, . . . , ρℓ+1;nb); 8: Xℓ+1 ≡[xℓ+1;1, xℓ+1;2, . . . , xℓ+1;nb] = ZW; 9: end for 10: return approximate eigenpairs to (λj, uj) for 1 ≤j ≤nb.
• Block Extended Steepest Descent method (BESD): all preconditioners Kℓ;j = I; • Block Preconditioned Extended Steepest Descent method (BPESD).
This framework is essentially the one implied in [40, section 4].
There are four important implementation issues to worry about in turning Algo-rithm 2.2 into a piece of working code.
1. In (40), a different preconditioner is used for each and every approximate eigenpair (ρℓ;j, xℓ;j) for 1 ≤j ≤nb. While, conceivably, doing so will speed up convergence for each approximate eigenpair because each preconditioner can be constructed to make that approximate eigenpair converge faster, but the cost in constructing these preconditioners may likely be too heavy to bear. A more practical approach would be to use one pre-conditioner Kℓfor all Kℓ;j aiming at speeding up the convergence of (ρℓ;1, xℓ;1) (or the first few approximate eigenpairs for tightly clustered eigenvalues). Once it (or the first few in the case of a tight cluster) is determined to be sufficiently accurate, the converged eigenpairs are locked up and deflated and a new preconditioner is computed to aim at the next non-converged eigenpairs, and the process continues. We will come back to discuss the deflation issue, i.e., Line 4 of Algorithm 2.2.
2. Consider implementing Line 6, i.e., generating a basis matrix for the subspace (40). In the most general case, Z can be gotten by packing the basis matrices of all Km(Kℓ;j(A −ρℓ;jB), xℓ;j) for 1 ≤j ≤nb together. There could be two problems with this: 1) such Z could be ill-conditioned, i.e., the columns of Z may not be sufficiently numerically linearly independent, and 2) the arithmetic operations in building a basis for each Km(Kℓ;j(A −ρℓ;jB), xℓ;j) are mostly matrix-vector multiplications, straying from one of the purposes: performing most arith-metic operations through matrix-matrix multiplications in order to achieve high perfor-mance on modern computers. To address these two problems, we do a tradeoffof using Kℓ;j ≡Kℓfor all j. This may likely degrade the effectiveness of the preconditioner per step in terms of rate of convergence for all approximate eigenpairs (ρℓ;j, xℓ;j) but may achieve 16 overall gain in using less time because then the code will run much faster in matrix-matrix operations, not to mention the saving in constructing just one preconditioner Kℓinstead of nb different ones. To simplify our discussion below, we will drop the subscript ℓfor readability. Since Kℓ;j ≡K for all j, (40) is the same as Km(KR, X) = span{X, KR(X), . . . , [KR]m−1(X)}, (41) where [KR]i( · ) is understood as successively applying the operator KR i times, e.g., [KR]2(X) = KRℓ(KR(X)). A basis matrix Z = [Z1, Z2, . . . , Zm] can be computed by the following block Arnoldi-like process in the B-inner product [40, Algorithm 5].
1: Z1 = X (recall XHBX = Inb already); 2: for i = 2 to m do 3: Y = K(AZi−1 −BΩZi−1); 4: for j = 1 to i −1 do 5: T = ZH j BY , Y = Y −ZjT; 6: end for 7: ZiT = Y (MGS in the B-inner product); 8: end for (42) There is a possibility that at Line 7 of (42), Y is numerically not of full column rank. If it happens, it poses no difficulty at all. In running MGS on Y ’s columns, anytime if a column is deemed linearly dependent on previous columns, that column should be deleted, along with the corresponding ρj from Ωto shrink its size by 1 as well. At the completion of MGS, Zi will have fewer columns than Y and the size of Ωis shrunk accordingly. Finally, at the end, the columns of Z are B-orthonormal, i.e., ZHBZ = I (of apt size) which may fail to an unacceptably level due to roundoff; so some form of re-orthogonalization should be incorporated.
4. At Line 4, a test for convergence are required. The same criteria (21) can be used: (ρℓ;j, xℓ;j) is considered acceptable if ∥rℓ;j∥2 ∥Axℓ;j∥2 + |ρℓ;j| ∥Bxℓ;j∥2 ≤rtol where rtol is a pre-set relative tolerance. Usually the eigenvalues λj are converged to in order, i.e., the smallest eigenvalues emerge first. All acceptable approximate eigenpairs should be locked in, say, a kcvgd × kcvgd diagonal matrix3 D D D for converged eigenvalues and an n × kcvgd tall matrix U U U for eigenvectors such that AU U U ≈BU U UD D D, U U UHBU U U ≈I to an acceptable level of accuracy. Every time a converged eigenpair is detected, delete the converged ρℓ;j and xℓ;j from Ωℓand Xℓ, respectively, and expand D D D and U U U to lock up the pair, accordingly. At the same time, either reduce nb by 1 or append a (possibly random) B-orthogonal column to X to maintain nb unchanged. There are two different ways to avoid recomputing any of the converged eigenpairs – a process called deflation.
3In actual programming code, it is likely an 1-D array. But we use a diagonal matrix for the sake of presentation.
17 1. At Line 7 in the above block Arnoldi-like process (42), each column of Zj+1 is B-orthogonalized against U U U.
2. Modify A −λB in form, but not explicitly, to (A + ζBU U UU U U HB) −λB, where ζ is a real number intended to move λj for 1 ≤j ≤kcvgd to λj +ζ; so it should be selected such that ζ + λ1 ≥λkcvgd+nb+1.
But if there is a good way to pick a ζ such that ζ+λ1 ≥λkcvgd+nb+1, the second approach is easier to use in implementation than the first one for which, if not carefully implemented, rounding errors can make R(Z) crawl into R(U U U) unnoticed.
2.4 Locally Optimal CG Methods As is well-known, the slow convergence of the plain steepest descent method is due to the extreme flat contours of the objective function near (sometimes local) optimal points. The nonlinear conjugate gradient method is another way, besides preconditioning technique, to move the searching direction away from the steepest descent direction. Originally, the conjugate gradient (CG) method was invented in 1950s by Hestenes and Stiefel [17, 1952] for solving linear system Hx = b with Hermitian and positive definite H, and later was interpreted as an iterative method for large scale linear systems. This is so-called the linear CG method [9, 13, 35]. In the 1960s, it was extended by Fletcher and Reeves [11, 1964] as an iterative method for solving nonlinear optimization problems (see also [35, 48]). We shall call the resulting method the nonlinear CG method. Often we leave out the word “linear” and “nonlinear” and simply call either method the CG method when no confusion can arise from this.
Because of the optimality properties (10) of the Rayleigh quotient ρ(x), it is natural to apply the nonlinear CG method to compute the first eigenpair and, with the aid of deflation, the first few eigenpairs of A −λB.
The article [7, 1966] by Bradbury and Fletcher seems to be the first one to do just that.
However, it is suggested that the local optimal CG (LOCG) method [39, 49] is more suitable for the symmetric eigenvalue problem. In its simplest form, LOCG for our eigenvalue problem A −λB is obtained by simply modifying the line-search (26) for the SD method to ρ ρ ρnew = min x∈span{x x x,x x xold,r r r} ρ(x), (43) where x x xold is the approximate eigenvector to u1 from the previous iterative step.
The three ideas we explained in the previous subsection to improve the plain SD method can be introduced to improve the approximation given by (43), too, upon noticing the search space in (43) is K2(A −ρ ρ ρB,x x x) + R(x x xold), making it possible for us to 1) extend the search space, 2) precondition the search direction r r r, and 3) introduce block implementation, in the same way as we did for the plain SD method.
All things considered, we now present an algorithmic framework: Algorithm 2.3, Locally Optimal Block Preconditioned Extended Conjugate Gradient method (LOBPECG) which has implementation choices: • block size nb; • preconditioners varying with iterative steps, with approximate eigenpairs, or not; 18 • the dimension m of Krylov subspaces in extending the search subspace at each iterative step. It may also vary with iterative steps, too.
Algorithm 2.3 Locally Optimal Block Preconditioned Extended Conjugate Gradient method (LOBPECG) Given an initial approximation X0 ∈Cn×nb with rank(X0) = nb, and an integer m ≥2, the algorithm attempts to compute approximate eigenpairs to (λj, uj) for 1 ≤j ≤nb.
1: compute the eigen-decomposition: (XH 0 AX0)W = (XH 0 BX0)WΩ0, where W H(XH 0 BX0)W = I, Ω0 = diag(ρ0;1, ρ0;2, . . . , ρ0;nb); 2: X0 ≡[x0;1, x0;2, . . . , x0;nb] = X0W, X−1 = 0; 3: for ℓ= 0, 1, . . . do 4: test convergence and lock up the converged; 5: construct preconditioners Kℓ;j for 1 ≤j ≤nb; 6: compute a basis matrix Z of the subspace nb ∑ j=1 Km(Kℓ;j(A −ρℓ;jB), xℓ;j) + R(Xℓ−1); (44) 7: compute the nb smallest eigenvalues and corresponding eigenvectors of ZH(A − λB)Z to get (ZHAZ)W = (ZHBZ)WΩℓ, where W H(ZHBZ)W = I, Ωℓ+1 = diag(ρℓ+1;1, ρℓ+1;2, . . . , ρℓ+1;nb); 8: Xℓ+1 ≡[xℓ+1;1, xℓ+1;2, . . . , xℓ+1;nb] = ZW; 9: end for 10: return approximate eigenpairs to (λj, uj) for 1 ≤j ≤nb.
The four important implementation issues we discussed for Algorithm 2.2 (BPESD) after its introduction essentially apply here, except some changes are needed in the com-putation of Z at Line 6 of Algorithm 2.3.
First Xℓ−1 can be replaced by something else while the subspace (44) remains the same. Specifically, we modify Lines 2, 6, and 8 of Algorithm 2.3 to 2: X0 ≡[x0;1, x0;2, . . . , x0;nb] = X0W, and Y0 = 0; 6: compute a basis matrix Z of the subspace nb ∑ j=1 Km(Kℓ;j(A −ρℓ;jB), xℓ;j) + R(Yℓ) (45) such that R(Z(:,1:nb)) = R(Xℓ), and let nZ be the number of the columns of Z; 8: Xℓ+1 ≡[xℓ+1;1, xℓ+1;2, . . . , xℓ+1;nb] = ZW, Yℓ+1 = Z(:,nb+1:nZ)W(nb+1:nZ,:); This idea is basically the same as the one in [18, 23]. Next we will compute a basis matrix for the subspace (45) (or (44)). For better performance (by using more matrix-matrix multiplications), we will assume Kℓ;j ≡Kℓfor all j for simplification.
Dropping the subscript ℓfor readability, we see (45) is the same as Km(KR, X) + R(Y ) = span{X, KR(X), . . . , [KR]m−1(X)} + R(Y ).
(46) We will first compute a basis matrix [Z1, Z2, . . . , Zm] for Km(KR, X) by the Block Arnoldi-like process in the B-inner product (42). In particular, Z1 = X. Then B-orthogonalize 19 Y against [Z1, Z2, . . . , Zm] to get Zm+1 satisfying ZH m+1BZm+1 = I. Finally take Z = [Z1, Z2, . . . , Zm+1].
3 Min-Max Principles for a Positive Semidefinite Pencil Let A −λB be an n × n positive semidefinite pencil, i.e., A and B are Hermitian and there is a real scalar λ0 such that A−λ0B is positive semidefinite. Note that this does not demand anything on the regularity of A −λB, i.e., a positive semidefinite matrix pencil can be either regular (meaning det(A −λB) ̸≡0) or singular (meaning det(A −λB) ≡0 for all λ ∈C).
Let the integer triplet (n−, n0, n+) be the inertia of B, meaning B has n−negative, n0 0, and n+ positive eigenvalues, respectively. Necessarily r := rank(B) = n+ + n−.
(47) We say µ ̸= ∞is a finite eigenvalue of A −λB if rank(A −µB) < max λ∈C rank(A −λB), (48) and x ∈Cn is a corresponding eigenvector if 0 ̸= x ̸∈N(A) ∩N(B) satisfies Ax = µBx, (49) or equivalently, 0 ̸= x ∈N(A −µB)(N(A) ∩N(B)). Let k+ and k−be two nonnegative integers such that k+ ≤n+, k−≤n−, and k+ + k−≥1, and set Jk = [Ik+ −Ik− ] ∈Ck×k, k = k+ + k−.
(50) Theorem 3.1 (). If A −λB is positive semidefinite, then A −λB has r = rank(B) finite eigenvalues all of which are real.
In what follows, if A −λB is positive semidefinite, we will denote its finite eigenvalues by λ± i arranged in the order: λ− n−≤· · · ≤λ− 1 ≤λ+ 1 ≤· · · ≤λ+ n+.
(51) For the case of a regular Hermitian pencil A−λB (i.e., det(A−λB) ̸≡0), Theorem 3.2 is a special case of the ones considered in [6, 34]. For a diagonalizable positive semidefinite Hermitian pencil A −λB with nonsingular B, Theorem 3.2 was implied in [26, 53]. Recall that a positive semidefinite Hermitian pencil A −λB can be possibly a singular pencil; so the condition of Theorem 3.2 does not exclude a singular pencil A −λB which was not considered before [29, 2013], not to mention that B may possibly be singular.
Theorem 3.2. Let A −λB be a positive semidefinite Hermitian pencil. Then for 1 ≤i ≤ n+ λ+ i = sup X codim X=i−1 inf x∈X xHBx=1 xHAx = sup X codim X=i−1 inf x∈X xHBx>0 xHAx xHBx, (52a) λ+ i = inf X dim X=i sup x∈X xHBx=1 xHAx = inf X dim X=i sup x∈X xHBx>0 xHAx xHBx, (52b) 20 and for 1 ≤i ≤n−, λ− i = − sup X codim X=i−1 inf x∈X xHBx=−1 xHAx = inf X codim X=i−1 sup x∈X xHBx<0 xHAx xHBx, (52c) λ− i = − inf X dim X=i sup x∈X xHBx=−1 xHAx = sup X dim X=i inf x∈X xHBx<0 xHAx xHBx.
(52d) In particular, setting i = 1 in (52) gives λ+ 1 = inf xHBx>0 xHAx xHBx, λ− 1 = sup xHBx<0 xHAx xHBx.
(53) All “ inf” and “ sup” can be replaced by “ min” and “ max” if A −λB is positive definite or positive semidefinite but diagonalizable 4 The following theorem for the case when B is also nonsingular is due to Kovaˇ c-Striko and Veseli´ c [25, 1995]. But in this general form, it is due to .
Theorem 3.3 (). Let A −λB be a Hermitian pencil of order n 1. Suppose A−λB is positive semidefinite. Let X ∈Cn×k satisfying XHBX = Jk, and denote by µ± i the eigenvalues of XHAX −λXHBX arranged in the order: µ− k−≤· · · ≤µ− 1 ≤µ+ 1 ≤· · · ≤µ+ k+.
(54) Then λ+ i ≤µ+ i ≤λ+ i+n−k, for 1 ≤i ≤k+, (55) λ− j+n−k ≤µ− i ≤λ− i , for 1 ≤j ≤k−, (56) where we set λ+ i = ∞for i > n+ and λ− j = −∞for j > n−.
2. If A −λB is positive semidefinite, then inf XHBX=Jk trace(XHAX) = k+ ∑ i=1 λ+ i − k− ∑ i=1 λ− i .
(57) (a) The infimum is attainable, if there exists a matrix Xmin that satisfies XH minBXmin = Jk and whose first k+ columns consist of the eigenvectors associated with the eigenvalues λ+ j for 1 ≤j ≤k+ and whose last k−columns consist of the eigen-vectors associated with the eigenvalues λ− i for 1 ≤i ≤k−.
(b) If A −λB is positive definite or positive semidefinite but diagonalizable, then the infimum is attainable.
(c) When the infimum is attained by Xmin, there is a Hermitian A0 ∈Ck×k whose eigenvalues are λ± i , i = 1, 2, . . . , k± such that XH minBXmin = Jk, AXmin = BXminA0.
4Hermitian pencil A −λB is diagonalizable if there exists a nonsingular matrix W such that both W HAW and W HBW are diagonal.
21 3. A −λB is a positive semidefinite pencil if and only if inf XHBX=Jk trace(XHAX) > −∞.
(58) 4. If trace(XHAX) as a function of X subject to XHBX = Jk has a local minimum, then A −λB is a positive semidefinite pencil and the minimum is global.
4 Linear Response Eigenvalue Problem We are interested in solving the standard eigenvalue problem of the form: [ 0 K M 0 ] [y x ] = λ [y x ] , (59) where K and M are n × n real symmetric positive semidefinite matrices and one of them is definite. We referred to it as a linear response (LR) eigenvalue problem because it is equivalent to the original LR eigenvalue problem [ A B −B −A ] [u v ] = λ [u v ] (60) via a simple orthogonal similarity transformation , where A and B are n × n real sym-metric matrices such that the symmetric matrix [A B B A ] is symmetric positive definite5 [41, 51]. In computational physics and chemistry literature, it is this eigenvalue problem that is referred to as the linear response eigenvalue problem (see, e.g., ), or random phase approximation (RPA) eigenvalue problem (see, e.g., ).
While (59) is not a symmetric eigenvalue problem, it has the symmetric structure in its submatrices and many optimization principles that are similar to those one usually find in the symmetric eigenvalue problem. For example, (59) has only real eigenvalues. But more can be said: its eigenvalues come in ±λ pairs. Denote its eigenvalues by −λn ≤· · · ≤−λ1 ≤+λ1 ≤· · · ≤+λn.
In practice, the first few positive eigenvalues and their corresponding eigenvectors are needed. In 1961, Thouless obtained a minimization principle for λ1, now known as Thouless’ minimization principle, which equivalently stated for (59) is λ1 = min x,y xTKx + yTMy 2|xTy| , (61) provided both K ≻0 and M ≻0. This very minimization principle, reminiscent of the first equation in (10), has been seen in action recently in, e.g., [8, 31, 33], for calculating λ1 and, with aid of deflation, other λj.
Recently, Bai and Li obtained Ky Fan trace min type principle, as well as Cauchy interlacing inequalities.
Theorem 4.1 (Bai and Li ). Suppose that one of K, M ∈Rn×n is definite.
5This condition is equivalent to that both A±B are positive definite. In [2, 3] and this article, we focus on very much this case, except that one of A ± B is allowed to be positive semidefinite.
22 1. We have k ∑ i=1 λi = 1 2 inf UTV =Ik trace(U TKU + V TMV ).
(62) Moreover, “ inf” can be replaced by “ min” if and only if both K and M are definite.
When they are definite and if also λk < λk+1, then for any U and V that attain the minimum can be used to recovered λj for 1 ≤j ≤k and their corresponding eigenvectors.
2. Let U, V ∈Rn×k such that U TV is nonsingular. Write W = U TV = W T 1 W2, where Wi ∈Rk×k are nonsingular, and define HSR = [ 0 W −T 1 UTKUW −1 1 W −T 2 V TMV W −1 2 0 ] .
(63) Denote by ±µi (1 ≤i ≤k) the eigenvalues of HSR, where 0 ≤µ1 ≤· · · ≤µk. Then λi ≤µi ≤ √ min{κ(K), κ(M)} cos ∠(U, V) λi+n−k for 1 ≤i ≤k, (64) where U = R(U) and V = R(V ), and κ(K) = ∥K∥2∥K−1∥2 and κ(M) = ∥M∥2∥M−1∥2 are the spectral condition numbers.
