content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
How can you rewrite the following sentence from the first-person perspective of someone in the sentence?: They bought a coat and gave it to the homeless man on the corner. | Socratic
How can you rewrite the following sentence from the first-person perspective of someone in the sentence?: They bought a coat and gave it to the homeless man on the corner.
1 Answer
I believe that the sentence already could be in 1st person
"They bought a coat and gave it to the homeless man on the corner" could be a statement made by someone from the 1st person perspective as long as the narrator was not a part of subject pronoun
Impact of this question
2365 views around the world | {"url":"https://socratic.org/questions/how-can-you-rewrite-the-following-sentence-from-the-first-person-perspective-of--13","timestamp":"2024-11-05T12:01:33Z","content_type":"text/html","content_length":"33302","record_id":"<urn:uuid:ef8dd50a-12a8-4ed4-807b-226a1cbc264c>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00736.warc.gz"} |
Unlimited Register Machine/Program/Basic Instruction
The basic instructions of a URM program form a finite sequence and hence can be considered a set indexed by the (positive) integer $1, 2, 3, \ldots$.
The basic instructions are as follows:
│Name │Notation │Effect │Description │
│Zero │$\map Z n$ │$0 \to R_n$ │Replace the number in $R_n$ by $0$. │
│Successor│$\map S n$ │$r_n + 1 \to R_n$ │Add $1$ to the number in $R_n$. │
│Copy │$\map C {m, n}$ │$r_m \to R_n$ │Replace the number in $R_n$ by the number in $R_m$ (leaving the one in $R_m$ as it was). │
│Jump │$\map J {m, n, q}$│$r_m = r_n ? \Rightarrow q$│If the numbers in $R_m$ and $R_n$ are equal, go to instruction number $q$, otherwise go to the next instruction.│
The operation of carrying out a basic URM instruction is referred to as execution.
Basic URM instructions are also (and more commonly) known as commands, because the word is shorter and quicker to say.
During the course of the exposition, the term instruction will also be found, which again means the same thing.
Also see
• Results about unlimited register machines can be found here. | {"url":"https://proofwiki.org/wiki/Definition:Unlimited_Register_Machine/Program/Basic_Instruction","timestamp":"2024-11-03T13:44:36Z","content_type":"text/html","content_length":"44541","record_id":"<urn:uuid:e0c2d38b-8ff5-4685-ac00-995e8cb8e2dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00549.warc.gz"} |
The true and predicted redshifts of 102,798 SDSS galaxies, using a simple decision tree regressor. Notice the presece of catastrophic outliers: those galaxies whose predicted redshifts are extremely
far from the true value.
Later, in Exercise #2, we will attempt to improve on this by optimizing the parameters of the decision tree.
In practice, the solutions to the photometric redshift problem can benefit from approaches that use physical intuition as well as machine learning tools. For example, some solutions involve the use
of libraries of synthetic galaxy spectra which are known to be representative of the true galaxy distribution. This extra information can be used either directly, in a physically motivated analysis,
or can be used to generate a larger suite of artificial training instances for a pure machine learning approach. | {"url":"https://notebook.community/diego0020/va_course_2015/AstroML/notebooks/08_regression_example","timestamp":"2024-11-05T01:22:09Z","content_type":"text/html","content_length":"49290","record_id":"<urn:uuid:d60a7d1b-35b0-4205-9584-0eb3c500298b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00652.warc.gz"} |
Theoretical Climate Dynamics
Copyright Notice: Many of the links included here are to publications copyrighted by the American Meteorological Society (AMS). Permission to place copies of these publications on this server has
been provided by the AMS. The AMS does not guarantee that the copies provided here are accurate copies of the published work. Permission to use figures, tables, and brief excerpts from these works in
scientific and educational works is hereby granted provided that the source is acknowledged. Any use of material in this work that is determined to be "fair use" under Section 107 of the U.S.
Copyright Act or that satisfies the conditions specified in Section 108 of the U.S. Copyright Act (17 USC §108, as revised by P.L. 94-553) does not require the AMS's permission. Republication,
systematic reproduction, posting in electronic form on servers, or other uses of this material, except as exempted by the above statement, requires written permission or a license from the AMS.
Additional details are provided in the AMS Copyright Policy, available on the AMS Web site located at (http://www.ametsoc.org/AMS) or from the AMS at 617-227-2425 or copyright@ametsoc.org. | {"url":"https://dept.atmos.ucla.edu/tcd/publications/author/49","timestamp":"2024-11-14T10:50:52Z","content_type":"text/html","content_length":"70048","record_id":"<urn:uuid:b55b9284-5fea-47f9-98db-77511c332a78>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00434.warc.gz"} |
MCCXVII in Hindu Arabic Numerals
MCCXVII = 1217
M C X I
MM CC XX II
MMM CCC XXX III
CD XL IV
D L V
DC LX VI
DCC LXX VII
DCCC LXXX VIII
CM XC IX
MCCXVII is valid Roman numeral. Here we will explain how to read, write and convert the Roman numeral MCCXVII into the correct Arabic numeral format. Please have a look over the Roman numeral table
given below for better understanding of Roman numeral system. As you can see, each letter is associated with specific value.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
How to write Roman Numeral MCCXVII in Arabic Numeral?
The Arabic numeral representation of Roman numeral MCCXVII is 1217.
How to convert Roman numeral MCCXVII to Arabic numeral?
If you are aware of Roman numeral system, then converting MCCXVII Roman numeral to Arabic numeral is very easy. Converting MCCXVII to Arabic numeral representation involves splitting up the numeral
into place values as shown below.
M + C + C + X + V + I + I
1000 + 100 + 100 + 10 + 5 + 1 + 1
As per the rule highest numeral should always precede the lowest numeral to get correct representation. We need to add all converted roman numerals values to get our correct Arabic numeral. The Roman
numeral MCCXVII should be used when you are representing an ordinal value. In any other case, you can use 1217 instead of MCCXVII. For any numeral conversion, you can also use our roman to number
converter tool given above.
Current Date and Time in Roman Numerals
The current date and time written in roman numerals is given below. Romans used the word nulla to denote zero because the roman number system did not have a zero, so there is a possibility that you
might see nulla or nothing when the value is zero. | {"url":"https://romantonumber.com/mccxvii-in-arabic-numerals","timestamp":"2024-11-10T14:27:18Z","content_type":"text/html","content_length":"89760","record_id":"<urn:uuid:9d298b67-6c55-43b2-a42e-0b394b7da0c8>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00131.warc.gz"} |
How to Calculate the Height of a Ridge Board
Landing Page
How to Calculate the Height of a Ridge Board
in today's video we're learning how to calculate the height of a ridge board for a common rafter roof today's maka may look familiar to you it's the same rafters that used in a previous video the
only difference is today I added a lot more markings on them and if you missed that video I'll do a quick overview now to get you all caught up and also leave a link up here and at the end of this
video so you can watch it what we have here and what we built in the beginner rafter video is a 912 common rafter roof meaning that for every 12 inches of horizontal run the roof rises 9 inches
vertically the overall span of the roof is 4 feet and the total run is half that distance or 2 feet we use the unit length method represented by these two triangles to determine our rough rafter
length after that we determine the bird's mouth cut length of the overhang and remove half the thickness of the ridge board to arrive here with 2 complete rafters overhangs and a ridge now that y'all
caught up I do want to add 3 things that I didn't have time to talk about in the first video that's important to know first make sure to sight down all rafter and ridge material making sure to
install with the crown facing up second it's normal to size the ridge board up a size or more from the rafters our rafters are 2 by 6 so you can use a 2 by 8 a 10 or 12 for a ridge in third I want to
talk a little bit more about the measuring line the measuring line is a theoretical line and it's really important because it's where all of our initial math is based from and if you remember this is
the line that our unit triangles followed to determine our rough rafter length the measuring line starts at the corner of the bird's mouth and continues all the way up to the top of the rafter
maintaining an equal distance from the bottom of the rafter lastly it's important to note that this line can either move up or down depending on the size of the seat cut so never assume that the
measuring line is always the same distance from the bottom of the rafter moving on during the process of laying out for the rafters we uncovered a few numbers that we can now use to determine the
height of the ridge the first measurement that we uncovered was labeled B or the unit rise which in our example is 9 inches the other measurement we found was the total run which was 2 feet or 24
inches but that measurement for the thickness of the ridge so for this time around we need to calculate the total run minus half the thickness of the ridge in order to get our total Ridge height so
in this case that's 24 minus 3/4 of an inch again because our ridge board that we're using is an inch and a half thick with a total round of twenty three and a quarter inches and a rise of nine
inches it's time to look at our first equation which is f equals B times D divided by 12 so for our mock up the F is the theoretical Ridge height B is the unit run and D is the total run therefore
the math looks like this 9 times 23 and 1/4 divided by 12 for a height of 17 and a half as you can see 17 and 1/2 gets us up to the theoretical Ridge height which is at the intersection of the ridge
and the measuring line so the next step is to find the height above plate which is the final number we need to solve for our total Ridge height to get the height above plate simply extend the heel
line of the birdsmouth up and then measure from the corner of that bird's mouth to the top of the rafter in case you're wondering why they call it height above plate it'll make better sense if you
look on the backside of the rafter here you can see I've actually drawn the same extended heel line but this time you can actually see the top of the plate so the height above plate is exactly that
it's the distance from the top of the plate to the top of the rafter and this measurement is really important when we start working with hip and Valley rafters well the height above plate of 4 and 1/
8 and with a theoretical Ridge height of 17 and a half it's time to look at our last equation which is G equals F plus E for our mock-up the G is the total Ridge height F is the theoretical Ridge
height and E is the height of a plate therefore the math looks like this 17 and 1/2 plus 4 and 1/8 equals a total Ridge height of 21 and five-eighths as a summary you need to know four things in
order to calculate for the total Ridge height first you need to know the ridge thickness second you need to know the total run – the ridge thickness and third you need to know the unit rise or pitch
which is often noted as a ratio like our roof which is a 9 12 pitch 9 being the rise and lastly you need to know the height above plate if you have any questions leave them in a comment section below
be sure to check out this video right now if you didn't already watch it thank you for joining me today thank you for subscribing and I'll see everybody next week bye | {"url":"https://bywoodworking.com/how-to-calculate-the-height-of-a-ridge-board/","timestamp":"2024-11-06T01:42:56Z","content_type":"text/html","content_length":"87271","record_id":"<urn:uuid:8df2751a-ee86-4427-9a17-bec47837d507>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00100.warc.gz"} |
Configure XGBoost "tree_method" Parameter
Choosing the appropriate “tree_method” parameter in XGBoost is crucial for optimizing both the speed of training and the performance of the model, especially when dealing with large datasets. This
tip explores how to select the best tree construction algorithm based on your data size and computational resources.
from sklearn.datasets import make_classification
from sklearn.model_selection import train_test_split
from xgboost import XGBClassifier
# Generate synthetic data
X, y = make_classification(n_samples=1000, n_features=20, n_informative=2, n_redundant=10, random_state=42)
# Split the dataset into training and test sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
# Configure the XGBoost model with a specific tree method
model = XGBClassifier(tree_method='hist', eval_metric='logloss')
# Fit the model
model.fit(X_train, y_train)
# Make predictions
predictions = model.predict(X_test)
Understanding the “tree_method” Parameter
The “tree_method” parameter in XGBoost specifies the algorithm used to construct the trees. It has several options, including:
• ‘auto’: XGBoost selects the most appropriate method based on the dataset.
• ’exact’: Utilizes an exact greedy algorithm. Best for small to medium datasets where precision is paramount.
• ‘approx’: Employs a histogram-based approximation of the greedy algorithm. Ideal for larger datasets to balance performance and speed.
• ‘hist’: Uses a faster histogram optimized algorithm, suitable for most datasets due to its effective balance of memory usage and speed.
Choosing the Right “tree_method” Value
Selecting the correct “tree_method” depends on your dataset and available resources:
• Use ’exact’ when your dataset is not extremely large and model accuracy is the critical factor.
• Opt for ‘approx’ or ‘hist’ for larger datasets, where training speed becomes a more significant consideration.
Practical Tips
• Begin with the ‘auto’ setting to allow XGBoost to automatically determine the best method for your data.
• Experiment with different methods to see which provides the best trade-off between training speed and model accuracy.
• Use cross-validation to evaluate the impact of different tree methods on your model’s performance, particularly to guard against overfitting.
• Always consider the hardware environment when choosing a tree method, especially if transitioning from a development to a production setting where different computational resources might be | {"url":"https://xgboosting.com/configure-xgboost-tree_method-parameter/","timestamp":"2024-11-02T18:04:44Z","content_type":"text/html","content_length":"8213","record_id":"<urn:uuid:0662a266-86e1-4d6b-ac2e-7af0cbee1d3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00787.warc.gz"} |
Simplify radical expression online
simplify radical expression online Related topics: pre algebra combining like term questions
intermediate algebra,16
divide rational expressions involving polynomials
5th grade math common factors
cube elementary school 2math games"
root mean square ti-89
number word poems
Author Message
Wo1f Posted: Tuesday 10th of Jan 20:33
Heya guys! Is someone here familiar with simplify radical expression online? I have this set of questions about it that I just can’t understand. Our class was tasked to solve it and
understand how we came up with the solution. Our Math professor will select random students to answer it as well as show solutions to class so I require comprehensive explanation about
simplify radical expression online. I tried answering some of the questions but I guess I got it completely wrong . Please help me because it’s a bit urgent and the deadline is close
already and I haven’t yet understood how to solve this.
Back to top
kfir Posted: Wednesday 11th of Jan 12:17
I don’t think I know of any website where you can get your solutions of simplify radical expression online checked within hours. There however are a couple of websites which do offer
assistance, but one has to wait for at least 24 hours before expecting some response.What I know for sure is that, this program called Algebrator, that I used during my college career was
really good and I was quite happy with it. It almost gives the type of results you need.
From: egypt
Back to top
Admilal Posted: Thursday 12th of Jan 16:41
`Leker It would really be great if you could tell us about a software that can provide both. If you could get us a home tutoring software that would give a step-by-step solution to our problem,
it would really be great . Please let us know the authentic websites from where we can get the tool .
From: NW
AR, USA
Back to top
lutidodi Posted: Friday 13th of Jan 09:08
To begin with, thanks for replying guys ! I really want to buy this program. Can you please tell me how to order this software? Can we order it through the web, or do we buy it from some
retail store?
From: Nova
Back to top
Gools Posted: Saturday 14th of Jan 07:07
You can download this program from https://softmath.com/algebra-software-guarantee.html. There are some demos available to see if it is really what want and if you are interested , you
can get a licensed version for a nominal amount.
From: UK
Back to top
Svizes Posted: Saturday 14th of Jan 14:18
function domain, converting fractions and adding functions were a nightmare for me until I found Algebrator, which is really the best math program that I have ever come across. I have
used it frequently through several math classes – Algebra 1, Pre Algebra and Remedial Algebra. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step
solution to the problem, and my math homework would be ready. I really recommend the program.
Back to top | {"url":"https://www.softmath.com/algebra-software-6/simplify-radical-expression.html","timestamp":"2024-11-09T02:39:20Z","content_type":"text/html","content_length":"42997","record_id":"<urn:uuid:9d3a2b1d-c5c3-4bd6-bba8-1525768a7efe>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00409.warc.gz"} |
Thunder Chicken CCT
This thread didn’t import into Discourse properly and include the whitepaper, so I am inserting it here …@brandon_martus
This zip file contains all the information about the 2002 ThunderChickens CCT. It contains a Word document describing the theory and math, an Excel document with the Bill of Materials, AutoCad 2000
drawings, a STEP solid model of the assembly, SolidWorks 2001Plus native solid models, assemblies, and drawings. This is the version of the CCT that we used at the Championship and has all the bugs
worked out. I hope this information will prove useful to anyone who tries to use it.
Have fun with it!!
-Paul Copioli
Team 217
EDIT: The CCT is now U.S. Patent Pending as of 2/20/2003
100262342342342002_thunderchickens_cct.zip (8.7 MB)
Originally created 07-22-2002 10:32 PM by @Paul_Copioli
This is specifically directed towards Paul on team 217, but if anyone else can help, I’d appreciate it!
I’m very interested in understanding and perhaps building a version of the CCT (the one that’s in the white papers) I’ve been doing quite a bit of analysis of the CCT setup, and I have some
1. I calculate the last stage of the bosch gearbox to have a 4.31:1 ratio, with a sun of 13 teeth, and a ring of 43 teeth. However, I counted 14 teeth on the planets, which seems to be one tooth
short… why do they allow such an amount of backlash?
2. Given the above gear ratio, and the ratio of the initial gear train between the chip motor and the sun, I end up with a total gear ratio of almost 61.6:1, with a 276 max RPM (if you run the chip
motor at 30Amp max RPM of 4048) Correct?
3. I assume you don’t exceed 276 RPM for the angular speed of the ring gear (the one driven by the worm). Thus your maximum speed ratio goes from 1:1 up to 4.31:1. Could you actually run the ring
faster than the sun, to get a output of faster than the max RPM? That doesn’t seem like a good idea…
4. OK, my main question:
Given the above ratio, and the torque that the system puts out at the max torque of the input motor (I get almost 36 NM @ 30A), your reaction torque (ring torque) is about 28.5 NM. Yes?
With a worm rato of 30:1 (60 teeth / 2 thread worm), you need 0.95 NM of torque @ up to 8280 RPM to drive the worm gear assembly. Yes?
At 30A each, the drill motor gives you 0.157 NM, and the FP motor 0.174 NM, which doesn’t add up to enough torque to spin the worm. Furthermore, at that load, the FP will only spin at 7700 RPM.
How do you get this to work? It seems to me that you would be tripping the breakers on the drill and FP motors at high loads, as you’re asking too much of them. Or do you have a further gear
reduction outside of this box, such that you don’t need as much reaction torque?
Sorry for the long message, but I’m truely enthralled by your design, and I’d like to understand it further. It’s quite a neat idea.
Simon G.
Mentor, Cheshire HS
Sikorsky Aircraft
I apologize for taking so long to get back to you, but I have been on vacation. I will answer in the order you asked your questions.
1. The reason is interference. There are 3 different interference conditions to deal with when designing planetary gears. Bosch chose to allow more backlash to avoid the interference, because
backlash is not that big of a deal for the kind of drill it was designed for.
2. Your gear ratio is slightly off, but here is the exact number:
(44/9) * (48/16) * (43/13 +1) = 63.18:1
That ratio is the speed ratio of the output if the ring gear is not spinning. At max speed of Chip that would be 88 RPM. The sun gear will spin at 375 RPM at max chip speed. The speed at 30 amps
is important, because that is the speed you will be going at max pulling time. That speed is 64 RPM at output and, as you said, 276 at the sun gear.
3. Oh, but we do spin the ring gear faster than the sun. Our design speed for the ring gear is 600 RPM. That would make our total speed ratio range from 63.18:1 to ~8:1. You are right that there are
some limitations when spinning the ring gear that fast.
4. I will try to address all your concerns here. All of the things you said in question 4 are true, however, we never spun the ring gear while at design load!! The worm gear setup I chose gave us
pretty good backdriving resistance. What I mean is that the FP and Drill did not have to work when sitting still (most of the time). The point at which a worm gear set is backdriveable is highly
dependent on 2 things: lead angle (number of threads) and Mu between the gear mesh. Every once in a while we would start backdriving the worm gear and it would look like we were in neutral. We
weren’t in neutral, but we were in a no move situation. You could stop us with just a pinky finger, but you could not move us back with 200 lbs pushing force. We did not like this condition, so
we had a “pulsing” button on one of the joysticks that would pulse power to the FP and drill to stop backdriving. After a few trials (and burnt motors, as you eluded to), we got the pulsing
routine to work. We did not have any further gear reduction, just used the backdriving resistance of worm gears.
Now, our design intent for the CCT had 2 main objectives: Max pushing force of 280 lbs and max speed of 10ft/sec. We knew you had to get to the goals fast and once you had them it did not matter.
What we found is that with 1 goal, we could spin the ring gear at 600 rpm and move quite easily (drained the battery), but once we had 2 goals and transferred the weight … no more ring gear spinning
except for anti-backdrive. We had it designed for a specific purpose, but if I were to design it so I could use the spinning gears at all times no matter what, then I would spin the ring gear at
about 75% of the sun gear speed at max speed (contrary to our 200% of sun gear speed). This only gives you a 3:1 jump in gear ratio as opposed to the 4.31:1 jump you talked about. The advantage of
the 3:1 jump is that you get max torque available at all times, even at high speed; but you do not get the variation in gear ration that you require.
The one thing I will do differently this year is LOCK the ring gear in place when I am slow/torque mode, probably with pneumatics. You can do it with the CCT as designed, but need to add some holes.
Another thing, do not directly couple the motors to the worm shaft as I did. Put the motors below the worm shaft and add another gear stage (1.5 or 2:1), it will make replacement easier.
I hope I helped.
Thanks for the detailed reply. It was very helpful. We actually looked at a similar design here at Sikorsky a while back for large scale helicopter applications…!
One more question. Your current worm lead angle is very close to 10 degrees (9 deg, 24’ to be exact), which would give you backdriving in extreme conditions, especially with your smooth, ground worm…
Why didn’t you use a single thread worm and a 30 tooth gear instead? (thus cutting your lead angle)
Was it a geometry (packaging) issue, or am I missing something critical? Your worm shaft thrust would go up with a smaller worm gear, but the torque requirements would be the same, right? Does the
efficiency drop with a smaller lead angle? Is there a limitation on RPM with a single thread worm too?
Sorry, but one more quick question…
It’s unclear how you actually attach the drill ring gear to your housing that holds the worm gear. I saw there was a note ref. on your 111-000 assembly drawing, but I didn’t see the actual note. It’s
all steel… are you welding it or brazing it together?
Thanks again
PS… did you actually spend ~ $130 per bearing for the Kaydon reali-slim part?! (we have a very slim budget) Or do you have a contact who might donate something like that?
The worm lead angle has a huge impact on efficiency. I would really want to use a quad thread worm, but I needed some backdriving resistance, so I chose dual. A single lead was just too inefficient.
The ring gear is press fit into a steel part (TC-2002-111-005)
that the worm gear (TC-2002-111-001) is bolted to. We used liquid Nitrogen (just because it is fun to play with) to shrink the ring gear and then pressed it into the steel part.
FANUC Robotics (the company I work for) is the largest non-military customer for KAYDON. I know our sales representative personally and he gives me any standard KAYDON bearing for free. In return,
KAYDON is placed as a sponsor to our team. So, to answer your question; no we do not pay one penny for the KAYDON bearings. If you decide to use the CCT and need a couple of those bearings, I can
probably get you a couple. | {"url":"https://www.chiefdelphi.com/t/thunder-chicken-cct/39311","timestamp":"2024-11-08T01:25:19Z","content_type":"text/html","content_length":"32264","record_id":"<urn:uuid:3b467bdb-73c7-4d80-b0c3-17124cbcdacb>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00318.warc.gz"} |
Maths Pop Up Quiz
Questions and Answers
• 1.
27 is a prime number.
Correct Answer
B. False
A prime number is a number that is only divisible by 1 and itself. However, 27 is divisible by 1, 3, 9, and 27, so it is not a prime number. Therefore, the correct answer is False.
• 2.
A composite number has more than two factors.
Correct Answer
A. True
A composite number is a positive integer that has more than two distinct positive divisors. This means that a composite number can be divided evenly by more than just 1 and itself. For example,
the number 6 is a composite number because it can be divided evenly by 1, 2, 3, and 6. Therefore, the statement "A composite number has more than two factors" is true.
• 3.
All prime numbers are odd.
Correct Answer
B. False
The statement is false because not all prime numbers are odd. The only even prime number is 2. All other prime numbers are odd.
• 4.
The number 9 has one prime factor.
Correct Answer
A. True
A prime factor is a prime number that divides another number evenly without leaving a remainder. The number 9 can be expressed as 3 x 3, where both factors are prime numbers. Therefore, the
statement that the number 9 has one prime factor is true.
• 5.
To find the product of number, We must divide.
Correct Answer
B. False
This statement is incorrect. To find the product of numbers, we must multiply them, not divide. Division is used to find the quotient or the result of dividing one number by another.
• 6.
To square a number, I must multiply the number by itself.
Correct Answer
A. True
The statement is true because squaring a number means multiplying the number by itself. For example, if we square the number 2, we get 2 x 2 = 4. So, in order to square any number, we need to
multiply it by itself.
• 7.
Average is sum divided by number.
Correct Answer
A. True
The statement is true because the average of a set of numbers is calculated by summing all the numbers together and then dividing that sum by the total number of values in the set. This is a
fundamental concept in mathematics and statistics.
• 8.
The area of a square whose perimeter is 36 cm is 81 square cm.
Correct Answer
A. True
The perimeter of a square is the sum of all its sides. In this case, the perimeter is 36 cm. Since a square has four equal sides, each side of the square would be 36 cm divided by 4, which is 9
cm. The area of a square is calculated by multiplying the length of one side by itself. Therefore, the area of this square would be 9 cm multiplied by 9 cm, which equals 81 square cm. Hence, the
statement is true.
• 9.
A hexagon has 5 angles.
Correct Answer
B. False
A hexagon has 6 angles, not 5. Each angle in a regular hexagon measures 120 degrees, so the total sum of all angles in a hexagon is 720 degrees. Therefore, the statement that a hexagon has 5
angles is false.
• 10.
All quadrilateral have four equal sides.
Correct Answer
B. False
The statement "All quadrilaterals have four equal sides" is false. Quadrilaterals can have sides of different lengths. Some examples of quadrilaterals with unequal sides include rectangles,
squares, parallelograms, and trapezoids. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=maths-pop-up-quiz","timestamp":"2024-11-10T15:45:43Z","content_type":"text/html","content_length":"447790","record_id":"<urn:uuid:5f93bbf2-5176-4f8a-882b-11153d1c47b7>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00281.warc.gz"} |
15.2 HMC algorithm parameters | Stan Reference Manual
15.2 HMC algorithm parameters
The Hamiltonian Monte Carlo algorithm has three parameters which must be set,
• discretization time \(\epsilon\),
• metric \(M\), and
• number of steps taken \(L\).
In practice, sampling efficiency, both in terms of iteration speed and iterations per effective sample, is highly sensitive to these three tuning parameters Neal (2011), Hoffman and Gelman (2014).