Armed with these minimization principles, we can work out extensions of the previously discussed steepest descent methods in subsection 2.3 and conjugate gradient methods in subsection 2.4 for the linear response eigenvalue problem (59). In fact, some extensions have been given in [2, 3, 42].
5 Hyperbolic Quadratic Eigenvalue Problem It was argued in that the hyperbolic quadratic eigenvalue problem (HQEP) is the closest analogue of the standard Hermitian eigenvalue problem Hx = λx when it comes to the quadratic eigenvalue problem (λ2A + λB + C)x = 0.
(65) In many ways, both problems share common properties: the eigenvalues are all real, and for HQEP there is a version of the min-max principles that is very much like the Courant-Fischer min-max principles.
When (65) is satisfied for a scalar λ and nonzero vector x, we call λ a quadratic eigenvalue, x an associated quadratic eigenvector, and (λ, x) a quadratic eigenpair.
One source of HQEP (65) is dynamical systems with friction, where A, C are associ-ated with the kinetic-energy and potential-energy quadratic form, respectively, and B is associated with the Rayleigh dissipation function. When A, B, and C are Hermitian, and A and B are positive definite and C positive semidefinite, we say the dynamical system is overdamped if (xHBx)2 −4(xHAx)(xHCx) > 0 for any nonzero vector x.
(66) An HQEP is slightly more general than an overdamped QEP in that B and C are no longer required positive definite or positive semidefinite, respectively. However, a suitable shift in λ can turn an HQEP into an overdamped HQEP .
23 In what follows, A, B, C ∈Cn×n are Hermitian, A ≻0, and (66) holds. Thus (65) is a HQEP for Q Q Q(λ) = λ2A + λB + C ∈Cn×n. Denote its quadratic eigenvalues by λ± i and arrange them in the order of λ− 1 ≤· · · ≤λ− n < λ+ 1 ≤· · · ≤λ+ n .
(67) Consider the following equation in λ f(λ, x) := xHQ Q Q(λ)x = λ2(xHAx) + λ(xHBx) + (xHCx) = 0, (68) given x ̸= 0. Since Q Q Q(λ) is hyperbolic, this equation always has two distinct real roots (as functions of x) ρ±(x) = −xHBx ± [ (xHBx)2 −4(xHAx)(xHCx) ]1/2 2(xHAx) .
(69) We shall call ρ+(x) the pos-type Rayleigh quotient of Q Q Q(λ) on x, and ρ−(x) the neg-type Rayleigh quotient of Q Q Q(λ) on x.
Theorem 5.1 below is a restatement of [32, Theorem 32.10, Theorem 32.11 and Re-mark 32.13]. However, it is essentially due to Duffin whose proof, although for over-damped Q Q Q, works for the general hyperbolic case. They can be considered as a general-ization of the Courant-Fischer min-max principles (see [38, p.206], [47, p.201]).
Theorem 5.1 (). We have λ+ i = max X⊆Cn codim X=i−1 min x∈X x̸=0 ρ+(x), λ+ i = min X⊆Cn dim X=i max x∈X x̸=0 ρ+(x), (70a) λ− i = max X⊆Cn codim X=i−1 min x∈X x̸=0 ρ−(x), λ− i = min X⊆Cn dim X=i max x∈X x̸=0 ρ−(x).
(70b) In particular, λ+ 1 = min x̸=0 ρ+(x), λ+ n = max x̸=0 ρ+(x), (71a) λ− 1 = min x̸=0 ρ−(x), λ− n = max x̸=0 ρ−(x).
(71b) To generalize Ky Fan trace min/max principle and Cauchy’s interlacing inequalities, we introduce the following notations. For X ∈Cn×k with rank(X) = k, XHQ Q Q(λ)X is a k × k hyperbolic quadratic matrix polynomial. Hence its quadratic eigenvalues are real.
Denote them by λ± i,X arranged as λ− 1,X ≤· · · ≤λ− k,X ≤λ+ 1,X ≤· · · ≤λ+ k,X.
Theorem 5.2.
1. We have min rank(X)=k k ∑ j=1 λ± j,X = k ∑ j=1 λ± j , max rank(X)=k k ∑ j=1 λ± j,X = k ∑ j=1 λ± n−k+j .
(72) 2. For X ∈Cn×k with rank(X) = k, λ+ i ≤λ+ i,X ≤λ+ i+n−k, i = 1, · · · , k, (73a) λ− j ≤λ− j,X ≤λ− j+n−k, j = 1, · · · , k.
(73b) Armed with these minimization principles, we can work out extensions of the previously discussed steepest descent methods in subsection 2.3 and conjugate gradient methods in subsection 2.4 for the HQEP Q Q Q(λ)x = 0. Details, among others, can be found in .
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189779 | https://www.quora.com/What-s-is-the-6th-term-in-the-following-geometric-sequence-3-15-75 | What’s is the 6th term in the following geometric sequence 3, 15, 75? - Quora
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What’s is the 6th term in the following geometric sequence 3, 15, 75?
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The 6th term in the geometric sequence 3, 15, 75 is 1,875.
The sequence is a geometric sequence with a common ratio of 5, as each term is 5 times the previous term. To find the 6th term, we can use the formula for the nth term of a geometric sequence:
a_n = a_1 r^(n-1)
Where:
a_n is the nth term
a_1 is the first term
r is the common ratio
n is the term number
Plugging in the values for this sequence:
a_1 = 3
r = 5
n = 6
a_6 = 3 5^(6-1) = 3 5^5 = 1,875.
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Matthew Boriack
Studied Engineering&Mathematics at Cinco Ranch High School (Graduated 2020) · Author has 124 answers and 256.3K answer views
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Originally Answered: What’s is the 7th term in the following geometric sequence 3, 15, 75? ·
Well, before we find the 7th term in the sequence is, we must find the pattern of the sequence. To find the pattern of the sequence, we must find the difference between two adjacent terms. First let's see if addition is the difference. The two numbers obtained will be the same if this is true.
15−3=12 15−3=12 and 75−15=60 75−15=60 The sequence is not an addition sequence because 12≠60 12≠60 so let's check multiplication.
15 3=5 15 3=5 and 75 3=5 75 3=5 5=5 5=5 so this is indeed a multiplication sequence! Now, let's find the n t h n t h term. n 1=3,n 2=15,n 3=75 n 1=3,n 2=15,n 3=75. So the n t h n t h term is 3(5)n 3(5)n. Whe
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Well, before we find the 7th term in the sequence is, we must find the pattern of the sequence. To find the pattern of the sequence, we must find the difference between two adjacent terms. First let's see if addition is the difference. The two numbers obtained will be the same if this is true.
15−3=12 15−3=12 and 75−15=60 75−15=60 The sequence is not an addition sequence because 12≠60 12≠60 so let's check multiplication.
15 3=5 15 3=5 and 75 3=5 75 3=5 5=5 5=5 so this is indeed a multiplication sequence! Now, let's find the n t h n t h term. n 1=3,n 2=15,n 3=75 n 1=3,n 2=15,n 3=75. So the n t h n t h term is 3(5)n 3(5)n. Where n=7 n=7. So 3(5)7 3(5)7 is the 7th term in the sequence.
5 7=78,125 5 7=78,125
3×78,125=234,375 3×78,125=234,375
So n 7=234,375 n 7=234,375.
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999 485
999 103
99 17
Penny Pan
Knows English
·3y
3(5 to the power of 0) = 3
3(5 to the power of 1) = 15
3(5 to the power of 2) = 75
……
3（5 to the power of 5 ) = 9, 375
Upvote ·
9 1
E Fact World
op
·3y
Sn = a(rn - 1)/r - 1
Where
S
n
= Sum of the first n terms of the Geometric Ratio
a = First Term of the geometric series
r = common ratio of the geometric series
For the series given in the problem statement
a = 3, r = 5
s6=sum of 6 terms
s6=3(5^6–1)/5–1
46874/4
11718.5
Upvote ·
9 1
Related questions
More answers below
What is the sequence of sequence 1, 3, 7, 15?
What is the sixth term of the geometric sequence 24, 16, 32/3?
How do you get the next number in this sequence: 3, 15, 75?
What is the next three terms in this geometric sequence, -5, -15, -75, -225?
What is the 6th term of the pattern 3, 3, 6, 9, 15?
Murali Krishna
Former Retired Senior Lecturer at GOVT DIET VOMRAVALLI (1989–2004) · Author has 6.5K answers and 6.7M answer views
·3y
GP
3 15 75
a=3
r is 15/3 =5
T6 = a r^5 =3 ×5^5=3×625×5 =3× 3125 = 9 375
Upvote ·
9 1
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Ella Diebolt Jackson
MA in Art History, University of Kentucky (Graduated 2011) · Author has 113 answers and 101.1K answer views
·7y
Originally Answered: What’s is the 7th term in the following geometric sequence 3, 15, 75? ·
while 3 seems arbitrary, the other two numbers seem to be a result of multiplication of preceding number by 5. So, just multiply 75 by 5, get 375, and so on, until you get to the 7th.
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Aaditya Dhami
7y
Originally Answered: What’s is the 7th term in the following geometric sequence 3, 15, 75? ·
35=15
155=75
755=375
3755=1875
18755=9375
93755=46875
Therefore the seventh term is 46875
Upvote ·
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Dave Palamar
I've written many papers on the topic of Math · Author has 2.3K answers and 2.7M answer views
·7y
Originally Answered: What’s is the 7th term in the following geometric sequence 3, 15, 75? ·
In this sequence we are multiplying by 5 per each step:
3, 15, 75, 375, 1 875, 9 375, 46 875
Upvote ·
9 1
Srinivasan-The Maths Tutor
Answered by
Srinivasan
· Author has 4K answers and 4.9M answer views
·Dec 9, 2021
The 6th term is ar^5=35^5=33125=9375
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Janez Vidmar
Former Retired · Author has 1.7K answers and 388.7K answer views
·3y
Geometric progression - Wikipedia
an=a1(r^(n-1))
r=an/a(n-1) an may be any term except first.
In this case
r=a3a2=75/15=5
a6=a1(r^(n-1))=3(5^(6–1))=3(5^5)=9 375
Upvote ·
Higa Michinao
Family Home Owner (2010–present) · Author has 535 answers and 603.3K answer views
·6y
Originally Answered: What’s is the 7th term in the following geometric sequence 3, 15, 75? ·
The 7th term is 46875.46875.
f(n)=3×5 n−1 f(n)=3×5 n−1
f(1)=3×5 1−1=3×1=3 f(1)=3×5 1−1=3×1=3
f(2)=3×5 2−1=3×5=15 f(2)=3×5 2−1=3×5=15
f(3)=3×5 3−1=3×25=75 f(3)=3×5 3−1=3×25=75
・
・
・
f(7)=3×5 7−1=3×5 6=3×15625=46875 f(7)=3×5 7−1=3×5 6=3×15625=46875
Upvote ·
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views
·1y
Related
What is the 6th term of the sequence of the 2/3, 7/154/15?
Write your three functions all with the same common denominator.
What is the common denominator of 2 3,7 15,4 15 2 3,7 15,4 15
Obviously, 15 is the common denominator, so multiply both numerator and denominator of the first fraction by the same number to convert 2/3 to fifteenths:
2 3×5 5=10 15 2 3×5 5=10 15
You now have three fractions with a common denominator:
10 15,7 15,4 15 10 15,7 15,4 15
What is the common difference?
Hint: Look at the numerators: 10, 7, 4
I bet you can answer your own question, now.
HINT: Your answer will be a negative fraction.
Upvote ·
Be Hustler
BSC in Mathematics&Fitness, Rani Durgawati Vishwavidhalaya (Graduated 2018)
·2y
Related
What is the 7th term in the geometric sequence 48, 24, 12, and 6?
So firstly let's see what is geometric sequence so a sequence is nothing but a taking one place to next place by always multiplying or dividing by the same number. And this number called common ratio.
So here is given geometric sequence is 48 ,24 ,12 ,6 so let's see
If we divide 48 by 2 then the result is 24
and again if we divide 24 by 2 then the result is 12
So we find the common ratio of 2 which is divide in each number for getting next number.
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
And next is
6 ÷ 2 = 3
3 ÷ 2 = 1.5
1.5 ÷ 2 = 7.5
7.5 ÷ 2 = 37.5 that's the 7th term
So answer is 37.5 which is 7th term .
If thi
Continue Reading
So firstly let's see what is geometric sequence so a sequence is nothing but a taking one place to next place by always multiplying or dividing by the same number. And this number called common ratio.
So here is given geometric sequence is 48 ,24 ,12 ,6 so let's see
If we divide 48 by 2 then the result is 24
and again if we divide 24 by 2 then the result is 12
So we find the common ratio of 2 which is divide in each number for getting next number.
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
And next is
6 ÷ 2 = 3
3 ÷ 2 = 1.5
1.5 ÷ 2 = 7.5
7.5 ÷ 2 = 37.5 that's the 7th term
So answer is 37.5 which is 7th term .
If this answer will helps you
Then please upvote .
Upvote ·
9 1
Related questions
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189781 | https://www.academia.edu/11025449/Inequalities_A_Mathematical_Olympiad_Approach | (PDF) Inequalities A Mathematical Olympiad Approach
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Inequalities A Mathematical Olympiad Approach
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Abstract
This book is intended for the Mathematical Olympiad students who wish to prepare for the study of inequalities, a topic now of frequent use at various levels of mathematical competitions. In this volume we present both classic inequalities and the more useful inequalities for confronting and solving optimization problems. An important part of this book deals with geometric inequalities and this fact makes a big difference with respect to most of the books that deal with this topic in the mathematical olympiad.
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Radmila Bulajich Manfrino José Antonio G6émez Ortega Rogelio Valdez Delgado
Now, we will present a geometric and a visual proof of the following inequal-
where we have used twice the statement that P,, is true. Now, by using the continuity of f and taking the limit, we get
The area can be positive or negative, this will depend on whether the triangle XYZ is positively oriented (anticlockwise oriented) or negatively oriented. For a convex function, we have that A > 0 and for a concave function, A < 0, as shown in the following graphs. As an application of this we get the following.
Using Holder’s inequality we obtain The inequality L < R follows by dividing both sides of L¢ < L9-? RP by LY? and taking the p-th root.
Both inequalities are direct consequence of inequality (1.11), as we can see as follows. and a simple inductive argument shows that
where the last inequality follows in the same way as in the previous example. Exercise 1.98. (South Africa, 1995) For a, b, c, d positive real numbers, prove that Exercise 1.99. Let a and 6 be positive real numbers. Prove that
Now applying the inequality between the arithmetic mean and the quadratic mean (see Exercise 1.68), we get The last inequality follows from Exercise 1.26.
Let BC = a. If AD is the altitude of the triangle at A, its length h can be expressed ash = Bq + y, where y measures its difference in comparison with the length of the altitude of the equilateral triangle. We also set d= $—ax ande=$+2, where x can be interpreted as the difference that the projection of A on BC has with respect to the projection of A on BC in an equilateral triangle, which in this case is the midpoint of BC. We obtain
Moreover, the equality holds if and only if = y = 0, that is, when the triangle is equilateral.
Using the cosine law we can deduce that
The equality holds if and only ifa+b=b+c=c-+a, or equivalently, if a=b=c. The previous inequality can be proved using inequality (2.3) as follows:
Set a = AC, b = CE andc = EA. Ptolemy’s inequality (see Exercise 2.11), applied to the quadrilateral ACE F’,, guarantees that AE- FC < FA-CE+AC-EF-. Since EF = FA, we have that c- FC < FA-b+FA-a. Therefore, 2.3 The use of inequalities in the geometry of the triangle Similarly, we can deduce the inequalities
Let M be the midpoint of BC and let Q be the orthogonal projection of J on the radius OD. Then
Proof. Let us use Stewart’s theorem which states!! that if L is a point on the side BC of a triangle ABC and if AL = 1, BL = m, LC =n, then a(l?+mn) = b?m + c?n.
11For a proof see [6, page 96] or [9, page 6]. 12See [6, page 83] or [9, page 10].
Exercise 2.55. Let a, b and c be the lengths of the sides of a triangle. Ifa+b+c = 1,
Exercise 2.59. On every side of a square with sides measuring 1, choose one point. The four points will form a quadrilateral of sides of length a, b, c and d, prove that
The sum of the areas of these parallelograms is cpg + ap. and this is equal to the area of the parallelogram A’ P’P’C’, where A’P’ is parallel to BP and of the same length. The area of A’P’P”C’ is at most b- PB. Moreover, the areas are equal if BP is perpendicular to A’C’ and this happens if and only if P is on BO, where O is the circumcenter of ABC.' Then,
Example 2.7.5. Using the notation of the Erdés-Mordell theorem, prove that As in the previous example, we have that aP.A > bp, + cp. Hence
Example 2.7.6. Using the notation of the Erdés-Mordell theorem, prove that of A,, By, Cy, we can deduce that
Thus, one of the following inequalities will be satisfied:
Since the triangles DEF and DMF are congruent, they have the same circumra- dius; moreover, since X M is the diameter of the circumcircle of triangle DMF, then XM = 2Repg. Similarly, YN = 2Ry and ZP = 2Rc. Thus, the inequality that needs to be proven can be written as lines through B, D, F and perpendicular to FA, BC, DE, respectively, where B ison YZ, D on ZX and F on XY. Observe that MNP and XYZ are similar triangles.
Exercise 2.71. Using the notation of the Erdés-Mordell theorem, prove that Exercise 2.72. Let ABC be a triangle, P be an arbitrary point in the plane and let pa, Pb Y Pe be the distances from P to the sides of a triangle of lengths a, b and c, respectively. If, for example, P and A are on different sides of the segment BC, then p, is negative, and we have a similar situation for the other cases. Prove that
If L, M and WN are the feet of the perpendiculars of P on the sides, it is clear that PM = NM’, where M’ is the intersection of PN with A’B’. Moreover, PM’ is the altitude of the equilateral triangle AP’ P. If A’P” is the altitude of the triangle AP'P from A’, it is clear that PM’ = A’P”. Let L’ be a point on B’C" such that A'L’ is the altitude of the triangle A’B’C’ from A’. Thus,
These perpendiculars determine a triangle DEF which is equilateral. This is so because the quadrilateral PBDC is cyclic, having angles of 90° in B and C. Now, since 7BPC' = 120°, we can deduce that 72BDC = 60°. This argument can be repeated for each angle. Therefore DEF is indeed equilateral.
Now let us go back to Steiner’s solution of the Fermat-Steiner problem. A point P that makes the sum PA+PB+PC a minimum can be one of the vertices A, B, C or a point of the triangle different from the vertices. In the first case, if P is one of the vertices, then one term of the sum PA+ PB + PC is zero and the other two are the lengths of the sides of the triangle ABC that have in common the chosen vertex. Hence, the sum will be minimum when the chosen vertex is opposite to the longest side of the triangle.
Proof. It is sufficient to see that the first two triangles are similar, since the othe cases are proved in a similar way. Schwarz’s solution. The German mathematician Hermann Schwarz provided the following solution to this problem for which he took as starting point two ob- servations that we present as lemmas. These lemmas will demonstrate that the inscribed triangle with the minimum perimeter is the triangle formed using the feet of the altitudes of the triangle. Such a triangle is known as the ortic triangle.
Proof. It follows directly from the previous lemma.
Reflect the complete figure on the side BC’, so that the resultant triangle is re- flected on C'A, then on AB, on BC and finally on CA.
Thus, we can conclude that if we fix the point L, the points M and N that make the minimum perimeter LM WN are the intersections of L’L” with CA and AB, respectively. Now, let us see which is the best option for the point DL. We already know that the perimeter of DMN is L’L”, thus L should make this quantity a minimum.
16A function f(a, b,...) is homogeneous if f(ta, tb,...) = tf(a,b,...) for each t € R. Then, ar inequality of the form f(a,b,...) > 0, in the case of a homogeneous function, is equivalent t« f (ta, tb,...) > 0 for any t > 0.
Solution 1.52. The inequality is equivalent to (+2) (4¢) (442) > 8. Now, we use the AM-GM inequality for each term of the product and the inequality follows immediately. Solution 1.53. Notice that
Solution 1.68. Apply Example 1.4.11.
Solution 1.75. Apply the rearrangement inequality to
is convex in each variable, therefore its maximum is attained at the endpoints. Second solution. The AM-GM inequality and the fact that abc = 1 imply that
4.1 Solutions to the exercises in Chapter 1 Solution 1.115. Observe that where Muirhead’s theorem has been used.
[
he last inequality can be written in the terminology of Muirhead’s theorem as 4[4, 0, 0] + 4[3, 0, 0] > [0, 0, 0] + 3[1, 0, 0] + 3[1, 1,0] + [1,1, 1). Now, note that
Solution 2.9. Construct a parallelogram ABDC, with one diagonal BC and the 17See [6, page 136] or [1, page 128]. 18See [6, page 97] or [9, page 13].
Second solution. Apply Ptolemy’s inequality (see Exercise 2.11) to the quadrilat- erals ABCP, ABPC and APBC; after cancellation of common terms we obtain that PB < PC + PA, PA < PC + PB and PC < PA+ PB, respectively, which establish the existence of the triangle.
For the last identity, see the end of the proof of Example 2.2.4. Solution 2.28. The part (i) follows from the following equivalences:
Solution 2.59. If 21, 1 — 21, 2, 1 — xg, ... are the lengths into which each side is divided for the corresponding point, we can deduce that a? + 6? + c? + d? = >> (a? + (1 — 2;)?). Prove that $ < 2(a; — $)?+$ =a? 4+ (1-2)? <1. For part (ii), the inequality on the right-hand side follows from the triangle inequality. For the one on the left-hand side, use reflections on the sides, as you can see in the figure. Solution 2.60. This is similar to part (ii) of the previous problem.
the last two inequalities follow from the fact that R > 2r (which implies that —r> =) and from s > 3V3r, respectively.
a < min {c,d}.
Since BQOP is cyclic (ZB = ZO = 90°), it follows that ZQBO = ZQPO = 45°, then O belongs to the internal bisector of 7B. Let T be the intersection of BO with RS, then ZQBT = ZQST = 45°, therefore BQT'S is cyclic and the center O' of the circumcircle of BQT'S is the intersection of the perpendicular bisectors of SQ and BT. But the perpendicular bisector of SQ is PR, hence the point O’ belongs to PR, and if V is the midpoint of BT, we have that VOO’ is a right triangle. Since O’O > O’V, then the chords SQ and BT satisfy SQ < BT, and the lemma follows. Through the vertices of the square draw parallel lines to the sides of the rectangle in such a way that those lines enclose the square as in the figure. Since the parallel lines form a square inside the rectangle and such a square contains the original square, we have the result.
4.2 Solutions to the exercises in Chapter 2
and N coincide with B and C, respectively. Therefore the maximum chord MN is BC. Solution 2.79. The quadrilateral APM N is cyclic and it is inscribed in the circle of diameter AP. The chord MN always opens the angle A (or 180° — ZA), therefore the length of MN will depend proportionally on the radius of the circumscribed circle to APM N. The biggest circle will be attained when the diameter AP is the biggest possible. This happens when P is diametrally opposed to A. In this case IV
The proof follows from the fact that the left sides of the equalities have the same sign.
Solution 3.45.
In the first two inequalities we applied inequality (1.11), and in the last inequality we used the AM-GM inequality.
Second solution. Using the Cauchy-Schwarz inequality we get econd solution. Use inequality (1.11) and the Cauchy-Schwarz inequality Solution 3.75. Applying the AM-GM inequality to each denominator, one obtains
The last inequality follows after using the AM-GM inequality for six numbers.
Similarly for the other two elements of the sum; then Then, it is enough to prove that
which in turn is equivalent to the inequality
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189782 | http://cc.kangwon.ac.kr/~kimoon/me/me-132/Nicholson-ch01.pdf | C H A P T E R 1 Economic Models The main goal of this book is to introduce you to the most important models that economists use to explain the behavior of consumers, firms, and markets. These models are central to the study of all areas of economics.
Therefore, it is essential to understand both the need for such models and the basic framework used to develop them. The goal of this chapter is to begin this process by outlining some of the conceptual issues that determine the ways in which economists study practically every question that interests them.
THEORETICAL MODELS A modern economy is a complicated entity. Thousands of firms engage in producing millions of different goods. Many millions of people work in all sorts of occupations and make decisions about which of these goods to buy. Let’s use peanuts as an example. Peanuts must be harvested at the right time and shipped to processors who turn them into peanut butter, peanut oil, peanut brittle, and numerous other peanut delicacies. These processors, in turn, must make certain that their products arrive at thousands of retail outlets in the proper quantities to meet demand.
Because it would be impossible to describe the features of even these peanut markets in complete detail, economists have chosen to abstract from the complexities of the real world and develop rather simple models that capture the “essentials.” Just as a road map is helpful even though it does not record every house or every store, economic models of, say, the market for peanuts are also useful even though they do not record every minute feature of the peanut economy. In this book we will study the most widely used economic models. We will see that, even though these models often make heroic abstractions from the complexities of the real world, they nonetheless capture essential features that are common to all economic activities.
The use of models is widespread in the physical and social sciences. In physics, the notion of a “perfect” vacuum or an “ideal” gas is an abstraction that permits scientists to study real-world phenomena in simplified settings. In chemistry, the idea of an atom or a molecule is actually a simplified model of the structure of matter. Architects use mock-up models to plan buildings.
Television repairers refer to wiring diagrams to locate problems. Economists’ models perform similar functions. They provide simplified portraits of the way individuals make decisions, the way firms behave, and the way in which these two groups interact to establish markets.
VERIFICATION OF ECONOMIC MODELS Of course, not all models prove to be “good.” For example, the earth-centered model of planetary motion devised by Ptolemy was eventually discarded because it proved incapable of accurately explaining how the planets move around the sun. An important purpose of scientific investigation is to sort out the “bad” models from the “good.” Two general methods have 3 Therefore, it is essential to understand both the need for such models and the basic framework used to develop them.
A modern economy is a complicated entity.
economists have chosen to abstract from the complexities of the real world and develop rather simple models that capture the essentials.” “e even though these models often make heroic abstractions from the complexities of the real world, they nonetheless capture essential features that are common to all economic activities.
Two general methods have been used for verifying economic models: (1) a direct approach, which seeks to establish the validity of the basic assumptions on which a model is based; and (2) an indirect approach, which attempts to confirm validity by showing that a simplified model correctly predicts real-world events. To illustrate the basic differences between the two approaches, let’s briefly examine a model that we will use extensively in later chapters of this book—the model of a firm that seeks to maximize profits.
The profit-maximization model The model of a firm seeking to maximize profits is obviously a simplification of reality. It ignores the personal motivations of the firm’s managers and does not consider conflicts among them. It assumes that profits are the only relevant goal of the firm; other possible goals, such as obtaining power or prestige, are treated as unimportant. The model also assumes that the firm has sufficient information about its costs and the nature of the market to which it sells to discover its profit-maximizing options. Most real-world firms, of course, do not have this information readily available. Yet, such shortcomings in the model are not necessarily serious.
No model can exactly describe reality. The real question is whether this simple model has any claim to being a good one.
Testing assumptions One test of the model of a profit-maximizing firm investigates its basic assumption: Do firms really seek maximum profits? Some economists have examined this question by sending ques-tionnaires to executives, asking them to specify the goals they pursue. The results of such studies have been varied. Businesspeople often mention goals other than profits or claim they only do “the best they can” to increase profits given their limited information. On the other hand, most respondents also mention a strong “interest” in profits and express the view that profit maximization is an appropriate goal. Testing the profit-maximizing model by testing its assumptions has therefore provided inconclusive results.
Testing predictions Some economists, most notably Milton Friedman, deny that amodel can be tested by inquiring into the “reality” of its assumptions.1 They argue that all theoretical models are based on “unrealistic” assumptions; the very nature of theorizing demands that we make certain ab-stractions. These economists conclude that the only way to determine the validity of a model is to see whether it is capable of predicting and explaining real-world events. The ultimate test of an economic model comes when it is confronted with data from the economy itself.
Friedman provides an important illustration of that principle. He asks what kind of a theory one should use to explain the shots expert pool players will make. He argues that the laws of velocity, momentum, and angles from theoretical physics would be a suitable model. Pool players shoot shots as if they follow these laws. But most players asked whether they precisely understand the physical principles behind the game of pool will undoubtedly answer that they do not. Nonetheless, Friedman argues, the physical laws provide very accurate predictions and therefore should be accepted as appropriate theoretical models of how experts play pool.
A test of the profit-maximization model, then, would be provided by predicting the behavior of real-world firms by assuming that these firms behave as if they were maximizing profits. (See Example 1.1 later in this chapter.) If these predictions are reasonably in accord with reality, we may accept the profit-maximization hypothesis. However, we would reject 1See M. Friedman, Essays in Positive Economics (Chicago: University of Chicago Press, 1953), chap. 1. For an alternative view stressing the importance of using “realistic” assumptions, see H. A. Simon, “Rational Decision Making in Business Organizations,” American Economic Review 69, no. 4 (September 1979): 493–513.
4 Part 1 Introduction been used for verifying economic models: (1) a direct approach, which seeks to establish the validity of the basic assumptions on which a model is based; d (2) an indirect approach, which attempts to confirm validity by showing that a simplified model correctly predicts real-world events. T the model if real-world data seem inconsistent with it. Hence, the ultimate test of either theory is its ability to predict real-world events.
Importance of empirical analysis The primary concern of this book is the construction of theoretical models. But the goal of such models is always to learn something about the real world. Although the inclusion of a lengthy set of applied examples would needlessly expand an already bulky book,2 the Ex-tensions included at the end of many chapters are intended to provide a transition between the theory presented here and the ways in which that theory is actually applied in empirical studies.
GENERAL FEATURES OF ECONOMIC MODELS The number of economic models in current use is, of course, very large. Specific assumptions used and the degree of detail provided vary greatly depending on the problem being addressed.
The models employed to explain the overall level of economic activity in the United States, for example, must be considerably more aggregated and complex than those that seek to interpret the pricing of Arizona strawberries. Despite this variety, however, practically all economic models incorporate three common elements: (1) the ceteris paribus (other things the same) assumption; (2) the supposition that economic decision makers seek to optimize something; and (3) a careful distinction between “positive” and “normative” questions. Because we will encounter these elements throughout this book, it may be helpful at the outset to briefly describe the philosophy behind each of them.
The ceteris paribus assumption As in most sciences, models used in economics attempt to portray relatively simple rela-tionships. A model of the market for wheat, for example, might seek to explain wheat prices with a small number of quantifiable variables, such as wages of farmworkers, rainfall, and consumer incomes. This parsimony in model specification permits the study of wheat pricing in a simplified setting in which it is possible to understand how the specific forces operate.
Although any researcher will recognize that many “outside” forces (presence of wheat diseases, changes in the prices of fertilizers or of tractors, or shifts in consumer attitudes about eating bread) affect the price of wheat, these other forces are held constant in the construction of the model. It is important to recognize that economists are not assuming that other factors do not affect wheat prices; rather, such other variables are assumed to be unchanged during the period of study. In this way, the effect of only a few forces can be studied in a simplified setting. Such ceteris paribus (other things equal) assumptions are used in all economic modeling.
Use of the ceteris paribus assumption does pose some difficulties for the verification of economic models from real-world data. In other sciences, such problems may not be so severe because of the ability to conduct controlled experiments. For example, a physicist who wishes to test a model of the force of gravity probably would not do so by dropping objects from the Empire State Building. Experiments conducted in that way would be subject to too many extraneous forces (wind currents, particles in the air, variations in temperature, and so forth) to permit a precise test of the theory. Rather, the physicist would conduct experiments in a laboratory, using a partial vacuum in which most other forces could be controlled or elim-inated. In this way, the theory could be verified in a simple setting, without considering all the other forces that affect falling bodies in the real world.
2For an intermediate-level text containing an extensive set of real-world applications, see W. Nicholson and C. Snyder, Intermediate Microeconomics and Its Application, 10th ed. (Mason, OH: Thomson/Southwestern, 2007).
Chapter 1 Economic Models 5 With a few notable exceptions, economists have not been able to conduct controlled experiments to test their models. Instead, economists have been forced to rely on various statistical methods to control for other forces when testing their theories. Although these statistical methods are as valid in principle as the controlled experiment methods used by other scientists, in practice they raise a number of thorny issues. For that reason, the limitations and precise meaning of the ceteris paribus assumption in economics are subject to greater con-troversy than in the laboratory sciences.
Optimization assumptions Many economic models start from the assumption that the economic actors being studied are rationally pursuing some goal. We briefly discussed such an assumption when investigating the notion of firms maximizing profits. Example 1.1 shows how that model can be used to make testable predictions. Other examples we will encounter in this book include consumers maxi-mizing their own well-being (utility), firms minimizing costs, and government regulators attempting to maximize public welfare. Although, as we will show, all of these assumptions are unrealistic, all have won widespread acceptance as good starting places for developing economic models. There seem to be two reasons for this acceptance. First, the optimization assumptions are very useful for generating precise, solvable models, primarily because such models can draw on a variety of mathematical techniques suitable for optimization problems. Many of these techniques, together with the logic behind them, are reviewed in Chapter 2. A second reason for the popularity of optimization models concerns their apparent empirical validity. As some of our Extensions show, such models seem to be fairly good at explaining reality. In all, then, opti-mization models have come to occupy a prominent position in modern economic theory.
EXAMPLE 1.1 Profit Maximization The profit-maximization hypothesis provides a good illustration of how optimization as-sumptions can be used to generate empirically testable propositions about economic behavior. Suppose that a firm can sell all the output that it wishes at a price of p per unit and that the total costs of production, C, depend on the amount produced, q. Then, profits are given by profits ¼ π ¼ pq CðqÞ: (1:1) Maximization of profits consists of finding that value of q which maximizes the profit ex-pression in Equation 1.1. This is a simple problem in calculus. Differentiation of Equation 1.1 and setting that derivative equal to 0 give the following first-order condition for a maximum: dπ dq ¼ p C0ðqÞ ¼ 0 or p ¼ C 0ðqÞ: (1:2) In words, the profit-maximizing output level (q) is found by selecting that output level for which price is equal to marginal cost, C0ðqÞ. This result should be familiar to you from your introductory economics course. Notice that in this derivation the price for the firm’s output is treated as a constant because the firm is a price taker.
Equation 1.2 is only the first-order condition for a maximum. Taking account of the second-order condition can help us to derive a testable implication of this model. The second-order condition for a maximum is that at q it must be the case that d2π dq2 ¼ C00ðqÞ < 0 or C00ðqÞ > 0: (1:3) 6 Part 1 Introduction That is, marginal cost must be increasing at q for this to be a true point of maximum profits.
Our model cannow be used to “predict”how afirm will react to a change inprice. To do so, we differentiate Equation 1.2 with respect to price (p), assuming that the firm continues to choose a profit-maximizing level of q: d½p C0ðqÞ ¼ 0 dp ¼ 1 C 00ðqÞ dq dp ¼ 0: (1:4) Rearranging terms a bit gives dq dp ¼ 1 C00ðqÞ > 0: (1:5) Here the final inequality again reflects the fact that marginal cost must be increasing if q is to be a true maximum. This then is one of the testable propositions of the profit-maximization hypothesis—if other things do not change, a price-taking firm should respond to an increase in price by increasing output. On the other hand, if firms respond to increases in price by reducing output, there must be something wrong with our model.
Although this is a very simple model, it reflects the way we will proceed throughout much of this book. Specifically, the fact that the primary implication of the model is derived by calculus, and consists of showing what sign a derivative should have, is the kind of result we will see many times.
QUERY: In general terms, how would the implications of this model be changed if the price a firm obtains for its output were a function of how much it sold? That is, how would the model work if the price-taking assumption were abandoned?
Positive-normative distinction A final feature of most economic models is the attempt to differentiate carefully between “positive” and “normative” questions. So far we have been concerned primarily with positive economic theories. Such theories take the real world as an object to be studied, attempting to explain those economic phenomena that are observed. Positive economics seeks to determine how resources are in fact allocated in an economy. A somewhat different use of economic theory is normative analysis, taking a definite stance about what should be done. Under the heading of normative analysis, economists have a great deal to say about how resources should be allocated. For example, an economist engaged in positive analysis might investigate how prices are determined in the U.S. health-care economy. The economist also might want to measure the costs and benefits of devoting even more resources to health care. But when he or she specifically advocates that more resources should be allocated to health care, the analysis becomes normative.
Some economists believe that the only proper economic analysis is positive analysis.
Drawing an analogy with the physical sciences, they argue that “scientific” economics should concern itself only with the description (and possibly prediction) of real-world economic events. To take moral positions and to plead for special interests are considered to be outside the competence of an economist acting as such. Other economists, however, believe strict application of the positive-normative distinction to economic matters is inappropriate. They believe that the study of economics necessarily involves the researchers’ own views about ethics, morality, and fairness. According to these economists, searching for scientific “objectivity” in such circumstances is hopeless. Despite some ambiguity, this book adopts a mainly positivist tone, leaving normative concerns for you to decide for yourself.
Chapter 1 Economic Models 7 DEVELOPMENT OF THE ECONOMIC THEORY OF VALUE Because economic activity has been a central feature of all societies, it is surprising that these activities were not studied in any detail until recently. For the most part, economic phenomena were treated as a basic aspect of human behavior that was not sufficiently interesting to deserve specific attention. It is, of course, true that individuals have always studied economic activities with a view toward making some kind of personal gain. Roman traders were not above making profits on their transactions. But investigations into the basic nature of these activities did not begin in any depth until the eighteenth century.3 Because this book is about economic theory as it stands today, rather than the history of economic thought, our discussion of the evolution of economic theory will be brief. Only one area of economic study will be examined in its historical setting: the theory of value.
Early economic thoughts on value The theory of value, not surprisingly, concerns the determinants of the “value” of a commodity.
This subject is at the center of modern microeconomic theory and is closely intertwined with the fundamental economic problem of allocating scarce resources to alternative uses. The logical place tostart iswithadefinitionofthe word “value.” Unfortunately, the meaning ofthistermhas not been consistent throughout the development of the subject. Today we regard value as being synonymous with the price of a commodity.4 Earlier philosopher-economists, however, made a distinction between the market price of a commodity and its value. The term “value” was then thought of as being, in some sense, synonymous with “importance,” “essentiality,” or (at times) “godliness.” Because “price” and “value” were separate concepts, they could differ, and most early economic discussions centered on these divergences. For example, St. Thomas Aquinas believed value to be divinely determined. Since prices were set by humans, it was possible for the price of a commodity to differ from its value. A person accused of charging a price in excess of a good’s value was guilty of charging an “unjust” price. For example, St. Thomas believed the “just” rate of interest to be zero. Any lender who demanded a payment for the use of money was charging an unjust price and could be—and sometimes was—prosecuted by church officials.