If \(\epsilon\) is too large, the leapfrog integrator will be inaccurate and too many proposals will be rejected. If \(\epsilon\) is too small, too many small steps will be taken by the leapfrog
integrator leading to long simulation times per interval. Thus the goal is to balance the acceptance rate between these extremes.
If \(L\) is too small, the trajectory traced out in each iteration will be too short and sampling will devolve to a random walk. If \(L\) is too large, the algorithm will do too much work on each
If the inverse metric \(M^{-1}\) is a poor estimate of the posterior covariance, the step size \(\epsilon\) must be kept small to maintain arithmetic precision. This would lead to a large \(L\) to
Integration time
The actual integration time is \(L \, \epsilon\), a function of number of steps. Some interfaces to Stan set an approximate integration time \(t\) and the discretization interval (step size) \(\
epsilon\). In these cases, the number of steps will be rounded down as
\[ L = \left\lfloor \frac{t}{\epsilon} \right\rfloor. \]
and the actual integration time will still be \(L \, \epsilon\).
Automatic parameter tuning
Stan is able to automatically optimize \(\epsilon\) to match an acceptance-rate target, able to estimate \(M\) based on warmup sample iterations, and able to dynamically adapt \(L\) on the fly during
sampling (and during warmup) using the no-U-turn sampling (NUTS) algorithm Hoffman and Gelman (2014).
Warmup Epochs Figure. Adaptation during warmup occurs in three stages: an initial fast adaptation interval (I), a series of expanding slow adaptation intervals (II), and a final fast adaptation
interval (III). For HMC, both the fast and slow intervals are used for adapting the step size, while the slow intervals are used for learning the (co)variance necessitated by the metric. Iteration
numbering starts at 1 on the left side of the figure and increases to the right.
When adaptation is engaged (it may be turned off by fixing a step size and metric), the warmup period is split into three stages, as illustrated in the warmup adaptation figure, with two fast
intervals surrounding a series of growing slow intervals. Here fast and slow refer to parameters that adapt using local and global information, respectively; the Hamiltonian Monte Carlo samplers, for
example, define the step size as a fast parameter and the (co)variance as a slow parameter. The size of the the initial and final fast intervals and the initial size of the slow interval are all
customizable, although user-specified values may be modified slightly in order to ensure alignment with the warmup period.
The motivation behind this partitioning of the warmup period is to allow for more robust adaptation. The stages are as follows.
1. In the initial fast interval the chain is allowed to converge towards the typical set,^20 with only parameters that can learn from local information adapted.
2. After this initial stage parameters that require global information, for example (co)variances, are estimated in a series of expanding, memoryless windows; often fast parameters will be adapted
here as well.
3. Lastly, the fast parameters are allowed to adapt to the final update of the slow parameters.
These intervals may be controlled through the following configuration parameters, all of which must be positive integers:
Adaptation Parameters Table. The parameters controlling adaptation and their default values.
parameter description default
initial buffer width of initial fast adaptation interval 75
term buffer width of final fast adaptation interval 50
window initial width of slow adaptation interval 25
Discretization-interval adaptation parameters
Stan’s HMC algorithms utilize dual averaging Nesterov (2009) to optimize the step size.^21
This warmup optimization procedure is extremely flexible and for completeness, Stan exposes each tuning option for dual averaging, using the notation of Hoffman and Gelman (2014). In practice, the
efficacy of the optimization is sensitive to the value of these parameters, but we do not recommend changing the defaults without experience with the dual-averaging algorithm. For more information,
see the discussion of dual averaging in Hoffman-Gelman:2014.
The full set of dual-averaging parameters are
Step Size Adaptation Parameters Table The parameters controlling step size adaptation, with constraints and default values.
parameter description constraint default
delta target Metropolis acceptance rate [0, 1] 0.8
gamma adaptation regularization scale (0, infty) 0.05
kappa adaptation relaxation exponent (0, infty) 0.75
t_0 adaptation iteration offset (0, infty) 10
By setting the target acceptance parameter \(\delta\) to a value closer to 1 (its value must be strictly less than 1 and its default value is 0.8), adaptation will be forced to use smaller step
sizes. This can improve sampling efficiency (effective sample size per iteration) at the cost of increased iteration times. Raising the value of \(\delta\) will also allow some models that would
otherwise get stuck to overcome their blockages.
Step-size jitter
All implementations of HMC use numerical integrators requiring a step size (equivalently, discretization time interval). Stan allows the step size to be adapted or set explicitly. Stan also allows
the step size to be “jittered” randomly during sampling to avoid any poor interactions with a fixed step size and regions of high curvature. The jitter is a proportion that may be added or
subtracted, so the maximum amount of jitter is 1, which will cause step sizes to be selected in the range of 0 to twice the adapted step size. The default value is 0, producing no jitter.
Small step sizes can get HMC samplers unstuck that would otherwise get stuck with higher step sizes. The downside is that jittering below the adapted value will increase the number of leapfrog steps
required and thus slow down iterations, whereas jittering above the adapted value can cause premature rejection due to simulation error in the Hamiltonian dynamics calculation. See Neal (2011) for
further discussion of step-size jittering.
Euclidean metric
All HMC implementations in Stan utilize quadratic kinetic energy functions which are specified up to the choice of a symmetric, positive-definite matrix known as a mass matrix or, more formally, a
metric Betancourt (2017).
If the metric is constant then the resulting implementation is known as Euclidean HMC. Stan allows a choice among three Euclidean HMC implementations,
• a unit metric (diagonal matrix of ones),
• a diagonal metric (diagonal matrix with positive diagonal entries), and
• a dense metric (a dense, symmetric positive definite matrix)
to be configured by the user.
If the metric is specified to be diagonal, then regularized variances are estimated based on the iterations in each slow-stage block (labeled II in the warmup adaptation stages figure). Each of these
estimates is based only on the iterations in that block. This allows early estimates to be used to help guide warmup and then be forgotten later so that they do not influence the final covariance
If the metric is specified to be dense, then regularized covariance estimates will be carried out, regularizing the estimate to a diagonal matrix, which is itself regularized toward a unit matrix.
Variances or covariances are estimated using Welford accumulators to avoid a loss of precision over many floating point operations.
Warmup times and estimating the metric
The metric can compensate for linear (i.e. global) correlations in the posterior which can dramatically improve the performance of HMC in some problems. This requires knowing the global correlations.
In complex models, the global correlations are usually difficult, if not impossible, to derive analytically; for example, nonlinear model components convolve the scales of the data, so standardizing
the data does not always help. Therefore, Stan estimates these correlations online with an adaptive warmup. In models with strong nonlinear (i.e. local) correlations this learning can be slow, even
with regularization. This is ultimately why warmup in Stan often needs to be so long, and why a sufficiently long warmup can yield such substantial performance improvements.
The metric compensates for only linear (equivalently global or position-independent) correlations in the posterior. The hierarchical parameterizations, on the other hand, affect some of the nasty
nonlinear (equivalently local or position-dependent) correlations common in hierarchical models.^22
One of the biggest difficulties with dense metrics is the estimation of the metric itself which introduces a bit of a chicken-and-egg scenario; in order to estimate an appropriate metric for
sampling, convergence is required, and in order to converge, an appropriate metric is required.
Dense vs. diagonal metrics
Statistical models for which sampling is problematic are not typically dominated by linear correlations for which a dense metric can adjust. Rather, they are governed by more complex nonlinear
correlations that are best tackled with better parameterizations or more advanced algorithms, such as Riemannian HMC.
Warmup times and curvature
MCMC convergence time is roughly equivalent to the autocorrelation time. Because HMC (and NUTS) chains tend to be lowly autocorrelated they also tend to converge quite rapidly.
This only applies when there is uniformity of curvature across the posterior, an assumption which is violated in many complex models. Quite often, the tails have large curvature while the bulk of the
posterior mass is relatively well-behaved; in other words, warmup is slow not because the actual convergence time is slow but rather because the cost of an HMC iteration is more expensive out in the
Poor behavior in the tails is the kind of pathology that can be uncovered by running only a few warmup iterations. By looking at the acceptance probabilities and step sizes of the first few
iterations provides an idea of how bad the problem is and whether it must be addressed with modeling efforts such as tighter priors or reparameterizations.
NUTS and its configuration
The no-U-turn sampler (NUTS) automatically selects an appropriate number of leapfrog steps in each iteration in order to allow the proposals to traverse the posterior without doing unnecessary work.
The motivation is to maximize the expected squared jump distance (see, e.g., Roberts, Gelman, and Gilks (1997)) at each step and avoid the random-walk behavior that arises in random-walk Metropolis
or Gibbs samplers when there is correlation in the posterior. For a precise definition of the NUTS algorithm and a proof of detailed balance, see Hoffman and Gelman (2014).
NUTS generates a proposal by starting at an initial position determined by the parameters drawn in the last iteration. It then generates an independent standard normal random momentum vector. It then
evolves the initial system both forwards and backwards in time to form a balanced binary tree. At each iteration of the NUTS algorithm the tree depth is increased by one, doubling the number of
leapfrog steps and effectively doubles the computation time. The algorithm terminates in one of two ways, either
• the NUTS criterion (i.e., a U-turn in Euclidean space on a subtree) is satisfied for a new subtree or the completed tree, or
• the depth of the completed tree hits the maximum depth allowed.
Rather than using a standard Metropolis step, the final parameter value is selected via multinomial sampling with a bias toward the second half of the steps in the trajectory Betancourt (2016b).^23
Configuring the no-U-turn sample involves putting a cap on the depth of the trees that it evaluates during each iteration. This is controlled through a maximum depth parameter. The number of leapfrog
steps taken is then bounded by 2 to the power of the maximum depth minus 1.
Both the tree depth and the actual number of leapfrog steps computed are reported along with the parameters in the output as treedepth__ and n_leapfrog__, respectively. Because the final subtree may
only be partially constructed, these two will always satisfy
\[ 2^{\mathrm{treedepth} - 1} - 1 \ < \ N_{\mathrm{leapfrog}} \ \le \ 2^{\mathrm{treedepth} } - 1. \]
Tree depth is an important diagnostic tool for NUTS. For example, a tree depth of zero occurs when the first leapfrog step is immediately rejected and the initial state returned, indicating extreme
curvature and poorly-chosen step size (at least relative to the current position). On the other hand, a tree depth equal to the maximum depth indicates that NUTS is taking many leapfrog steps and
being terminated prematurely to avoid excessively long execution time. Taking very many steps may be a sign of poor adaptation, may be due to targeting a very high acceptance rate, or may simply
indicate a difficult posterior from which to sample. In the latter case, reparameterization may help with efficiency. But in the rare cases where the model is correctly specified and a large number
of steps is necessary, the maximum depth should be increased to ensure that that the NUTS tree can grow as large as necessary.
Hoffman, Matthew D., and Andrew Gelman. 2014. “The No-U-Turn Sampler: Adaptively Setting Path Lengths in Hamiltonian Monte Carlo.” Journal of Machine Learning Research 15: 1593–1623. http://jmlr.org/
Neal, Radford. 2011. “MCMC Using Hamiltonian Dynamics.” In Handbook of Markov Chain Monte Carlo, edited by Steve Brooks, Andrew Gelman, Galin L. Jones, and Xiao-Li Meng, 116–62. Chapman; Hall/CRC.
Nesterov, Y. 2009. “Primal-Dual Subgradient Methods for Convex Problems.” Mathematical Programming 120 (1): 221–59.
Roberts, G.O., Andrew Gelman, and Walter R. Gilks. 1997. “Weak Convergence and Optimal Scaling of Random Walk Metropolis Algorithms.” Annals of Applied Probability 7 (1): 110–20. | {"url":"https://mc-stan.org/docs/2_26/reference-manual/hmc-algorithm-parameters.html","timestamp":"2024-11-04T07:19:44Z","content_type":"text/html","content_length":"105735","record_id":"<urn:uuid:0007c88f-fbce-4e7a-8530-1bce3268d10e>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00633.warc.gz"} |
Conservation of Momentum - Physics 1 Impulse and Momentum
Conservation of Momentum
A champagne bottle of mass 90 g has a 2 g cork in it. The pressure built up in the bottle causes the cork to shoot out, as shown below. If after the explosion, the bottle is moving directly to the
left at 4 m/s, what is the velocity of the cork?
Posted by Kim Miller a year ago
Related Problems
A 2kg car is traveling east at 3 m/s and a 4kg car is traveling west at 6 m/s. Now assume the cars bounce off each other in a perfectly elastic collision. The final velocity of the 2kg car is -4 m/s.
Find the direction and velocity of the other car after the collision.
The red ball (4 kg) hits the blue ball (5 kg) moving at 2 m/s, and the balls move as shown in the figure below. $\theta$ = 30 degrees. After the impact, the red ball is measured moving at 1.8 m/s.
What is the velocity of the blue ball, and at what angle relative to the horizontal is the ball moving after the impact?
The mass shown below is blown apart into three pieces. Find the direction and velocity of the pink piece.
A crash test car of mass 1,000 kg moving at a constant speed of 12 m/s collides completely inelastically with an object of mass M at time t = 0. The ojbect was initally at rest. The speed v in m/s of
the car object system after the collision is given as a function of time t in seconds by the expression
$v = \frac{8}{1 + 5t}$
A. Calculate the mass M of the object
B. Assuming an initial position of x = 0, determine an expression for the position of the car object system after the collision as a function of time t.
C. Determine an expression for the resisting force on the car object system after the collision as a function of time t.
D. Determine the impulse delivered to the car object system from t = 0 to t = 2.0 s. | {"url":"https://www.practiceproblems.org/problem/Physics_1/Impulse_and_Momentum/Conservation_of_Momentum","timestamp":"2024-11-02T23:39:29Z","content_type":"text/html","content_length":"36989","record_id":"<urn:uuid:f739cbcb-e6bf-41c0-ba75-77292706dec8>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00878.warc.gz"} |
2X2 Between Anova | Zach Shipstead
top of page
2 x 2 ANOVA, between subjects variables
ds = dataset you are currently using.
DV = dependent variable of interest
IV = independent variable of interest
XYXY = dummy name for a variable, matrix, or data frame into which you are moving information.
Balanced Design
XYXY<-aov(DV ~ IV1 * IV2, data=ds)
Unbalanced Design
I'm going to approach the unbalanced design from the SPSS perspective. I understand some don't agree with this. But if you're so above the SPSS approach, I'm not sure what you're doing here.
The problem: R defaults to type 1 sum of squares, which can give wildly varying answers, depending on the order in which variables are entered.
The solution: (1) Default contrasts that R runs need to be changed. (2) The "Anova()" function from the "car" package, which allows you to specify type III sum of squares.
XYXY <- aov(DV ~ IV1 * IV2,
contrasts=list(IV1 = 'contr.sum', IV2 = 'contr.sum'),
data = ds)
Anova(XYXY, type = 3)
bottom of page | {"url":"https://www.zshipstead.com/copy-of-one-way-anova","timestamp":"2024-11-09T11:00:59Z","content_type":"text/html","content_length":"428295","record_id":"<urn:uuid:933076c3-8e49-4be4-8a3d-97f291547ed7>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00366.warc.gz"} |
MCQ Questions Class 10 Mathematics Circles with Answer
MCQs for Mathematics Class 10 with Answers Chapter 10 Circles
Students of class 10 Mathematics should refer to MCQ Questions Class 10 Mathematics Circles with answers provided here which is an important chapter in Class 10 Mathematics NCERT textbook. These MCQ
for Class 10 Mathematics with Answers have been prepared based on the latest CBSE and NCERT syllabus and examination guidelines for Class 10 Mathematics. The following MCQs can help you to practice
and get better marks in the upcoming class 10 Mathematics examination
Chapter 9 Circles MCQ with Answers Class 10 Mathematics
MCQ Questions Class 10 Mathematics Circles provided below have been prepared by expert teachers of grade 10. These objective questions with solutions are expected to come in the upcoming Standard 10
examinations. Learn the below provided MCQ questions to get better marks in examinations.
Question. Two circle touch each other externally at C and AB is a common tangent to the circles. Then,∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Question. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at anangle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Question. A circle with centre O is inscribed in a right angle triangle ABC, right angled at B. If BC = 15 cm, AC = 17 cm, the radius ‘r’ of the circle is :
(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 5 cm
Question. If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equalto
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Question. The length of a tangent drawn from a point at a distance of 10 cm of circle is 8 cm. The radius ofthe circle is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 7 cm
Question. Tangents from an external point to a circle are
(a) equal
(b) not equal
(c) parallel
(d) perpendicular
Question. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that <POR = 120°, then <OPQ is :
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Question. Two parallel lines touch the circle at points A and B respectively. If area of the circle is 25 n cm 2 ,then AB is equal to
(a) 5 cm
(b) 8 cm
(c) 10 cm
(d) 25 cm
Question. A line through point of contact and passing through centre of circle is known as
(a) tangent
(b) chord
(c) normal
(d) segment
Question. Segment joining the points of contact of two parallel tangents
(a) may or may not pass through the centre.
(b) will pass through the centre.
(c) will not pass through the centre.
(d) none of these
Question. At point A on a diameter AB of a circle of radius 10 cm, tangent XAY is drawn to the circle. Thelength of the chord CD parallel to XY at a distance 16 cm from A is
(a) 8 cm
(b) 10 cm
(c) 16 cm
(d) 18 cm
Question. Number of tangents drawn at a point of the , circle is/are
(a) one
(b) two
(c) none
(d) infinite
Question. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point P. The length of TP is :
(a) 5 cm
(b) 5 2/ 3 cm
(c) 6 cm
(d) 6 2/3 cm
Question. A circle touches all the four sides of a quadrilateral ABCD, with AB = 7 cm, BC = 8 cm and CD = 5 cm. The length of AD is :
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Question. A tangent is drawn from a point at a distance of 17 cm of circle C(0, r) of radius 8 cm. The lengthof its tangent is
(a) 5 cm
(b) 9 cm
(c) 15 cm
(d) 23 cm
Question. The maximum number of common tangents that can be drawn to two circles intersecting at twodistinct point is
(a) 2
(b) 4
(c) 1
(d) 3
Question. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q
(a) √119 cm
(b) 13 cm
(c) 12 cm
(d) 8.5 cm
Question. In a circle of radius 7cm, tangent PT is drawn from point P such that PT = 24cm. If O is the centreof the circle, then the length of OP is:
(a) 30cm
(b) 31cm
(c) 28cm
(d) 25cm
Question. The length of a tangent from a point 25 cm from the centre of a circle is 24 cm. The radius of the circle is:
(a) 5 cm
(b) 7 cm
(c) 10 cm
(d) 12 cm
Question. If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Question. A tangent PQ at a point P of a circle of radius 8 cm meets a line through the centre O at a point Q such that OQ = 15 cm. Length PQ is :
(a) 17 cm
(b) 10 cm
(c) 125 cm
(d) none of these
Question. The tangents drawn at the extremities of the diameter of a circle are
(a) perpendicular
(b) parallel
(c) equal
(d) none of these
Question. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, thepair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR
(a) 60 cm^ 2
(b) 65 cm^ 2
(c) 30 cm^2
(d) 32.5 cm^ 2
Question. A circle touches all the four sides of a quadrilateral ABCD. Then,
(a) AB + CD = AC + BC
(b) AB + CD = BC + AD
(c) AC + AD = BD + CD
(d) AC + BD = BC + DB
Question. Two circle touch each other externally at C and AB is a common tangent to the circles. Then,∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Question. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circlesuch that ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Question. Two circles touch each other externally at C and AB is a common tangent to the circles. Then, <ACB is:
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Question. Two circles of radii 10 cm and 8 cm are concentric. The length of a chord of the larger circle which touches the smaller is :
(a) 6 cm
(b) 11 cm
(c) 12 cm
(d) 13 cm
Question. The length of tangents drawn from an external point to the circle
(a) are equal
(b) are not equal
(c) sometimes are equal
(d) are not defined
Question. If TP and TQ are tangents to a circle with centre O such that <POQ = 110°, then <PTQ is equal to :
(a) 60°
(b) 70°
(c) 80°
(d) 90°
We hope the above multiple choice questions for Class 10 Mathematics for Chapter 9 Circles provided above with answers based on the latest syllabus and examination guidelines issued by CBSE, NCERT
and KVS are really useful for you. Circles is an important chapter in Class 10 as it provides very strong understanding about this topic. Students should go through the answers provided for the MCQs
after they have themselves solved the questions. All MCQs have been provided with four options for the students to solve. These questions are really useful for benefit of class 10 students. Please go
through these and let us know if you have any feedback in the comments section. | {"url":"https://dkgoelsolutions.com/mcqs-for-mathematics-class-10-with-answers-chapter-10-circles/","timestamp":"2024-11-07T16:20:55Z","content_type":"text/html","content_length":"140962","record_id":"<urn:uuid:ad145cd8-dacc-4d1f-bb43-a6f7327e1827>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00492.warc.gz"} |
What is kinetic energy? - Learn all about the energy of movement
What is kinetic energy? – Learn all about the energy of movement
Image Source: FreeImages
Have you ever wondered why a speeding car can cause so much damage? Or why a rolling ball can knock over a stack of blocks? The answer lies in kinetic energy – the energy of movement. Kinetic energy
is a fundamental concept in physics, describing the energy that an object possesses due to its motion. Understanding kinetic energy is crucial for engineers, scientists, and anyone interested in how
things work. In this article, we’ll explore the basics of kinetic energy, including its definition, formula, and examples. So, whether you’re a student, a science enthusiast, or someone who simply
wants to learn more about the world around you, read on to discover all about the fascinating energy of movement.
Understanding the concept of energy
Before we dive into the specifics of kinetic energy, it’s important to understand the concept of energy itself. Energy can be defined as the ability of an object to do work. Work, in turn, is defined
as the application of force over a distance. Essentially, energy is what makes things happen – it’s what allows us to move, to think, and to create.
However, energy comes in many different forms, each of which has its own unique properties and characteristics. One of the most fundamental distinctions is between potential and kinetic energy.
Potential energy is the energy that an object possesses due to its position or configuration, while kinetic energy is the energy that an object possesses due to its motion. In other words, potential
energy is the energy that could be released if an object were to move, while kinetic energy is the energy that is actually released when it does.
Difference between potential and kinetic energy
To understand the difference between potential and kinetic energy, let’s consider a simple example. Imagine that you’re holding a ball at the top of a hill. The ball has potential energy because it’s
in a position where it could roll down the hill and gain kinetic energy. However, until it actually starts moving, the ball doesn’t have any kinetic energy. Once you release the ball and it starts
rolling down the hill, it begins to gain kinetic energy as it picks up speed. The faster the ball goes, the more kinetic energy it has. At the same time, it loses potential energy as it moves farther
away from its original position at the top of the hill.
It’s important to note that potential and kinetic energy are not mutually exclusive – in fact, they’re often closely related. For example, a roller coaster at the top of a hill has potential energy
due to its height, but it also has kinetic energy because it’s moving. As the roller coaster starts to descend the hill, it begins to lose potential energy and gain kinetic energy. By the time it
reaches the bottom of the hill, it has converted all of its potential energy into kinetic energy.
Types of kinetic energy
Now that we’ve established what kinetic energy is and how it differs from potential energy, let’s take a closer look at the different types of kinetic energy. While all forms of kinetic energy
involve the motion of an object, there are several specific types that are worth noting.
One of the most common types of kinetic energy is translational kinetic energy, which is the energy that an object possesses due to its linear motion. This is the type of energy that a car has when
it’s moving down the highway, or that a person has when they’re running.
Another type of kinetic energy is rotational kinetic energy, which is the energy that an object possesses due to its rotational motion. This is the type of energy that a spinning top has, or that a
planet has as it orbits its star.
Finally, there is also vibrational kinetic energy, which is the energy that an object possesses due to its vibrational motion. This is the type of energy that a guitar string has when it’s plucked,
or that a tuning fork has when it’s struck.
Examples of kinetic energy in everyday life
Now that we’ve covered the basics of kinetic energy, let’s take a look at some examples of how it manifests in everyday life. In fact, kinetic energy is all around us – it’s what makes things move,
and what allows us to do work.
One of the most obvious examples of kinetic energy is in the movement of cars and other vehicles. When a car is moving down the road, it has a significant amount of kinetic energy due to its
translational motion. This is why a speeding car can cause so much damage – it has a lot of energy behind it.
Another example of kinetic energy is in sports. When a basketball player jumps to make a shot, they’re converting potential energy into kinetic energy as they move upwards. When they reach the peak
of their jump and start to come back down, they’re converting kinetic energy back into potential energy.
In fact, any time we move our bodies, we’re using kinetic energy. When we walk, run, or dance, we’re converting potential energy stored in our muscles into kinetic energy that propels us forward.
How kinetic energy is calculated
So, how do we calculate the amount of kinetic energy that an object has? The formula for kinetic energy is relatively simple:
KE = 1/2mv^2
In this formula, KE represents kinetic energy, m represents mass, and v represents velocity. The formula tells us that the amount of kinetic energy an object has is proportional to both its mass and
its velocity. Specifically, the kinetic energy of an object increases as its mass and velocity increase.
To see how this formula works in practice, let’s consider an example. Imagine that we have a baseball with a mass of 0.145 kilograms (the average weight of a baseball). If the baseball is thrown at a
velocity of 40 meters per second (roughly 90 miles per hour), we can calculate its kinetic energy as follows:
KE = 1/2 * 0.145 kg * (40 m/s)^2 KE = 1/2 * 0.145 kg * 1600 m^2/s^2 KE = 116 Joules
So, the baseball has a kinetic energy of 116 Joules. This may not seem like a lot, but it’s enough to knock over a stack of blocks or cause a bruise if it hits someone.
The conservation of kinetic energy
One of the fundamental principles of physics is the conservation of energy, which states that energy cannot be created or destroyed – it can only be converted from one form to another. This principle
applies to kinetic energy as well. In a closed system (i.e. a system with no external forces acting upon it), the total amount of kinetic energy remains constant.
This means that if one object loses kinetic energy, another object must gain an equal amount of kinetic energy. For example, if a ball is rolling across a table and starts to slow down, the kinetic
energy of the ball is being converted into other forms of energy – such as heat and sound – but the total amount of kinetic energy in the system remains constant.
Real-life applications of kinetic energy
Now that we’ve covered the basics of kinetic energy, let’s take a look at some real-life applications. Kinetic energy is used in a wide variety of fields, from engineering to medicine to
One of the most important applications of kinetic energy is in transportation. Cars, trains, planes, and other vehicles all rely on the energy of movement to get from one place to another. Engineers
and designers use their knowledge of kinetic energy to create vehicles that are efficient, safe, and reliable.