The founding of modern economics During the latter part of the eighteenth century, philosophers began to take a more scientific approach to economic questions. The 1776 publication of The Wealth of Nations by Adam Smith (1723–1790) is generally considered the beginning of modern economics. In his vast, all-encompassing work, Smith laid the foundation for thinking about market forces in an ordered and systematic way. Still, Smith and his immediate successors, such as David Ricardo (1772–1823), continued to distinguish between value and price. To Smith, for example, the value of a commodity meant its “value in use,” whereas the price represented its “value in exchange.” The distinction between these two concepts was illustrated by the famous water-diamond paradox. Water, which obviously has great value in use, has little value in exchange (it has a low price); diamonds are of little practical use but have a great value in exchange. The paradox with which early economists struggled derives from the observation that some very useful items have low prices whereas certain nonessential items have high prices.
3For a detailed treatment of early economic thought, see the classic work by J. A. Schumpeter, History of Economic Analysis (New York: Oxford University Press, 1954), pt. II, chaps. 1–3.
4This is not completely true when “externalities” are involved and a distinction must be made between private and social value (see Chapter 19).
8 Part 1 Introduction Labor theory of exchange value Neither Smith nor Ricardo ever satisfactorily resolved the water-diamond paradox. The con-cept of value in use was left for philosophers to debate, while economists turned their attention to explaining the determinants of value in exchange (that is, to explaining relative prices). One obvious possible explanation is that exchange valuesof goodsare determined by what it coststo produce them. Costs of production are primarily influenced by labor costs—at least this was so in the time of Smith and Ricardo—and therefore it was a short step to embrace a labor theory of value. For example, to paraphrase an example from Smith, if catching a deer takes twice the number of labor hours as catching a beaver, then one deer should exchange for two beavers. In other words, the price of a deer should be twice that of a beaver. Similarly, diamonds are relatively costly because their production requires substantial labor input.
To students with even a passing knowledge of what we now call the law of supply and demand, Smith’s and Ricardo’s explanation must seem incomplete. Didn’t they recognize the effects of demand on price? The answer to this question is both yes and no. They did observe periods of rapidly rising and falling relative prices and attributed such changes to demand shifts.
However, they regarded these changes as abnormalities that produced only a temporary divergence of market price from labor value. Because they had not really developed a theory of value in use, they were unwilling to assign demand any more than a transient role in deter-mining relative prices. Rather, long-run exchange values were assumed to be determined solely by labor costs of production.
The marginalist revolution Between 1850 and 1880, economists became increasingly aware that to construct an adequate alternative to the labor theory of value, they had to come to devise a theory of value in use.
During the 1870s, several economists discovered that it is not the total usefulness of a commodity that helps to determine its exchange value, but rather the usefulness of the last unit consumed. For example, water is certainly very useful—it is necessary for all life. But, because water is relatively plentiful, consuming one more pint (ceteris paribus) has a relatively low value to people. These “marginalists” redefined the concept of value in use from an idea of overall usefulness to one of marginal, or incremental, usefulness—the usefulness of an additional unit of a commodity. The concept of the demand for an incremental unit of output was now contrasted to Smith’s and Ricardo’s analysis of production costs to derive a comprehensive picture of price determination.5 Marshallian supply-demand synthesis The clearest statement of these marginal principles was presented by the English economist Alfred Marshall (1842–1924) in his Principles of Economics, published in 1890. Marshall showed that demand and supply simultaneously operate to determine price. As Marshall noted, just as you cannot tell which blade of a scissors does the cutting, so too you cannot say that either demand or supply alone determines price. That analysis is illustrated by the famous Marshallian cross shown in Figure 1.1. In the diagram the quantity of a good purchased per period is shown on the horizontal axis and its price appears on the vertical axis. The curve DD represents the quantity of the good demanded per period at each possible price. The curve is negatively sloped to reflect the marginalist principle that as quantity increases, people are 5Ricardo had earlier provided an important first step in marginal analysis in his discussion of rent. Ricardo theorized that as the production of corn increased, land of inferior quality would be used and this would cause the price of corn to rise. In his argument Ricardo implicitly recognized that it is the marginal cost—the cost of producing an additional unit—that is relevant to pricing. Notice that Ricardo implicitly held other inputs constant when discussing diminishing land productivity; that is, he employed one version of the ceteris paribus assumption.
Chapter 1 Economic Models 9 willing to pay less for the last unit purchased. It is the value of this last unit that sets the price for all units purchased. The curve SS shows how (marginal) production costs rise as more output is produced. This reflects the increasing cost of producing one more unit as total output expands.
In other words, the upward slope of the SS curve reflects increasing marginal costs, just as the downward slope of the DD curve reflects decreasing marginalvalue. The two curves intersect at p, q. This is an equilibrium point—both buyers and sellers are content with the quantity being traded and the price at which it is traded. If one of the curves should shift, the equilibrium point would shift to a new location. Thus price and quantity are simultaneously determined by the joint operation of supply and demand.
Paradox resolved Marshall’s model resolves the water-diamond paradox. Prices reflect both the marginal eval-uation that demanders place on goods and the marginal costs of producing the goods. Viewed in this way, there is no paradox. Water is low in price because it has both a low marginal value and a low marginal cost of production. On the other hand, diamonds are high in price because they have both a high marginal value (because people are willing to pay quite a bit for one more) and a high marginal cost of production. This basic model of supply and demand lies behind much of the analysis presented in this book.
General equilibrium models Although the Marshallian model is an extremely useful and versatile tool, it is a partial equilibrium model, looking at only one market at a time. For some questions, this narrowing of perspective gives valuable insights and analytical simplicity. For other, broader questions, such a narrow viewpoint may prevent the discovery of important relationships among markets. To answer more general questions we must have a model of the whole economy that suitably mirrors the connections among various markets and economic agents. The French economist Leon Walras (1831–1910), building on a long Continental tradition in such analysis, created the basis for modern investigations into those broad questions. His method of representing the FIGURE 1.1 The Marshallian Supply-Demand Cross Marshall theorized that demand and supply interact to determine the equilibrium price (p) and the quantity (q) that will be traded in the market. He concluded that it is not possible to say that either demand or supply alone determines price or therefore that either costs or usefulness to buyers alone determines exchange value.
Quantity per period Price S S D D q p 10 Part 1 Introduction economy by a large number of simultaneous equations forms the basis for understanding the interrelationships implicit in general equilibrium analysis. Walras recognized that one cannot talk about a single market in isolation; what is needed is a model that permits the effects of a change in one market to be followed through other markets.
EXAMPLE 1.2 Supply-Demand Equilibrium Although graphical presentations are adequate for some purposes, economists often use algebraic representations of their models to both clarify their arguments and make them more precise. As an elementary example, suppose we wished to study the market for peanuts and, on the basis of statistical analysis of historical data, concluded that the quantity of peanuts demanded each week (q, measured in bushels) depended on the price of peanuts (p, measured in dollars per bushel) according to the equation quantity demanded ¼ qD ¼ 1,000 100p: (1:6) Because this equation for qD contains only the single independent variable p, we are implicitly holding constant all other factors that might affect the demand for peanuts. Equation 1.6 indicates that, if other things do not change, at a price of $5 per bushel people will demand 500 bushels of peanuts, whereas at a price of $4 per bushel they will demand 600 bushels. The negative coefficient for p in Equation 1.6 reflects the marginalist principle that a lower price will cause people to buy more peanuts.
To complete this simple model of pricing, suppose that the quantity of peanuts supplied also depends on price: quantity supplied ¼ qS ¼ 125 þ 125p: (1:7) Here the positive coefficient of price also reflects the marginal principle that a higher price will call forth increased supply—primarily because (as we saw in Example 1.1) it permits firms to incur higher marginal costs of production without incurring losses on the additional units produced.
Equilibrium price determination. Equation 1.6 and 1.7 therefore reflect our model of price determination in the market for peanuts. An equilibrium price can be found by setting quantity demanded equal to quantity supplied: qD ¼ qS (1:8) or 1,000 100p ¼ 125 þ 125p (1:9) or 225p ¼ 1,125, (1:10) so p ¼ 5: (1:11) At a price of $5 per bushel, this market is in equilibrium: at this price people want to purchase 500 bushels, and that is exactly what peanut producers are willing to supply. This equilibrium is pictured graphically as the intersection of D and S in Figure 1.2.
A more general model. In order to illustrate how this supply-demand model might be used, let’s adopt a more general notation. Suppose now that the demand and supply functions are given by (continued) Chapter 1 Economic Models 11 EXAMPLE 1.2 CONTINUED FIGURE 1.2 Changing Supply-Demand Equilibria The initial supply-demand equilibrium is illustrated by the intersection of D and S (p ¼ 5, q ¼ 500).
WhendemandshiftstoqD 0 ¼ 1,450 100p (denotedasD0),theequilibriumshiftstop ¼ 7,q ¼ 750.
0 Quantity per period (bushels) Price ($) S S D′ D′ D D 14.5 10 7 5 500 750 1000 1450 qD ¼ a þ bp and qS ¼ c þ dp (1:12) where a and c are constants that can be used to shift the demand and supply curves, respectively, and b (<0) and d (>0) represent demanders’ and suppliers’ reactions to price.
Equilibrium in this market requires qD ¼ qS or a þ bp ¼ c þ dp: (1:13) So, equilibrium price is given by6 p ¼ a c d b : (1:14) 6Equation 1.14 is sometimes called the “reduced form” for the supply-demand structural model of Equations 1.12 and 1.13. It shows that the equilibrium value for the endogenous variable p ultimately depends only on the exogenous factors in the model (a and c) and on the behavioral parameters b and d. A similar equation can be calculated for equilibrium quantity.
12 Part 1 Introduction Notice that, in our prior example, a ¼ 1,000, b ¼ 100, c ¼ 125, and d ¼ 125, so p ¼ 1,000 þ 125 125 þ 100 ¼ 1,125 225 ¼ 5: (1:15) With this more general formulation, however, we can pose questions about how the equi-librium price might change if either the demand or supply curve shifted. For example, differentiation of Equation 1.14 shows that dp da ¼ 1 d b > 0, dp dc ¼ 1 d b < 0: (1:16) That is, an increase in demand (an increase in a) increases equilibrium price whereas an in-crease in supply (an increase in c) reduces price. This is exactly what a graphical analysis of supply and demand curves would show. For example, Figure 1.2 shows that when the con-stant term, a, in the demand equation increases to 1450, equilibrium price increases to p ¼ 7 ½¼ ð1,450 þ 125Þ=225.
QUERY: How might you use Equation 1.16 to “predict” how each unit increase in the constant a affects p? Does this equation correctly predict the increase in p when the constant a increases from 1,000 to 1,450?
For example, suppose that the demand for peanuts were to increase. This would cause the priceofpeanutstoincrease.Marshalliananalysiswouldseektounderstandthesizeofthisincrease by looking at conditions of supply and demand in the peanut market. General equilibrium analysiswouldlooknotonlyatthatmarketbutalsoatrepercussionsinothermarkets.Ariseinthe price of peanuts would increase costs for peanut butter makers, which would, in turn, affect the supply curve for peanut butter. Similarly, the rising price of peanuts might mean higher land prices for peanut farmers, which would affect the demand curves for all products that they buy.
The demand curves for automobiles, furniture, and trips to Europe would all shift out, and that might create additional incomes for the providers of those products. Consequently, the effects of the initial increase in demand for peanuts eventually would spread throughout the economy.
Generalequilibriumanalysisattemptstodevelopmodelsthatpermitustoexaminesucheffectsin a simplified setting. Several models of this type are described in Chapter 13.
Production possibility frontier Here we briefly introduce some general equilibrium ideas by using another graph you should remember from introductory economics—the production possibility frontier. This graph shows the various amounts of two goods that an economy can produce using its available resources during some period (say, one week). Because the production possibility frontier shows two goods, rather than the single good in Marshall’s model, it is used as a basic building block for general equilibrium models.
Figure 1.3 shows the production possibility frontier for two goods, food and clothing.
The graph illustrates the supply of these goods by showing the combinations that can be produced with this economy’s resources. For example, 10 pounds of food and 3 units of clothing could be produced, or 4 pounds of food and 12 units of clothing. Many other combinations of food and clothing could also be produced. The production possibility frontier shows all of them. Combinations of food and clothing outside the frontier cannot be produced because not enough resources are available. The production possibility frontier Chapter 1 Economic Models 13 reminds us of the basic economic fact that resources are scarce—there are not enough resources available to produce all we might want of every good.
This scarcity means that we must choose how much of each good to produce. Figure 1.3 makes clear that each choice has its costs. For example, if this economy produces 10 pounds of food and 3 units of clothing at point A, producing 1 more unit of clothing would “cost” 1 2 pound of food—increasing the output of clothing by 1 unit means the production of food would have to decrease by 1 2 pound. So, the opportunity cost of 1 unit of clothing at point A is 1 2 pound of food.
On the other hand, if the economy initially produces 4 pounds of food and 12 units of clothing at point B, it would cost 2 pounds of food to produce 1 more unit of clothing. The opportunity cost of 1 more unit of clothing at point B has increased to 2 pounds of food. Because more units of clothing are produced at point B than at point A, both Ricardo’s and Marshall’s ideas of increasing incremental costs suggest that the opportunity cost of an additional unit of clothing will be higher at point B than at point A. This effect is shown by Figure 1.3.
The production possibility frontier provides two general equilibrium insights that are not clear in Marshall’s supply and demand model of a single market. First, the graph shows that producing moreof one goodmeans producing less ofanothergood becauseresourcesare scarce.
Economists often (perhaps too often!) use the expression “there is no such thing as a free lunch” to explain that every economic action has opportunity costs. Second, the production possibility frontier shows that opportunity costs depend on how much of each good is produced. The frontier is like a supply curve for two goods: it shows the opportunity cost of producing more of one goodasthedecrease intheamountof thesecond good.Theproductionpossibility frontieris therefore a particularly useful tool for studying several markets at the same time.
FIGURE 1.3 Production Possibility Frontier The production possibility frontier shows the different combinations of two goods that can be produced from a certain amount of scarce resources. It also shows the opportunity cost of producing more of one good as the amount of the other good that cannot then be produced. The opportunity cost at two different levels of clothing production can be seen by comparing points A and B.
Quantity of food per week B A 0 2 4 9.5 10 3 4 12 13 Quantity of clothing per week Opportunity cost of clothing = 2 pounds of food Opportunity cost of clothing = pound of food 1 2 14 Part 1 Introduction EXAMPLE 1.3 The Production Possibility Frontier and Economic Inefficiency General equilibrium models are good tools for evaluating the efficiency of various economic arrangements. As we will see in Chapter 13, such models have been used to assess a wide variety of policies such as trade agreements, tax structures, and environmental regulation. In this simple example, we explore the idea of efficiency in its most elementary form.
Suppose that an economy produces two goods, x and y, using labor as the only input.The production function for good x is x ¼ l0:5 x (where lx is the quantity of labor used in x production) and the production function for good y is y ¼ 2l0:5 y . Total labor available is constrained by lx þ ly 200. Construction of the production possibility frontier in this economy is extremely simple: lx þ ly ¼ x2 þ 0:25y2 200 (1:17) if the economy is to be producing as much as possible (which, after all, is why it’s called a “frontier”). Equation 1.17 shows that the frontier here has the shape of a quarter ellipse—its concavity derives from the diminishing returns exhibited by each production function.
Opportunity cost. Assuming this economy is on the frontier, the opportunity cost of good y in terms of good x can be derived by solving for y as y2 ¼ 800 4x2 or y ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 800 4x2 p ¼ ½800 4x20:5 (1:18) and then differentiating this expression: dy dx ¼ 0:5½800 4x20:5ð8xÞ ¼ 4x y : (1:19) Suppose, for example, labor is equally allocated between the two goods. Then x ¼ 10, y ¼ 20, and dy=dx ¼ 4ð10Þ=20 ¼ 2. With this allocation of labor, each unit increase in x output would require a reduction in y of 2 units. This can be verified by considering a slightly different allocation, lx ¼ 101 and ly ¼ 99. Now production is x ¼ 10:05 and y ¼ 19:9. Moving to this alternative allocation would have Dy Dx ¼ ð19:9 20Þ ð10:05 10Þ ¼ 0:1 0:05 ¼ 2, which is precisely what was derived from the calculus approach.
Concavity. Equation 1.19 clearly illustrates the concavity of the production possibility frontier.
The slope of the frontier becomes steeper (more negative) as x output increases and y output falls.
For example, if labor is allocated so that lx ¼ 144 and ly ¼ 56, then outputs are x ¼ 12 and y 15 and so dy=dx ¼ 4ð12Þ=15 ¼ 3:2. With expanded x production, the opportunity cost of one more unit of x increases from 2 to 3.2 units of y.
Inefficiency. If an economy operates inside its production possibility frontier, it is operating inefficiently. Moving outward to the frontier could increase the output of both goods. In this book we will explore many reasons for such inefficiency. These usually derive from a failure of some market to perform correctly. For the purposes of this illustration, let’s assume that the labor market in this economy does not work well and that 20 workers are permanently unemployed.
Now the production possibility frontier becomes x2 þ 0:25y2 ¼ 180, (1:20) (continued) Chapter 1 Economic Models 15 EXAMPLE 1.3 CONTINUED and the output combinations we described previously are no longer feasible. For example, if x ¼ 10 then y output is now y 17:9. The loss of about 2.1 units of y is a measure of the cost of the labor market inefficiency. Alternatively, if the labor supply of 180 were allocated evenly between the production of the two goods then we would have x 9:5 and y 19, and the inefficiency would show up in both goods’ production—more of both goods could be pro-duced if the labor market inefficiency were resolved.
QUERY: How would the inefficiency cost of labor market imperfections be measured solely in terms of x production in this model? How would it be measured solely in terms of y production? What would you need to know in order to assign a single number to the efficiency cost of the imperfection when labor is equally allocated to the two goods?
Welfare economics Inadditiontotheiruseinexaminingpositivequestionsabouthowtheeconomyoperates,thetools used in general equilibrium analysis have also been applied to the study of normative questions about the welfare properties of various economic arrangements. Although such questions were a major focus of the great eighteenth- and nineteenth-century economists (Smith, Ricardo, Marx, Marshall, and so forth), perhaps the most significant advances in their study were made by the British economist Francis Y. Edgeworth (1848–1926) and the Italian economist Vilfredo Pareto (1848–1923) in the early years of the twentieth century. These economists helped to provide a precise definition for the concept of “economic efficiency” and to demonstrate the conditions under which markets will be able to achieve that goal. By clarifying the relationship between the allocation pricing of resources, they provided some support for the idea, first enunciated by Adam Smith, that properly functioning markets provide an “invisible hand” that helps allocate resources efficiently. Later sections of this book focus on some of these welfare issues.
MODERN DEVELOPMENTS Research activity in economics expanded rapidly in the years following World War II. A major purpose of this book is to summarize much of this research. By illustrating how economists have tried to develop models to explain increasingly complex aspects of economic behavior, this book seeks to help you recognize some of the remaining unanswered questions.