Another important application of kinetic energy is in medicine. Many medical devices, such as MRI machines and X-ray machines, rely on the principles of kinetic energy to function properly. Doctors
and scientists also use their understanding of kinetic energy to study the movement of cells and molecules within the body, which can help to diagnose and treat diseases.
Finally, kinetic energy is also used in entertainment. From roller coasters to water slides to bungee jumping, many forms of entertainment rely on the thrill of movement and the release of kinetic
Fun experiments to demonstrate kinetic energy
If you’re interested in learning more about kinetic energy, there are plenty of fun experiments you can do to see it in action. Here are a few ideas to get you started:
• Roll a ball down a ramp and measure how far it travels. Then, try rolling the same ball down a steeper ramp and see how much farther it goes. This demonstrates how the kinetic energy of the ball
increases as its velocity increases.
• Hold a pendulum at rest and then release it. Watch as it swings back and forth, converting potential energy into kinetic energy and back again.
• Fill a balloon with air and then release it. Watch as it flies across the room, propelled by the kinetic energy of the escaping air.
Conclusion: Importance of kinetic energy in the world around us
In conclusion, kinetic energy is a fundamental concept in physics that describes the energy of movement. Understanding kinetic energy is crucial for engineers, scientists, and anyone interested in
how things work. By learning about the different types of kinetic energy, how it’s calculated, and how it’s conserved, we can gain a deeper appreciation for the world around us and the role that
energy plays in our lives. Whether we’re designing new technologies, studying the human body, or simply enjoying a day at an amusement park, kinetic energy is always at play – and knowing how it
works can help us to better understand and appreciate the world we live in.
Leave a Comment | {"url":"https://wondersc.com/what-is-kinetic-energy-learn-all-about-the-energy-of-movement/","timestamp":"2024-11-03T18:29:19Z","content_type":"text/html","content_length":"91418","record_id":"<urn:uuid:831b3581-a64a-4b51-8601-af2e8268fa5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00660.warc.gz"} |
THE BRITISH SUBMARINE FLEET: 1992 AND BEYOND - NSL Archive
Like the other Western powers that saw the Cold War through to its highly successful conclusion, Britain is now reassessing its future force requirements. This far-reaching process is affecting every
element of all three Services — the Royal Navy’s Submarine service as much as any.
During the period of the Cold War the Submarine Service evolved into a force of three elements, all under the functional command of Aag Officer Submarines (FOSM), based at Headquarters
Commander-in-Chief Aeet (CINCFLEET) at Northwood, near London. The most important single element of the submarine flotilla was the Polaris force of four RESOLUTION class SSBNs. These were the outcome
of the Kennedy-MacMillan talks in Nassau in 1962, in which the President agreed to provide Polaris A3 missiles for installation in British-built submarines. Commissioned between 1967 and 1969 the
four boats took over sole responsibility for Britain’s independent strategic deterrent from the Royal Air Force.
The second element was a force of SSNs. Since the commissioning of the first British SSN, DREADNOUGHT, in 1963, a regular programme resulted in continuous production. In general terms, a new class of
5-7 boats joined the fleet in each decade and by 1990, with DREADNOUGHT stricken, there were eighteen boats in service. The third element was a small number of SSKs, all of which had been completed
in the 1960s.
The four current SSBNs displace 8,500 tons submerged and are generally similar to the U.S. Navy’s LAFAYEITE class. They have been regularly refitted and modernised, but are now approaching the end of
their service lives and there have been recent press reports of reactor problems in certain boats.
These SSBNs have been armed throughout their service by U.S.-supplied Polaris A3 SLBMs. These were rebuilt and remotored by Lockheed in the 1980s and the original British-built warheads were also
replaced between 1982 and 1988. The new warheads (code-named Chevaline ), are described by Norman Friedman in World Naval Weapons Systems, (USNI, 1991), as a “compromise between the multiple, but
co-targeted warheads of U.S. Polaris systems and the fully-independently targeted warheads of Poseidon;” i.e., a cross between MRV and MIRV. Officially designated Polaris A3TK, each missile carries
six lSOKT Chevaline warheads.
Over the years there has been some vociferous and widely reported, domestic political agitation to get rid of the nuclear deterrent This view was, in reality, confined to some fringe elements of
British society, and the maintenance of the SSBN force was never in serious doubt. Thus, when the question of replacing the RESOLUTION class was addressed in the 1970s, the question was not one of
principle, but rather of the most cost-effective way of achieving it.
Initially the government announced (1980) that it would purchase Trident I (C-4) SLBMs for a new class of SSBNs, to be commissioned during the 1990s. In 1982, however, it was decided that it would be
more sensible to purchase Trident IT (D-5). Firm orders were then placed for the boats; #1 in 1986, #2 in 1987 and #3 in 1990. The fourth has still not been ordered, although the Minister for Defence
Procurement announced on 1 July 1991 that the government “stood by its intention” to place such an order.
The new submarines will be much larger than their predecessors, displacing some 15,000 tons submerged. They will each carry sixteen Trident II (D-5) missiles, but, as with the Polaris, these will
carry an entirely British front-end, consisting of 1SOKT MIRVed warheads and decoys/penaids. It has also been publicly stated by the Ministry of Defence, that, although the missiles are capable of
carrying up to twelve warheads, the British will never mount more than eight. The first-of-class, VANGUARD, will be launched in February 1992 and commissioned later in the year, followed by the
others in 1994, 1995 and 1997, respectively.
The first British, new-build, post-war submarines were eight PORPOISE class boats, launched in 1956-59, which were quickly followed by thirteen of the very similar OBERON class. Fourteen OBERONs were
also exported to Australia (6), Brazil (3), Canada (3) and Chile (2). The Royal Navy’s OBERONs have served well, with no losses, and several took part in both the Falklands and Gulf wars, although
their role has been shrouded in secrecy.
The PORPOISE class was discarded between 1976 and 1987 after a relatively short operational life. The OBERONs have served longer and five underwent a major modification programme in the early 1980s,
and may remain in service for a few more years. Others of the class are being stricken at a slow rate, with the last due to strike in 1994. Egypt has already purchased one of the PORPOISE class boats
and the first of the OBERONs to be offered for sale, and has expressed a desire to buy six more as they become available.
There was a long debate throughout the 1970s about the value of building a successor to the OBERONs and the RN considered an all-nuclear submarine force, as exists in the United States Navy. It was
eventually decided, however, that diesel-electric submarines continue to have substantial advantages in some operational areas and are also much cheaper. The UPHOLDER class boats, for example, are
reported to cost some £150-200 million (U.S. $262.6- 350 million) to build compared with about £300 million (U.S. $525.5 million) for a TRAFALGAR class SSN, while the life-cycle cost of an SSN is
reported by Jane’s Defense Weekly (April27, 1991) to be some three times that of an SSK. Another important consideration is that SSKs require much smaller crews (although they are, admittedly,
smaller boats): UPHOLDER, for example, has a crew of 7 officers and 37 enlisted men compared to 12 officers and 97 enlisted men for TRAFALGAR.
These cost factors, coupled with the excellent operational performance of the SSKs, led to the new UPHOLDER class being ordered in 1983. The design was based on that of the Vickers Type 2400, which
was being marketed at that time by Vickers Shipbuilding to foreign navies.
The class has not been without its problems, which have included time delays, cost overruns and design faults. The delays on entry to service have been considerable; #1 was 3 years late and #2 18
months late, while #3 and #4 are estimated to be 6 and 3 months behind schedule, respectively. Part of the initial delay was due to a power-loss problem, and later a design fault was found in the
torpedo doors, which requires the first three to be docked for rectification, although the fourth will be modified during construction.
The VALIANT class comprises five boats completed between 1966 and 1971, and followed on from the first British SSN, DREADNOUGHT. These five boats were due to reach the end of their operational
careers in the mid-1990s, but a combination of reactor problems and the need to cut expenditure has led to the deletion of three in 1990-91, leaving just two (VALIANT and COURAGEOUS}, which, despite
recent refits, are also likely to be stricken in the near future. It has been a successful class. CHURCHILL carried out the UK trials for Sub-Harpoon and was also one of the first Western submarines
to be fitted with anechoic tiles to reduce the acoustic signature. CONQUEROR remains the only nuclear-powered submarine in any navy to have sunk a hostile surface warship (ARA GENERAL BELGRANO; May
2, 1982).
The six SWIFI’SURE class boats were built in the 1970s, introducing a new pressure-hull which maintains its diameter for a greater proportion of its length than in the earlier classes, giving much
greater usable internal volume. The forward hydroplanes are fitted in the bow below the waterline and retract into the outer casing. They have a very quiet hull form and all were given elastomeric
acoustic tile coatings during their first refits. They are powered by PWR-1 reactors with a core which gives a theoretical life of 12 years, although the refueling cycle will probably be about 8
years. They are fitted with five torpedo tubes, one less than in the earlier SSNs. Each boat is undergoing a 30-month mid-life refit, the first being completed in 1987, the second in 1989 and the
third in 1991, with the remainder following at two-year intervals. Assuming the usual 25-years operational life, the SWJFfSURE will be due for replacement between 1998 and 2006.
The TRAFALGAR class was ordered in 1977, the first-ofclass joining the fleet in 1983; the seventh and last will be commissioned in 1992. These boats incorporate yet further improvements, including a
new type of conformal anechoic tiling on both the pressure and outer hulls. All have strengthened fins and retractable bow hydroplanes for under-ice operations. TRAFALGAR is fitted with a
conventional 7-bladed propeller, but al! subsequent boats have a shrouded, pump-jet propulsor — a major British breakthrough in underwater technology.
THE PLAN IN 1990
The plan for the future of the submarine force as the Cold War drew to its close was fairly straightforward. The first three VAN GUARD class SSBNs had been ordered and a contract had been placed with
Vickers in 1987 for design work on the new W (SSN-20) class SSNs, with project definition having started in 1989. The plan was for a class of six (possibly seven), with the first being ordered in
1993 for commissioning in 2000. Also, construction of the first four UPHOLDER class SSKs was well in hand, to be followed by five (possibly eight) of a larger Batch 2 design.
All this was thrown into jeopardy by the end of the Cold War and the consequential reassessment of defence needs carried out by the Government and the Ministry of Defence in 1990/91. After
considerable discussion, much of it behind closed doors, the new plan is now becoming clear. The Royal Navy will reduce from some 31 submarines to 20, of which four will be SSBNs, four will be SSKs
and the balance of 12 will be SSNs. This, as always, will be the fleet total, and of the 20, those available immediately or at very short notice will be 2 SSBNs, 3 SSKs and 7-9 SSNs, while with
adequate notice the number of SSNs might increase to 10.
With three of the VAN GUARD class already under construction and the fourth and last of the UPHOLDER class launched and fitting out, speculation about the future can be limited to the SSNs. It has
already been officially declared that development of the W (SSN-20) class has ended. Thus the replacement for the SWIFrSURE class, which must join the fleet between 1998 and 2006, could either be a
development of the TRAFALGAR class (which is variously reported as an Improved TRAFALGAR, TRAFALGAR Batch 2, or even SSN-19% !), or a scaled-down VAN GUARD design. Whichever of the designs is
selected, the aim must be to construct two to four boats in the mid-1990s.
There will then, however, be a need to replace the TRAFALGAR class in the 2005-2010 time-frame, which fortuitously coincides with the French Navy’s requirement to replace their RUBIS class SSNs.
Tentative moves are thus being made towards a collaborative programme, with the UK using development work already done on the SSN-20 project and the French their work on the AMETHYSTE and l..e
TRIOMPHANT classes.
The history of European naval collaborative projects has not been particularly good; the collapse of the NATO Frigate programme being a recent example. However, there have been some good examples of
Anglo-French programmes; several sonar projects have been successful and the current AngloFrench Future Frigate programme has gone well so far.
The end of the Cold War and the subsequent coiJapse of Soviet power has necessitated a fundamental review of Western military forces and it is not surprising that reductions should be sought in
expenditure, manpower and commitments in all areas of defence. However, there comes a time when reductions are so deep that they threaten the viability of what remains and it is this writer’s view
that planned reductions in the British submarine force have reached that point.
The SSBN force of four VANGUARD class is the bare minimum to achieve a guarantee of one boat always at sea. However, one such boat with sixteen Trident II (D-5), each with eight warheads, packs
sufficient power to serve as a deterrent for the foreseeable future. Apart from the Soviet Navy, there is no naval force likely to have the capability to find, let alone destroy, such a vessel while
it is on patrol, at least for the foreseeable future.
The SSK force of four boats is also at the absolute minimum. It is unlikely that in a sudden crisis more than two will be available, although a third should normally be available at short notice. In
such a small force, however, a mechanical problem or a minor collision could make one boat unavailable for several months, with disproportionate effects of front-line availability.
The most serious worry, however, is with the SSNs. Government policy is to have a force of about 12 boats, of which 7-9 should be available at any one time, which with adequate rwtice might be
increased to ten. The qualifications are emphasised, since experience indicates that British governments take full advantage of lower limits. ” The SSN has proved to be one of •the most powerful,
flexible and influential of all modem weapons systems. The use of CONQUEROR in the Falldands War showed that since the Argentine Navy had no way of detecting such an SSN, it had to assume that she
(and maybe at least one more SSN) could be anywhere in South Atlantic waters. As a result, once CONQUEROR had sunk the GENERAL BELGRANO, the Argentine surface fleet was effectively prevented from any
further operations which could have seriously threatened the Royal Navy task force.
The British series of SSNs has been particularly successful, even though built in small numbers. But even smaller numbers in the future will exacerbate the problem of the industrial base. There is
only one British shipyard capable of building nuclear boats: Vickers (VSEL) at Barrow-in-Fumess. VSEL has already suffered from lack of continuity in orders, but the future will be worse. There is
unlikely to be another SSK order from the Royal Navy for a decade and once the fourth SSBN has been completed there will be no more orders for such boats for some twenty years. Thus, without export
orders (and the British have not exported any new-build submarines since the OBERON class) the work is likely to be very sporadic and even when they do have such work it will be at a low intensity.
Thus, the position of the British Submarine Service is that it remains firmly in the business and that the quality of men and materiel will be as high as ever. But, the quantities will be less even
than now and thus the ability to deal with sudden and unexpected crises will also be reduced. Will it be sufficient to meet the new and unpredictable threats in a highly uncertain world? Only time
will tell.
1991 NSL FACT BOOK
There were several administrative errors made in the preparation of the 1991 Fact Book. The correct data is summarized below:
• Page 13, change to Captain Thomas J. Flanagan.
• Page 14, change to Admiral Harold E. Shear.
• Page 66, change location of Submarine Squadron TWO to Groton. cr.
• Page 67, change location of Submarine Squadron TWENTYTWO to La Maddalena, Italy. Change USS Tecumseh’s hull number to SSBN 628.
• Page 68, under Submarine Squadron THREE, add: USS Haddock (SSN 621), USS Pogy (SSN 647) and USS Houston (SSN 713). Delete USS Houston (SSN 713) from Submarine Squadron SEVEN.
• Page 69, change Greenling’s hull number to SSN 614, location to Groton. Change Gate’s hull number to SSN 615 and location to Groton. Add SSN 621 Haddock. San Diego.
• Page 70, Change L Mendell Rivers’ hull number to SSN 686. Change City of Corpus Christi’s location to Groton.
• Page 107, change PERS OOW toPERS 003.
[The homeport assignments for submarines change frequently. The Fact Book lists the current assignments as each issue goes to press. No further attempt will be made to keep the list up-to-date. The
local submarine area commander’s office should be consulted for a current listing of submarines assigned to a particular homeport] | {"url":"https://archive.navalsubleague.org/1992/the-british-submarine-fleet-1992-and-beyond","timestamp":"2024-11-13T19:38:10Z","content_type":"text/html","content_length":"55858","record_id":"<urn:uuid:da520203-e42d-45c3-9300-babddb626c5b>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00675.warc.gz"} |
Foundations and Applications of Quantum Science
The research of the SFB F40 (FoQuS) encompasses theoretical and experimental investigations in the field of quantum optics and quantum information with particular emphasis on the foundations and
applications of quantum science. Throughout the last 14 years, the consortium has learned to control and manipulate increasingly complex quantum systems. Thus, the intended research will continue to
employ a large variety of quantum systems ranging form single photons from parametric down-conversion in crystals as well as from single atoms, in some cases together with high-finesse cavities, over
single trapped and laser-cooled ions in electrodynamic traps, and samples of laser-cooled atoms or molecules in traps, to ultra-cold gases, mesoscopic quantum systems and various combinations of
The goal of this research programme is the focussed and collaborative research that addresses fundamental questions in quantum information, in quantum optics with atoms and photons as well as their
application for computation, communication, and metrological problems. Furthermore, one of the general goals of the intended SFB is the investigation of what is usually called the quantum-classical
boundary. While the laws of classical physics are known to describe the behaviour of large systems and the laws of quantum mechanics govern the realm of quantum systems, it is hitherto unknown and
widely unexplored where and how the transition between the quantum and classical world appears.
Increasingly complex systems exhibit a wealth of new phenomena and features that can be applied for the solution of technical questions and problems. For example, the ability to control and
manipulate large registers of quantum objects allows one to build a quantum computer, or more generally, quantum devices for advanced metrology and sensor technology. While large-scale number
crunching with such machines still seems a long way off, scaling up systems comprised of building blocks that can be individually controlled can lead to very practical devices such as quantum
repeaters for long-distance quantum communication, advanced atomic clocks, highly sensitive detectors and more. Moreover, the theoretical understanding and the experimental use of small to mesoscopic
quantum arrangements can be used to simulate systems that are virtually impossible to compute on classical computers. | {"url":"https://www.uibk.ac.at/foqus/","timestamp":"2024-11-12T02:14:49Z","content_type":"application/xhtml+xml","content_length":"9356","record_id":"<urn:uuid:cfae51a6-5ec0-42c4-baf0-c1da0e3bcc08>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00787.warc.gz"} |
Annihilator Gun
CPU usage time limit is 1 second
Runtime memory usage limit is 256 megabytes
In the Link-Cut system, there are planets, each with its own type. The type of a planet is an integer ranging from to . You are given an array of integers representing the types of each planet. Your
goal is to make all planets of the same type or determine that it is impossible.
To achieve this, Lucy has constructed an annihilator gun for you, which can change the types of each planets to any other types (from to ) that are not equal to their previous types in one use. The
new types of planets do not necessarily have to be distinct.
More formally, in one use of the gun, you can create an array of integers from to such that for all , and then assign each element of the array a new value .
Since using the annihilator gun is quite resource-intensive, and you need to save resources for capturing the evil Chmyaaax, you need to achieve the goal in the minimum number of uses of the gun.
Your task is to determine this number or to determine that it is impossible to make all planets of one type.
The first line contains two integers and — the number of planets and the number of possible types for each planet.
The second line contains integers — the types of the planets.
In a single line, you need to output the minimum number of uses of the annihilator gun required to make all planets of the same type or if it is impossible.
In the first example, you can do the only use of the gun, changing the types of both planets to . Since their initial types are different, the answer cannot be .
In the second example, it can be shown that it is not possible to make both planets of the same type.
In the third example, all the planets have the same type initially, thus you do not need to do anything.
In the fourth example, you can use the gun two times:
It can be shown that the condition cannot be satisfied in one use of the gun.
1. ( points): ;
2. ( points): ;
3. ( points): ;
4. ( points): ;
5. ( points): ;
6. ( points): no additional constraints.
Submissions 228
Acceptance rate 14% | {"url":"https://basecamp.eolymp.com/en/problems/11727","timestamp":"2024-11-06T14:02:25Z","content_type":"text/html","content_length":"303944","record_id":"<urn:uuid:7eaf4782-7284-4bb9-9457-853611ff54a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00773.warc.gz"} |
Electricity And Energy Class 10th Science Tamilnadu Board Solution
Class 10^th Science Tamilnadu Board Solution
Question 1.
The potential difference required to pass a current 0.2 A in a wire of resistance 20 ohm is ________.
A. 100 V
B. 4 V
C. 0.01 V
D. 40 V
As by looking at the problem we know that we have to find V - potential difference. And the data we are given with is of, current - I which is 0.2 A (ampere) and of resistance - R which is 20 ohm
So, by using Ohm’s law relation we can find the potential difference across the conductor with the help of the values of current and resistance.
Mathematically, V (potential difference) = I(current) × R(resistance)
V = 0.2A× 20ohm = 4 V.
Question 2.
Two electric bulbs have resistances in the ratio 1 : 2. If they are joined in series, the energy consumed in these are in the ratio _________. (1 : 2, 2 : 1, 4 : 1, 1 : 1)
We are given with the ratio of resistance of the two bulbs, let it be R[1]: R[2] = 1 : 2 , the bulbs are connected in series , so same amount of current will pass through both of the bulbs. The
energy consumed by a bulb is given by,
E(electric energy) = V(potential difference) X I(current) X T(time)
As we do not know about the potential difference in this question but we know about the current (same in both bulbs) , so in order to solve this question we need to change the equation in the form of
current , resistance and time only.
Put V = IR form Ohm’s law relation in this electric energy equation.
E = (IR) × I × T =
Let the electric energies of bulb[1] be E[1] and bulb[2] be E[2].
As time taken by the current to pass through both bulbs is always same ,
The ratio of the electric energies is same as the ratio of resistances of the corresponding bulbs.
Question 3.
Kilowatt-hour is the unit of __________.
A. potential difference
B. electric power
C. electric energy
D. charge
As we know that the unit of power is watt or Kilowatt, and the units Kilowatt hour are representing power multiplied by time(hour). And we know that,
So, E = P× T and in the form of units ,
units of E = units of P× units of T = Kilowatt× hour.
Question 4.
________ surface absorbs more heat than any other surface under identical conditions.
A. White
B. Rough
C. Black
D. Yellow
Black body’s are known for their ability to absorb every radiation, there are theories on black body radiations which you will study in higher classes, due to these reasons black body are used in
solar cookers and solar heaters because they absorb all the heat radiation and do not let them escape easily, the other colors white or any other color cannot be more efficient in absorbing
radiations than black body or black color.
Question 5.
The atomic number of natural radioactive element is _________.
A. greater than 82
B. less than 82
C. not defined
D. atleast 92
After atomic number 82 which is Lead(Pb) , there are the radioactive elements , which are highly unstable and humans cannot go near it as it is. We cannot choose less than 82 and less than 92 because
they include simple metals and non-metals too.
Question 6.
Which one of the following statements does not represents Ohm’s law?
(i) current / potential difference = constant
(ii) potential difference / current = constant
(iii) current = resistance x potential difference
As the resistance of the conductor is a constant quantity at constant temperature and same dimensions and also in Ohm’s law resistance is specified as the constant of proportionality, so the first
two options are correct,
and the statement of Ohm’s law is V = I× R , not I = R× V through which the third statement is wrong but is a correct answer.
Question 7.
What is the major fuel used in thermal power plants?
Thermal power plants generally use fossil fuels as their major fuel to produce heat by burning them few examples are petroleum, coal, charcoal. But the majorly used in industries is coal.
Question 8.
Which is the ultimate source of energy?
Sun is the ultimate source of energy because with the help of sun all the processes goes on Earth, like plants making food etc.
Also Energy is produced by the sun because sun keeps on giving radiations of heat and light without having any intake, but there are certain nuclear reactions (FUSION) are going on inside it which
make this all happen. Also many of the sources derive their energy from the sun.
Question 9.
What must be the minimum speed of wind to harness wind energy by turbines?
The minimum speed of wind to harness wind energy by turbines is 15 km/h because at this speed the turbines manage to maintain the required speed. Also, the wind energy is set up at those places where
there is wind going on for maximum time of the year.
Not only the speed but time is also the factor governing the production of electricity by using wind energy.
Question 10.
What is the main raw material used in the production of biogas?
Cow dung-cakes are the main material used in the formation of biogas that is why biogas is also called gobar-gas, but animal waste and plant residue and sewage are also used in the formation of
biogas. It is the improved conventional source of energy.
Question 1.
Fill in the blanks
i) Potential difference: voltmeter; then current __________.
ii) Hydro power plant: Conventional source of energy; then solar energy: _________.
(i) As the question’s pattern is telling us a quantity: and it’s measuring instrument, as the quantity is potential difference: so, it’s measuring instrument will be voltmeter, and in the second case
the quantity is current: so it’s measuring instrument will be ammeter.
The units of potential difference are volt so it’s measuring instrument is named voltmeter and the units of current is ampere so it’s measuring instrument is named ammeter.
(ii). Alternative or Non-Conventional source of energy.
EXPLANATION – As in the question’s pattern it is first giving the name of a method of source of energy and: then asking the type of method whether it is a conventional or non-conventional source of
Question 2.
In the list of sources of energy given below, find out the odd one.
(wind energy, solar energy, hydroelectric power)
Out of all the three options available, if we compare the type of method of these sources of energy have, the wind energy and the hydroelectric power are the Conventional methods of energy whereas
the solar energy is the Non-Conventional method of energy.
Question 3.
Correct the mistakes, if any, in the following statements.
i) A good source of energy would be one which would do a small amount of work per unit volume of mass.
ii) Any source of energy we use to do work is consumed and can be used again.
(i). A good source of energy is the one which would do a large amount of work per unit volume or mass.
When we want to use something, we give it input and we desire considerable output from that thing and of course we want large amount of output or work done by that thing not a small amount of output
or work done.
(ii). Any source of energy we use to do work is consumed and cannot be used again.
When we use some energy, it is in the usable form and when we apply it some energy is used and some energy is dissipated to the surroundings but the amount of energy we used is fully gone with the
law of conservation of mass.
Question 4.
The schematic diagram, in which different components of the circuit are represented by the symbols conveniently used, is called a circuit diagram. What do you mean by the term components?
It is difficult to draw the actual parts of the circuit which are in real life so, there is a way to draw the circuit with some suitable symbols. Electric components or simply components of the
circuit means, the parts with which our circuit is made. For e.g.
A plug key, connecting wires, cell or battery, resistance or bulb, etc.
It’s the simplest circuit.
Components are generally the parts through which our circuit is made.
Here are the symbols of some components used in making circuits.
The simple circuit contains only the main components which should be in every circuit.
Question 5.
The following graph was plotted between V and I values. What would be the values of V / I ratios when the potential difference is 0.5 V and 1 V?