The mathematical foundations of economic models A major postwar development in microeconomic theory was the clarification and formalization of the basic assumptions that are made about individuals and firms. The first landmark in this development was the 1947 publication of Paul Samuelson’s Foundations of Economic Analysis, in which the author (the first American Nobel Prize winner in economics) laid out a number of models of optimizing behavior.7 Samuelson demonstrated the importance of basing behav-ioral models on well-specified mathematical postulates so that various optimization techniques from mathematics could be applied. The power of his approach made it inescapably clear that mathematics had become an integral part of modern economics. In Chapter 2 of this book we review some of the mathematical concepts most often used in microeconomics.
7Paul A. Samuelson, Foundations of Economic Analysis (Cambridge, MA: Harvard University Press, 1947).
16 Part 1 Introduction New tools for studying markets A second feature that has been incorporated into this book is the presentation of a number of new tools for explaining market equilibria. These include techniques for describing pricing in single markets, such as increasingly sophisticated models of monopolistic pricing or models of the strategic relationships among firms that use game theory. They also include general equilibrium tools for simultaneously exploring relationships among many markets. As we shall see, all of these new techniques help to provide a more complete and realistic picture of how markets operate.
The economics of uncertainty and information A final major theoretical advance during the postwar period was the incorporation of uncertainty and imperfect information into economic models. Some of the basic assumptions used to study behavior in uncertain situations were originally developed in the 1940s in connection with the theory of games. Later developments showed how these ideas could be used to explain why individuals tend to be adverse to risk and how they might gather information in order to reduce the uncertainties they face. In this book, problems of uncertainty and information enter the analysis on many occasions.
Computers and empirical analysis One final aspect of the postwar development of microeconomics should be mentioned—the increasing useofcomputerstoanalyzeeconomicdataandbuildeconomicmodels. Ascomputers have become able to handle larger amounts of information and carry out complex mathematical manipulations, economists’ ability to test their theories has dramatically improved. Whereas previous generations had to be content with rudimentary tabular or graphical analyses of real-world data, today’s economists have available a wide variety of sophisticated techniques together with extensive microeconomic data with which to test their models. To examine these tech-niques and some of their limitations would be beyond the scope and purpose of this book. But, Extensionsattheend of most chaptersare intended tohelpyou startreading about some ofthese applications.
SUMMARY This chapter provided background on how economists ap-proach the study of the allocation of resources. Much of the material discussed here should be familiar to you from intro-ductory economics. In many respects, the study of economics represents acquiring increasingly sophisticated tools for ad-dressing the same basic problems. The purpose of this book (and, indeed, of most upper-level books on economics) is to provide you with more of these tools. As a starting place, this chapter reminded you of the following points: • Economics is the study of how scarce resources are al-located among alternative uses. Economists seek to develop simple models to help understand that process.
Many of these models have a mathematical basis be-cause the use of mathematics offers a precise shorthand for stating the models and exploring their consequences.
• The most commonly used economic model is the supply-demand model first thoroughly developed by Alfred Marshall in the latter part of the nineteenth century. This model shows how observed prices can be taken to represent an equilibrium balancing of the production costs incurred by firms and the willingness of demanders to pay for those costs.
• Marshall’s model of equilibrium is only “partial”—that is, it looks only at one market at a time. To look at many markets together requires an expanded set of general equilibrium tools.
• Testing the validity of an economic model is perhaps the most difficult task economists face. Occasionally, a model’s validity can be appraised by asking whether it is based on “reasonable” assumptions. More often, however, models are judged by how well they can explain economic events in the real world.
Chapter 1 Economic Models 17 SUGGESTIONS FOR FURTHER READING On Methodology Blaug, Mark, and John Pencavel. The Methodology of Econom-ics: Or How Economists Explain, 2nd ed. Cambridge: Cam-bridge University Press, 1992.
A revised and expanded version of a classic study on economic meth-odology. Ties the discussion to more general issues in the philosophy of science.
Boland, Lawrence E. “A Critique of Friedman’s Critics.” Journal of Economic Literature (June 1979): 503–22.
Good summary of criticisms of positive approaches to economics and of the role of empirical verification of assumptions.
Friedman, Milton. “The Methodology of Positive Econom-ics.” In Essays in Positive Economics, pp. 3–43. Chicago: University of Chicago Press, 1953.
Basic statement of Friedman’s positivist views.
Harrod, Roy F. “Scope and Method in Economics.” Economic Journal 48 (1938): 383–412.
Classic statement of appropriate role for economic modeling.
Hausman, David M., and Michael S. McPherson. Economic Analysis, Moral Philosophy, and Public Policy, 2nd ed.
Cambridge: Cambridge University Press, 2006.
The authors stress their belief that consideration of issues in moral philosophy can improve economic analysis.
McCloskey, Donald N. If You’re So Smart: The Narrative of Economic Expertise. Chicago: University of Chicago Press, 1990.
Discussion of McCloskey’s view that economic persuasion depends on rhetoric as much as on science. For an interchange on this topic, see also the articles in the Journal of Economic Literature, June 1995.
Sen, Amartya. On Ethics and Economics. Oxford: Blackwell Reprints, 1989.
The author seeks to bridge the gap between economics and ethical studies. This is a reprint of a classic study on this topic.
Primary Sources on the History of Economics Edgeworth, F. Y. Mathematical Psychics. London: Kegan Paul, 1881.
Initial investigations of welfare economics, including rudimentary notions of economic efficiency and the contract curve.
Marshall, A. Principles of Economics, 8th ed. London: Mac-millan & Co., 1920.
Complete summary of neoclassical view. A long-running, popular text. Detailed mathematical appendix.
Marx, K. Capital. New York: Modern Library, 1906.
Full development of labor theory of value. Discussion of “transforma-tion problem” provides a (perhaps faulty) start for general equilibrium analysis. Presents fundamental criticisms of institution of private property.
Ricardo, D. Principles of Political Economy and Taxation.
London: J. M. Dent & Sons, 1911.
Very analytical, tightly written work. Pioneer in developing careful analysis of policy questions, especially trade-related issues. Discusses first basic notions of marginalism.
Smith, A. The Wealth of Nations. New York: Modern Library, 1937.
First great economics classic. Very long and detailed, but Smith had the first word on practically every economic matter. This edition has helpful marginal notes.
Walras, L. Elements of Pure Economics. Translated by W. Jaffé.
Homewood, IL: Richard D. Irwin, 1954.
Beginnings of general equilibrium theory. Rather difficult reading.
Secondary Sources on the History of Economics Backhouse, Roger E.The Ordinary Businessof Life: The History of Economics from the Ancient World to the 21st Century.
Princeton, NJ: Princeton University Press, 2002.
An iconoclastic history. Quite good on the earliest economic ideas, but some blind spots on recent uses of mathematics and econometrics.
Blaug, Mark. Economic Theory in Retrospect, 5th ed.
Cambridge: Cambridge University Press, 1997.
Very complete summary stressing analytical issues. Excellent “Readers’ Guides” to the classics in each chapter.
Heilbroner, Robert L. The Worldly Philosophers, 7th ed. New York: Simon & Schuster, 1999.
Fascinating, easy-to-read biographies of leading economists. Chap-ters on Utopian Socialists and Thorstein Veblen highly recommended.
Keynes, John M. Essays in Biography. New York: W. W.
Norton, 1963.
Essays on many famous persons (Lloyd George, Winston Churchill, Leon Trotsky) and on several economists (Malthus, Marshall, Edgeworth, F. P.
Ramsey, and Jevons). Shows the true gift of Keynes as a writer.
Schumpeter, J. A. History of Economic Analysis. New York: Oxford University Press, 1954.
Encyclopedictreatment. Coversallthefamousandmanynot-so-famous economists. Also briefly summarizes concurrent developments in other branches of the social sciences.
18 Part 1 Introduction Blaug, Mark, and John Pencavel. The Methodology of Econom-ics: Or How Economists Explain, 2nd ed. Cambridge: Cam-bridge University Press, 1992.
Friedman, Milton.
The Methodology of Positive Econom-“T ics.” In Essays in Positive Economics, pp. 3–43. Chicago: University of Chicago Press, 1953.
Blaug, Mark. Economic Theory in Retrospect, 5th ed.
Cambridge: Cambridge University Press, 1997. |
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Bar Graph Worksheets
Grade 3
Bar Graph Worksheets
Jo-ann Caballes
Updated on July 23, 2025
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Subjects & Topics:
Geometry Worksheets
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Bar graph worksheets are a compilation of questions that train kids’ ability to use bar graphs to represent data in the form of rectangular bars. They teach children how to use this form of graphing to compare data, which they will encounter pretty often while working with information. Brighterly worksheets offer a comfortable way to start.
Bar graph worksheet: Examples
Brighterly’s free printable bar graph worksheets are designed to show kids the bar chart basics in an entertaining form.
Exercises in making bar graph worksheets are aimed at illustrating the difference in frequency between different data. For example, they can use the bar graph to show the subjects a child has the highest grades in. The colorful charts with single and double bars become familiar and interesting to analyze, create, and compare.
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Bar graph worksheet PDF [download for free]
Get the bar graph worksheets grade 3 PDF for free from Brighterly tutors under the link below.
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More 3rd grade math worksheets:
Subtraction Without Regrouping Worksheets
3rd Grade Subtraction Worksheets
Counting Money Worksheets 3rd Grade
3 Digit Multiplication Worksheets
Addition And Subtraction With Regrouping Worksheets
Combine exercises from the bar graph worksheets PDF with other 3rd grade level worksheets to tackle all the math problems your kid struggles with in the current year.
Easy bar graph worksheets: Benefits
Sequential learning
Appropriate for different age groups
Plotting practice
Improves data interpretation skills
Engaging yet effective
Reading a bar graph worksheet – benefit #1: Sequential learning
The worksheets teach children about bar graphs by following a step-by-step approach. This helps them learn better as they begin with the simpler concepts before moving to the more technical ones.
Note: Purposeful pacing boosts reflection, repetition, and personalization, which fosters meaningful integration of the learned concept.
Make a bar graph worksheet printable – benefit #2: Appropriate for different age groups
Brighterly’s simple bar graph worksheets can be used for children in different grades. Thanks to this, the parent or supervising adult will be the one to determine which exercises are most appropriate for the student, depending on their skill level and grade.
Blank bar graph worksheets – benefit #3: Plotting practice
The bar graph worksheet grade 3 offers space for the child to practice their illustration and plotting skills. They can put their skills to the test by plotting the data given to see how applicable bar, line, and pie graphs can be in the real world.
Note:The PubMed research shows that spatial visualizations, when incorporated in math learning, can significantly improve math performance and the quality of spatial reasoning.
Double bar graph worksheets – benefit #4: Improves data interpretation skills
With the bar graph practice worksheet, students get the tool to interpret, understand, and analyze data. Given that the exercises are focused on working with bar charts, the worksheets help build the foundation of data sets analysis.
Free bar graph worksheets – benefit #5: Engaging yet effective
Reading bar graph worksheets are designed to look easy, entertaining, and appealing to kids, but their key goal is to boost math knowledge. Their fun nature makes it easier for kids to engage in dedicated practice, which is the foundation for any skill development.
Note: Deliberate practice is the type of focused repetition that helps to achieve excellence in a chosen domain.
Bar graph worksheet PDF: Practice with ease!
Get your kid these 3rd grade bar graph worksheets PDF from Brighterly so they can understand what bar graphs are and how to represent data with them. Students can also learn more about bar graph word problems from Graph Worksheets and Picture Graph Worksheets that contain extra practices that they can find helpful.
See the complete test with its answers on the Bar Graph Practice Problems page. Make the math practice easy — browse the exercises we have on the Geometry Worksheets page.
Jo-ann Caballes
14 articles
As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable.
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189784 | https://www.math.uri.edu/~mcomerford/math141/Fall11/lesson12.pdf | Lesson 2.5: The Second Derivative: Consider a function f with derivative f . We can also differentiate f to get the so-called second derivative f of f, ie f = (f )(the derivative of the derivative).
Sometimes written d2f dx2 = d dx( d f dx) or d2y dx2 = d dx( dy dx) if y = f(x).
What do derivative tell us?
Recall that we have the following: • If f > 0 on an interval, then f is increasing on that interval.
• If f < 0 on an interval, then f is decreasing on that interval.
• If f = 0 on an interval, then f is constant on that interval.
If we apply this result to f instead of f, we get, for example, in the first case that if f > 0 on an interval then f is increasing on that interval. Since f represents the slope of the tangent line of the graph, it is plausible that this means that the graph of f must bend upwards - ie f is concave up.
Similar reasoning applies when f < 0.
1 The full result is: • If f > 0 on an interval, then f is increasing on that interval, and f is concave up on that interval.
• If f < 0 on an interval, then f is decreasing on that interval, and f is concave down on that interval.
• If f = 0 on an interval, then f is constant on that interval, and f is linear on that interval.
TRAP: The converse of the first two parts is not (quite) true.
f is concave up on all of R, but f (0) = 0 - check this!!
What we can say is as follows: • If the graph of f is concave up on an interval, then f ≥0.
• If the graph of f is concave down on an interval, then f ≤0.
2 Example 1: What do the following 3 graphs tell us about the second derivative?
a) Concave up everywhere, so f ≥0.
b) Concave down everywhere so f ≤0.
c) Concave down for x < 0 so f ≤0 for x < 0. Concave up for x > 0 so f > 0 for x > 0.
In the last example, the function changed from concave down to concave up as we passed through x = 0. A point where the concavity changes from up to down or vice versa is called a point of inflection.
Interpretation of the second derivative as a rate of change. Mention defence budged in 1985.
3 Example 2: Logistic Growth Supposed a population of animals P grows with time t according the to the following graph: We have a point of inflection at t = t0 where the graph changes from concave up to concave down.
For t < t0, d2P dt2 ≥0 and the population growth rate dP dt appears to be growing.
For t > t0, d2P dt2 ≤0 and the population growth rate dP dt appears to be decreasing.
dP dt is actually at a maximum when t = t0. As t →∞, P →L and the population appears to stabilize.
4 Example 3: Tests on the C5 Chevy Corvette sports car gave the following table of data: Time t (sec) 0 3 6 9 12 Velocity v (m/sec) 0 20 33 43 51 a) Estimate dv dx for the time intervals shown.
b) What can you say about d2v dt2 over the period shown?
a) For each (short) time interval, dv dt is approximated by the average velocity: From t = 0 to t = 3 dv dt ≈Average rate of change of velocity = v(3) −v(0) 3 −0 = 20 −0 3 −0 = 6.67m/s2 The full table is: Time interval t (sec) 0-3 3-6 6-9 9-12 Average acceleration v (m/s2) 6.67 4.33 3.33 2.67 b) Since the values of dv dt are decreasing, we expect d2v dt2 ≤0 and the graph of v against t should be concave down. That dv dt > 0 means that the car is accelerating, but d2v dt2 ≤0 means the acceleration is decreasing.
5 Velocity and Acceleration: When a car is speeding up, we say it is accelerating. We define acceleration as the rate of change of velocity with respect to time. For an object with displacement s(t) and velocity v(t) = ds dt, the average acceleration over a time interval [t, t + h] is Average Acceleration from t to t + h = v(t + h) −v(t) h while the instantaneous acceleration or acceleration at t is a(t) where a(t) = v(t) = lim h→0 v(t + h) −v(t) h = s(t) Example 5: A particle is moving in a straight line, its acceleration is zero only once.
Its displacement s(t) has the following graph: a) What is the particle moving to the right and when is it moving to the left?
b) When is the acceleration zero, negative, positive?
a) The particle is moving to the right whenever s is increasing. From the graph, this is for 0 < t < 1 and t > 2. Similarly for 1 < t < 2 s is decreasing, and the particle is moving the the left. Note that at t = 2, we are back where we started.
b) Particle is accelerating when the graph is concave up, decelerating when the graph is concave down. From the graph, this happens for t > 1.5 and 0 < t < 1.5 respectively. Particle is not accelerating when concavity changes at the point of inflection located at t = 1.5.
6 |
189785 | https://www.khanacademy.org/math/geometry-home/right-triangles-topic/law-of-cosines-geo/v/law-of-cosines | Use of cookies
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189786 | https://math.stackexchange.com/questions/4287635/take-away-game-problem-find-the-winning-strategy | contest math - Take-away game problem - find the winning strategy - Mathematics Stack Exchange
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Take-away game problem - find the winning strategy
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There are 200 coins in a pile. During a players turn they must take at least one coin and cannot take more than half of the coins. The person who takes the last coin wins. Devise a strategy so you always win.
What strategy/method should I use to solve this problem and related problems. Would proof by induction be sufficient or is there some kind of formula. I’ve played around with this question for a while but haven’t had much luck. Any suggestions/pointers would be helpful.
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asked Oct 26, 2021 at 6:30
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Commented Oct 26, 2021 at 6:35
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Under this rule set, if there is a single coin you have to take it (rule 1) but are banned from taking it because 1>12×11>12×1 (rule 2).
– Mark Bennet
Commented Oct 26, 2021 at 6:46
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Commonly, with a game like this one (two-player, symmetric, zero-sum games of perfect information and no possibility of a tie), you can start at the end of the game and work your way backwards, marking each board position as either a winning position or a losing position, defined recursively by the two rules:
A board position is a winning position if you either immediately win or you can move so that the board position on your opponent's turn is a losing position.
A board position is a losing position if you either immediately lose or every option you have to move will put the board in a winning position on your opponent's turn.
For instance:
If on my turn there is one coin remaining, then I win. Therefore one coin is a winning position.
If there are two coins remaining, then I must take one coin, leaving one left, which is a winning position. Therefore two coins is a losing position.
With three coins, I can remove one to put the board in a losing position, so three coins is a winning position.
After you do this a few more times, you might notice that the losing positions become pretty rare, and we can further investigate what pattern they follow. In this case, we find that the losing positions are all the positions with 3(2n)−13(2n)−1 coins for n≥0n≥0 an integer.
Once you have that, the winning strategy is fairly apparent. Whatever position the board is in on your turn, you must take away the right number of coins to leave 3(2n)−13(2n)−1 coins on your opponent's turn. Proving that this is a winning strategy amounts to an inductive proof that having 3(2n)−13(2n)−1 coins at the start of your turn is a losing position as defined by the two rules I gave above, which effectively is just noting
12(3(2n)−1−1)≤3(2n−1)−1<12(3(2n)−1).12(3(2n)−1−1)≤3(2n−1)−1<12(3(2n)−1).
In a game starting with 200200 coins, if we go first, we should take 99 coins, leaving 3(26)−1\=1913(26)−1\=191 coins. Following this strategy each turn will result in our win.
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edited Oct 26, 2021 at 8:10
answered Oct 26, 2021 at 7:58
Brian MoehringBrian Moehring
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Note that I have assumed you may take one coin if only one coin is remaining on your turn, even though as Mark Bennet pointed out in a comment, the rules do not actually allow this. Otherwise all games end in a stalemate as nobody can take the final coin.
– Brian Moehring
Commented Oct 26, 2021 at 8:11
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189787 | https://www.youtube.com/watch?v=J-Gow6SfG2c | An Introduction to "sequences" and discrete calculus | Algebraic Calculus One | Wild Egg
Wild Egg Maths
88 likes
3028 views
27 May 2018
One of the serious challenges for any rigorous calculus course is finding suitable definitions of "sequences". These are the main objects of the Discrete Calculus, also sometimes called the Difference Calculus.