Ohm’s law was given by experimenting and making graph of V vs. I that is V on y axis and I on X axis and by the observations Ohm deduced that V/I is coming out to be nearly constant, and as we
increase V(potential difference) I(current) increases simultaneously and linearly that is current I increases linearly with potential difference V.
He told that V/I is a constant and is named as resistance(R).
Now we know that V/I is constant, so let’s calculate the ratio.
When the potential difference is 0.5 V then the value of current comes out to be 0.2 A , and when the potential difference is 1 V the value of the current comes out to be 0.4 V.
Hence the Ohm’s law is also verified that is,
Question 6.
We know that γ – rays are harmful radiations emitted by natural radioactive substances.
i) Which are other radiations from such substances?
ii) Tabulate the following statements as applicable to each of the above radiations
(They are electromagnetic radiation. They have high penetrating power. They are electrons. They contain neutrons)
(i). There are three types of radiations which emitted from natural Radioactive substances or elements, which are: -
Alpha rays or α-rays, beta rays or β-rays, gamma rays or γ-rays.
The first two alpha and beta rays consist of charges with them but the third one the gamma rays is made of energy only.
(ii). The table is here,
The first two alpha and beta rays consist of charges with them but the third one the gamma rays is made of energy only.
Question 7.
Draw the schematic diagram of an electric circuit consisting of a battery of two cells of 1.5V each, three resistances of 5-ohm, 10 ohm and 15 ohms respectively and a plug key all connected in
This is the circuit having one plug key, one 5 ohm resistor, one 10 ohm resistor, one 15 ohm resistor and a battery of 2 cells each 1.5 V and everything is connected in series, so the battery is
total of 3 V. As you can see there is open plug key, so, there is no current flowing in the circuit and hence we say that it is an open circuit.
As you can see it is the similar circuit as we have drawn the above one, but the only difference is in the plug key, this time the plug key is closed through which the circuit gets closed and gets
completed connection, so, there is flow of current and remember the direction of flow of current is always in opposite direction of the flow of electrons, that is from positive to negative terminal
Question 8.
Fuse wire is made up of an alloy of ___________ which has high resistance and_______.
Accordingly Low melting point.
Nickel, Chromium, Manganese, Iron, Zinc, Tin type of metals and the most common Domestically used fuse wire is of an alloy Nichrome .
The alloy of fuse wire is made and tested that it should have low melting point accordingly, so that it is able to break the circuit by melting when high ampere of current passes through it which can
destroy our electrical appliances.
Question 9.
Observe the circuit given and find the resistance across AB.
As we can see there are two sets of resistors, one is at left and the other is at right, both have 2 resistors in parallel each of 1 ohm. As the sets have parallel resistors, so, we can simplify them
and make them one resistor to get the equivalent resistance of the circuit.
We will solve one set first and in the similar way the second set, we know that when resistors are parallel we add the reciprocals of the resistors and we equate it to the reciprocal of the total
resistance after cross-multiplying we get the total resistance of the circuit.
Let the total resistance of the set A be R[A] and the total resistance of the set B be R[B].
Similarly, R[B] = [A] in set A and R[B] in set B which are in series with each other because when we resolved the 1 ohm resistors which were in parallel they got merged and attached to the main
circuit wire.
Now when resistors are in series we add them algebraically or simply like we add two integer numbers.
R = r[1] + r[2] + …………………. add all the resistors
As R[A] and R[B] are 0.5 ohm calculated by us, so when we add them we get 1 ohm as the answer.
Question 10.
Complete the table choosing the right terms within the brackets.
(zinc, copper, carbon, lead, lead dioxide, aluminum.)
This is the complete table and the answers are in the third column.
These are the answers because, the lead acid battery or accumulator has positive electrode of lead dioxide and leclanche cell has negative electrode of zinc, so we just need to look out for the
patterns of questions to solve them.
Question 11.
How many electrons flow through an electric bulb every second, if the current that passes through the bulb is 1.6 A.
As we know the relation
We have studied the relation q = n× e
Where q is total amount of charge; n is the number of electrons; e is the amount of charge on 1 electron which has value 1.6 X 10^-19 C.
And is always constant(n).
In this problem we have to find n that is number of electrons, so we have the value of current I = 1.6 A and time t = 1 sec , and time will be 1 sec if q = 1.6C , as ,
Question 12.
Vani’s hair dryer has a resistance of 50 Ω when it is first turned on.
i) How much current does the hair dryer draw from the 230 V – line in Vani’s house?
ii) What happens to the resistance of the hair dryer when it runs for a long time?
(Hint: As the temperature increases the resistance of the metallic conductor increases.)
At time t = 0, the given initial resistance(R) of the hair dryer is 50 ohm
i). We are given with the input voltage(V) = 230 V and we have to find the amount of current withdrawn by the dryer which is actually the initial current drawn by the dryer, so,
By using ohm’s law, we will calculate the current,
V = I× R,
ii). As we know that the dryer is used to dry our hair and it does by throwing hot air on us and as time passes when we are using it, it becomes hot and when it’s temperature is increased by becoming
hot, the resistance of the metallic parts increases , the resistance of the metals increases with the increase in temperature till a certain temperature or within a specific range. The longer we use
the hair dryer the more hot it becomes and the more resistance is offered by it.
Question 13.
In the given network, find the equivalent resistance between A and B.
To find the effective resistance of any typical network , first we should look out for small structures of the main structure which can be resolved and we should solve the question diagram by diagram
provided each diagram is simplified from the previous one and don’t just interpret and solve some structures in between structures because there are no resistors in parallel or in series there they
have more connections and it becomes more complex to solve from between, so try to solve from outside if it does not solve then go inside to simplify the whole structure, so, first solve the outer
structures then go to the inner ones if the outer ones can be solved first.
ALWAYS REMEMBER THE MAJOR RULE :-
By using this rule you can create a passage of current from any point where there is potential difference given in the question
In this question the point can be any but from only A and B the two points given in the circuit because between these points there is potential difference , THESE ARE CALLED GIVEN POINTS between
which we have to calculate the effective resistance.
First the extreme left resistors will be solved as they are in series, so we will get, they will get added normally as integers.
Now again the left most resistors are to be solved but now they are parallel and to be calculated by parallel method of evaluation.
Now again we have to solve the left most two resistors at the top which are in series and will be added like integers.
Now again solve the left most resistors which are parallel, we will get,
Now solve the resistors in series, which will be,
Now solve the resistors in parallel which are at the top, which are evaluated to,
Now add the two 5 ohm resistors like integers because they are in series, for the last time,
Solve the two remaining 10 ohm resistors which are in parallel combination to each other,
Thus, we get the resultant or the final calculated resistance.
Question 14.
Old – fashioned serial lights were connected in a series across a 240V household line.
If a string of these lights consists of 12 bulbs, what is the potential difference across each bulb?
We should not get afraid if there is no number data or less number data as there is everything in the question itself, As in the question there is given that the 12 bulbs are identical and there is
the input voltage of 240 V, and the bulbs are connected in series which means the current is same in each and every bulb.
Suppose that each bulb has resistance ‘R’ and ‘I’ is the current passing through each bulb,
We have to find the potential difference across each bulb, which will come out to be same because the current passing and resistance of each bulb is same, and V = I× R, so V is same which is
potential difference across one bulb.
By adding the resistance of all bulbs, we get,
Potential difference across all bulbs = (current through one bulb) × (resistance of all bulbs)
Which is, 240 = I× 12R = 12 IR,
Now potential difference through one bulb is,
Question 15.
Old – fashioned serial lights were connected in a series across a 240V household line.
If the bulbs were connected in parallel, what would be the potential difference across each bulb?
We know that potential difference across each bulb is same when they are connected in parallel and in this case the bulbs too are identical that is they have same resistance and so the current will
be divided in same proportion in each branch (to each resistor).
The conditions are same as in the previous case,
Effective resistance of the network is,
We get,
By Ohm’s law, 240 = I× R[t]
As current through circuit is I, so current through each bulb will be
By Ohm’s law,
Hence now we know the rule works always.
Question 16.
The figure is a part of a closed circuit. Find the currents i[1], i[2] and i[3]
If we know the main current one partition we can find other by using simple maths.
For example- 3A is the main current for forward going 1A and i[2] , and 1A and i[2] are the partitions of 3A current.
As there is law of conservation of energy.
There is always conservation of current(charge) ,
So 3 = 1 + i[2] , by which we can get i[2] which is = 3-1 = 2A.
Now to calculate i[1] we must obtain the equation of it , which is
i[1] + 2 = 3 , i[1] = 3-2 = 1A.
Now to calculate i[3] we will look at the point where it is getting divided and by whom that is , what is the main current of i[3] and the second partition of the main current which is companion with
As i[2] = 2A is the main current and it’s one partition is 1.5A and second partition is i[3] , so ,
i[2] = i[3] + 1.5 , 2 = i[3] + 1.5 , i[3] = 2 - 1.5 = 0.5A
Question 17.
If the reading of the Ideal voltmeter (V) in the given circuit is 6V, then find the reading of the ammeter (A).
We are given with potential difference across 15 ohm resistor and we have to calculate the amount of current flowing in the circuit which will be the reading of ammeter.
Let the potential difference across 15 ohm resistor be V and the current flowing in the circuit be I. As we are given that the voltmeter is ideal which means that no current or very less current
passes through it and hence nearly all the current passes through the 15 ohm resistor.
V = 6 V , R = 15 ohm , by using Ohm’s law , we get
V = I× R ,
Always remember current passing through ideal voltmeter is always zero , Mathematically
I = 0 A, through an ideal voltmeter.
Question 18.
A wire of resistance 8 Ω is bent into a circle. Find the resistance across the diameter.
As in this question we have to calculate net resistance across the diameter of the circle and we are given with the resistance of the wire across it’s free ends which is 8 ohm.
If we remember our previous questions in which we have calculate net resistance, we will know that the given points are whom across there is potential difference and these two points will not merge
and form one-point, potential difference at one point is different from another point that is A and B have different potential.
But the potential difference across each resistor is same as they have same ending points A and B.
So, we are at conclusion, that the two resistors are parallel to each other and we can calculate the net resistance now, as the resistance of the full wire across it’s free ends is 8 ohm, So by the
We can say that by dividing the length to half of the original keeping other things constant which they are in this question, the resistance will reduce to half of the original, that is now the
resistance of the two parallel wires is 4 ohm instead of 8 ohm.
R[t] = 2 ohm, is the resistance across A and B (the diameter).
Question 19.
A wire is bent into a circle. The effective resistance across the diameter is 8 Ω. Find the resistance of the wire.
In this question we are given with the resistance across the diameter AB which 8 ohms is, and we have to calculate the resistance of the wire formed from the circle across its free ends.
The situation of resistors is same as of previous question, So,
We need to find the resistance of the two parallel wires and then we can find the resistance of the total full wire.
Let the resistance of the parallel wires be R ohm each.
So, [t]
We are given that R[t] = 8 ohm, so, R = 2 R[t] = 2× 8 = 16 ohm.
So, the total resistance of the full straight wire across it’s ends will be 16 + 16 = 2× 16 = 32 ohm, because R was the resistance of half wire.
Question 20.
Two bulbs of 40 W and 60 W are connected in series to an external potential difference. Which bulb will glow brighter? Why?
Remember one thing whenever there is this type of question we will check the power dissipation in each bulb, the bulb dissipating more power will glow brighter than the bulb with dissipating less
power. The dissipation of power depends upon the resistance of the bulb that is more the resistance of the bulb more is the dissipation of the power.
As the bulbs are connected in series that means there is same amount of current flowing through them, so,
Power(P) = potential difference(V)× current(I)
By Ohm’s law, V = I× R
Where V is the rating voltage of the bulbs which a very important concept in is these types of questions.
We will have to take input voltage less than or equal to the V volt or the rating voltage of the bulbs otherwise our bulb will fuse out.
Now we have to calculate R[1] of bulb 1, so
Now we have to calculate R[2] of bulb 2, so
Now as we have got the resistance of bulb 1 and 2, so we will find the current flowing in the circuit , which is
Now let’s calculate power dissipation which is ,
P[1] =
P[2] =
As we can see that power dissipated by P[1] is higher than that of P[2] ,
So, bulb 1 with 40 W power will glow brighter than bulb 2 with 60 W power.
Question 21.
Two bulbs of 70 W and 50 W are connected in parallel to an external potential difference. Which bulb will glow brighter? Why?
Remember in these type of questions we will have to calculate the amount of power dissipated by the bulbs , the bulbs dissipating more power will glow more brightly.
Again we will first evaluate the resistance of each bulb and after that the current flowing in the circuit , then we will put both the values of current and resistance to evaluate the power
dissipation by each bulb.
Let the rating voltage of both the bulbs be same as the input voltage which is V volts.
P[1] = [1] = [1] = 50 W
P[2] = [2] = [2] = 70 W
Now to calculate the current flowing in the circuit,
The net resistance is R,
R =
Now let’s calculate power dissipation which is ,
P[1] =
P[2] =
Hence power dissipated by the bulb 2 which has 70 W power will glow more brighter than the 50 W bulb 1.
Hence, we got the same answers when the bulbs are connected in parallel combination.
Question 22.
Write about ocean thermal energy?
OCEAN THERMAL ENERGY :-
The water at the surface of the sea or ocean which, has considerable depth or deepness, is heated by the Sun while as we go deeper and deeper in the ocean or sea the upcoming sections become
relatively colder than the previous one. This difference in temperature is used to obtain energy in ocean-thermal-energy conversion plants. These plants can operate if the temperature difference
between the water at the surface and water at depths up to 2 km is 293 K (20°C) or more. The warm surface-water is used to boil a volatile liquid like ammonia. The vapours of the liquid are then used
to run the turbine of generator. The cold water from the depth of the ocean is pumped up and condense vapour again to liquid.
Question 23.
In a hydroelectric power plant, more electrical power can be generated if water falls from a greater height. Give reasons.
In hydroelectric power plants, the electric energy is produced by potential energy of the water at a certain height , and potential energy increases with the increase of height of the water.
When water is taken at some height it gains potential energy and when water from this height is thrown the potential energy of the water starts converting into kinetic energy and when the water hits
the turbine the kinetic energy rotates the turbine and hence the electricity or electric energy is generated by conversion.
The more is the height of the water, the more is the potential energy, the more is the kinetic energy and , hence the more is the electric energy.
Question 24.
What measures would you suggest to minimize environmental pollution caused by burning of fossil fuel?
The oxides of carbon, nitrogen and sulphur that are released on burning fossil fuels are acidic oxides. These lead to acid rain which affects our water and soil resources. To reduce these kind of
hazardous incidents, the following measures will help you to reduce these incidents:-
1. The pollution caused by burning fossil fuels can be reduced by increasing the efficiency of the combustion process.
2. Using various techniques to reduce the escape of harmful gases and convert them into non-harmful gases.
3. Use that fuel which could create less ashes, so that the ashes do not get into surroundings and pollute it.
4. Use that fuel which produces high level of heat and less level of smoke and gases.
Question 25.
What are the limitations in harnessing wind energy?
Wind energy is an environment-friendly and efficient source of
renewable energy. It requires no recurring expenses for the production of electricity. But there are many limitations in harnessing wind energy, that are:-
1. Wind energy farms can be established only at those places where wind blows for the greatest part of a year.
2. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine.
3. There should be some back-up facilities (like storage cells) to take care of the energy needs during a period when there is no wind.
4. Establishment of wind energy farms requires large area of land and also the initial cost of establishment of the farm is quite high.
Question 26.
What is biomass? What can be done to obtain bioenergy using biomass?
The fuels which are plants and animal’s products, the source of these fuels is said to be bio-mass. Bio means living things and mass means the matter that is matter obtained from the living.
As the bio-mass when burned do not provide much heat and give out a lot of smoke or pollution through them, so there is a method called bio gas plant. The plant has a dome-like structure built with
bricks. Anaerobic micro-organisms that do not require oxygen decompose or break down complex compounds of the cow-dung or sewage or dead plants, slurry. It takes a few days for the decomposition
process to be complete and generate gases like methane, carbon dioxide, hydrogen and hydrogen sulphide.
After the above process the formed biogas is stored in the plant in a tank and can be used through pipes.
Question 27.
Which form of energy leads to the least amount of environmental pollution in the process of harnessing and utilization? Justify your answer.
Solar energy is the energy which leads to the least amount of environmental pollution in the process of harnessing and utilizing it.
Because :-
1. The input of solar energy is the sun which is the ultimate source of energy and will never die and is nonpolluting.
2. There is no much activities or harming of nature when harnessing it like it is in wind energy, we have to consider large area of land and place windmills there that might include cutting of trees
or digging of earth too deep.
3. There is no release of harmful gases as there is in geothermal energy when the melted rocks come to the surface of the earth from the core, then they release harmful gases when they form contact
with water.
4. There is no burning of any fossil fuels like there is in thermal power plants and no radioactive waste as it is in nuclear power plants.
5. There is no disturbance caused to the wildlife, because we can place it on top of our home and use it while we are in our house , but there might be some disturbance caused to the aquatic life in
energy from the sea methods and by hydroelectric power plant also.
Question 1.
Veena’s car radio will run from a 12 V car battery that produces a current of 0.20 A even when the car engine is turned off. The car battery will no longer operate when it has lost 1.2 x 10^6 J of
energy. If Veena gets out of the car, leaving the radio on by mistake, how long will it take for the car battery to go completely dead, i.e. lose all energy?
(1 day = 86400 second)
In this question we need to find the time taken (T) by the car to go dead when the radio is left on by Veena , We are given with the amount of energy car can supply (E) which is 1.2 X 10^6 .
We are also given with the car battery’s potential difference (V) which it can give which is 12 V , and the amount of current produced (I) is 0.2 A .
By using the equation , E = V× I× T ,
1 day = 86400 seconds, 5 X 10^5 seconds =
Question 2.
Find the total current that passes through the circuit. Find the heat generated across the each resistor.
First we need to find the total current (I) flowing in the circuit, that we can find by using Ohm’s law, because we are given with the battery’s potential difference (V) and we need to find the net
resistance of the circuit (R).
As we can see there are 3 resistors 4 ohm, 6 ohm and 12 ohm, and there are two resistors in parallel the 6 and 12 ohm ones, sowe will first solve the parallel ones and get the parallel resistance R
R[p] = 4 ohm
Now the net resistance will come by simply adding the R[p] = 4 ohm and the other 4 ohm resistor because they both are now in series,
R = R[p] + 4 = 4 + 4 = 8 ohm
R = 8 ohm
By using Ohm’s law, V = I× R,
Now we need to find the heat generated across each resistor, which we will do by using Joule’s law of heating,
Where I is current passing through conductor; T is time; R is resistance of the conductor; and H is heat generated across the conductor.
As the 4 ohm resistor is alone and in series so current I[1] passing through it will be equal to the total current passing through the circuit ,
Which is 2 A and the potential difference across the 4 ohm resistor is
V = I× R = 2× 4 = 8 V
Now the two resistors 6 and 12 ohm are in parallel that is they have same potential difference across them which is 8 V because the battery has total potential difference of 16 V and 8 V is consumed
by the 4 ohm resistor which is in series and alone so , 16-8 = 8 V the remaining potential difference is 8 V which is consumed by both 6 and 12 ohm resistors in parallel.
Let the current passing through 6-ohm resistor be I[2] and current passing through 12 ohm resistor be I[3] .
By Ohm’s law,
V = I[2]× R[2],
V = I[3]× R[3] ,
Heat generated by 4-ohm resistor,
Heat generated by 6-ohm resistor,
Heat generated by 12-ohm resistor,
Time will be same for all the resistors always, in this case as no time is given so we will calculate for one second of heat generated or we can leave the answer in terms of T.
Question 3.
Find the total current that passes through the circuit given in the diagram. Also find the potential difference across 1Ω resistor.
To find the total current(I) passing through the circuit, we must find the total resistance(R) and also we know the battery’s potential difference(V) which is 1.5 V , so by Ohm’s law we can find the
net current flowing in the circuit.
As we can see by the figure that two resistors 1 ohm and 2 ohm are in series to each other, so we can directly add them , 1 + 2 = 3 ohm,
Also we can see that 3 resistors 4 ohm , 6 ohm , and 12 ohm are parallel to each other , so , let the net parallel resistance of these three resistors be R[p] ,
Therefore, R[p] = 2 ohm
Hence by Ohm’s law, V = I× R
As the current passing through 1 ohm resistor is the net current flowing in the circuit, so
V = I× R = 0.75× 1 = 0.75 V. It is the potential difference(V) across the 1 ohm resistor.
Question 4.
Raman’s air-conditioner consumes 2160 W of power, when a current of 9.0 A passes through it.
i) What is the voltage drop when the air-conditioner is running?
ii) How does this compare to the usual household voltage?
iii) What would happen if Raman tried connecting his air-conditioner to a 120V line?
We are given the amount of power consumed by Raman’s air-conditioner when a current of 9 A flows through it.
(i). The meaning of voltage drop simply means the potential difference across Raman’s air conditioner when it is running which we have to find ,
Now to find the potential difference by knowing power and amount of electric current passing through passing through it we can use this relation ,
P = V× I ,
(ii). In India, we have a household voltage of 220 V current supply,
which changes time to time and in very short time intervals.
As there is no major difference in 220 V and 240 V A.C. supply , which we will study in higher classes not in this class in detail , So yes these are comparable voltages and this air conditioner can
be used in houses.
(iii). If raman tries to connect his air conditioner to a 120 V supply line , then his air conditioner would do less amount of work and deliver less power(rate of doing work) , also as it’s potential
difference is reduced to half and of course it’s resistance is constant , so the current passing through it would also become less and will be reduced to half itself , so basically the power would
become 1/4^th of the original power , also this situation might destroy your appliance or the air conditioner in Raman’s case.
Question 5.
The effective resistance of three resistors connected in parallel is 60/47 Ω. When one wire breaks, the effective resistance becomes 15/8 ohms. Find the resistance of the wire that is broken.
As the three resistors are in parallel combination to each other , so , let the total resistance of the 3 resistors be R[3] and the total resistance of the two resistors be R[2] ,
Let the three resistors be R[a] , R[b] , R[c] .
First we will calculate the total resistance of 3 resistors,
We are given with the values of R[3] = 60/47 ohm and R[2] = 15/8 ohms ,
When one wire is broken down , let the wire be of resistance R[a] , so
Put the value of two resistor equation in three resistor equation ,
We get ,
Hence this is the required resistance.
Question 6.
Find the resistance across
A and D
In order to find the total resistance, first we need to memorize the rule which we have learned which is , THE POINTS WHO’S ACROSS WE HAVE TO EVALUATE THE TOTAL RESISTANCE ARE THE GIVEN POINTS AND WE
And the rest are the normal points of the circuit.
A and D are the given points in this part of the question and the rest of the points are the simple points (that is B and C).
As the resistors in the line BC are in series and the resistors in the line AD are also in series with each other , so they will get added like integers or they will be simply added.
Now merge the points B into A and C into D , because they have same potential , all the three resistors are parallel to each other ,
By solving we will get ,
These two resistors are in parallel combination , so when we solve these two we will get ,
This is the resultant resistance.
Question 7.
Find the resistance across
B and D.
In order to find the total resistance, first we need to memorize the rule which we have learned which is , THE POINTS WHOSE ACROSS WE HAVE TO EVALUATE THE TOTAL RESISTANCE ARE THE GIVEN POINTS AND WE
B and D are the given points in this part and the remaining are the simple points to deal with (that is A and C).
The resistors in the line BC are in series as well as the resistors in the line AD are also in series with each other, so by solving we get
Merge the point A into point B and also merge the point C into point D, all three resistors are parallel to each other, so,
These two resistors are parallel to each other, so by solving them we get,
Thus, it is the resultant or total resistance.
Question 8.
Explain the two different ways of harnessing energy from the ocean.
Basically there are three ways of harnessing energy from the oceans and seas, but we are today going to discuss only two,
Which are :-
1. OCEAN THERMAL ENERGY :-
The water at the surface of the sea or ocean which, has considerable depth or deepness, is heated by the Sun while as we go deeper and deeper in the ocean or sea the upcoming sections become
relatively colder than the previous one. This difference in temperature is used to obtain energy in ocean-thermal-energy conversion plants. These plants can operate if the temperature difference
between the water at the surface and water at depths up to 2 km is 293 K (20°C) or more. The warm surface-water is used to boil a volatile liquid like ammonia. The vapours of the liquid are then used
to run the turbine of generator. The cold water from the depth of the ocean is pumped up and condense vapour again to liquid.
2. OCEAN TIDAL ENERGY :-
Due to the gravitational pull of mainly the moon on the spinning earth, the level of water in the sea rises and falls. This phenomenon is called high and low tides and the difference in sea-levels
gives us tidal energy by the same method as we saw in hydroelectric power plant , but in this method the gravitational energy by moon gives potential energy as well as kinetic energy to the water
thereby there is formation of tides or waves. Tidal energy is harnessed by constructing a dam across a narrow opening to the sea. A turbine fixed at the opening of the dam converts tidal energy to
There is also Ocean wave energy.
Question 9.
Five resistors of resistance ‘R’ are connected such that they form a letter ‘A’. Find the effective resistance across the free ends.
The simple resistors must be identified and after that they must be first solved and when they are solved look for again the simple resistors pattern solve till you end up with only one resistor.
We will solve diagram by diagram,
First we will add the two top most resistors because they are in series, we get ,
Now the two topmost resistors are parallel to each other because their free ends are attached to same points if we look it as wire.
We get,
Now these three are in series with each other, so we will add them simply like we are adding integers.
We will get,
Therefore, the total or the resultant resistance is = 8R/3 ohm. | {"url":"https://www.kalakadu.com/2021/01/electricity-and-energy-class-10th.html","timestamp":"2024-11-03T19:03:40Z","content_type":"application/xhtml+xml","content_length":"1002652","record_id":"<urn:uuid:e4647c17-0737-4ed9-aedd-11d5fa0d183f>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00441.warc.gz"} |
Thomas Kreuz - ISI- and SPIKE-distance
Source Code - SPIKE- and ISI-Distance
SPIKE- and ISI-Distance between two or more spike trains
For a detailed description of the methods please refer to:
Kreuz T, Chicharro D, Houghton C, Andrzejak RG, Mormann F:
Monitoring spike train synchrony
J Neurophysiol 109, 1457 (2013) [PDF]
Kreuz T:
Scholarpedia 7(12), 30652 (2012)
For an overview please refer to:
Mulansky M, Bozanic N, Sburlea A, Kreuz T:
A guide to time-resolved and parameter-free measures of spike train synchrony
IEEE Proceeding Event-based Control, Communication, and Signal Processing (EBCCSP), 1-8 and arXiv [PDF] (2015) [PDF]
Powerpoint presentation
Kreuz T:
Monitoring spike train synchrony
Further information on the SPIKE-distance can also be found on the SPIKY Facebook page.