Here we introduce the subject informally through the eyes of Thomas Harriot, an early 17th century pioneer in this kind of algebra. The difference and summation operators on sequences are the main topics, and difference and summation tables figure prominently.
There is an arithmetical mistake near the end of Slide 3: the sum 302+132=434, not 334. So that moves through to the result 7^4=2401, not 2301.
Video Contents:
0:00 Intro
0:13 Catalan numbers
1:56 Calculus of differences
4:06 Thomas Harriot
8:47 Harriot's triangular summation table
12:53 Triangular numbers and pyramidal numbers
16:17 Harriot's triangular/binomial "sequences"
20:30 Harriot's Difference theorem
Screenshot PDFs for my videos are available at the website These give you a concise overview of the contents of the lectures for various Playlists: great for review, study and summary.
My research papers can be found at my Research Gate page, at
My blog is at where I will discuss lots of foundational issues, along with other things.
Online courses will be developed at openlearning.com. The first one, already underway is Algebraic Calculus One at Please join us for an exciting new approach to one of mathematics' most important subjects!
If you would like to support these new initiatives for mathematics education and research, please consider becoming a Patron of this Channel at Your support would be much appreciated.
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Low Mach Number Derivation
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1. Low Mach Number Derivation¶
The low Mach number equations are a subset of the fully compressible equations of motion (momentum, continuity and energy), admitting large variations in gas density while remaining acoustically incompressible. The low Mach number equations are preferred over the full compressible equations for low speed flow problems as the accoustics are of little consequence to the overall simulation accuracy. The technique avoids the need to resolve fast-moving acoustic signals. Derivations of the low Mach number equations can be found in found in Rehm and Baum,[RB78], or Paolucci,[Pao82].
The equations are derived from the compressible equations using a perturbation expansion in terms of the lower limit of the Mach number squared; hence the name. The asymptotic expansion leads to a splitting of pressure into a spatially constant thermodynamic pressure and a locally varying dynamic pressure. The dynamic pressure is decoupled from the thermodynamic state and cannot propagate acoustic waves. The thermodynamic pressure is used in the equation of state and to determine thermophysical properties. The thermodynamic pressure can vary in time and can be calculated using a global energy balance.
1.1. Asymptotic Expansion¶
The asymptotic expansion for the low Mach number equations begins with the full compressible equations in Cartesian coordinates. The equations are the minimum set required to propagate acoustic waves. The equations are written in divergence form using Einstein notation (summation over repeated indices):
∂ρ∂t+∂ρ u j∂x j∂ρ u i∂t+∂ρ u j u i∂x j+∂P∂x i∂ρ E∂t+∂ρ u j H∂x j=0,=∂τ i j∂x j+ρ g i,=−∂q j∂x j+∂u i τ i j∂x j+ρ u i g i.∂ρ∂t+∂ρ u j∂x j=0,∂ρ u i∂t+∂ρ u j u i∂x j+∂P∂x i=∂τ i j∂x j+ρ g i,∂ρ E∂t+∂ρ u j H∂x j=−∂q j∂x j+∂u i τ i j∂x j+ρ u i g i.
The primitive variables are the velocity components, u i u i, the pressure, P P, and the temperature T T. The viscous shear stress tensor is τ i j τ i j, the heat conduction is q i q i, the total enthalpy is H H, the total internal energy is E E, the density is ρ ρ, and the gravity vector is g i g i. The total internal energy and total enthalpy contain the kinetic energy contributions. The equations are closed using the following models and definitions:
P E H τ i j q i=ρ R W T,=H−P/ρ,=h+1 2 u k u k,=μ(∂u i∂x j+∂u j∂x i)−2 3 μ∂u k∂x k δ i j,=−k∂T∂x i P=ρ R W T,E=H−P/ρ,H=h+1 2 u k u k,τ i j=μ(∂u i∂x j+∂u j∂x i)−2 3 μ∂u k∂x k δ i j,q i=−k∂T∂x i
The mean molecular weight of the gas is W W, the molecular viscosity is μ μ, and the thermal conductivity is k k. A Newtonian fluid is assumed along with the Stokes hypothesis for the stress tensor.
The equations are scaled so that the variables are all of order one. The velocities, lengths, and times are nondimensionalized by a characteristic velocity, U∞U∞, and a length scale, L L. The pressure, density, and temperature are nondimensionalized by P∞P∞, ρ∞ρ∞, and T∞T∞. The enthalpy and energy are nondimensionalized by C p,∞T∞C p,∞T∞. Dimensionless variables are noted by overbars. The dimensionless equations are:
∂ρ¯∂t¯+∂ρ¯u¯j∂x¯j∂ρ¯u¯i∂t¯+∂ρ¯u¯j u¯i∂x¯j+1 γ M a 2∂P¯∂x¯i∂ρ¯h¯∂t¯+∂ρ¯u¯j h¯∂x¯j=0,=1 R e∂τ¯i j∂x¯j+1 F r i ρ¯,=−1 P r 1 R e∂q¯j∂x¯j+γ−1 γ∂P¯∂t¯+γ−1 γ M a 2 R e∂u¯i τ¯i j∂x¯j+ρ¯u¯i γ−1 γ M a 2 F r i−γ−1 2 M a 2(∂ρ¯u¯k u¯k∂t¯+∂ρ¯u¯j u¯k u¯k∂x¯j).∂ρ¯∂t¯+∂ρ¯u¯j∂x¯j=0,∂ρ¯u¯i∂t¯+∂ρ¯u¯j u¯i∂x¯j+1 γ M a 2∂P¯∂x¯i=1 R e∂τ¯i j∂x¯j+1 F r i ρ¯,∂ρ¯h¯∂t¯+∂ρ¯u¯j h¯∂x¯j=−1 P r 1 R e∂q¯j∂x¯j+γ−1 γ∂P¯∂t¯+γ−1 γ M a 2 R e∂u¯i τ¯i j∂x¯j+ρ¯u¯i γ−1 γ M a 2 F r i−γ−1 2 M a 2(∂ρ¯u¯k u¯k∂t¯+∂ρ¯u¯j u¯k u¯k∂x¯j).
The groupings of characteristic scaling terms are:
R e P r F r i M a=ρ∞U∞L μ∞,x x x R e y n o l d s n u m b e r,=C p,∞μ∞k∞,x x x P r a n d t l n u m b e r,=u 2∞g i L,x x x x x x i F r o u d e n u m b e r,g i≠0,=u 2∞γ R T∞/W−−−−−−−−−√,M a c h n u m b e r,R e=ρ∞U∞L μ∞,x x x R e y n o l d s n u m b e r,P r=C p,∞μ∞k∞,x x x P r a n d t l n u m b e r,F r i=u∞2 g i L,x x x x x x i F r o u d e n u m b e r,g i≠0,M a=u∞2 γ R T∞/W,M a c h n u m b e r,
where γ γ is the ratio of specific heats.
For small Mach numbers, M a≪1 M a≪1, the kinetic energy, viscous work, and gravity work terms can be neglected in the energy equation since those terms are scaled by the square of the Mach number. The inverse of Mach number squared remains in the momentum equations, suggesting singular behavior. In order to explore the singularity, the pressure, velocity and temperature are expanded as asymptotic series in terms of the parameter ϵ ϵ:
P¯u¯i T¯=P¯0+P¯1 ϵ+P¯2 ϵ 2…=u¯i,0+u¯i,1 ϵ+u¯i,2 ϵ 2…=T¯0+T¯1 ϵ+T¯2 ϵ 2…P¯=P¯0+P¯1 ϵ+P¯2 ϵ 2…u¯i=u¯i,0+u¯i,1 ϵ+u¯i,2 ϵ 2…T¯=T¯0+T¯1 ϵ+T¯2 ϵ 2…
The zeroeth-order terms are collected together in each of the equations. The form of the continuity equation stays the same. The gradient of the pressure in the zeroeth-order momentum equations can become singular since it is divided by the characteristic Mach number squared. In order for the zeroeth-order momentum equations to remain well-behaved, the spatial variation of the P¯0 P¯0 term must be zero. If the magnitude of the expansion parameter is selected to be proportional to the square of the characteristic Mach number, ϵ=γ M a 2 ϵ=γ M a 2, then the P¯1 P¯1 term can be included in the zeroeth-order momentum equation.
1 γ M a 2∂P¯∂x i=∂∂x i(1 γ M a 2 P¯0+ϵ γ M a 2 P¯1+…)=∂∂x i(P¯1+ϵ P¯2+…1 γ M a 2)1 γ M a 2∂P¯∂x i=∂∂x i(1 γ M a 2 P¯0+ϵ γ M a 2 P¯1+…)=∂∂x i(P¯1+ϵ P¯2+…1 γ M a 2)
The form of the energy equation remains the same, less the kinetic energy, viscous work and gravity work terms. The P 0 P 0 term remains in the energy equation as a time derivative. The low Mach number equations are the zeroeth-order equations in the expansion including the P 1 P 1 term in the momentum equations. The expansion results in two different types of pressure and they are considered to be split into a thermodynamic component and a dynamic component. The thermodynamic pressure is constant in space, but can change in time. The thermodynamic pressure is used in the equation of state. The dynamic pressure only arises as a gradient term in the momentum equation and acts to enforce continuity. The unsplit dimensional pressure is
P=P t h+γ M a 2 P 1,P=P t h+γ M a 2 P 1,
where the dynamic pressure, p=P−P t h p=P−P t h, is related to a pressure coefficient
P¯1=P−P t h ρ∞u 2∞P t h.P¯1=P−P t h ρ∞u∞2 P t h.
The resulting unscaled low Mach number equations are:
∂ρ∂t+∂ρ u j∂x j∂ρ u i∂t+∂ρ u j u i∂x j+∂P∂x i∂ρ h∂t+∂ρ u j h∂x j=0,=∂τ i j∂x j+(ρ−ρ∘)g i,=−∂q j∂x j+∂P t h∂t,∂ρ∂t+∂ρ u j∂x j=0,∂ρ u i∂t+∂ρ u j u i∂x j+∂P∂x i=∂τ i j∂x j+(ρ−ρ∘)g i,∂ρ h∂t+∂ρ u j h∂x j=−∂q j∂x j+∂P t h∂t,
where the ideal gas law becomes
P t h=ρ R W T.P t h=ρ R W T.
The hydrostatic pressure gradient has been subtracted from the momentum equation, assuming an ambient density of ρ∘ρ∘. The stress tensor and heat conduction remain the same as in the original equations.
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HomeClass 12BIOLOGYGemmae are...
Gemmae are
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non green, multicellular, asexual buds
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green,multicellular, asexual buds
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non green, unicellular, asexual buds
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non green, multicellular, sexual buds
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Step-by-Step Solution: 1. Understanding Gemmae: Gemmae are asexual reproductive structures found in certain plants, particularly in liverworts and some mosses. They are small, multicellular, and can develop into new individuals. 2. Formation of Gemmae: Gemmae are produced in a cup-like structure known as the gemma cup, which is located on the thallus (the body of the plant). This structure aids in the asexual reproduction process. 3. Characteristics of Gemmae: - They are typically haploid, meaning they contain a single set of chromosomes. - They are capable of developing directly into a new organism when they come into contact with suitable soil conditions. 4. Examples of Plants with Gemmae: Gemmae are commonly found in plants like Marchantia (a type of liverwort) and various moss species. 5. Analyzing the Options: In the question, we need to determine the correct definition of gemmae from the provided options: - The first option describes them as non-green multicellular asexual buds, which is incorrect. - The second option describes them as green multicellular asexual buds, which is correct since gemmae are indeed green and multicellular. 6. Conclusion: The correct option regarding gemmae is that they are green multicellular asexual buds that can give rise to new plants.
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Generating function
The idea behind generating functions is to create a power series whose coefficients, , give the terms of a sequence which is of interest. Therefore the power series (i.e. the generating function) is and the sequence is .
Contents
[hide]
1 Simple Example
2 More Examples
3 Convolutions
4 Simple Exercises
5 See also
6 External Link
Simple Example
If we let , then we have .
This function can be described as the number of ways we can get heads when flipping different coins.
The reason to go to such lengths is that our above polynomial is equal to (which is clearly seen due to the Binomial Theorem). By using this equation, we can rapidly uncover identities such as (let ), also .
More Examples
Many generating functions can be derived using the sum formula for geometric series for .
For example, using a change of variables, for (if ) or (if ), generating the powers of for any real . The identity holds for all when , but the result is uninteresting (both the generating function and the desired power series are just ).
Taking the derivative and multiplying by gives a generating function for the nonnegative integers: for .
Convolutions
Suppose we have the sequences and . We can create a new sequence, called the convolution of and , defined by . Generating functions allow us to represent the convolution of two sequences as the product of two power series. If is the generating function for and is the generating function for , then the generating function for is .
Simple Exercises
There are three baskets on the ground: one has 2 purple eggs, one has 2 green eggs, and one has 3 white eggs. Eggs of the same color are indistinguishable. In how many ways can I choose 4 eggs from the baskets?
Solution for problem 1: The generating function for the first basket is , since there is one way to choose 0 purple eggs (do nothing), 1 way to choose 1 purple egg (since eggs are indistinguishable), and 1 way to choose 2 purple eggs. In a similar fashion the generating function for the green egg basket is and the generating function for the white egg basket is . Now to find the number of ways to pick eggs from multiple baskets, just simply multiply the functions together, getting . We want the number of ways to choose 4 eggs, so we just need to look at the coefficient of and see that there are 8 ways to choose 4 eggs.
See also
Combinatorics
Polynomials
Series
External Link
generatingfunctionology
Retrieved from "
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189792 | https://www.youtube.com/watch?v=XQBLV2VMj3E | [Production Theory Basics] Elasticity of Substitution | High & Low elasticity of substitution | 42 |
nishant mehra
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Posted: 7 Jun 2021
This video talks about
Basics of Production Theory
Elasticity of Substitution
High & Low elasticity of substitution
(REFERENCE : Nicholson and Snyder Chapter 9)
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1) Intermediate Microeconomics Course in their semesters
2) UGC Net Economics
3) Basics of MA Economics Entrance Exam (DSE/ISI/JNU/IGIDR/MSE)
4) Indian Economic Services
For any course related query please call 9999886629
or visit nishantmehra.com for IAS Economics Optional/ Indian Economic Services/ UGC Net Economics
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Transcript:
elasticity of substitution so in a sense it means how easy it is to substitute one input for the another right so along an isoquant how easy it is to substitute one input for the another there are many things related to it so let me just start writing it out it's you will understand how easy it is to substitute one input for another right so it measures how easy it is it is to substitute or it is to move along the points right between the different points along an isoquant how easy it is to move or how easy it is to move right between points along an isoquant along an isoquant so you mean this better so you have an isoquant like this and you have labor you have capital and let's say you're sitting at point a and here you are sitting at point b and data this is going to give you the amount of labor and capital input which you are going to use now so this is telling you this is the amount of labor input this is the amount of capital input at point a this is telling you this is the amount of labor input this is the amount of capital input at point b this is what it is right so you have labor input capital input labor input capital input right and you can also measure mrts at these points you know you can also measure mrts at these points so mrts is the slope of the isoquant at these points slope of the isoquant at point a and at point b right so when you move along an isoquant right so when you move along an isoquant from point a to point b what happens is that your k by l ratio is also changing and mrts is also changing right so how do you do this that is in moving from point a to point b on q is equal to q naught that is your q naught isopod let's say right q is equal to q naught isoquant both k by l is changing so here k by ratio is this k 1 by l 1 here k by ratio is this k by l ratio is this k 2 by l 2 right both k by l ratio and mrts so mrts at point a is given by the slope of the isoquant at point a and mrts at point b is given by the slope of the isoquant at point b mrts will change right so the elasticity of substitution which we denoted by this sigma this is defined as the ratio of the proportional change of k by l to mrts right so is defined as is sigma is defined to be the ratio of these proportional changes to be the ratio of these proportional changes right so let me also write a formal definition so that you understand this so for a production function for a production function q is equal to f of kl sigma measures the proportionate change in k by l relative to the proportionate change in mrts relative to the proportionate change in mrt so basically you mean this proportionate change in k by l upon proportionate change in mrts so of course you can write proportionate change in k by l as k by l at point b minus k by l at point a upon k by l at point a upon m mrts at point b minus mrts at point a upon mrts at point a so this could be written also as del log of t by l upon del log of mrts del log of k by l upon del log of mrts right i hope you know this i mean this is exactly like your elasticity so i don't know whether you remember this or not so if i have a function uh let's say p is equal to f of x right so that's a demand function which i have so i can actually write this also so i mean when i write the elasticity how will i write del x by x that is the proportional change in x upon del p by p that is the proportionate change in v this could be written as del x by del p into p by x that is what elasticity is but i can also write this in terms of log how i can write this as log of p is equal to log of f x or let me just write x let's say right so if i write if i take the total derivatives d log p is equal to d log x right so what is d log p that is 1 upon p that is the derivative of log p dp is equal to 1 upon x that is the derivative of log x into dx so dx by dp you can just take it up like this dx by dp into p by x this is also let's say equal to whatever and you could have also so either you can write your elasticity in this don't write one or you could have also written it like this d log x upon d log you can also write it like this so this is also the meaning this is also the same meaning so otherwise you can also write it like this d log of sorry not d log of d k by l upon dmrts into mrts upon k by l you can also write it like this right you can also write it like this so we generally use this definition it is easy so i've just told you how why we are writing it in this fashion and not in any other fashion right so along an isoquant when you are moving down from a to b right your cabinet ratio is also falling here you are using more capital less labor here you're using less capital more labor so k by ratio is is falling and not only that your slope is also becoming flatter so slope of the isoquant at point a is this which is steeper and slope of the isoquant at point b is this which is flatter so slope is also getting flatter so along one isoquant along one isoquant the mrts will decrease s k by l ratio will fall mrts will also decrease escape ratios right so i mean you can just think of like this so if mrts is not at all changing with a change in uh k by l ratio you will say that the substitution is very easy it is very easy to substitute right so you can just think of like this if mrts does not change at all for changes in kpil we might say substitution is easy right so the ratio of the marginal productivities is not at all changing as k by l is changing right as the input ratios are changing so this generally happens in case of perfect substitutes kind of the production function perfect substitutes kind of production function what about the other one so if mrts changes frequently or it it changes very rapidly for even a very very small change in the cable ratio then you say that substitution is very very difficult right if mrt is changes rapidly for small changes in kb we would say substitution is difficult we would say substitution is difficult so a very small minor variation in the input productivities will bring about a huge change in mrts right so this happens in case of perfect complement kind of production function right and this elasticity of substitution is a scale-free measure of how easy it is to substitute this right so this is a scale-free measure of responsiveness skill free measure of responsibilities right so let me just i mean can we just give some points to it and and make it more understandable so let us say i have five labor here 20 capital here 10 labor here and 10 capital here right okay so if i have this then how do i get the what is the cable ratio at point kebile ratio at point a it is 28.5 four k while ratio at point b is what 10 upon 10 which is equal to 1. mrts at point a is the slope of this line right at this particular point that is just 28.54 and mrts at point b right so only at this point so you can just assume it to be one right so elasticity of substitution would be what the change in k by l upon this guy your change in mrts proportionate change so it is what change in uh this mrt so it is one minus four so this is k by ratio at point b upon keyboard ratio point a upon 4 upon mrts at point b minus mrt s at point a upon mrts at 0.8 so this what you get as 1. so this is what elasticity of substitution is this is what elasticity of substitution is so in a sense i mean basically if if you get the positive value of sigma right then you say that there's yes there is a certain degree of substitutability between the production inputs a positive value of sigma indicates a certain degree of substitutability between production inputs between production engines right production and input so if sigma is high right then mrts will not change much relative to k by l and uh isoquant will be relatively flat so you can have a high sigma or you can have a low sigma so high sigma is this so if sigma is high mrts is not going to change much right mrts will not change much relative to keybl right and the i sequence will be relatively flat and the isoquant will be relatively flat right so when there are let's say many people to run machines i mean when there are many people to run machines uh then you can actually introduce machines right while there are still enough people to run machines when there are very few people then in which are involved in the production process then it is difficult to swap machines for people so there you will have low sigma right there you will have low sigma so you can write it like this so let me just define it in one way and then i'll come back to this particular point and a low value of sigma would mean what that you have a very curved isoquant you have a very curved isoquant so mrts is going to change hugely by a very very small changes in k by l a low value of sigma indicates a rather curved isocon so a high value would indicate a relatively flat isocone mrts will change by the substantial amount by l changes right this i'm giving you a picture for for the next class right where we will be talking about perfect substitutes kind of elasticity of substitution what happens to a large serious substitution in in in a kingd isoquant for example min of ak comma bl so those kinds of things i just think about it a high value so it is harder for example there are uh in in the production process you're using both machines and uh and labor so if there are very few people involved it is difficult to swap machines for people so low value of sigma so you can just think of like this it may be harder to swap machines for people means you cannot i mean substitutability is low low sigma right when only a few people are involved in the production process when only a few people are involved in the production process whereas it is easier that you can introduce machines if there are enough people involved in the production process so means you're trying to substitute machines easily in place of people whereas it is easier means your sigma is higher whereas it is easier to introduce machines that is high signal while there are still enough people to run machines while there are still enough people to run machines excuse me vita so i mean in this recording what you have learned you've learned what do you mean by elasticity of substitution graphically how do you think about it and also one more thing in general i mean your sigma is going to change along an isoquant right so sigma is going to be different at different points along an iso one let me write that point also so in general you can just write somewhere in general sigma changes different points along an iso sigma is going to change at different points along an isoquant right so what is that you've learned you learn what is the meaning of elasticity of substitution and although you have also a constant elasticity of substitution okay so but i'm just doing this simple case here you understood what is the meaning of high sigma or a low sigma although we have not seen the example yet a very uh theoretical example i've given out here what is the meaning of high and low sigma and a very very simple numerical here and here i've just told you i mean what is the definition or what is the formula for elasticity of substitution right so what is the formula for elasticity of substitution so this is what i wanted to do in this class thank you |
189793 | https://math.stackexchange.com/questions/339649/maximizing-distance-between-points | Skip to main content
Maximizing distance between points
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I asked a similar question on SciComp, but it is a little out of the domain, so I thought I'd give it a try here as well.
Give n points, I would like to place them in a periodic box (periodic such that the distance between two points "wraps around" to the other side) so that the minimum distance between any two points is as large as it can possibly be.
How do I do this? I imagine analytically this could be quite difficult, but is there at least a numerical procedure?
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asked Mar 24, 2013 at 14:46
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Place them on a hexagonal grid? I believe that is the densest packing. If the numbers don't fill up the points on the grid, it gets hairy. There are packing problems (some proposed by Kepler, and still unsolved) for spheres, perhaps 2D is simpler.
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Yeah, it won't fill up the grid evenly. I'm trying to place 361 points at the moment.
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This kind of problem is very hard. You might have a look at packomania which has solutions for many numbers of circles in squares and rectangles. Most of them are found experimentally and not proven to be maximal. You can almost incorporate the wraparound by shrinking the square or rectangle by the radius in each direction, but I don't think it gets the corners right.
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answered Mar 24, 2013 at 15:10
Ross MillikanRoss Millikan
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I came up with the following, relatively easy to implement algorithm based on forces/constraints. I'm not sure if it always converges though.
Consider a box with width w and height h filled with n particles represented as 2D vectors pi=(pi,x, pi,y)
Now for each step, calculate for each pair of points (pi, pj) calculate the distance vector r=pj−pi and use that to separate the points. First calculate a scaled version rs=F|r|Dr. F is a parameter that represents the amount of force and D represents how much force decreases with distance. For D=1, the force is constant (you get a unit vector) and for D=3 force decreases with the distances squared. Generally, a higher D yields better results. The force needs to be adjusted manually to the problem. Then move the points away from eachother by rs:
pi(new)=pi(old)−rs
pj(new)=pj(old)+rs
So now we have an algorithm to create the maximum distance between points, but the problem lies in the distance function. To find the shortest distance in a periodic box, imagine all the adjacent boxes point pj could be in. In 2D this wouldbe 9 boxes total and in 3D 27. The offset vector v could take any of the following values.
v={vx=−w,0,wvy=−h,0,h
Now calculate all 9 possible distances rk=(pj+vk)−pi and pick the rk with the smallest |rk|.
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answered May 16, 2016 at 13:59
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189794 | https://artofproblemsolving.com/wiki/index.php/Rearrangement_Inequality?srsltid=AfmBOoqGuXiESdO6IDBP33Q5ShQJpC-pUoUThFEoJGq7GIXBbQWW9dX4 | Art of Problem Solving
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Rearrangement Inequality
The Rearrangement Inequality states that, if is a permutation of a finiteset (in fact, multiset) of real numbers and is a permutation of another finite set of real numbers, the quantity is maximized when and are similarly sorted (that is, if is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of ). Conversely, is minimized when and are oppositely sorted (that is, if is less than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members ).
Contents
[hide]
1 Introductory
2 Intermediate
2.1 Proof of the Rearrangement Inequality
3 Uses
4 See Also
5 External Links
Introductory
Consider the following simple application: suppose you are involved in the hold-up of a convenience store. You note, as you are emptying the register, that there are different numbers of each denomination (penny, nickel, dime, quarter, dollar bill, five dollar bill, ten dollar bill and twenty dollar bill) in the register. When would your take be maximized? Certainly, you would hope that there would be the largest number of twenty dollar bills, then the next largest number of tens, etc. Meanwhile, you would find yourself very disappointed if there were more pennies than nickels, more nickels than dimes, and so on. This is a simple application of the rearrangement inequality. It is also an application of the greedy algorithm, so one possible interpretation of the rearrangement inequality is that sometimes, the greedy algorithm works.
Intermediate
Proof of the Rearrangement Inequality
The proof of the Rearrangment Inequality can be handled with proof by contradiction. Only the maximization form is proved here; the minimization proof is virtually identical.
Before we begin the proof properly, it is useful to consider the case where . Without loss of generality, sort and so that and . By hypothesis, . Expanding and taking some terms to the other side of the inequality, we get , as desired.
Now for the general case. Again, without loss of generality, set and ; and suppose the grouping that maximizes the desired sum of products is not the one that pairs with , with , and so on. This means that there is at least one instance where is paired with while is paired with , where and . However, using the technique seen above to prove the inequality for , we can see that the sum of products can only increase if we instead pair with and with (unless both a's or both b's are equal, in which case either we can choose another pair of products or note that the current arrangement is actually identical to the desired one), which contradicts our assumption that the arrangement we had was already the largest one.
Note: The minimization equality can be very easily proved by noting that if we have the set , ordered in decreasing order and the set , ordered in increasing order, then the maximum sum is just . Thus, by negating all values the inequality follows.
Uses
The Rearrangement Inequality has a wide range of uses, from MathCounts level optimization problems to Olympiad level inequality problems. A relatively simple example of its use in solving higher-level problems is found in the proof of Chebyshev's Inequality. It is particularly useful in that it does not require any terms of either sequence to be positive or negative, unlike the power-mean family of inequalities.
See Also
Chebyshev's Inequality
Power Mean Inequality
External Links
The Rearrangement Inequality by Dragos Hrimiuc
Retrieved from "
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189795 | https://math.stackexchange.com/questions/4800845/uniform-sampling-points-on-a-line-using-2-uniform-distributions | Uniform Sampling points on a line using 2 Uniform distributions - Mathematics Stack Exchange
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Uniform Sampling points on a line using 2 Uniform distributions
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I have been struggling with this problem for quite some time but I am not sure how to proceed.
So I am given a sampling algorithm : We would like to uniformly sample points on a line between A and B. Start with sampling x,y x,y from U([0,1])U([0,1]) (a uniform distribution),then the point P=x A+y B x+y P=x A+y B x+y.
For the sake of simplicity I let A=0 and B=1 and generated some results in R to see if the points are uniformly sampled, but they are not. So I would like to understand why this sampling algorithm does not yield uniform samples.
Some Work I have done: I tried looking into some prior questions but I could not find a match. I think finding the cdf and showing its not a straight line (like the one for uniform distribution) would suffice to show that the algorithm does not yield uniform sampling, however I am struggling to find such CDF. Any help would be appreciated.
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asked Nov 5, 2023 at 6:26
kuuhakukuuhaku
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Without loss of generality A=0 A=0 and B=1 B=1. We have to verify that Q=X/(X+Y)Q=X/(X+Y) is not uniform. To this end, we compute Pr(X1 y r>1 and Pr(X<Y r|Y=y)=y r Pr(X<Y r|Y=y)=y r if y r<1 y r<1 which implies Pr(X<Y r)=∫1 0 y r d y=r 2 Pr(X<Y r)=∫0 1 y r d y=r 2 if r<1 r<1 and Pr(X1.r>1.
This leads for 0<q<1 0<q<1 to F(q)=Pr(X/(X+Y)<q)=Pr(X/Y≤q/(1−q))F(q)=Pr(X/(X+Y)<q)=Pr(X/Y≤q/(1−q)) which is
F(q)=q 2(1−q)F(q)=q 2(1−q) and F′(q)=1 2(1−q)2 F′(q)=1 2(1−q)2 if 0<q<1/2 0<q<1/2,
and also F(q)=2 q−1 q F(q)=2 q−1 q and F′(q)=1 2 q 2 F′(q)=1 2 q 2 if 1/2<q<1.1/2<q<1.
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edited Nov 5, 2023 at 8:58
answered Nov 5, 2023 at 8:52
Gérard LetacGérard Letac
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Thank you for the response ! It might be a bit dumb but I do not understand what the variable yr is. I am also unsure as to how you make the transition from P r(X/(X+Y)<q)=P r(X/Y≤q/(1−q))P r(X/(X+Y)<q)=P r(X/Y≤q/(1−q)). Appreciate the response !kuuhaku –kuuhaku 2023-11-05 10:13:00 +00:00 Commented Nov 5, 2023 at 10:13
Sorry I got that the transition P r(X/(X+Y)<q)=P r(X/Y≤q/(1−q))P r(X/(X+Y)<q)=P r(X/Y≤q/(1−q)) is just algebra so thats all good. Just confused about the y r y r variable and the first part !kuuhaku –kuuhaku 2023-11-05 10:22:42 +00:00 Commented Nov 5, 2023 at 10:22
Well, if Y=y,Y=y, then Y r=y r Y r=y r, I cannot explain more. Also Pr(X<Y r|Y=y)=Pr(X/Y<r|Y=y).Pr(X<Y r|Y=y)=Pr(X/Y<r|Y=y).Gérard Letac –Gérard Letac 2023-11-05 10:47:24 +00:00 Commented Nov 5, 2023 at 10:47
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It doesn't work because the sum x+y x+y will not be uniformly distributed. You should be able to sample uniformly if you use one uniform variable; e.g.,
P=(1−x)A+x B,P=(1−x)A+x B,
where x x is uniform on [0,1][0,1]. Then when x=0 x=0, P=A P=A, and when x=1 x=1, P=B P=B.
It's also worth mentioning that A A and B B need not be limited to R R. They can be arbitrary points in R n R n, and you will still get a uniform sampling on the line segment joining A A and B B.
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answered Nov 5, 2023 at 6:32
heropupheropup
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Thank you so much for the response ! So basically if I were to calculate the cdf of the x+y x+y then it will not be constant right? I think I can use convolution to get the pdf and then cdf to show that.kuuhaku –kuuhaku 2023-11-05 06:37:31 +00:00 Commented Nov 5, 2023 at 6:37
@kuuhaku I've already shown you that you only need one uniform variable. Try it out for yourself. Let x x be uniform on [0,1][0,1], and let A=(2,3)A=(2,3), and B=(5,7)B=(5,7). Then calculate P=(1−x)A+x B=(2(1−x)+5 x,3(1−x)+7 x)=(3 x+2,4 x+3).P=(1−x)A+x B=(2(1−x)+5 x,3(1−x)+7 x)=(3 x+2,4 x+3). These will be uniformly distributed on the line segment joining A A and B B. You can also try to calculate the PDF of P P when A,B∈R A,B∈R.heropup –heropup 2023-11-05 06:41:39 +00:00 Commented Nov 5, 2023 at 6:41
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189796 | https://www.quora.com/How-does-a-frameshift-mutation-prove-that-our-genetic-code-is-a-triplet | How does a frameshift mutation prove that our genetic code is a triplet? - Quora
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Genetics and Heredity
Codon Table
Frameshift Mutation
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Genetic Code
5
How does a frameshift mutation prove that our genetic code is a triplet?
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James Knight
Graduate student in Evolutionary Biology
·7y
Insertion and deletions of 1 or 2 bases cause a frameshift, while the insertion or deletion of a triplet does not. In fact, an insertion or deletion of a number of bases that isn’t a multiple of 3 will cause a frame shift, while any multiple of 3 will not. This demonstrates that the genetic code is comprised of triplet codons.
We knew before then that it had to be at least triplet (not singlet or doublet) because at least 20 amino acids had to be coded for, and since there are 4 nucleotide bases, a singlet code could only potentially code for 4 amino acids (4^1 = 4), a doublet code could only c
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Insertion and deletions of 1 or 2 bases cause a frameshift, while the insertion or deletion of a triplet does not. In fact, an insertion or deletion of a number of bases that isn’t a multiple of 3 will cause a frame shift, while any multiple of 3 will not. This demonstrates that the genetic code is comprised of triplet codons.
We knew before then that it had to be at least triplet (not singlet or doublet) because at least 20 amino acids had to be coded for, and since there are 4 nucleotide bases, a singlet code could only potentially code for 4 amino acids (4^1 = 4), a doublet code could only code for up to 16 (4^2 = 44 = 16). A triplet code can has 64 distinct codons (4^3 = 444 = 64), a quadruplet code would have 256 distinct codons (4^4 = 4444 = 256), etc.
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How is it proved that genetic code is triplet?
What is an attempted experiment to prove that genetic code is triplet?
Why is the genetic code a triplet code?
How many types of tRNA are necessary if the genetic code is triplet code?
Which type of mutation helped lead to the understanding that the genetic code is based on triplets?
Richard Rombouts
Freelance C++ Software Engineer (1998–present) · Author has 3.6K answers and 2.6M answer views
·7y
Since we have A, T, C and G = 4 symbols, there are 4 x 4 x 4 = 64 possible combinations for a triplet. Some are start or stop codons, the others map on to 20 amino acids. We know the mapping of triplets onto amino acids. We also know what sequences there are in normal human DNA.
If there is a large enough stretch of DNA letters affected, say we miss the first letter, then instead of for instance (ATC)(GAT)(CGA)(TCG) this would be read as: (TCG)(ATC)(GAT)(CG?)
Now if this sequence does not match any human, but we can make a match by adding (reinserting) the presumed missing letter and we find a m
Continue Reading
Since we have A, T, C and G = 4 symbols, there are 4 x 4 x 4 = 64 possible combinations for a triplet. Some are start or stop codons, the others map on to 20 amino acids. We know the mapping of triplets onto amino acids. We also know what sequences there are in normal human DNA.
If there is a large enough stretch of DNA letters affected, say we miss the first letter, then instead of for instance (ATC)(GAT)(CGA)(TCG) this would be read as: (TCG)(ATC)(GAT)(CG?)
Now if this sequence does not match any human, but we can make a match by adding (reinserting) the presumed missing letter and we find a match, we can be pretty sure we had a frame shift.
The shift will usually produce different proteins that were not ‘intended’. This can be a start of a great new mutation, a super intelligent human or so, but more often it is harmful. Maybe the newly unwanted proteins that are formed are not so bad, but the carrier might miss the ones that were supposed to be encoded. Note that lots of proteins work as enzyme to speed up or slow down reactions.
I can imagine frame-shifting can not only happen in DNA, but also in m-RNA, so when copying instructions from our DNA.
Upvote ·
Logan R. Kearsley
MA in Linguistics from BYU, 8 years working in research for language pedagogy. · Author has 8.7K answers and 8.2M answer views
·4y
Related
How is it proved that genetic code is triplet?
Well, there's the hard way, or…
Count the types of nucleotides. Count the types of amino acids.
There are 4 nucleotides, and 21 amino acids. 4^2 is less than 21. 4^3 is more than 21. And there you go, codons must be triplets of nucleotides.
Upvote ·
Matthew Funk
A biochemistry student at CSU.
·5y
Related
Why is the genetic code a triplet code?
mRNA only has four bases, A,U,G, and C. There are 20 amino acids used in protein synthesis.
A single nucleotide cannot directly code for an amino acids, because there are two many of them and any code would be nonspecific. Keep in mind it is very important that the code is specific, because small changes in protein sequence will destroy the entire tertiary structure of the protein. 4^2=16, so there are still not enough nucleotides in a doublet code to select for the right amino acid. With a triple code, you get 64 different possibilities. This is more than enough to code for the 20 amino acids.
Continue Reading
mRNA only has four bases, A,U,G, and C. There are 20 amino acids used in protein synthesis.
A single nucleotide cannot directly code for an amino acids, because there are two many of them and any code would be nonspecific. Keep in mind it is very important that the code is specific, because small changes in protein sequence will destroy the entire tertiary structure of the protein. 4^2=16, so there are still not enough nucleotides in a doublet code to select for the right amino acid. With a triple code, you get 64 different possibilities. This is more than enough to code for the 20 amino acids. Some of the amino acids are coded for with more than one base triplet, allowing for some flexibility in the genetic code.
This is the obvious, pragmatic reason for having a triplet genetic code. I don’t really know much about the evolution of the code, but it likely started very early in the history of life since it is the same across all organisms today. Keep in mind that DNA takes up a lot of space and has to be wound down extremely tightly. A quadruplet code would require more DNA to get to the same amino acids, so it would be evolutionarily favorable for organisms to adopt a three-letter code.
So, to summarize, it is a very practical, pragmatic reason.
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Which is a point mutation and not a frameshift mutation? Is mutagen insertion deletion silent?
Which triplet in DNA codes for valine?
Which is a point mutation and not a frameshift mutation?
Does the genetic code triplet of nucleotides bases?
Why is the triplet of codons used instead of 4?