The best demonstration of the SPIKE-distance can be seen in the following movie. This and many more movies can now also be seen on the SPIKY Youtube Channel.
Movie: How the SPIKE-distance traces spike train clustering
Here we explain how this time-resolved and parameter-free measure traces spike train clustering with both its normal and its realtime version.
Top: Rasterplot with artificially generated spike trains whose clustering changes every 500 ms.
Middle: Dissimilarity matrices for the regular (1st column) and the real-time (2nd column) SPIKE-distance. The spike train group matrices (3rd and 4th column) are obtained by averaging over the
respective submatrices of the original dissimilarity matrices.
Bottom: Corresponding dendrograms.
In the first part of the movie the instantaneous dissimilarities and the clustering dendrograms of both measures are updated as the green time bar moves from left to right. The second part comprises
several examples of selective temporal averaging of individual or combined intervals as well as triggered averaging.
A second example:
Movie: Demonstration of externally triggered averaging.
Here we explain how the instantaneous nature of the SPIKE-distance can be very useful when this method is combined with triggered averaging.
In this simulated setup a population of 50 neurons is recorded upon presentation of a time-varying non-periodic stimulus, in this case a chirp function. It is assumed that half of the neurons are
sensitive to negative amplitudes and accordingly exhibit higher (lower) reliability for local minima (maxima) of the chirp function.
As the amplitude of the chirp function varies, so does the spike train synchrony of half of the neurons, but due to the non-periodicity of the stimulus this is quite difficult to detect. Detection is
facilitated by externally triggered averaging where the triggering is performed on common stimulus features, here the amplitude of the chirp function. As can be seen in the dissimilarity matrices of
the SPIKE-distance and even better in the dendrogram, a decrease of this stimulus amplitude leads to the emergence of a spike train cluster consisting of the modulated spike trains which indicates
their increase in reliability.
The definiton of the SPIKE-distance builds on these earlier papers:
Kreuz T, Chicharro D, Greschner M, Andrzejak RG:
Time-resolved and time-scale adaptive measures of spike train synchrony
J Neurosci Methods 195, 92 (2011) [PDF]
Kreuz T, Chicharro D, Andrzejak RG, Haas JS, Abarbanel HDI:
Measuring multiple spike train synchrony
J Neurosci Methods 183, 287 (2009) [PDF]
Kreuz T, Haas JS, Morelli A, Abarbanel HDI, Politi A:
Measuring spike train synchrony
J Neurosci Methods 165, 151 (2007) [PDF]
The Matlab code is now complemented by PySpike, an open source Python library written by Mario Mulansky and available on github. Its core functionality is the implementation of the bivariate ISI- and
SPIKE-distance. Additionally, it provides functions to compute multivariate ISI- and SPIKE-distances, as well as averaging and general spike train processing. All computation intensive parts are
implemented in C (via cython) to reach a competitive performance (factor 100-200 over plain Python). There is also documentation, currently consisting of a brief introduction and the API reference.
See also:
Python-Implementation of the pairwise SPIKE-distance (written by Jeremy Fix)
Python-Implementation of the pairwise ISI-distance (maintained by Michael Chary)
Matlab code to calculate and visualize the SPIKE- and the ISI-Distance between two or more spike trains:
(Last updated version 1.2: April 29, 2013; Copyright: Thomas Kreuz, Nebojsa Bozanic)
Important note: This code will not be updated anymore. In principle it is replaced by the program 'SPIKY_loop' which is included in the zip-file of the new graphical user interface SPIKY which makes
use of the same MEX-files. The Beta-version of these source codes can now be found here.
This version uses MEX- instead of m-files for the most time-consuming calculations. This reduces the computational cost considerably (on my notebook the code running all the examples below was faster
by a factor of 85). It also uses a new output-structure 'results' which allows easy access not only to the overall dissimilarity value but also to the measure profiles and the pairwise distance
matrices for all the selected measures.
Note of warning:
This program calculates the SPIKE- and the ISI-Distance between two or more spike trains. It also allows the time-resolved visualization via measure profiles and the spatial visualization via
dissimilarity matrices (plots and figures). This, however, makes it a rather complex program so it might happen initially that there are some bugs or that the program is not as error-tolerant as it
should be. Whenever you encounter any problem please provide feedback and we will try to resolve the problem and improve the source codes accordingly.
The name of the zip-file will always include the date of release which will allow users to ensure that they have the most recent version. As the development of the measure evolved, more and more
applications came to mind and were integrated, and this might not be the end of it. Thus it might be worthwhile to check for updates now and then, in particular in these early alpha development
stages. If you want to be informed about possible future releases, please send an email to Thomas Kreuz at "thomas.kreuz (at) cnr.it". You will then be included in a mailing list.
The zip-file contains many m- and eight MEX-files which all should be stored in the same directory. Before the first run of the program the MEX-files should be compiled. To do so please run from this
directory the program 'Compile_MEX.m'.
Distances_Main_Demo: Demo program, used to create all the figures below
f_get_data_demo: Main function (sets spike-matrix as well as the parameters describing these data), called by 'Distances_Main_Demo'
Distances_Main: Main program; loads or creates data, sets parameters, and calls main function 'f_distances'
f_get_data: Main function (sets spike-matrix as well as the parameters describing these data), called by 'Distances_Main'
f_distances_MEX: Main function (calculates distances and does the plotting), called by 'Distances_Main' or 'Distances_Main_Demo', uses MEX-functions for computationally expensive loops
This function calls the following functions (included):
f_measure_profiles: Function that creates individual measure profiles, called by 'f_distances_MEX'
f_moving_average (& variants): Functions to generate the (weighted) moving average of a measure profile, depending on the measure profile the non-causal or the causal variant is used
f_compute_gauss_smooth: Function that creates a smoothed version of the PSTH
(Copyright: Jude Mitchell, http://www.snl.salk.edu/~jude/sfn2008/index.html)
as well as a few more functions (some of them from Matlab toolboxes)
To get started, try either Distances_Main_Demo or Distances_Main. However, before the first start of the program a few variables (see also below) defining the work environment should be set:
f_para.imagespath: Path where images (ps) will be stored
f_para.moviespath: Path where movies (avi) will be stored
f_para.matpath: Path where Matlab files (mat) will be stored
Main function f_distances_MEX:
Matrix spikes with two or more spike trains (if trains have different numbers of spikes, fill with zeros)
Structure d_para that describe the data (see below)
Structure f_para that determines the appearance of the figures and the movie
Structure 'results' with these fields:
all of these variables have one entry for each measure selected via the variable 'f_para.subplot_posi' (see below):
measures: Name of selected measures (helps to identify the order within all other variables)
dissimilarity_profiles: cell with overall dissimilarity profiles obtained by averaging over spike train pairs (either piecewise constant (size 'num_isi') or sampled (length 'len'), see additional
figure 1 below for an example)
distance_matrices: pairwise distance matrices, obtained by averaging over time
overall_dissimilarities: overall dissimilarities, obtained by averaging over both spike trains and time
In case a sampled measure was selected:
time: sampling values of length 'len' (defines x-axis for sampled measures)
In case a piecewise constant measure was selected:
isi: length of successive isi within the pooled spike train, length 'num_isi' (defines x-axis for piecewise constant measures)
For the piecewise constant profiles (here for the first) the function f_pico can be used to obtain the average value as well as x- and y-vectors for plotting:
[overall_dissimilarity,plot_x_values,plot_y_values] = f_pico(results.isi,results.dissimilarity_profiles{1},d_para.dts,d_para.tmin);
Example call:
After all other parameters have been set to standard values, i.e.
d_para.tmin=0; d_para.tmax=101; d_para.dts=0.001;
num_trains=4; num_spikes=100;
for rc=2:num_trains
The program uses three sets of parameters : Each of these are stored in a Matlab-struct.
1. d_para: Parameters that describe the data (set within the function f_get_data and then used in the call of the main function f_distances_MEX from the main program)
2. f_para: Parameters that determine the appearance of the figures and the movie (set in the main program and then used in the call of the main function f_distances_MEX from the main program)
3. s_para: Parameters that determine the appearance of the individual subplots in the figure (set in the function f_distances_MEX and then used in the call of the function f_measure_profiles)
Here is a description of the individual structures:
1. Structure 'd_para': parameters that describe the data
tmin: Beginning of recording
tmax: End of recording
dts: Sampling interval, precision of spike times [!!! Please take care that this value is not larger than the actual sampling size, otherwise two spikes can occur at the same time instant and this
can lead to problems in the algorithm !!!]
all_train_group_names: Names of spike train groups
all_train_group_sizes: Sizes of respective spike train groups
select_train_mode: Selection of spike trains (1-all,2-selected groups,3-selected trains)
select_train_groups: Selected spike train groups (if 'select_train_mode==2')
select_trains: Selected spike trains (if 'select_train_mode==3')
select_averages: One or more continuous intervals for selective temporal averaging
trigger_averages: Non-continuous time-instants for triggered temporal averaging, external (e.g. certain stimulus properties) but also internal (e.g. certain event times)
markers: Relevant time instants
markers2: Even more relevant time instants
separators: Relevant seperations between groups of spike trains
separators2: Even more relevant seperations between groups of spike trains
interval_separators: Edges of subsections
interval_strings: Captions for subsections
comment_string: Comment on the data, will be used in file and figure names
2. Structure 'f_para': parameters that determine the appearance of the figures (and the movie)
imagespath: Path where images (postscript) will be stored
moviespath: Path where movies (avi) will be stored
matpath: Path where Matlab file (mat) will be stored
filename: Name under which images, movies and Matlab files will be stored
title_string: Appears in the figure titles
saving: Saving to postscript file? (0-no,1-yes)
print_mode: Saving to postscript file? (0-no,1-yes)
publication: Omits otherwise helpful information, such as additional comments (0-no,1-yes)
comment_string: Additional comment on the example, will be used in file and figure names
num_fig: Number of figure
pos_fig: Position of figure
font_size: Font size of labels (and headlines)
multi_figure: Open many figures (0-no,1-yes)
timeunit_string: string with time unit, used in labels
xfact: Conversion of time unit
ma_mode: Moving average mode: (0-no,1-only,2-both)
spike_mao: Order of moving average (pooled ISI)
time_mao: Order of moving average (time)
dtm: Sampling of measure profile, setting this to larger values allows downsampling in order to facilitate memory management
mov_step: Step size for movie frames
mov_frames_per_second: Well, frames per second
mov_num_average_frames: Number of mov_frames the averages are shown at the end of the movie (if this is too small they are hardly visable)
mov_frames: Selection of individual frames to be shown in the movie (replaces use of mov_step as step size if non-empty)
plot_mode: +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie (binary addition allows all combinations)
norm_mode: normalization of averaged bivariate measure profiles (1-Absolute maximum value 'one',2-Overall maximum value,3-Individual maximum value)
color_norm_mode: normalization of pairwise color matrices (1-Absolute maximum value,2-Overall maximum value,3-Each one individually)
block_matrices: Allows tracing the overall synchronization among groups of spike trains (0-no,1-yes)
dendrograms: Cluster trees from distance matrices (0-no,1-yes)
dendro_color_mode: coloring of dendrograms (0-no,1-groups,2-trains)
subplot_size: Vector with relative size of subplots
subplot_posi: Vector with order of subplots
3. Structure 's_para': parameters that describe the appearance of the individual subplots in the figures (measure time profiles)
window_mode: time interval selection (1-all (recording limits),2:outer spikes,3-inner spikes,4-smaller analysis window)
colors: Colors for 1:Stim/STs/ISI,2:Stim/STs/ISI-outside,3:Mean values,4:Mean values-outside,5:Moving averages,6:Moving averages-outside
font_size: Font size of labels (and headlines)
line_width: Line width
line_style: Line style (1-,2:,3-.,4--,5:none)
nma: Selected moving averages (depends on f_para.ma_mode)
spike_mao: Order of moving average (pooled ISI)
time_mao: Order of moving average (time)
causal: determines kind of moving average, set automatically for each measure (0-no,1-yes)
itmin: Beginning of recording
itmax: End of recording
wmin: Beginning of selected window (if window_mode==4)
wmax: End of selected window (if window_mode==4)
num_subplots: depends on f_para.subplot_posi, set automatically
xl: x-limits for plotting, set automatically
yl: y-limits for plotting, set automatically
dtm: equals f_para.dtm
plot_mode: equals f_para.plot_mode
Some of the more important parameters are here explained in more detail:
f_para.plot_mode - This variable selects the types of plots (and movie)
+1: time-resolved measure profile (see Figs. 2 and 3 of the 2013 paper and below)
+2: different measures and cuts (see Figs. 5 and 6 of the 2013 paper and below)
+4: different measures (creates one figures of different cuts for each measure)
+8: different cuts (basically screenshots of the movie, see Fig. 9 in the 2013 paper and below)
+16: different cuts-Movie (basically a succession of plots, see Fig. 9 in the 2013 paper and below as well as the supplementary movie)
Binary addition allows all combinations of these plots/movies.
For the measure profiles (created by adding +1 to f_para.plot_mode):
f_para.subplot_posi and f_para.subplot_size - These variables allow you to choose the selection and order the subplots as well as their relative size.
Use f_para.subplot_posi to select the order of the subplots. Use 0 if a subplot is not needed. Make sure that the range from 1 to the desired number of subplots is covered, no (positive) number
should appear twice.
Use f_para.subplot_size to select the relative size of the selected subplots.
The first variable f_para.subplot_posi refers to the standard order of the subplots which is defined below.
For each of the following variables there will be an entry in the output variable (in the order of the list below.
The second variable f_para.subplot_size then refers to the subplots selected by "f_para.subplot_posi" and gives their relative sizes from top to bottom (its length should be equal to the maximum
number in "f_para.subplot_posi").
The subplot "Stimulus" allows you to add any representation of the stimulus that triggered the spike trains (put 0 if not needed).
In f_distances_MEX the function f_measure_profiles is called once for each selected measure profile. This is a list of the measures provided.
Stimulus, Spike trains, Classic, Rate, ISI:
Stimulus Spikes PSTH GPSTH ISI <ISI> Rate <Rate> Ia I^a
Sa S^a Sa_r S_r^a Sa_f S_f^a Sta St^a Sta_r St_r^a Sta_f St_f^a
From 9 to 22 these are six pairs where the first entry (Xa) refers to all pairwise measure profiles and the second entry (X^a) to the averaged bivariate measure profiles.
Piecewise constant and sampled variants:
The dissimilarity profile S (t) of the SPIKE-distance is piecewise linear rather than piecewise constant (as is the case for the ISI-distance). Therefore, when the localized visualization is desired,
a new value has to be calculated for each sampling point and not just once per each interval in the pooled spike train.
In cases where the distance value itself is sufficient, the short computation time can be even further decreased by representing each interval by the value of its center and weighting it by its
length. This is not only faster, but it actually gives the exact result, whereas the time-resolved calculation is a very good approximation only for sufficiently small sampling intervals dt (imagine
the example of a rectangular function, at some point any sampled representation has to cut the right angle). The dissimilarity profile S_r (t) of the real-time SPIKE-distance is hyperbolic and not
linear but here also the exact result can be obtained by piecewise integration over all intervals of the pooled spike train.
Measures 9-16 are piecewise constant (one value for each interval in the pooled spike train), while measures 17-22 are sampled with the sampling interval given by the variable f_para.dtm.
9-10: ISI-distance (per definition piecewise constant)
11-12: SPIKE-distance (piecewise constant)
13-14: Real-time SPIKE-distance (piecewise constant)
15-16: Future SPIKE-distance (piecewise constant)
17-18: SPIKE-distance (sampled with variable 'time')
19-20: Real-time SPIKE-distance (sampled with variable 'time')
21-22: Future SPIKE-distance (sampled with variable 'time')
For each selected variable a value in the variable 'mean_values' is returned (the order is defined by f_para.subplot_posi).
There exist a number of different possibilities how the first and the last inter-spike interval are defined. For longer spike trains with sufficient statistics these differences become rather
s_para.window_mode (1-all, 2-outer spikes, 3-inner spikes, 4-chosen analysis window)
1. All - time profiles are calculated within this interval (which should cover the spike train such that in each spike train the first interval is from d_para.tmin to the first spike, the last
interval is from the last spike to d_para.tmax). This corresponds to the case where spike trains were recorded for a certain time. The variable d_para.min then denotes the onset of the recording,
d_para.tmax its end.
2. Outer spikes - time profiles are averaged from the very first spike till the very last spike.
3. Inner spikes - time profiles are calculated on the common support, i.e., from the latest first spike till the earliest last spike. This option does not make much sense if the spike trains do not
have overlapping support (i.e., if one spike train starts after the other has already ended).
4. Window - restricts the averaging to a smaller window [s_para.wmin s_para.wmax] to be defined. These values should be within the range [tmin tmax].
s_para.wmin and s_para.wmax - beginning and end of the analysis window (for window_mode=4).
s_para.colors - colors used to display the stimulus/spike trains/ISI, the mean values and the moving averages both for the whole time profile and the selected analysis window (for window_mode>1)
f_para.publication (0-no, 1-yes) - If yes, the figure will have a title with all the selected distance values and the postscript-file will be a large landscape print, otherwise no title and a smaller
portrait file.
For the movies (created by adding +16 to f_para.plot_mode):
The movies will always append individual figures in this order:
1. Individual frames (either individually selected or sampled with a certain step size)
2. Selected averages
3. Triggered averages
There are two possibilities to select individual frames in the first part of the movie.
1. f_para.mov_frames - Selection of individual frames to be shown in the movie (replaces use of mov_step as step size if non-empty)
2. f_para.mov_step - Step size for movie frames (will only be used if f_para.mov_frames is empty).
f_para.mov_num_average_frames - Number of frames the selected and triggerad averages are shown at the end of the movie (in case the first part of the movie contains a sequence of snapshots with high
temporal resolution and f_para.mov_frames_per_second is rather large this number has to be rather large (e.g., of the order of f_para.mov_frames_per_second) in order to make the averages visible)
f_para.block_matrices: Allows tracing the overall synchronization among groups of spike trains (0-no,1-yes) (see Fig. 9 of the 2013 paper and below)
f_para.dendrograms: Cluster trees from distance matrices (0-no,1-yes), (see Figs. 8 and 9 of the 2013 paper and below)
d_para.select_averages: Continuous intervals for selective temporal averaging (see Fig. 7 of the 2013 paper and below)
d_para.trigger_averages: Non-continuous time-instants for triggered temporal averaging, external (e.g. certain stimulus properties) but also internal (e.g. certain event times) (see Fig. 8 of the
2013 paper and below)
To control the position and size of the matrix or dendrogram subplots the variable supos is set in f_distances_MEX. Standard settings for up to 8 different subplots are provided, however, further
adjustments might be necessary.
In case you want to create your own plots the pairwise results are stored in the four-dimensional variable 'mov_mat' with dimensions [#(selected measures)] * [#(selected frames) + #(selected
averages) + #(selected triggers)] *[#(spike trains)[ * [#(spike trains)]. The respective variables for the second dimensions are num_frame_select, num_select_averages, and num_trigger_averages.
Memory management:
For large numbers of spike trains with many spikes the interval is cut into smaller segments, and the averaging over all pairs of spike trains is performed for each segment separately. During
calculations this is indicated on the screen by 'run_info = #run #(overall runs)'.
In the beginning of the function the variable 'max_memo_init' is set. This variable should be big enough to hold the basic matrices for one run of the segment loop but small enough to not run out of
In the following you will find an explanation of the most relevant parameter settings needed in order to reproduce some of the plots from the 2013 paper:
To reproduce these figures run 'Distance_Main_Demo' and select the figures by setting the variable 'examples' accordingly
Fig. 2a: SPIKE-distance applied to a bivariate example with varying spike match
(example 21)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.comment_string=['Fig2a']; % Additional comment on the example, will be used in file and figure names
Fig. 2b: SPIKE-distance applied to a multivariate example
(example 22)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.comment_string=['Fig2b']; % Additional comment on the example, will be used in file and figure names
Fig. 3a: Real-time SPIKE-distance applied to a bivariate example with varying spike match
(example 31)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.ma_mode=2; % Moving average mode: (0-no,1-only,2-both)
f_para.time_mao=120; % Order of moving average (time)
f_para.comment_string=['Fig3a_Realtime']; % Additional comment on the example, will be used in file and figure names
Fig. 3b: SPIKE-distance applied to a multivariate example
(example 32)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.ma_mode=2; % Moving average mode: (0-no,1-only,2-both)
f_para.time_mao=190; % Order of moving average (time)
f_para.comment_string=['Fig3b_Realtime']; % Additional comment on the example, will be used in file and figure names
Fig. 4: Real-time SPIKE-distance: Peaks during reliable spiking events are not spurious
(example 41)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.comment_string=['Fig4']; % Additional comment on the example, will be used in file and figure names
Fig. 5A: Instantaneous clustering for artificially generated spike trains
(example 51)
d_para.interval_separators=500:500:d_para.tmax-500; % Edges of subsections
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 3 0 0]; % Vector with order of subplots
f_para.plot_mode=2; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.mov_frames=[250 750 1190 1510]; % Selection of individual frames
f_para.xfact=1000; % Conversion of time unit
f_para.timeunit_string='[s]'; % Time unit, used in labels
f_para.comment_string=['Fig5a']; % Additional comment on the example, will be used in file and figure names
Fig. 5B: Further examples of instantaneous clustering for artificially generated spike trains
(example 61)
d_para.interval_separators=500:500:d_para.tmax-500; % Edges of subsections
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 3 0 0]; % Vector with order of subplots
f_para.plot_mode=2; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.mov_frames=[250 750 1250 1750]; % Selection of individual frames f_para.xfact=1000; % Conversion of time unit
f_para.timeunit_string='[s]'; % Time unit, used in labels
f_para.comment_string=['Fig5b']; % Additional comment on the example, will be used in file and figure names
Fig. 6: Selective temporal averaging
(example 71)
d_para.interval_separators=500:500:d_para.tmax-500; % Edges of subsections
d_para.select_averages={[0 500];[1000 2000];[2000 2500 3000 3500];[0 4000]}; % Selected average over different intervals
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 3 0 0]; % Vector with order of subplots
f_para.plot_mode=2; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.xfact=1000; % Conversion of time unit
f_para.timeunit_string='[s]'; % Time unit, used in labels
f_para.comment_string=['Fig6']; % Additional comment on the example, will be used in file and figure names
Fig. 7a: Triggered temporal averaging 1: Whole Interval
(example 81)
d_para.select_averages={[d_para.tmin d_para.tmax]}; % One continuous intervals for selective temporal averaging
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=8; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.dendrograms=1; % Cluster trees from distance matrices (0-no,1-yes)
f_para.comment_string=['Fig7b,c1']; % Additional comment on the example, will be used in file and figure names
Fig. 7b: Triggered temporal averaging: Firings of first spike train
(example 81)
d_para.trigger_averages{1}=spikes(1,1:num_spikes(1)); % Triggered averaging over all time instants when a certain neuron fires
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=8; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.dendrograms=1; % Cluster trees from distance matrices (0-no,1-yes)
f_para.comment_string=['Fig7b,c2']; % Additional comment on the example, will be used in file and figure names
Fig. 8: Screenshot from the supplementary movie
(example 91)
d_para.markers=[500:500:d_para.tmax-500]; % Relevant time instants
d_para.all_train_group_names={'G1';'G2';'G3';'G4'}; % Names of spike train groups
d_para.all_train_group_sizes=[10 10 10 10]; % Sizes of respective spike train groups
d_para.select_averages={500+[0 500 1000 1500]}; % Two continuous intervals for selective temporal averaging
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 3 0 0]; % Vector with order of subplots
f_para.plot_mode=8; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.block_matrices=1; % Allows tracing the overall synchronization among groups of spike trains (0-no,1-yes)
f_para.dendrograms=1; % Cluster trees from distance matrices (0-no,1-yes)
f_para.dendro_color_mode=1; % Coloring of dendrograms (0-no,1-groups,2-trains)
f_para.comment_string=['Fig8']; % Additional comment on the example, will be used in file and figure names
Additional Figure 1: Comparison of piecewise constant and sampled dissimilarity profile
(example 21)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0]; % Vector with order of subplots
f_para.subplot_size=[1 1.5]; % Vector with relative size of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.publication=0; % Omits otherwise helpful information, such as additional comments (0-no,1-yes)
[The dissimilarity profile with the normal line width is sampled (with one value each sample point), whereas the bold dissimilarity profile is piecewise constant (with one value for each interval in
the pooled spike train). This profile is obtained by representing each interval by the value of its center and weighting it by its length. This is not only faster, but it actually gives the exact
result, whereas the sampled calculation is a very good approximation only for sufficiently small sampling intervals (imagine the example of a rectangular function, at some point any sampled
representation has to cut the right angle). On the other hand, the sampled profile has a higher temporal resolution in the visualization. If this is not essential or if the averaged values are
sufficient, the calculation of the piecewise constant distances is preferable.
9-10: ISI-distance (per definition piecewise constant)
11-12: SPIKE-distance (piecewise constant)
13-14: Real-time SPIKE-distance (piecewise constant)
15-16:Future SPIKE-distance (piecewise constant)
17-18: SPIKE-distance (sampled with variable 'time')
19-20: Real-time SPIKE-distance (sampled with variable 'time')
21-22: Future SPIKE-distance (sampled with variable 'time')
In the source codes the sampled variables can be recognized by the ending '_t'.
Regarding the plotting parameters: The sampled profile was generated because f_para.subplot_posi(18)>0 and the piecewise constant profile was generated because f_para.subplot_posi(12)>0. Since both
values were 2 they were plotted in the same subplot. And since f_para.subplot_size was set to [1 1.5], the second subplot is 1.5 times larger than the first.]