Jon Deriel
Author has 6.6K answers and 1.4M answer views
·Updated 4y
Related
How come genetic code is a triplet code? From Crick and Brenner's experiment, I don't understand why it was for sure a triplet code when they found out it didn't modify the virus that much
Acridine dyes intercalate into the DNA double helix (they are exactly the same width as a base-pair) and cause miscopying, resulting in insertion or deletion of one nucleotide at a time. A single one-nucleotide insertion mutation changes the reading frame and messes up the whole amino acid sequence downstream from the insertion — so the protein is non-functional. If you combine two closely linked insertions, the reading frame is still wrong and the amino acid sequence is still messed up downstream from the two insertions. Still non-functional. But if you combine three closely linked insertions
Continue Reading
Acridine dyes intercalate into the DNA double helix (they are exactly the same width as a base-pair) and cause miscopying, resulting in insertion or deletion of one nucleotide at a time. A single one-nucleotide insertion mutation changes the reading frame and messes up the whole amino acid sequence downstream from the insertion — so the protein is non-functional. If you combine two closely linked insertions, the reading frame is still wrong and the amino acid sequence is still messed up downstream from the two insertions. Still non-functional. But if you combine three closely linked insertions, presto, you get a functional protein. That must mean that combining three single-nucleotide insertions restores the original reading frame and restores the downstream amino acid sequence, which in turn means that the nucleotide acid sequence is being read in groups of three nucleotides.
Ditto for single nucleotide deletions.
Mark Twain once said science is fascinating because you get such a wholesale return of conjecture from such a trifling investment of fact. He would have loved this example.
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Rich Hochstim
BS in Chemistry&Biology, University of Miami (Graduated 1983) · Author has 9.4K answers and 3.9M answer views
·4y
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How come genetic code is a triplet code? From Crick and Brenner's experiment, I don't understand why it was for sure a triplet code when they found out it didn't modify the virus that much
Q:
How come genetic code is a triplet code? From Crick and Brenner's experiment, I don't understand why it was for sure a triplet code when they found out it didn't modify the virus that much
Since evolution arrived at 20 different amino acids and 4 different nucleotides a coding schema that was not to small and not to big was desirable. Goldilocks tried one base per codon, but that could only code for four amino acids. Then Goldilocks tried two bases per codon, but that only coded for 16 amino acids. Then Goldilocks tried three bases for a codon and that gave 64 possible (words) amino acids, which left room for punctuation and redundancy, and that was just right!
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Keith Robison
Professional in bioinformatics/genomics since 1991. Blogs at Omics! Omics! · Upvoted by
Johannes Soellner
, PhD Molecular Genetics, University of Vienna · Author has 7.2K answers and 6.6M answer views
·6y
Related
If we have two copies of each gene, both of which are made of double helices, does that mean that when there is a single point mutation the base-pairing with that mutation changes as well?
Yes, with exceptions such as when a base conversion occurs while a DNA segment is truly single stranded due to DNA repair associated exonuclease activity, there will be mispairing.
Mismatch repair enzymes recognize the associated bulges and engage DNA repair processes. A mutation in mismatch repair is responsible for Lynch Syndrome, an autosomal dominant hereditary predisposition to colorectal cancer
Hereditary nonpolyposis colorectal cancer - Wikipedia
Cytosine can spontaneously deaminate to uracil; the enzyme Uracil N-Glycosylase (UNG) clips out the uracil base to create an abasic site, which t
Continue Reading
Yes, with exceptions such as when a base conversion occurs while a DNA segment is truly single stranded due to DNA repair associated exonuclease activity, there will be mispairing.
Mismatch repair enzymes recognize the associated bulges and engage DNA repair processes. A mutation in mismatch repair is responsible for Lynch Syndrome, an autosomal dominant hereditary predisposition to colorectal cancer
Hereditary nonpolyposis colorectal cancer - Wikipedia
Cytosine can spontaneously deaminate to uracil; the enzyme Uracil N-Glycosylase (UNG) clips out the uracil base to create an abasic site, which triggers other repair enzymes. Note that when 5-methylcytosine deaminates thymidine is firmed, so that is again a mismatch repair problem.
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TJ Berens
Retired Aerospace Defense Consultant at United States Armed Forces · Author has 24.8K answers and 12.6M answer views
·4y
Related
How do mutations appear out of nowhere and pass down genetically?
How do mutations appear out of nowhere and pass down genetically?
A mutation is just a mistake in making the genetic materials.
It is a chemical reaction, and, chemical reactions require certain conditions.
When the conditions are off a bit, say, the pH, or temperature, or some other chemical is present to interfere, or radiation causes an interference, and so forth…
… a mutation is caused.
If the mutation is in a cell that is part of the body, but, not in a sperm or egg, the cell can become cancerous, or function differently, etc… but, it won;t be passed on to offspring.
If the mutation is in a ger
Continue Reading
How do mutations appear out of nowhere and pass down genetically?
A mutation is just a mistake in making the genetic materials.
It is a chemical reaction, and, chemical reactions require certain conditions.
When the conditions are off a bit, say, the pH, or temperature, or some other chemical is present to interfere, or radiation causes an interference, and so forth…
… a mutation is caused.
If the mutation is in a cell that is part of the body, but, not in a sperm or egg, the cell can become cancerous, or function differently, etc… but, it won;t be passed on to offspring.
If the mutation is in a germ cell, a sperm or egg, then, it can be passed on to offspring, as, they are products of sperm and eggs, etc.
:D
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Dhruti Shah
Visualiser | Graphic Designer (2018–present)
·Sep 12
What are the best professional affordable and comfortable shoes for women?
I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours.
Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them.
They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the
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I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours.
Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them.
They’re professional enough for work wear but don’t feel heavy and keep me comfortable throughout the day, even when I’m commuting to the office. I got mine for around ₹999 from Bata, which felt like a steal compared to some other brands I looked at. They’ve held up really well, and I can easily pair them with trousers, skirts for my work outfits. If you’re on a budget but still want something that is comfortable and follows fashion trends, Ballerinas by Bata are the perfect choice. I picked up mine from a Bata store near me, you can grab yours too.
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Israel Ramirez
Retired Biopsychologist · Author has 4.4K answers and 31.7M answer views
·4y
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How do gene mutations work when we have two copies of each chromosome?
Dominant beneficial genes become common more easily than recessive ones.
Suppose a new dominant mutation appears. Anyone having one copy of that gene benefits from it and passes on that gene to its descendants. Over time, that mutated version of the gene becomes more and more common.
If a brand-new mutation is recessive, it won’t have any immediate effect on the individual. That mutation might get passed on to that individuals, children, grandchildren, great grandchildren, etc. Eventually, two descendants having that mutated gene might mate and produce some children with two copies of that mutat
Continue Reading
Dominant beneficial genes become common more easily than recessive ones.
Suppose a new dominant mutation appears. Anyone having one copy of that gene benefits from it and passes on that gene to its descendants. Over time, that mutated version of the gene becomes more and more common.
If a brand-new mutation is recessive, it won’t have any immediate effect on the individual. That mutation might get passed on to that individuals, children, grandchildren, great grandchildren, etc. Eventually, two descendants having that mutated gene might mate and produce some children with two copies of that mutation. If that mutation is beneficial, that individual might have more children than others but, because the mutation is still rare, most children will only have just one copy of that mutation. So, a beneficial mutation spreads slowly.
The reverse happens with a harmful mutation. A harmful dominant mutation is immediately removed by natural selection, but a harmful one can persist for a long time.
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Howard Landman
General science nerd, studied biochemistry, member NCSE. · Author has 1.6K answers and 1.8M answer views
·1y
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How do scientists determine that mutations do not add new information to the genetic code?
First, your question appears somewhat garbled. The genetic code is the translation table from RNA to amino acids.
Mutations in normal DNA don’t change the genetic code. The code CAN change, and has changed, but only extremely rarely and only in very simple single-celled organisms.
So I will assume that you really meant to ask “Is it impossible for mutations to increase the information in an organism’s genome?”. No. It’s quite easy to mathematically prove that it isn’t, for almost any possible measure of information that you care to propose.
Let I()I() be a measure of information. That is, it’s a mat
Continue Reading
First, your question appears somewhat garbled. The genetic code is the translation table from RNA to amino acids.
Mutations in normal DNA don’t change the genetic code. The code CAN change, and has changed, but only extremely rarely and only in very simple single-celled organisms.
So I will assume that you really meant to ask “Is it impossible for mutations to increase the information in an organism’s genome?”. No. It’s quite easy to mathematically prove that it isn’t, for almost any possible measure of information that you care to propose.
Let I()I() be a measure of information. That is, it’s a math function that takes in a genome g g and spits out a number I(g)I(g) that tells you how much information g g contains.
If I()I() is constant - that is, if it gives the same answer for every possible g g - we call it trivial. Trivial measures don’t tell us anything about anything, so they’re useless. In what follows, we ignore them.
So let I()I() be a non-trivial measure of information. That means it gives 2 different answers A=I(a)A=I(a) and B=I(b)B=I(b) for some pair of genomes a a and b b. Let’s assume that a a has more information than b b (if not, we can just swap them). So A=I(a)>I(b)=B A=I(a)>I(b)=B.
Any genome can be converted into any other genome by some sequence of mutations. So construct such a sequence (there are many). At least one mutation along that path must decrease the information. That is, there are genomes x x and y y that differ by a single mutation and I(x)>I(y)I(x)>I(y).
But every mutation is reversible. So if the mutation x→y x→y decreases information, then the mutation y→x y→x must increase information. QED.
Thus we see that saying “some mutations decrease information” is logically equivalent to saying “some mutations increase information”. You can’t have one without the other also being true.
In addition to this mathematical proof, empirical evidence from biology supports the idea that mutations can and do increase genetic information. New genes, new functions, and increased complexity in genomes arise through various types of mutations, such as gene duplications, insertions/deletions, and point mutations. These changes contribute to the diversity and adaptability of life.
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A Hammouda
Member of Bioinformatics Hub Net, Professor at Medical Biochemistry · Author has 756 answers and 142.8K answer views
·2y
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Discuss the notion of the genetic code, and how do you read the triplets of genetic code?
An aminoacid is encoded by a sequence of three bases (triplet) in DNA or mRNA. This triplet is called a codon. Since we have four bases, there are 64 codons, a state of codon redundancy. The twenty aminoacids are encoded by 61 codons. Three (UAA, UGA, UAG) are stop codons (nonsense codons), which terminate translation. They are all written in the 5' to 3' direction. There is one start codon (initiation codon), AUG, coding for methionine. Protein synthesis begins with methionine (Met) in eukaryotes, and formyl-methionine (fmet) in prokaryotes.
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An aminoacid is encoded by a sequence of three bases (triplet) in DNA or mRNA. This triplet is called a codon. Since we have four bases, there are 64 codons, a state of codon redundancy. The twenty aminoacids are encoded by 61 codons. Three (UAA, UGA, UAG) are stop codons (nonsense codons), which terminate translation. They are all written in the 5' to 3' direction. There is one start codon (initiation codon), AUG, coding for methionine. Protein synthesis begins with methionine (Met) in eukaryotes, and formyl-methionine (fmet) in prokaryotes.
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Mike Lieberman
SVP, Medical Science at Klick Health (2018–present) · Author has 1.3K answers and 10.2M answer views
·7y
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How are point mutation and frameshift mutation similar?
They’re both mutations involving individual nucleotides.
Point mutations are typically base substitutions - so one nucleotide is substituted for another (A becomes T or G becomes C, just for example.). The net result of a point mutation with respect to protein coding can be a premature stop codon, an amino acid change, or nothing. The genetic code is somewhat tolerant of point mutations since there are 64 possible combinations of nucleotide triplets, but only 20 amino acids coded for (plus two stop codons).
(Image credit: whyevolutionistrue.wordpress.com)
Frameshift mutations are insertions or de
Continue Reading
They’re both mutations involving individual nucleotides.
Point mutations are typically base substitutions - so one nucleotide is substituted for another (A becomes T or G becomes C, just for example.). The net result of a point mutation with respect to protein coding can be a premature stop codon, an amino acid change, or nothing. The genetic code is somewhat tolerant of point mutations since there are 64 possible combinations of nucleotide triplets, but only 20 amino acids coded for (plus two stop codons).
(Image credit: whyevolutionistrue.wordpress.com)
Frameshift mutations are insertions or deletions of bases - these change the reading frame, which often impacts every downstream amino acid, which is why they’re often called nonsense mutations.
But this is all a bit theoretical - it’s easier to actually try them out and see what happens.
Take the sentence: “Molecular biology is a fun and interesting subject.”
Now we’re going to mutate it a bit and see what happens:
Point mutation:
Molecular bioloty is a fun and interesting subject.
See? A single mutation - changing the “g” in “biology” to a “t” but you can still read the sentence and get the gist of it. And most proteins are similar - while certain mutations can change the functionality of the protein, many are tolerated with little overall impact.
Frameshift mutation:
Molecular iologyi sa f una ndi terestings ubject.
All I did was delete the “b” in biology but keep the number of letters in subsequent words consistent. That’s much, much harder to parse out - and molecular machinery is no different. In practical terms, a frameshift mutation will often make entire proteins, or at least large portions of them, unusable.
But the question asked about what is similar between the two, and my point was, they’re individual base mutations. There are other types of mutations altogether. We can see how they play out.
Duplications
Molecular biology is a biology is a fun and interesting subject.
Here I’ve duplicated the “biology is a “ portion of the sentence. This would actually be a very small duplication, in reality they’re often considerably larger. Sometimes the resulting protein makes sense, sometimes it doesn’t, sometimes it takes on a new function.
Deletions
Molecular biology is a subject.
Deletions are, rather obviously, when a portion of the genome is deleted. They can be large or small deletions, but obviously can have a large impact on the protein, as you see here with the sentence itself. Whole functional domains can be eliminated through a deletion.
Transpositions
Molecular biology is a fun Andy Warhol paintings are overrated and expensive interesting subject.
Sections of DNA can move around the genome, by “transposing” from one region to another. Sometimes a transposon lands right in the middle of a gene and all of the sudden the protein can get all new functions.
These last three types of mutation are fundamentally different from the first two though- they’re about large changes in DNA sequence. Whereas point and frameshift mutations are about individual changes in single base pairs.
The two general types of mutation can work in tandem though in ways that make evolution quite plausible at the molecular level. Once whole genes are duplicated, the new genes start leading to proteins found in whole new regions of the body. And because they’re new, they’re not required for cells to function, so they’re very tolerant of other mutations. “Molecular biology is a fun and interesting subject.” Can become, after a duplication and a short series of point mutations, a deletion and a transposition, “Mole people can be a fundamental pillar of modern government.” A whole new meaning and function from the same starting material.
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Related questions
How is it proved that genetic code is triplet?
What is an attempted experiment to prove that genetic code is triplet?
Why is the genetic code a triplet code?
How many types of tRNA are necessary if the genetic code is triplet code?
Which type of mutation helped lead to the understanding that the genetic code is based on triplets?
Which is a point mutation and not a frameshift mutation? Is mutagen insertion deletion silent?
Which triplet in DNA codes for valine?
Which is a point mutation and not a frameshift mutation?
Does the genetic code triplet of nucleotides bases?
Why is the triplet of codons used instead of 4?
How are point mutation and frameshift mutation similar?
Who first suggested the triplet nature of genetic code?
What are the effects of point and frameshift mutations?
Is it possible for non-coding DNA to become coding DNA through mutations, such as exonization?
How does an insertion mutation affect the protein coded by the gene with which it occurs? What about deletions and frameshift mutations?
Related questions
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Which type of mutation helped lead to the understanding that the genetic code is based on triplets?
Which is a point mutation and not a frameshift mutation? Is mutagen insertion deletion silent?
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189797 | https://pubmed.ncbi.nlm.nih.gov/28140341/ | Estimating the minimum stimulation frequency necessary to evoke tetanic progression based on muscle twitch parameters - PubMed
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. 2017 Mar;38(3):466-476.
doi: 10.1088/1361-6579/aa5bd1. Epub 2017 Jan 31.
Estimating the minimum stimulation frequency necessary to evoke tetanic progression based on muscle twitch parameters
Shogo Watanabe1,Shinichi Fukuhara,Takeshi Fujinaga,Hisao Oka
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1 Department of Medical Technology, Graduate School of Health Sciences, Okayama University, 2-5-1, Shikata-cho, Kita-ku, Okayama 700-8558, Japan.
PMID: 28140341
DOI: 10.1088/1361-6579/aa5bd1
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Estimating the minimum stimulation frequency necessary to evoke tetanic progression based on muscle twitch parameters
Shogo Watanabe et al. Physiol Meas.2017 Mar.
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. 2017 Mar;38(3):466-476.
doi: 10.1088/1361-6579/aa5bd1. Epub 2017 Jan 31.
Authors
Shogo Watanabe1,Shinichi Fukuhara,Takeshi Fujinaga,Hisao Oka
Affiliation
1 Department of Medical Technology, Graduate School of Health Sciences, Okayama University, 2-5-1, Shikata-cho, Kita-ku, Okayama 700-8558, Japan.
PMID: 28140341
DOI: 10.1088/1361-6579/aa5bd1
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The summation of the muscle force caused by an increase in the firing rate is named a tetanic contraction (tetanus), and the minimum stimulation frequency necessary to evoke an unfused/fused tetanus is related to the contraction time (CT) and relaxation time (RT) of the twitch. In particular, the fusion index (FI) is a very useful indicator, and it is used to evaluate the change in the muscle fiber component ratio. However, the measurement of the FI is invasive, because most patients experience pain during the electrical stimulation for tetanus. We expect that the twitch parameters CT and RT can substitute for the FI in the future. We found that the minimum stimulation frequency necessary to evoke the unfused/fused tetanus can be estimated from the twitch parameters as a first step. The results showed that (1) the minimum stimulation frequencies calculated from twitch parameters during unfused/fused tetanus were very similar to those calculated from FI parameters, and (2) they were also strongly correlated with FI parameters regardless of fiber components. The basic characteristics of tetanic progression in different fiber types could be estimated from twitch parameters.
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189799 | https://www.ixl.com/math/grade-5/multiply-a-decimal-by-a-one-digit-whole-number-using-the-distributive-property | IXL | Multiply a decimal by a one-digit whole number using the distributive property | 5th grade math
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CC.4
Multiply a decimal by a one-digit whole number using the distributive property
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Learn with an example
Use the distributive property to find the product.
3
×
5.3
=
3
×
i
+
0.3
i
=
i
3
×
i
+
i
3
×
i
=
+
=
Submit
Back to practice
ref_doc_title.
Back to practice
Learn with an example
Learn with an example
question
Use the distributive property to find the product.
3
×
7.3
=
3
×
i
7
+
i
=
i
3
×
i
+
i
3
×
i
=
+
=
key idea
Distributive property
You can multiply a sum by multiplying each addend separately.
Example:
2
×
(
5
+
7
)
=
(
2
×
5
)
+
(
2
×
7
)
solution
You can use the distributive property to find the product
3
×
7.3
.
First, write 7.3 as 7 + 0.3.
3
×
(
7
+
0.3
)
Use the distributive property to rewrite the problem as the sum of two products.
(
3
×
7
)
+
(
3
×
0.3
)
Find the products.
21
+
0.9
Finally, find the sum.
21.9
So,
3
×
7.3
=
21.9
.
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Lesson: The distributive property Learn about the distributive property! Using the distributive property with numbers Using the distributive property with variables Showing the distributive property with area models Lesson: The distributive property
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Work it out Not feeling ready yet? These can help:D.16 Multiply using propertiesD.16 Multiply using properties - Fifth grade 6PNCC.2 Multiply a decimal by a one-digit whole number: tenths or hundredthsCC.2 Multiply a decimal by a one-digit whole number: tenths or hundredths - Fifth grade 2TZD.7 Multiply by 1-digit numbersD.7 Multiply by 1-digit numbers - Fifth grade 7H4Lesson: The distributive property Learn about the distributive property! Using the distributive property with numbers Using the distributive property with variables Showing the distributive property with area models Lesson: The distributive property
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