Additional Figure 2: Poisson with different firing rate - Distance value only depends on distance from diagonal (distances are scale-free)
(example 121)
d_para.select_averages={[d_para.tmin d_para.tmax]}; % Continuous intervals for selective temporal averaging
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 2 0 3 0 0 0 0 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=8; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.color_norm_mode=3; % normalization of pairwise color matrices (1-Absolute maximum value,2-Overall maximum value,3-Each one individually)
Additional Figure 3: Poisson with different firing rate - Distance value only depends on distance from diagonal (distances are scale-free)
(example 121)
d_para.select_averages={[d_para.tmin d_para.tmax]}; % Continuous intervals for selective temporal averaging
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 2 0 3 0 0 0 0 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=8; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.color_norm_mode=3; % normalization of pairwise color matrices (1-Absolute maximum value,2-Overall maximum value,3-Each one individually)
Additional Figure 4: Expectation values for Poisson spike trains with identical rate (but these values do not depend on the rate)
(example 131)
f_para.subplot_posi=[0 1 0 0 0 0 0 0 0 0 0 2 0 3 0 0 0 0 0 0 0 0]; % Vector with order of subplots
f_para.plot_mode=1; % +1:vs time,+2-different measures and cuts,+4-different measures,+8-different cuts,+16:different cuts-Movie
f_para.publication=0; % Omits otherwise helpful information, such as additional comments (0-no,1-yes)
[The firing rate of the Poisson spike trains is 1000 Hz so these are almost 1.000.000 spikes. It takes quite some time (~1h on my notebook) to run this example. The interval is cut into smaller
segments, and the averaging over all pairs of spike trains is performed for each segment separately.] | {"url":"https://www.thomaskreuz.org/source-codes/isi-and-spike-distance","timestamp":"2024-11-07T23:23:19Z","content_type":"text/html","content_length":"385788","record_id":"<urn:uuid:ee9efba7-647a-4fa0-9f6b-5eafdb291da2>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00054.warc.gz"} |
The Stacks project
Proposition 47.15.11. Let $A$ be a Noetherian ring which has a dualizing complex. Then any $A$-algebra essentially of finite type over $A$ has a dualizing complex.
Comments (2)
Comment #1693 by Sándor Kovács on
Typo: "dualixing" should be "dualizing"
Comment #1741 by Johan on
Thanks, fixed here.
There are also:
• 2 comment(s) on Section 47.15: Dualizing complexes
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder,
this is tag 0A7K. Beware of the difference between the letter 'O' and the digit '0'.
The tag you filled in for the captcha is wrong. You need to write 0A7K, in case you are confused. | {"url":"https://stacks.math.columbia.edu/tag/0A7K","timestamp":"2024-11-08T11:37:26Z","content_type":"text/html","content_length":"15271","record_id":"<urn:uuid:5f976628-db21-4167-b054-88a2a6d2474a>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00368.warc.gz"} |
Re: Crypto and new computing strategies
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: Crypto and new computing strategies
> Jim Choate writes:
> > In the latest issue of Scientific American there is an article...
On Seth Lloyd's grain-of-salt computer, actually. I didn't know he was
going to build one. Anyway, his technique *may* be useful to make quantum
computers, but it's more likely to be useful for making regular
deterministic massive single-instruction-multiple-data computers out of
fairly simple crystals--"maybe even a grain of salt."
His technique would make every repeating unit of the 3D crystal into a
computing unit. You lose a couple factors of 10 for addressing,
making higher-level modules, and error-correction. Still, that's a lot
of compute power.
Tim May says-
> No need to worry just yet.
> There is no convincing evidence that "quantum computers" can calculate
> in any way differently from "ordinary" computers.
Right. This is just a large power increase using deterministic stuff.
It's based on electrons in the shells of atoms in crystals responding
to different frequencies of photons depending on their own and
neighboring atoms' shells' states.
> Devices that are built on a size scale where quantum effects are
> important, such as quantum-well devices, don't use QM as a
> computational mechanism per se. The devices are just real small. But
> not small enough to matter for large RSA moduli--the computations
> required to factor a 1000-decimal-digit number swamp even a universe
> _made_ of computers!
Which is what a naive guess would have said about 129-digit numbers.
I would love to see some sort of curve of factoring algorithm
efficiencies over time. You could show the log of the difficulty for
a selection of number sizes over the past hundred years, say. The
experts say it's flattening out and will probably stay that way.
A sudden jump in the high end of computer power would mean that we
would need to use larger keys sooner than we thought. A key length
requiring a little bit more work on the user's part means a
lot more work on the cracker's part, but I don't know how many more
bits of key compensate for a 10^9 increase in cracking power, say.
quote me
- - - - - - - - - - - - - - -
blue pill, Pharm. a pill of blue mass, used as an alterative...
alterative, adj. tending to alter...
-----BEGIN PGP SIGNATURE-----
Version: 2.3a
-----END PGP SIGNATURE----- | {"url":"https://cypherpunks.venona.com/date/1994/03/msg01147.html","timestamp":"2024-11-13T18:25:08Z","content_type":"text/html","content_length":"6983","record_id":"<urn:uuid:4c26af4a-b183-43f4-90ad-bee6cf8aaf84>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00115.warc.gz"} |
Match Charting Project Tactics Stats: Glossary
I’m in the process of rolling out more stats based on Match Charting Project data across Tennis Abstract. This is one of several glossaries intended to explain those stats and point interested
visitors to further reading.
At the moment, the following tactics-related stats can be seen at a variety of leaderboards.
• SnV Freq% – Serve-and-volley frequency. The percentage of service points (excluding aces) on which the server comes in behind the serve. I exclude aces because serve-and-volley attempts are less
clear (and thus less consistently charted) if the server realizes immediately that he or she has hit an unreturnable serve. I realize this is a minority opinion and thus an unorthodox way to
calculate the stat, but I’m sticking with it.
• SnV W% – Serve-and-volley winning percentage. The percentage of (non-ace) serve-and-volley attempts that result in the server winning the point.
• Net Freq – Net point frequency. The percentage of total points in which the player comes to net, including serve-and-volley points. I include points in which the player doesn’t hit any net shots
(such as an approach shot that leads to a lob winner), but I do not count points ended by a winner that appears to be an approach shot.
• Net W% – Net point winning percentage. The percentage of net points won by this player.
• FH Wnr% – Forehand winner percentage. The percentage of topspin forehands (excluding forced errors) that result in winners or induced forced errors.
• FH DTL Wnr% – Forehand down-the-line winning percentage. The percentage of topspin down-the-line forehands (excluding forced errors) that result in winners or induced forced errors. Here, I
define “down-the-line” a bit broadly. The Match Charting Project classifies the direction of every shot in one of three categories. If a forehand is hit from the middle of the court or the
player’s forehand corner and hit to the opponent’s backhand corner (or a lefty’s forehand corner), it counts as a down-the-line shot. Thus, some shots that would typically be called “off”
forehands end up in this category.
• FH IO Wnr% – Forehand inside-out winning percentage. The percentage of topspin inside-out forehands (excluding forced errors) that result in winners or induced forced errors. This one is defined
more strictly, only counting forehands hit from the player’s own backhand corner to the opponent’s backhand corner (or a lefty’s forehand corner).
• BH Wnr% – Backhand winner percentage. The percentage of topspin backhands (excluding forced errors) that result in winners or induced forced errors.
• BH DTL Wnr% – Backhand down-the-line winner percentage. The percentage of topspin down-the-line backhands (excluding forced errors) that result in winners or induced forced errors. As with the
forehand down-the-line stat, I define these a bit broadly, catching some “off” backhands as well.
• Drop Freq – Dropshot frequency. The percentage of groundstrokes that are dropshots. This excludes dropshots hit at the net and those hit in response to an opponent’s dropshot (re-drops).
• Drop Wnr% – Dropshot winner percentage. The percentage of dropshots that result in winners or induced forced errors. Note that this number itself isn’t a verdict on the dropshot tactic, as it
doesn’t count extended points that the player who hit the dropshot went on to win.
• RallyAgg – Rally Aggression Score. A variation of Aggression Score, a stat invented by MCP contributor Lowell West. At its simplest, any member of this family of aggression metrics is the
percentage of shots that end the point–winners, unforced errors, and shots that induce forced errors. RallyAgg excludes serves and is a bit more complex, following the logic that I outlined for
Return Aggression by separating winners from unforced errors. For each match, the player’s unforced error rate and winner rate are normalized relative to tour average and expressed in standard
deviations above or below the mean. RallyAgg is the average of those two numbers, multiplied by 100 for the sake of readability. The higher the score, the more aggressive the player. Tour average
is zero.
• ReturnAgg – Return Aggression Score. Another variation of Aggression score, considering only return winners and return errors. As with RallyAgg, winners and errors are separated, and each rate is
normalized relative to tour average. ReturnAgg is the average of those two normalized rates, multiplied by 100 for the sake of readability. The higher the number, the more aggressive the
returner, and tour average is zero. | {"url":"http://www.tennisabstract.com/blog/2019/08/17/match-charting-project-tactics-stats-glossary/","timestamp":"2024-11-03T23:05:30Z","content_type":"text/html","content_length":"61868","record_id":"<urn:uuid:41c22e92-ff91-4c69-87b7-30323fe5a007>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00069.warc.gz"} |
MATH 127C University of Nairobi Cauchy Sequence Mathematics Homework 1 - Course Help Online
please write the answer and logic clearly.
here is hw and some handout.
hope nice writing
Math 127C – Homework 1
Write complete solutions to each of the following problems. Submit your
answers in PDF format to Gradescope by Monday, June 29 at 11:59 PM.
1. Show that the 1-norm/taxicab norm on Rn , defined by k(x1 , . . . , xn )k1 =
|x1 | + . . . + |xn | for all (x1 , . . . , xn ) ∈ Rn , is a norm.
2. Let V be any nontrivial (V 6= {0}) vector space over R, and let k · k be
a norm on V . Show that for any positive real number r > 0 there exists
x ∈ V such that kxk = r.
3. Let (M, d) be a metric space, and let M be finite. Show that (M, d) is
4. Let {xn }∞
n=1 be a sequence in a metric space, and suppose the sequence
satisfies the following property: for every > 0, there exists N ≥ 1 and a
ball B of radius such that
n > N =⇒ xn ∈ B.
Show that {xn }∞
n=1 is a Cauchy sequence.
5. Show that
S for any family {Uα : α ∈ I} of open sets in a metric space, the
union α∈I Uα is open.
6. Show that if U and V are open sets in a metric space, then U ∩ V is open.
7. Show that a finite subset of a metric space is connected if and only if it
contains exactly one point. | {"url":"https://coursehelponline.com/math-127c-university-of-nairobi-cauchy-sequence-mathematics-homework-1/","timestamp":"2024-11-13T20:54:16Z","content_type":"text/html","content_length":"42013","record_id":"<urn:uuid:f85d41e7-1e56-4836-8db4-9352946bd398>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00462.warc.gz"} |
The Department of Statistics offers courses in the basic principles and techniques of probability and statistics, advanced theory and methods courses, courses in stochastic processes and methods, and
courses statistical methods in finance.
For questions about specific courses, contact the department.
For questions about specific courses, contact the department.
A friendly introduction to statistical concepts and reasoning with emphasis on developing statistical intuition rather than on mathematical rigor. Topics include design of experiments, descriptive
statistics, correlation and regression, probability, chance variability, sampling, chance models, and tests of significance.
Course Number
3 pts
Spring 2025
Mo 14:40-15:55We 14:40-15:55
Section/Call Number
0 of 86
Victor de la Pena
A friendly introduction to statistical concepts and reasoning with emphasis on developing statistical intuition rather than on mathematical rigor. Topics include design of experiments, descriptive
statistics, correlation and regression, probability, chance variability, sampling, chance models, and tests of significance.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 86
Anthony Donoghue
A friendly introduction to statistical concepts and reasoning with emphasis on developing statistical intuition rather than on mathematical rigor. Topics include design of experiments, descriptive
statistics, correlation and regression, probability, chance variability, sampling, chance models, and tests of significance.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 86
Ashley Datta
Prerequisites: intermediate high school algebra. Designed for students in fields that emphasize quantitative methods. Graphical and numerical summaries, probability, theory of sampling distributions,
linear regression, analysis of variance, confidence intervals and hypothesis testing. Quantitative reasoning and data analysis. Practical experience with statistical software. Illustrations are taken
from a variety of fields. Data-collection/analysis project with emphasis on study designs is part of the coursework requirement.
Course Number
3 pts
Spring 2025
Th 10:10-11:25Tu 10:10-11:25
Section/Call Number
0 of 160
Wayne Lee
Prerequisites: intermediate high school algebra. Designed for students in fields that emphasize quantitative methods. Graphical and numerical summaries, probability, theory of sampling distributions,
linear regression, analysis of variance, confidence intervals and hypothesis testing. Quantitative reasoning and data analysis. Practical experience with statistical software. Illustrations are taken
from a variety of fields. Data-collection/analysis project with emphasis on study designs is part of the coursework requirement.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 86
Banu Baydil
Prerequisites: one semester of calculus. Designed for students who desire a strong grounding in statistical concepts with a greater degree of mathematical rigor than in STAT W1111. Random variables,
probability distributions, pdf, cdf, mean, variance, correlation, conditional distribution, conditional mean and conditional variance, law of iterated expectations, normal, chi-square, F and t
distributions, law of large numbers, central limit theorem, parameter estimation, unbiasedness, consistency, efficiency, hypothesis testing, p-value, confidence intervals, maximum likelihood
estimation. Serves as the pre-requisite for ECON W3412.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 86
Hammou El Barmi
Prerequisites: one semester of calculus. Designed for students who desire a strong grounding in statistical concepts with a greater degree of mathematical rigor than in STAT W1111. Random variables,
probability distributions, pdf, cdf, mean, variance, correlation, conditional distribution, conditional mean and conditional variance, law of iterated expectations, normal, chi-square, F and t
distributions, law of large numbers, central limit theorem, parameter estimation, unbiasedness, consistency, efficiency, hypothesis testing, p-value, confidence intervals, maximum likelihood
estimation. Serves as the pre-requisite for ECON W3412.
Course Number
3 pts
Spring 2025
Mo 08:40-09:55We 08:40-09:55
Section/Call Number
0 of 85
Joyce Robbins
Prerequisites: one semester of calculus. Designed for students who desire a strong grounding in statistical concepts with a greater degree of mathematical rigor than in STAT W1111. Random variables,
probability distributions, pdf, cdf, mean, variance, correlation, conditional distribution, conditional mean and conditional variance, law of iterated expectations, normal, chi-square, F and t
distributions, law of large numbers, central limit theorem, parameter estimation, unbiasedness, consistency, efficiency, hypothesis testing, p-value, confidence intervals, maximum likelihood
estimation. Serves as the pre-requisite for ECON W3412.
Course Number
3 pts
Spring 2025
Th 10:10-11:25Tu 10:10-11:25
Section/Call Number
0 of 86
Joyce Robbins
Prerequisites: one semester of calculus. Designed for students who desire a strong grounding in statistical concepts with a greater degree of mathematical rigor than in STAT W1111. Random variables,
probability distributions, pdf, cdf, mean, variance, correlation, conditional distribution, conditional mean and conditional variance, law of iterated expectations, normal, chi-square, F and t
distributions, law of large numbers, central limit theorem, parameter estimation, unbiasedness, consistency, efficiency, hypothesis testing, p-value, confidence intervals, maximum likelihood
estimation. Serves as the pre-requisite for ECON W3412.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 86
Corequisites: An introductory course in statistic (STAT UN1101 is recommended). This course is an introduction to R programming. After learning basic programming component, such as defining variables
and vectors, and learning different data structures in R, students will, via project-based assignments, study more advanced topics, such as recursion, conditionals, modular programming, and data
visualization. Students will also learn the fundamental concepts in computational complexity, and will practice writing reports based on their statistical analyses.
Course Number
3 pts
Spring 2025
Tu 16:10-17:25Th 16:10-17:25
Section/Call Number
0 of 120
Alex Pijyan
Prerequisites: An introductory course in statistics (STAT UN1101 is recommended). Students without programming experience in R might find STAT UN2102 very helpful. Develops critical thinking and data
analysis skills for regression analysis in science and policy settings. Simple and multiple linear regression, non-linear and logistic models, random-effects models. Implementation in a statistical
package. Emphasis on real-world examples and on planning, proposing, implementing, and reporting.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 85
Daniel Rabinowitz
Prerequisites: STAT UN2103 is strongly recommended. Students without programming experience in R might find STAT UN2102 very helpful. This course covers statistical models amd methods for analyzing
and drawing inferences for problems involving categofical data. The goals are familiarity and understanding of a substantial and integrated body of statistical methods that are used for such
problems, experience in anlyzing data using these methods, and profficiency in communicating the results of such methods, and the ability to critically evaluate the use of such methods. Topics
include binomial proportions, two-way and three-way contingency tables, logistic regression, log-linear models for large multi-way contingency tables, graphical methods. The statistical package R
will be used.
Course Number
3 pts
Spring 2025
Mo 08:40-09:55We 08:40-09:55
Section/Call Number
0 of 86
Ronald Neath
Prerequisites: STAT UN2103. Students without programming experience in R might find STAT UN2102 very helpful. This course is a machine learning class from an application perspective. We will cover
topics including data-based prediction, classification, specific classification methods (such as logistic regression and random forests), and basics of neural networks. Programming in homeworks will
require R.
Course Number
3 pts
Spring 2025
Tu 14:40-15:55Th 14:40-15:55
Section/Call Number
0 of 86
Wayne Lee
Prerequisites: the project mentors permission. This course provides a mechanism for students who undertake research with a faculty member from the Department of Statistics to receive academic credit.
Students seeking research opportunities should be proactive and entrepreneurial: identify congenial faculty whose research is appealing, let them know of your interest and your background and skills.
Course Number
3 pts
Spring 2025
Section/Call Number
0 of 2
Ronald Neath
Topics in Modern Statistics that provide undergraduate students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
field. Courses listed are reviewed and approved by the Undergraduate Advisory Committee of the Department of Statistics. A good working knowledge of basic statistical concepts (likelihood,
Bayes' rule, Poisson processes, Markov chains, Gaussian random vectors), including especially linear-algebraic concepts related to regression and principal components analysis, is necessary. No
previous experience with neural data is required.
Course Number
3 pts
Spring 2025
Tu 14:40-15:55Th 14:40-15:55
Section/Call Number
0 of 16
Joyce Robbins
Prerequisites: Calculus through multiple integration and infinite sums. A calculus-based tour of the fundamentals of probability theory and statistical inference. Probability models, random
variables, useful distributions, conditioning, expectations, law of large numbers, central limit theorem, point and confidence interval estimation, hypothesis tests, linear regression. This course
replaces SIEO 4150.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 100
Cristian Pasarica
Prerequisites: Calculus through multiple integration and infinite sums. A calculus-based tour of the fundamentals of probability theory and statistical inference. Probability models, random
variables, useful distributions, conditioning, expectations, law of large numbers, central limit theorem, point and confidence interval estimation, hypothesis tests, linear regression. This course
replaces SIEO 4150.
Course Number
3 pts
Spring 2025
Mo 13:10-14:25We 13:10-14:25
Section/Call Number
0 of 86
Sumit Mukherjee
Prerequisites: At least one semester, and preferably two, of calculus. An introductory course (STAT UN1201, preferably) is strongly recommended. A calculus-based introduction to probability theory. A
quick review of multivariate calculus is provided. Topics covered include random variables, conditional probability, expectation, independence, Bayes’ rule, important distributions, joint
distributions, moment generating functions, central limit theorem, laws of large numbers and Markov’s inequality.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 60
Marco Avella Medina
Prerequisites: At least one semester, and preferably two, of calculus. An introductory course (STAT UN1201, preferably) is strongly recommended. A calculus-based introduction to probability theory. A
quick review of multivariate calculus is provided. Topics covered include random variables, conditional probability, expectation, independence, Bayes’ rule, important distributions, joint
distributions, moment generating functions, central limit theorem, laws of large numbers and Markov’s inequality.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 5
Marco Avella Medina
Prerequisites: STAT GU4203. At least one semester of calculus is required; two or three semesters are strongly recommended. Calculus-based introduction to the theory of statistics. Useful
distributions, law of large numbers and central limit theorem, point estimation, hypothesis testing, confidence intervals maximum likelihood, likelihood ratio tests, nonparametric procedures, theory
of least squares and analysis of variance.
Course Number
3 pts
Spring 2025
Tu 13:10-14:25Th 13:10-14:25
Section/Call Number
0 of 45
Banu Baydil
Prerequisites: STAT GU4203. At least one semester of calculus is required; two or three semesters are strongly recommended. Calculus-based introduction to the theory of statistics. Useful
distributions, law of large numbers and central limit theorem, point estimation, hypothesis testing, confidence intervals maximum likelihood, likelihood ratio tests, nonparametric procedures, theory
of least squares and analysis of variance.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 35
Ashley Datta
Prerequisites: STAT GU4204 or the equivalent, and a course in linear algebra. Theory and practice of regression analysis. Simple and multiple regression, testing, estimation, prediction, and
confidence procedures, modeling, regression diagnostics and plots, polynomial regression, colinearity and confounding, model selection, geometry of least squares. Extensive use of the computer to
analyse data.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 35
Ronald Neath
Prerequisites: STAT GU4204 and GU4205 or the equivalent. Introduction to programming in the R statistical package: functions, objects, data structures, flow control, input and output, debugging,
logical design, and abstraction. Writing code for numerical and graphical statistical analyses. Writing maintainable code and testing, stochastic simulations, paralleizing data analyses, and working
with large data sets. Examples from data science will be used for demonstration.
Course Number
3 pts
Spring 2025
Fr 10:10-12:40
Section/Call Number
0 of 40
Yongchan Kwon
Prerequisites: STAT GU4203 and two, preferably three, semesters of calculus. Review of elements of probability theory. Poisson processes. Renewal theory. Wald's equation. Introduction to discrete and
continuous time Markov chains. Applications to queueing theory, inventory models, branching processes.
Course Number
3 pts
Spring 2025
Tu 16:10-17:25Th 16:10-17:25
Section/Call Number
0 of 50
Anne van Delft
Prerequisites: STAT GU4203 and two, preferably three, semesters of calculus. Review of elements of probability theory. Poisson processes. Renewal theory. Walds equation. Introduction to discrete and
continuous time Markov chains. Applications to queueing theory, inventory models, branching processes.
Course Number
3 pts
Spring 2025
Mo 11:40-12:55We 11:40-12:55
Section/Call Number
0 of 35
Mark Brown
Prerequisites: STAT GU4205 or the equivalent. Least squares smoothing and prediction, linear systems, Fourier analysis, and spectral estimation. Impulse response and transfer function. Fourier
series, the fast Fourier transform, autocorrelation function, and spectral density. Univariate Box-Jenkins modeling and forecasting. Emphasis on applications. Examples from the physical sciences,
social sciences, and business. Computing is an integral part of the course.
Course Number
3 pts
Spring 2025
Sa 10:10-12:40
Section/Call Number
0 of 25
Franz Rembart
Prerequisites: STAT GU4204 or the equivalent. Statistical inference without parametric model assumption. Hypothesis testing using ranks, permutations, and order statistics. Nonparametric analogs of
analysis of variance. Non-parametric regression, smoothing and model selection.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 25
Arian Maleki
This course introduces the Bayesian paradigm for statistical inference. Topics covered include prior and posterior distributions: conjugate priors, informative and non-informative priors; one- and
two-sample problems; models for normal data, models for binary data, Bayesian linear models; Bayesian computation: MCMC algorithms, the Gibbs sampler; hierarchical models; hypothesis testing, Bayes
factors, model selection; use of statistical software.
Prerequisites: A course in the theory of statistical inference, such as STAT GU4204 a course in statistical modeling and data analysis, such as STAT GU4205.
Course Number
3 pts
Spring 2025
Tu 19:40-20:55Th 19:40-20:55
Section/Call Number
0 of 25
Dobrin Marchev
Prerequisites: STAT GU4204 or the equivalent. Introductory course on the design and analysis of sample surveys. How sample surveys are conducted, why the designs are used, how to analyze survey
results, and how to derive from first principles the standard results and their generalizations. Examples from public health, social work, opinion polling, and other topics of interest.
Course Number
3 pts
Spring 2025
Tu 14:40-15:55Th 14:40-15:55
Section/Call Number
0 of 25
Rongning Wu
Prerequisites: STAT GU4206. The course will provide an introduction to Machine Learning and its core models and algorithms. The aim of the course is to provide students of statistics with detailed
knowledge of how Machine Learning methods work and how statistical models can be brought to bear in computer systems - not only to analyze large data sets, but to let computers perform tasks that
traditional methods of computer science are unable to address. Examples range from speech recognition and text analysis through bioinformatics and medical diagnosis. This course provides a first
introduction to the statistical methods and mathematical concepts which make such technologies possible.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 50
Samory Kpotufe
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will be covered using existing tools in
R. Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of
several core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
We 18:10-20:55
Section/Call Number
0 of 25
Alex Pijyan
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will be covered using existing tools in
R. Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of
several core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
Fr 18:10-20:55
Section/Call Number
0 of 25
Haiyuan Wang
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will be covered using existing tools in
R. Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of
several core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
Tu 16:10-18:40
Section/Call Number
0 of 30
Galen McKinley
Prerequisites: STAT GU4205 or the equivalent. A fast-paced introduction to statistical methods used in quantitative finance. Financial applications and statistical methodologies are intertwined in
all lectures. Topics include regression analysis and applications to the Capital Asset Pricing Model and multifactor pricing models, principal components and multivariate analysis, smoothing
techniques and estimation of yield curves statistical methods for financial time series, value at risk, term structure models and fixed income research, and estimation and modeling of volatilities.
Hands-on experience with financial data.
Course Number
3 pts
Spring 2025
Sa 10:10-12:40
Section/Call Number
0 of 25
Zhiliang Ying
Prerequisites: STAT GU4203. STAT GU4207 is recommended. Basics of continuous-time stochastic processes. Wiener processes. Stochastic integrals. Ito's formula, stochastic calculus. Stochastic
exponentials and Girsanov's theorem. Gaussian processes. Stochastic differential equations. Additional topics as time permits.
Course Number
3 pts
Spring 2025
Mo 16:10-17:25We 16:10-17:25
Section/Call Number
0 of 25
Steven Campbell
Prerequisites: STAT GU4264. Mathematical theory and probabilistic tools for modeling and analyzing security markets are developed. Pricing options in complete and incomplete markets, equivalent
martingale measures, utility maximization, term structure of interest rates. This is a core course in the MS program in mathematical finance.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 25
Graeme Baker
Prerequisites: STAT GU4205 and at least one statistics course numbered between GU4221 and GU4261. This is a course on getting the most out of data. The emphasis will be on hands-on experience,
involving case studies with real data and using common statistical packages. The course covers, at a very high level, exploratory data analysis, model formulation, goodness of fit testing, and other
standard and non-standard statistical procedures, including linear regression, analysis of variance, nonlinear regression, generalized linear models, survival analysis, time series analysis, and
modern regression methods. Students will be expected to propose a data set of their choice for use as case study material.
Course Number
3 pts
Spring 2025
Fr 10:10-12:40
Section/Call Number
0 of 25
Gabriel Young
Prerequisites: At least one semester of calculus. A calculus-based introduction to probability theory. Topics covered include random variables, conditional probability, expectation, independence,
Bayes rule, important distributions, joint distributions, moment generating functions, central limit theorem, laws of large numbers and Markovs inequality.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 35
Marco Avella Medina
Prerequisites: STAT GR5203 and GR5204 or the equivalent. Theory and practice of regression analysis, Simple and multiple regression, including testing, estimation, and confidence procedures,
modeling, regression diagnostics and plots, polynomial regression, colinearity and confounding, model selection, geometry of least squares. Extensive use of the computer to analyse data.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 50
Ronald Neath
Corequisites: STAT GR5204 and GR5205 or the equivalent. Introduction to programming in the R statistical package: functions, objects, data structures, flow control, input and output, debugging,
logical design, and abstraction. Writing code for numerical and graphical statistical analyses. Writing maintainable code and testing, stochastic simulations, paralleizing data analyses, and working
with large data sets. Examples from data science will be used for demonstration.
Course Number
3 pts
Spring 2025
Fr 10:10-12:40
Section/Call Number
0 of 50
Yongchan Kwon
Corequisites: GR5203 or the equivalent. Review of elements of probability theory. Poisson processes. Renewal theory. Wald's equation. Introduction to discrete and continuous time Markov chains.
Applications to queueing theory, inventory models, branching processes.
Course Number
3 pts
Spring 2025
Mo 11:40-12:55We 11:40-12:55
Section/Call Number
0 of 100
Mark Brown
Prerequisites: STAT GR5205 Least squares smoothing and prediction, linear systems, Fourier analysis, and spectral estimation. Impulse response and transfer function. Fourier series, the fast Fourier
transform, autocorrelation function, and spectral density. Univariate Box-Jenkins modeling and forecasting. Emphasis on applications. Examples from the physical sciences, social sciences, and
business. Computing is an integral part of the course.
Course Number
3 pts
Spring 2025
Sa 10:10-12:40
Section/Call Number
0 of 125
Franz Rembart
Prerequisites: STAT GR5205 Statistical inference without parametric model assumption. Hypothesis testing using ranks, permutations, and order statistics. Nonparametric analogs of analysis of
variance. Non-parametric regression, smoothing and model selection.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 86
Arian Maleki
Bayesian data analysis: building, fitting, evaluating and improving probability models. Prior information, hierachical models, and combining information. Linear and nonlinear models. Simulation of
fake data and evaluation of methods. Computing using R and Stan.
Course Number
3 pts
Spring 2025
Tu 19:40-20:55Th 19:40-20:55
Section/Call Number
0 of 125
Dobrin Marchev
Course Number
3 pts
Spring 2025
Tu 14:40-15:55Th 14:40-15:55
Section/Call Number
0 of 86
Rongning Wu
Prerequisites: STAT GR5206 or the equivalent. The course will provide an introduction to Machine Learning and its core models and algorithms. The aim of the course is to provide students of
statistics with detailed knowledge of how Machine Learning methods work and how statistical models can be brought to bear in computer systems - not only to analyze large data sets, but to let
computers perform tasks that traditional methods of computer science are unable to address. Examples range from speech recognition and text analysis through bioinformatics and medical diagnosis. This
course provides a first introduction to the statistical methods and mathematical concepts which make such technologies possible.
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 86
Genevera Allen
Prerequisites: STAT GR5206 or the equivalent. The course will provide an introduction to Machine Learning and its core models and algorithms. The aim of the course is to provide students of
statistics with detailed knowledge of how Machine Learning methods work and how statistical models can be brought to bear in computer systems - not only to analyze large data sets, but to let
computers perform tasks that traditional methods of computer science are unable to address. Examples range from speech recognition and text analysis through bioinformatics and medical diagnosis. This
course provides a first introduction to the statistical methods and mathematical concepts which make such technologies possible.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 86
Yisha Yao
Prerequisites: STAT GR5206 or the equivalent. The course will provide an introduction to Machine Learning and its core models and algorithms. The aim of the course is to provide students of
statistics with detailed knowledge of how Machine Learning methods work and how statistical models can be brought to bear in computer systems - not only to analyze large data sets, but to let
computers perform tasks that traditional methods of computer science are unable to address. Examples range from speech recognition and text analysis through bioinformatics and medical diagnosis. This
course provides a first introduction to the statistical methods and mathematical concepts which make such technologies possible.
Course Number
3 pts
Spring 2025
Mo 18:10-19:25We 18:10-19:25
Section/Call Number
0 of 86
Alberto Gonzalez Sanz
Prerequisites: STAT GR5206 or the equivalent. The course will provide an introduction to Machine Learning and its core models and algorithms. The aim of the course is to provide students of
statistics with detailed knowledge of how Machine Learning methods work and how statistical models can be brought to bear in computer systems - not only to analyze large data sets, but to let
computers perform tasks that traditional methods of computer science are unable to address. Examples range from speech recognition and text analysis through bioinformatics and medical diagnosis. This
course provides a first introduction to the statistical methods and mathematical concepts which make such technologies possible.
Course Number
3 pts
Spring 2025
Tu 13:10-14:25Th 13:10-14:25
Section/Call Number
0 of 86
Chenyang Zhong
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will covered using existing tools in R.
Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of several
core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
We 18:10-20:40
Section/Call Number
0 of 86
Alex Pijyan
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will covered using existing tools in R.
Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of several
core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
Th 18:10-20:40
Section/Call Number
0 of 86
Haiyuan Wang
Prerequisites: Pre-requisite for this course includes working knowledge in Statistics and Probability, data mining, statistical modeling and machine learning. Prior programming experience in R or
Python is required. This course will incorporate knowledge and skills covered in a statistical curriculum with topics and projects in data science. Programming will covered using existing tools in R.
Computing best practices will be taught using test-driven development, version control, and collaboration. Students finish the class with a portfolio of projects, and deeper understanding of several
core statistical/machine-learning algorithms. Short project cycles throughout the semester provide students extensive hands-on experience with various data-driven applications.
Course Number
3 pts
Spring 2025
Tu 16:10-18:40
Section/Call Number
0 of 30
Galen McKinleyTian Zheng
Prerequisites: STAT GR5204 or the equivalent. STAT GR5205 is recommended. A fast-paced introduction to statistical methods used in quantitative finance. Financial applications and statistical
methodologies are intertwined in all lectures. Topics include regression analysis and applications to the Capital Asset Pricing Model and multifactor pricing models, principal components and
multivariate analysis, smoothing techniques and estimation of yield curves statistical methods for financial time series, value at risk, term structure models and fixed income research, and
estimation and modeling of volatilities. Hands-on experience with financial data.
Course Number
3 pts
Spring 2025
Sa 10:10-12:40
Section/Call Number
0 of 150
Zhiliang Ying
Prerequisites: STAT GR5203 or the equivalent. Basics of continuous-time stochastic processes. Wiener processes. Stochastic integrals. Ito's formula, stochastic calculus. Stochastic exponentials and
Girsanov's theorem. Gaussian processes. Stochastic differential equations. Additional topics as time permits.
Course Number
3 pts
Spring 2025
Mo 16:10-17:25We 16:10-17:25
Section/Call Number
0 of 86
Steven Campbell
Prerequisites: STAT GR5264 Available to SSP, SMP. Mathematical theory and probabilistic tools for modeling and analyzing security markets are developed. Pricing options in complete and incomplete
markets, equivalent martingale measures, utility maximization, term structure of interest rates.
Course Number
3 pts
Spring 2025
Tu 18:10-19:25Th 18:10-19:25
Section/Call Number
0 of 135
Graeme Baker
Prerequisites: W4315 and either another statistics course numbered above the 4200 or permission of instructor. Required for the major in statistics. Data analysis using a computer statistical package
and selected exploratory data analysis subroutines. Topics include editing of data for errors, exploratory and standard techniques for one-way analysis of variance, linear regression, and two-way
analysis of variance. Material is presented in case-study format.
Course Number
3 pts
Spring 2025
Fr 10:10-12:40
Section/Call Number
0 of 225
Gabriel Young
Topics in Modern Statistics will provide MA Statistics students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
Course Number
3 pts
Spring 2025
Tu 14:40-15:55Th 14:40-15:55
Section/Call Number
0 of 86
Joyce Robbins
Topics in Modern Statistics will provide MA Statistics students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
Course Number
3 pts
Spring 2025
Tu 13:10-14:25Th 13:10-14:25
Section/Call Number
0 of 35
Philip Protter
Topics in Modern Statistics will provide MA Statistics students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
Course Number
3 pts
Spring 2025
Mo 16:10-17:25We 16:10-17:25
Section/Call Number
0 of 86
Parijat Dube
Topics in Modern Statistics will provide MA Statistics students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
Course Number
3 pts
Spring 2025
We 18:10-20:40
Section/Call Number
0 of 86
Lei Kang
Topics in Modern Statistics will provide MA Statistics students with an opportunity to study a specialized area of statistics in more depth and to meet the educational needs of a rapidly growing
Course Number
3 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 20
Andrew Gelman
This course is intended to provide a mechanism to MA students in Statistics who undertake on-campus project work or research. The course may be signed up with a faculty member from the Department of
Statistics for academic credit. Students seeking to enroll in the course should identify an on-campus project and a congenial faculty member whose research is appealing to them, and who are able to
serve as their mentor. Students should then submit an application to enroll in this course, which will be reviewed and approved by the Faculty Director of the MA in Statistics program.
Course Number
3 pts
Spring 2025
Section/Call Number
0 of 35
Demissie Alemayehu
Prerequisites: GR5203; GR5204 &GR5205 and at least 4 approved electives This course is an elective course for students in the M.A. in Statistics program that counts towards the degree requirements.
To receive a grade and academic credits for this course, students are expected to engage in approved off-campus internships that can be counted as an elective. Statistical Fieldwork should provide
students an opportunity to apply their statistical skills and gain practical knowledge on how statistics can be applied to solve real-world challenges.
Course Number
1 pts
Spring 2025
Section/Call Number
0 of 25
Demissie Alemayehu
Prerequisites: (STAT GR5701) working knowledge of calculus and linear algebra (vectors and matrices), STAT GR5701 or equivalent, and familiarity with a programming language (e.g. R, Python) for
statistical data analysis. In this course, we will systematically cover fundamentals of statistical inference and modeling, with special attention to models and methods that address practical data
issues. The course will be focused on inference and modeling approaches such as the EM algorithm, MCMC methods and Bayesian modeling, linear regression models, generalized linear regression models,
nonparametric regressions, and statistical computing. In addition, the course will provide introduction to statistical methods and modeling that addresses various practical issues such as design of
experiments, analysis of time-dependent data, missing values, etc. Throughpout the course, real-data examples will be used in lecture discussion and homework problems. This course lays the
statistical foundation for inference and modeling using data, preparing the MS in Data Science students, for other courses in machine learning, data mining and visualization.
Course Number
3 pts
Spring 2025
Tu 17:40-18:55Th 17:40-18:55
Section/Call Number
0 of 180
Dobrin Marchev
Prerequisites: STAT GR6101 Continuation of STAT GR6101.
Course Number
4 pts
Spring 2025
Mo 10:10-11:25We 10:10-11:25
Section/Call Number
0 of 25
Yuqi Gu
Course Number
4 pts
Spring 2025
Tu 14:10-16:00
Section/Call Number
0 of 25
Liam Paninski
Prerequisites: STAT GR6102 or instructor permission. The Deparatments doctoral student consulting practicum. Students undertake pro bono consulting activities for Columbia community researchers under
the tutelage of a faculty mentor.
Course Number
3 pts
Spring 2025
Mo 08:40-09:55We 08:40-09:55
Section/Call Number
0 of 15
Tian ZhengAshley Datta
Prerequisites: STAT GR6201 Continuation of STAT G6201
Course Number
4 pts
Spring 2025
Mo 14:40-15:55We 14:40-15:55
Section/Call Number
0 of 25
Cynthia Rush
Prerequisites: STAT GR6301. Conditional distributions and expectations. Martingales; inequalities, convergence and closure properties, optimal stopping theorems, Burkholder-Gundy inequalities,
Doob-Meyer decomposition, stochastic integration, Itos rule. Brownian motion: construction, invariance principles and random walks, study of sample paths, martingale representation results Girsanov
Theorem. The heat equation, Feynman-Kac formula. Dirichlet problem, connections with potential theory. Introduction to Markov processes: semigroups and infinitesimal generators, diffusions,
stochastic differential equations.
Course Number
4 pts
Spring 2025
Tu 10:10-11:25Th 10:10-11:25
Section/Call Number
0 of 25
Marcel Nutz
Course Number
3 pts
Spring 2025
Mo 10:10-12:00
Section/Call Number
0 of 25
Christopher Harshaw
Departmental colloquium in statistics.
Course Number
3 pts
Spring 2025
Mo 16:10-17:25
Section/Call Number
0 of 45
Yuqi GuBianca Dumitrascu
Departmental colloquium in probability theory.
Course Number
3 pts
Spring 2025
Fr 11:40-12:55
Section/Call Number
0 of 25
Ivan Corwin
A colloquiim in applied probability and risk.
Course Number
1 pts
Spring 2025
Th 13:10-14:25
Section/Call Number
0 of 25
Chenyang ZhongSumit Mukherjee
A colloquium on topics in mathematical finance
Course Number
3 pts
Spring 2025
Th 16:10-17:25
Section/Call Number
0 of 25
Marcel NutzPhilip Protter | {"url":"https://sps.columbia.edu/courses/professional-academic-development/statistics","timestamp":"2024-11-11T02:01:30Z","content_type":"text/html","content_length":"312368","record_id":"<urn:uuid:6cae615a-14df-4cd0-8fe3-8256a9dff22e>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00487.warc.gz"} |
Feature-product networks (FP-nets) are inspired by end-stopped cortical cells with FP-units that multiply the outputs of two filters. We enhance state-of-the-art deep networks, such as the ResNet and
MobileNet, with FP-units and show that the resulting FP-nets perform better on the Cifar-10 and ImageNet benchmarks. Moreover, we analyze the hyperselectivity of the FP-net model neurons and show
that this property makes FP-nets less sensitive to adversarial attacks and JPEG artifacts. We then show that the learned model neurons are end-stopped to different degrees and that they provide
sparse representations with an entropy that decreases with hyperselectivity.
For machine learning to work, one needs appropriate biases to constrain the solution for the problem at hand. Deep convolutional neural networks (CNNs), for example, are successful due to two
constraints that specialize them relative to more general networks such as the multilayer perceptron (MLP): sparse connections and shared weights. It is well known that biases cannot be learned from
the data or derived by logical deduction (
Watanabe, 1985
). In computer vision, appropriate biases can be obtained, as in the case of the CNNs, by studying biological vision (
LeCun et al., 2015
Majaj & Pelli, 2018
). Besides inspiring the use of localized (oriented) filters (the two CNN biases above) followed by a pointwise nonlinearity, biological vision can provide additional insight, an issue that currently
receives somewhat limited attention in the deep-learning community (
Majaj & Pelli, 2018
Paiton et al., 2020
We here focus on the principle of efficient coding (
Barlow, 1961
Simoncelli & Olshausen, 2001
) and the related neural phenomenon of end-stopping (
Hubel & Wiesel, 1965
). Statistical analysis shows that oriented linear filters reduce the entropy of natural images by encoding oriented straight patterns (one-dimensional [1
] regions) such as vertical and horizontal edges (
Zetzsche et al., 1993
). In cortical area V2, however, the majority of cells are end-stopped to different degrees (
Hubel & Wiesel, 1965
). End-stopped cells are thought to detect two-dimensional (2
) regions such as junctions and corners. Since 2
regions are unique and sparse in natural images (
Barth & Watson, 2000
Mota & Barth, 2000
Zetzsche et al., 1993
), they represent images efficiently, that is, with a high degree of sparseness and minimal information loss. A standard way of modeling end-stopped cells is to multiply outputs of
orientation-selective cells, resulting in an AND-combination of simple-cell outputs (
Zetzsche & Barth, 1990
). For example, a corner can be detected by the logical combination of “horizontal edge AND vertical edge.” In
Paiton et al. (2020)
, the authors argue convincingly that principles adopted from vision should be beneficial for deep networks and that the exploitation of multiplicative interactions between neurons has not been
sufficiently explored in this specific context. There is, nevertheless, a vast literature on sigma-pi networks in general (e.g.,
Mel & Koch, 1990
Rumelhart et al., 1986
), which is not surprising since such networks define a large class of possible systems.
It has been shown that end-stopping can emerge from the principle of predictive coding based on recursive connections (
Rao & Ballard, 1999
); the latter has also been observed in
Barth and Zetzsche (1998)
. Note that in
Rao and Ballard (1999)
, end-stopping emerges based on unsupervised learning with natural images and, in our case, on task-driven supervised learning in a natural vision task.
Feature-product networks (FP-nets) implement a network architecture that contains explicit multiplications of the feature maps obtained with pairs of linear filters. The main feature of these
networks is that they learn the appropriate filter pairs to be multiplied based on the task at hand. An early FP-net architecture has been presented as a preprint (
Grüning et al., 2020b
), and it has been shown in Grüning et al. (
Grüning & Barth, 2021
) that a similar network can predict subjective image quality well. Of course, we do not assume that neurons would compute ideal multiplications; the AND terms could be created in alternative ways,
for example, by using logarithms (
Grüning et al., 2020b
) or the minimum operation (
Grüning & Barth, 2021a
) instead of multiplications. AND terms could also be generated by traditional CNNs with linear filters followed by simple ReLU nonlinearities (
Barth & Zetzsche, 1998
), but this would require larger networks and would be limited in terms of the possible tuning properties of the resulting nonlinear functions (see also
Paiton et al., 2020
, regarding the limits of pointwise nonlinearities). Here, we present a novel FP-net architecture that is closer to vision models than the ones introduced previously in
Grüning and Barth (2021b)
Grüning et al. (2020b)
. We first demonstrate its performance and then analyze the learned units by relating them to biological vision.
Regarding the use of multiplicative terms in CNNs,
Zoumpourlis et al. (2017)
have shown that quadratic forms added to the first layer of a CNN can improve generalization. An FP-net can be interpreted as a special case of a network with an additional second-order Volterra
kernel, but it has much fewer parameters. However, CNNs are also special cases of MLPs and, as we have argued above, the challenge is to find the right biases that can take us from the general to the
more special case. For more comprehensive overviews on how FP-nets relate to various deep-network architectures, especially to bilinear CNNs (
Li et al., 2017
), see
Grüning et al. (2020a)
Grüning and Barth (2021b)
. In addition, we would like to mention recent work of
Chrysos et al. (2020)
, which illustrates that the Hadamard product of layers in deep network and the resulting higher-order polynomial representation can improve classification performance. Finally, in recurrent
networks, multiplications are used to implement useful gating mechanisms (
Collins et al., 2016
FP-nets as competitive deep networks
With FP-nets, we denote a deep-network architecture that contains one or several FP-blocks. Each
of a deep network implements a sequence of layers and operations that transforms an input tensor
\(\mathbf {T}_{0} \in \mathbb {R}^{h \times w \times d_{in}}\)
to an output tensor
\(\mathbf {T}_{out} \in \mathbb {R}^{\frac{h}{s} \times \frac{w}{s} \times d_{out}}\)
. A tensor consists of a number (e.g.,
) of feature maps, each with spatial width
and height
that may be altered by a factor
. The typical input tensor for a CNN is an image, the three color channels being the feature maps. The sequence of operations in an FP-block is shown in
Figure 1
and consists of three steps: (a) a first linear combination, (b) the feature product, (c) a second linear combination. In the first step, the
feature maps of an input tensor
\(\mathbf {T}_{0}\)
are linearly combined, followed by a ReLU, to yield the tensor
\(\mathbf {T}_{1}\)
\(q d_{out}\)
feature maps:
\begin{eqnarray} &&\mathbf {T}_{1}[i,j,m] = ReLU\left(\sum _{n=1}^{d_{in}} w_{m}^{n} \mathbf {T}_{0}[i,j,n]\right);\nonumber\\ && m= 1,..., q d_{out}. \end{eqnarray}
\(\mathbf {T}_{1}[i,j,m]\)
is the value of
\(\mathbf {T}_{1}\)
at pixel position
\((i, j)\)
and feature map
are learned weights and
is an expansion factor that controls the block size. By
\(\mathbf {T}_{1}^{m} \in \mathbb {R}^{h \times w}\)
, we denote the
th feature map of
\(\mathbf {T}_{1}\)
. The second step is the computation of feature products, the centerpiece of the FP-block. Each feature map
\(\mathbf {T}_{1}^{m}; m=1,..., q d_{out}\)
, is convolved with two learned filters
\(\mathbf {V}^{m}\)
\(\mathbf {G}^{m} \in \mathbb {R}^{k \times k}\)
. Filtering is followed by instance normalization (IN) (
Ulyanov et al., 2016
) and ReLU nonlinearity yielding two new feature maps. Subsequently, the product of the two filter outputs is computed. For any particular image patch
\(\mathbf {X} \in \mathbb {R}^{k \times k}\)
, with the center pixel being
, of a particular feature map
\(\mathbf {T}_{1}^{m}\)
, the filter operation for the vectorized image patch
\(\mathbf {x}= vect(\mathbf {X})\in \mathbb {R}^{k^{2}}\)
is the scalar product of the image patch with the vectorized filters
\(\mathbf {v} = vect(\mathbf {V}^{m})\)
\(\mathbf {g} = vect(\mathbf {G}^{m})\)
\begin{eqnarray} &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf {T}_{2}[i,j,m] \,{=}\, \frac{1}{\sigma _{v}\sigma _{g}} ReLU(\mathbf {x}^{T}\mathbf {v} \,{-}\, \mu _{v})ReLU(\mathbf {g}^{T}\mathbf {x} \,{-}
\, \mu _{g}). \;\; \end{eqnarray}
\(\mathbf {T}_{2}\in \mathbb {R}^{\frac{w}{s} \times \frac{h}{s} \times q d_{out}}\)
is the resulting tensor and
the stride of the filter operation. If
is greater than 1,
\(\mathbf {T}_{2}\)
’s width and height are subsampled.
are the mean value and standard deviation of
\(\mathbf {T}_{1}^{m}\)
after convolution with either
\(\mathbf {V}^{m}\)
\(\mathbf {G}^{m}\)
$$\mu _{v} = \frac{s^{2}}{hw} \sum _{i, j}{(\mathbf {T}_{1}^{m} * \mathbf {V}^{m})[i, j]},$$
$$\sigma _{v} = \frac{s^{2}}{hw} \sum _{i, j}{(\mathbf {T}_{1}^{m} * \mathbf {V}^{m} - \mu _{v})^{2}[i, j]},$$
\((\mathbf {T}_{1}^{m} * \mathbf {V})[i, j]\)
being the
th pixel of the filter result. In the third step, a second linear combination transforms
\(\mathbf {T}_{2} \in \mathbb {R}^{ \frac{h}{s} \times \frac{w}{s} \times qd_{out}}\)
\(\mathbf {T}_{3} \in \mathbb {R}^{\frac{h}{s} \times \frac{w}{s} \times d_{out}}\)
. To comply with the baseline architectures ResNet and MobileNet, a residual connection defines the final output as:
$$\mathbf {T}_{out} = \mathbf {T}_{0} + \mathbf {T}_{3}.$$
Using the above FP-block, we designed four different FP-nets based on different baseline architectures: an FP-net based on (a) the original ResNet, and (b) the PyrBlockNet trained on Cifar-10, (c) a
ResNet-50, and (d) a MobileNet-V2 both trained on ImageNet. A
is a larger segment of the network, consisting of several
. Except for the first stack that may have a stride of 1, each new stack starts with a block with a stride of 2 that reduces the size of each feature map. Within a stack, all blocks operate on
feature maps of the same size. Different network architectures may have different numbers and types of blocks. In our case, basic blocks, pyramid blocks, bottleneck blocks, and inverted residual
blocks define the ResNet-Cifar, PyrBlockNet, ResNet-50, and MobileNet-V2 architecture, respectively. The block is the core module of an architecture and contains several
. Layers are the smallest network building units such as convolution layers and max-pooling layers.
Figure 2
shows an example of a ResNet-Cifar architecture that has three stacks with five blocks each. Each first block of the second and third stacks contains a convolution layer with stride
that downsamples the input. The two other architectures that we used are similar: The ResNet-50 has four stacks with varying numbers of bottleneck blocks. The MobileNet-V2 has six stacks consisting
of inverted-residual blocks.
We transform the four baseline architectures defined above into FP-nets using a simple design rule: Substitute each stack’s first block with an FP-block. The input and output dimensions of the block
are kept equal; only the internal operations differ.
We developed this design rule to improve upon already well-established architectures, making FP-nets practical since only a few changes need to be done to create an FP-net. To be compatible with
state-of-the-art architectures, the FP-block has a structure similar to the MobileNet-V2 block (
Sandler et al., 2018
). We found that combinations of convolution blocks and FP-blocks work best and that larger kernel sizes do not improve performance. One way to view a stack is that it constitutes a visual processing
chain for a specific image scale. One would expect end-stopping to be more useful at the beginning of this chain. Thus, we replaced the first block of each stack. Note, however, that later stacks,
for example, the second and third stack in the Cifar-10 networks, already work with highly processed inputs coming from the previous stacks. Therefore, one would expect that there is a lower
necessity of extracting 2
regions in later stacks. Indeed, we will show, when analyzing the
values of FP-blocks, that highly selective neurons are more common in earlier stacks.
Due to the moderate size of the data set, Cifar-10 is often used to evaluate the potential of new architectures and designs. For our experiments on this data set, we used ResNets (
He et al., 2016
) as baseline; see
Figure 2
for an example. These networks have three stacks, each consisting of
blocks. We evaluated two types of the ResNet-20, ResNet-32, ResNet-44, and ResNet-56, with
\(N = 3\)
, 5, 7, and 9 blocks, respectively (the numbers after the names indicate the number of convolution or linear layers). Since the first publication of the ResNet architecture, several additional blocks
were proposed; see
Han et al. (2017)
for an overview. As two baselines on Cifar-10, we used the original ResNet and a variant using the pyramid block that we denote PyrBlockNet. For both variants, we created FP-nets by replacing
baseline blocks with FP-blocks according to our design rule. We used the same number of blocks, but note that an FP-block contains one additional convolution layer in each block. The FP-net-23,
FP-net-35, FP-net-47, and FP-net-59 are based on the PyrBlockNet: Each stack’s first block is an FP-block, and all other blocks are pyramid blocks. Analogously,
FP-net (basic)
denotes an FP-net based on the original ResNet: Each stack’s first block is an FP-block, and the remaining blocks are basic blocks.
Next, we evaluated the performance of FP-nets with the larger ImageNet data set that contains over 1.2 million training examples and 50,000 validation examples (we tested on the publicly available
validation set). With an input size of at least \(224 \times 224\) pixels and 1,000 classes, ImageNet poses a greater challenge than Cifar-10. We compared the ResNet-50 to two FP-net-50: one smaller
net with an expansion factor \(q=0.8\) and a slightly larger network with \(q=1\). In both cases, for each of the four stacks of the ResNet-50, the first block was replaced by an FP-block to obtain
the FP-net-50. Note that, if not explicitly mentioned, the term FP-net-50 refers to the \(q=1\) variant.
To further validate our approach, we evaluated an FP-net based on the popular MobileNet-V2 architecture. As with the ResNet, we replaced the first block of each stack with an FP-block, using \(q=3\)
The results of the Cifar-10 experiments are shown in
Figure 3
: The left side compares the original ResNet to the FP-net (basic), and the right side compares the PyrBlockNet to the FP-net. Each point of the two curves shows the best possible test error
occurring over all training epochs averaged over five runs and for one particular network (i.e., one particular number of blocks). The black line shows the baseline network, the green line the
resulting FP-net when substituting the first blocks of the baseline’s stacks. The
-axis displays the number of parameters, a number that increases with the number of blocks. Note, however, that the inclusion of FP-blocks reduces the number of parameters. Overall, the FP-nets are
more compact and perform better with a lower test error and only a small overlap in the standard deviations.
Table 1
shows the results on ImageNet. Note that the FP-net (
) performs better than the baseline ResNet-50, and the validation error is reduced by almost 0.4. When considering the already compact MobileNet architecture, the FP-net performs better than the
MobileNet with an error decreased by 0.2. We trained the MobileNet-V2 baseline network ourselves to obtain its validation error. For the ResNet-50, we report the value from the Tensorpack repository
Wu, 2016
). The performance depending on the number of parameters for the ResNet and FP-variants is illustrated in
Figure 4
FP-nets and visual coding
Hyperselectivity of FP-units
Vilankar and Field (2017)
used the term
to quantify how strongly a neuron is tuned to its optimal stimulus, that is, how quickly the response drops when the optimal stimulus changes. In the context of deep learning, hyperselectivity is
relevant because it can increase robustness, for example, robustness against adversarial attacks (
Paiton et al., 2020
). One way to quantify hyperselectivity is to measure the
iso-response contours
. Given an
-dimensional input to a function
, an
-dimensional surface may exist such that for all points
\(\mathbf {s}\)
on the surface, the output
\(f(\mathbf {s})\)
is a constant. As
can be a high dimension, 2
projections are used to analyze such iso-surfaces, which in two dimensions become iso-response contours
\(\mathbf {s} = \phi (t), t \in \mathbb {R}\)
The typical linear-nonlinear (LN) model neuron used in CNNs is a function
\(f_{LN}(\mathbf {x})\)
that involves a linear projection on a weight vector
\(\mathbf {w} \in \mathbb {R}^{n}\)
followed by a pointwise nonlinearity
\(\rho (x)\)
. To analyze the iso-response contour of such a neuron, one first projects the input on
\(\mathbf {w}\)
, the axis corresponding to the optimal stimulus
\(\mathbf {x}_{opt}\)
. To find a second axis, one searches for a vector orthogonal to
\(\mathbf {x}_{opt}\)
, for example, by picking
random values and using the Gram–Schmidt process (see
Equation 16
) to transform the random vector to one that is orthogonal to
\(\mathbf {{x}_{opt}}\)
. When looking at the output of an LN-neuron for
\(\mathbf {x_{opt}}\)
perturbed by any orthogonal vector
\(\mathbf {z}\)
\( \mathbf {{x}_{opt}}^{T} \mathbf {z} = \mathbf {w}^{T} \mathbf {z} = 0\)
, the iso-response contour is always a straight line parallel to
\(\mathbf {z}\)
, because
\(f_{LN}(\mathbf {x}_{opt} + \mathbf {z}) = \rho (\mathbf {w}^{T}(\mathbf {x}_{opt} + \mathbf {z})) = \rho (\mathbf {w}^{T} \mathbf {x}_{opt}) = f_{LN}(\mathbf {x}_{opt})\)
. Thus, for LN-neurons, the iso-response contours have zero curvature. For hyperselective neurons (
\(f_{HS}(\mathbf {x})\)
), there exist vectors
\(\mathbf {z}\)
that are orthogonal to
\(\mathbf {x}_{opt}\)
and decrease the neuron’s optimal response such that
\(f_{HS}(\mathbf {x}_{opt} + \mathbf {z}) \lt f_{HS}(\mathbf {x}_{opt})\)
. In this case, the exo-origin iso-response contour bends away from the origin of the basis defined by
\(\mathbf {x}_{opt}\)
\(\mathbf {z}\)
. A higher curvature of this bend indicates a more significant activation dropoff in regions that are different from the optimal stimulus (i.e., a greater hyperselectivity). One way to quantify the
curvature is to use the coefficient of the quadratic term obtained by fitting a second-order polynomial to the iso-response contour. FP-nets contain FP-blocks that consist of FP-units, or
, which yield the feature-product output for a pixel
in a feature map
as defined by
Equation 2
. As shown in the
, FP-neurons exhibit curved exo-origin iso-response contours with a curvature that depends on the angle
\(\gamma = \measuredangle (\mathbf {v}, \mathbf {g})\)
. Iso-response contours are shown in
Figure 5
for different values of
. Note that curvature, and thus hyperselectivity, increases with
. Accordingly, a large
leads to a lower entropy of the resulting feature maps; see
Figure 6
Entropy and degree of end-stopping
To further support the view that FP-neurons are hyperselective depending on
, we analyzed the entropy of the feature maps generated by different FP-neurons. The results in
Figure 6
show that the learned filters tend to have a
larger than zero, that is, the majority of FP-neurons are hyperselective and that a high
-value leads to a lower entropy. Details of how the entropy is computed are given in the
In order to analyze the end-stopping behavior of the model neurons that are learned in the FP-nets trained on Cifar-10 and ImageNet, we needed to quantify the degree of end-stopping. In order to
relate to physiological measurements, we started by analyzing the response of FP-neurons to straight lines and line ends, but this turned out to be problematic because the FP-nets use small
\(3 \times 3\)
filters and subsample the input. To keep the analogy, but with a more robust measure, we used a square as input and quantified the average responses to the uniform zero-dimensional (0
) regions, the straight 1
edges, and the 2
corners. The degree of end-stopping is then defined by the relation between 1
and 2
responses. In order to account for ON/OFF- type responses, we used both a bright and a dark square. The results are shown in
Figure 7
, and the details of the algorithm are given in the
Note that, as the real neurons in cortical areas V1 and V2, the model neurons in the FP-net are end-stopped to different degrees. Thus, end-stopping seems to be beneficial for both the ImageNet and
Cifar-10 tasks, since the emergence of end-stopping is here driven by the classification error. As expected, the multiplication in the FP-block shifts the distribution toward a higher degree of
end-stopping. However, the network could have learned filter pairs that do not lead to end-stopped FP-neurons. The bias that we introduce (i.e., the multiplication) just makes it easier for the
network to learn end-stopped representations.
The angle distributions in
Figure 8
show that indeed linear FP-neurons are learned as well since more than 15% of FP-neurons have a
-value near zero. With increasing network depth, the number of linear FP-neurons increases, indicating that hyperselectivity and especially end-stopping are more frequent in earlier stages of the
visual processing chain.
FP-neurons are more robust against adversarial attacks
Although outperforming almost all alternative approaches on many vision tasks, CNNs are surprisingly sensitive to barely visible perturbations of the input images (
Szegedy et al., 2013
). An adversarial attack on a classifier function
adds a noise pattern
\(\mathbf {\eta }\)
to an input image
\(\mathbf {x}\)
so that
\(f(\mathbf {x} + \mathbf {\eta })\)
does not return the correct class
\(y=f(\mathbf {x})\)
. Furthermore, the attacker ensures that some
-norm of
\(\mathbf {\eta }\)
does not exceed
. In many cases, including this work, the infinity-norm is chosen, and the
values are in the set
\(\lbrace {1}/{255}, {2}/{255}, ...\rbrace\)
. Thus, for example, for
\(\epsilon = {1}/{255}\)
, each 8-bit pixel value is at most altered by adding or subtracting the value 1.
Goodfellow et al. (2014)
argue that the main reason for the sensitivity to adversarial examples is due to the linearity of CNNs: With a high-dimensional input, one can substantially change a linear neuron’s output, even with
small perturbations. Consider the output of an LN-neuron for an input
\(\mathbf {x}\)
with dimension
perturbed by
\(\mathbf {\eta }\)
. We choose
\(\mathbf {\eta }\)
to be the sign function of the weight vector multiplied with
\(\mathbf {\eta }=sign(\mathbf {w}) \cdot \epsilon\)
. Thus,
\(\mathbf {\eta }\)
roughly points in the direction of the optimal stimulus (which is also the gradient), but its infinity-norm does not exceed
. Assuming that the mean absolute value of
\(\mathbf {w}\)
\(f_{LN}(\mathbf {\eta })\)
is approximately equal to
\(\epsilon n m\)
. Accordingly, a significant change of the LN-neuron’s output can be achieved by a small
value if the input dimension
is large, which is the case for many vision-related tasks. This gradient-ascent method can also be applied to nonlinear neurons. Within a local region, the output of almost any function
can be approximated by a linear function. To optimally increase the output, the input needs to be moved along the gradient direction. The fast gradient sign method (FGSM;
Goodfellow et al., 2014
) perturbs the original input image
\(\mathbf {x}\)
by adding
\(\mathbf {\eta }=\epsilon sign(\nabla f(\mathbf {x}))\)
. Another approach is to define
\(\mathbf {\eta }\)
to be the gradient times a positive step size
followed by clipping to
\(\eta \in [-\epsilon , +\epsilon ]^{n}\)
. The clipped iterative gradient ascent (CIGA) greedily moves along the direction of the highest linear increase,
\begin{eqnarray} \begin{array}{@{}r@{\;}c@{\;}l@{}} \mathbf {\eta }_{0} &=& \mathbf {0}; \tau \gt 0\\ \mathbf {q}_{i+1} &=& \mathbf {\eta }_{i} + \tau \nabla f(\mathbf {x} + \mathbf {\eta }_{i})\\ \
mathbf {\eta }_{i+1}^{j} &=& \min (\max (q_{i+1}^{j}, -\epsilon ), \epsilon ),\\ \end{array} \qquad \end{eqnarray}
being the
th entry of the unbounded result
\(\mathbf {q}_{i}\)
at the
th iteration step. In the following, we use CIGA in our illustrations of the principle, and in our experiments, we employ FGSM as it is a widely recognized adversarial attack method. When regarding
an iso-response contour plot, one can easily spot the direction of the gradient, which is orthogonal to an iso-response contour (
Paiton et al., 2020
). In
Figure 9
on the left, the gradient for an LN-neuron is parallel to the optimal stimulus (black line). As long as the initial input yields a nonzero gradient, each step of CIGA maximally increases the
LN-neuron output. Thus, the algorithm’s effectiveness is only bounded by
but widely independent of the initial input
\(\mathbf {x}\)
. For a step size larger than
, CIGA finds the optimal solution in one step. We now investigate the effects of CIGA on a simplified version of an FP-neuron:
$$F(\mathbf {x}) = \mathbf {x}^T \mathbf {v} \mathbf {g}^T \mathbf {x}.$$
Note that in the following particular example, the input is chosen to yield nonnegative projections on
\(\mathbf {v}\)
\(\mathbf {g}\)
; thus, we can remove the ReLUs. The resulting gradient is
$$abla _{F}(x) = (\mathbf {v}^{T}\mathbf {x}) \mathbf {g} + (\mathbf {g}^{T}\mathbf {x}) \mathbf {v}.$$
The effectiveness of an iteration step strongly depends on the current position. The highest possible increase would be obtained along the line defined by the optimal stimulus. In
Figure 9
on the right, this is the black line. If the initial input is located on this line, any step in the gradient direction yields an optimal increase of the FP-neuron output. However, for any other
position with a nonzero gradient, an unbounded iteration step would move toward the optimal stimulus line. The blue curve in
Figure 9
shows the path for several iterations of CIGA: Starting above the optimal stimulus line, each step slowly converges to the optimal stimulus line, eventually moving almost parallel to it. Once the
threshold of 1 is reached in the horizontal dimension, the (now bounded) path runs parallel to the vertical dimension to increase the neuron output further. The optimal solution is found once the
bound is also reached in the vertical dimension. The important difference when comparing with LN-neurons is that there are numerous conditions (depending on
\(\mathbf {x}\)
, and
) where CIGA would need several steps to find an optimal solution. This reduced effectiveness of the gradient ascent illustrates why hyperselective neurons are more robust against adversarial
attacks; for example, if
is too small, or
is chosen poorly, or with too few iterations, an attack might not increase the FP-neuron output by much. Note that single neurons are usually not the target of adversarial attacks; instead, the
gradient is determined on the classification loss function. Still, the argument holds that hyperselective neurons are harder to activate than LN-neurons, resulting in an increased robustness.
To test this hypothesis, we created new Cifar-10 test sets
\(\mathcal {S}_{\epsilon _{i}} = \lbrace FGSM(\mathbf {x}, \epsilon _{i}): \mathbf {x} \in \mathcal {X}_{C10} \rbrace\)
derived from the original test set
\(\mathcal {X}_{C10}\)
. Here, we focused on the most subtle adversarial attacks: we created one test set
\(\mathcal {S}_{{1}/{255}}\)
, where each test image was perturbed by using FGSM with
\(\epsilon ={1}/{255}\)
. Results for larger
-values are shown in the
Table 2
Table 3
). To exclude the hypothesis that the better accuracy (with perturbations) is due to the fact that the FP-nets already generalize better, we present results where we measure the percentage of changed
predictions of the classifier
\begin{eqnarray} &&Perc.\ of\ changed\ predictions (f, \Gamma , \theta )\nonumber\\ &&\qquad = \frac{1}{ |\mathcal {X}_{C10}| }\sum _{\mathbf {x} \in \mathcal {X}_{C10}} \mathbb {1}( f(\mathbf {x}) \
ne f(\Gamma (\mathbf {x}, \theta ) )), \quad \end{eqnarray}
\(\mathbb {1}\)
is the indicator function returning a 1 for a true statement and a zero otherwise.
is some function (here, FGSM) that perturbs the original image
\(\mathbf {x}\)
based on some parameter
. We evaluated this metric for each of the four architectures that we trained on the original Cifar-10 training set (see Section “
FP-nets as competitive deep networks
”); no additional adversarial training scheme was employed. As shown in
Figure 10
, 40% to 50% of the predictions did change. However, for both baseline models, substituting some of the LN-neurons with FP-neurons increased the robustness against FGSM attacks.
The results reiterate that CNN predictions can be significantly altered by deliberate and subtle attacks (we show some example images in the
). Unfortunately, this lack of robustness creates problems of practical relevance beyond such attacks. For example, JPEG-compression can create artifacts that have similar effects. To evaluate
robustness against JPEG artifacts, we created the Cifar-10 test sets
\(\mathcal {S}_{Q_i} = \lbrace JPEG(\mathbf {x}, Q_i): \mathbf {x} \in \mathcal {X}_{C10} \rbrace\)
, with
\(JPEG(\mathbf {x}, Q)\)
being the JPEG-compressed version of the original image
\(\mathbf {x}\)
with a quality rate
\(Q \in \lbrace 1,2,...,100 \rbrace\)
, 100 being the original image. A low quality indicates a high compression with stronger artifacts (example images are given in the
). In
Figure 11
, we show the results for the low compression test set
\(\mathcal {S}_{90}\)
and further results in the
Tables 4
Again, using FP-neurons increased the robustness against artifacts. However, even a moderate compression alters up to 10% of the CNNs’ predictions.
Example FP-unit
As shown above, the learned FP-neurons are hyperselective and end-stopped to different degrees. However, these two properties do not fully specify an FP-neuron. When analyzing the individual
FP-neurons in more detail, it is difficult to further specify them according to simple properties such as orientation or phase. Nevertheless, some FP-neurons look as if they were taken from a
textbook on “how to model end-stopped neurons,” and we show one example in
Figure 12
Discussion and conclusions
We have presented a novel FP-net architecture and have demonstrated its competitive performance. To do so, we have designed experiments with state-of-the-art deep networks and showed that we could
improve their performance by substituting original blocks in the network architecture with FP-blocks that implement an explicit multiplication of feature maps. Given this simple design rule, we can
expect our approach to be of practical use, since any traditional network can easily be transformed into an FP-net that will most likely perform better. We did not employ any hyperparameter tuning
specific to the FP-nets but just used the hyperparameters of the original networks; one may thus expect even better performance with additional tuning. We believe that the improvement that comes with
FP-nets is due to an appropriate bias, which allows the network to learn efficient representations based on units (model neurons) that are end-stopped to different degrees. The multiplications that
we introduce allow for AND rather than OR combinations and thus make the resulting units more selective than linear filters with pointwise nonlinearities. Note that the key feature of FP-nets is that
one learns pairs of linear filters, which are then AND combined. In case of FP-nets, the AND is implemented by multiplications. We could, however, show that logarithms (
Grüning et al., 2020b
) and the minimum operation (
Grüning & Barth, 2021a
) can also work as AND operation. We consider the improvements that bio-inspired FP-nets achieve over the baseline networks to be the main contribution of our article.
Moreover, we have analyzed the selectivity of the FP-units in an attempt to relate them to what is known about visual neurons. We could show that FP-units are indeed end-stopped to different degrees.
The emergence of end-stopping in a network that learns based on only the classification error demonstrates that end-stopping is beneficial for the task of object recognition. This finding is
supported by previously known mathematical results, according to which (a) 2
features such as corners and junctions are statistically rare in natural images, leading to sparse representations (
Zetzsche et al., 1993
), and (b) 2
features are still unique since there exists a mathematical proof that 0
(uniform) and 1
(straight) regions in images are redundant (
Mota & Barth, 2000
), although being statistically frequent.
Of course, the considerations above cannot be taken to imply that biological vision implements an FP-net architecture, especially as the FP-nets implement additional and typical deep-network
operations such as linear recombinations that increase the entropy of the representation. In other words, much of what well-performing deep networks do is not something one would necessarily consider
to be optimal.
It is known that sparse-coding units are more selective than typical CNN units, that is, than linear neurons with pointwise nonlinearities (
Paiton et al., 2020
), and thus less prone to certain adversarial attacks. This increased selectivity has been quantified with the curvature of the iso-response contours. We could show that the iso-response contours of
the FP-units are curved, with the degree of curvature depending on the angle between the multiplied feature vectors, and that a large number of hyperselective units emerge in FP-nets trained for
object recognition. Furthermore, our results show that FP-nets are indeed more robust against adversarial attacks and compression artifacts, and this is, again, due to the vision-inspired FP-units.
Commercial relationships: none.
Corresponding author: Philipp Grüning.
Email: gruening@inb.uni-luebeck.de.
Address: Institute for Neuro- and Bioinformatics, University of Lübeck, Germany.
Barlow, H. (1961). Possible principles underlying the transformation of sensory messages. Sensory Communication, 1(1), 217–234.
Barth, E., & Zetzsche, C. (1998). Endstopped operators based on iterated nonlinear center-surround inhibition. In Rogowitz, B. E. & Pappas, T. N. (Eds.),
Human vision and electronic imaging
(Vol. 3299, pp. 67–78). Bellingham, WA: Optical Society of America, Available from
Bradski, G. (2000). The openCV library. Dr. Dobb's Journal: Software Tools for the Professional Programmer, 25(11), 120–123.
Chrysos, G., et al. (2020). P-nets: Deep polynomial neural networks. In
2020 IEEE/CVF Conference on Computer Vision and Pattern Recognition (CVPR)
, Seattle, WA, USA, pp. 7323–7333, doi:
Collins, J., Sohl-Dickstein, J., & Sussillo, D. (2016). Capacity and trainability in recurrent neural networks. Stat, 1050, 29.
Deng, J., Dong, W., Socher, R., Li, L.-J., Li, K., & Fei-Fei, L. (2009). ImageNet: A large-scale hierarchical image database.
2009 IEEE Conference on Computer Vision and Pattern Recognition
, pp. 248–255, doi:
Goodfellow, I. J., Shlens, J., & Szegedy, C. (2015). Explaining and harnessing adversarial examples. In Bengio, Y., LeCun, Y. (Eds.),
3rd International Conference on Learning Representations, ICLR 2015, San Diego, CA, USA, Conference Track Proceedings
. Opgehaal van,
Grüning, P., & Barth, E. (2021b). Fp-nets for blind image quality assessment. Journal of Perceptual Imaging, 4(1), 10402-1–10402-13.
Grüning, P., Martinetz, T., & Barth, E. (2020a). Feature products yield efficient networks. arXiv preprint arXiv:2008.07930.
Grüning, P., Martinetz, T., & Barth, E. (2020b). Log-nets: Logarithmic feature-product layers yield more compact networks. In Farkaš, I., Masulli, P., & Wermter, S. (Eds.), Artificial Neural Networks
and Machine Learning – ICANN 2020 (pp. 79–91). Cham, Switzerland: Springer International Publishing.
Han, D., Kim, J., & Kim, J. (2017). Deep pyramidal residual networks. In
2017 IEEE Conference on Computer Vision and Pattern Recognition (CVPR)
, Honolulu, HI, USA, pp. 6307–6315, doi:
He, K., Zhang, X., Ren, S. & Sun, J. (2016). Deep residual learning for image recognition.
2016 IEEE Conference on Computer Vision and Pattern Recognition (CVPR)
, pp. 770–778, doi:
Hubel, D. H., & Wiesel, T. N. (1965). Receptive fields and functional architecture in two nonstriate visual areas (18 and 19) of the cat.
Journal of Neurophysiology,
28(2), 229–289.
Kim, H. (2020). Torchattacks: A pytorch repository for adversarial attacks. arXiv preprint arXiv:2010.01950.
Krizhevsky, A., Nair, V., & Hinton, G. (2021). Cifar-10 (Canadian Institute for Advanced Research).
Li, Y., Wang, N., Liu, J., & Hou, X. (2017). Factorized bilinear models for image recognition. In
2017 IEEE International Conference on Computer Vision (ICCV), Venice, Italy
, pp. 2098–2106, doi:
Lu, L. (2020). Dying ReLU and initialization: Theory and numerical examples.
Communications in Computational Physics,
28(5), 1671–1706, doi:
Majaj, N. J., & Pelli, D. G. (2018). Deep learning—Using machine learning to study biological vision.
Journal of Vision,
18(13), 2–2.
Mel, B. W., & Koch, C. (1990). Sigma-pi learning: On radial basis functions and cortical associative learning. In Touretzky, D. (Ed.), Advances in Neural Information Processing Systems, 2.
Paiton, D. M., Frye, C. G., Lundquist, S. Y., Bowen, J. D., Zarcone, R., & Olshausen, B. A. (2020). Selectivity and robustness of sparse coding networks.
Journal of Vision,
20(12), 10,
Paszke, A., Gross, S., Massa, F., Lerer, A., Bradbury, J., Chanan, G. et al. (2019). Pytorch: An imperative style, high-performance deep learning library. In Wallach, H., Larochelle, H., Beygelzimer,
A., d'Alché-Buc, F., Fox, E., Garnett, R., Paszke, A., Gross, S., Massa, F., Lerer, A., Bradbury, J., Chanan, G., Chintala, S. (Eds.), Advances in neural information processing systems (Vol. 32, pp.
8026–8037). Red Hook, NY: Curran Associates, Inc.
Rao, R. P., & Ballard, D. H. (1999). Predictive coding in the visual cortex: A functional interpretation of some extra-classical receptivefield effects.
Nature Neuroscience,
2(1), 79–87.
Rumelhart, D. E., Hinton, G. E., & McClelland, J. L. (1986). A general framework for parallel distributed processing. Parallel Distributed Processing: Explorations in the Microstructure of Cognition,
1(26), 45–76.
Sandler, M., Howard, A., Zhu, M., Zhmoginov, A., & Chen, L. (2018). MobileNetV2: Inverted residuals and linear bottlenecks. In
2018 IEEE/CVF Conference on Computer Vision and Pattern Recognition (CVPR)
, Salt Lake City, UT, USA, pp. 4510–4520, doi:
Simoncelli, E. P., & Olshausen, B. A. (2001). Natural image statistics and neural representation.
Annual Review of Neuroscience,
24(1), 1193–1216.
Srivastava, R. K., Greff, K., & Schmidhuber, J. (2015). Training very deep networks. Proceedings of the 28th International Conference on Neural Information Processing Systems - Volume 2, pp.
2377–2385. Presented at the Montreal, Canada. Cambridge, MA, USA: MIT Press.
Szegedy, C., Liu, W., Jia, Y., Sermanet, P., Reed, S., Anguelov, D. et al. (2014). Going deeper with convolutions.
Szegedy, C., Zaremba, W., Sutskever, I., Bruna, J., Erhan, D., Goodfellow, I., & Fergus, R. (2014). Intriguing properties of neural networks. Paper presented at 2nd International Conference on
Learning Representations, ICLR 2014, Banff, Canada.
Ulyanov, D., Vedaldi, A., & Lempitsky, V. (2016). Instance normalization: The missing ingredient for fast stylization. arXiv preprint arXiv:1607.08022.
Veit, A., Wilber, M. J., & Belongie, S. (2016). Residual networks behave like ensembles of relatively shallow networks. In Lee, D., Sugiyama, M., Luxburg, U., Guyon, I., Garnett, R. (Eds.), Advances
in Neural Information Processing Systems (Vol 29). Curran Associates, Inc.
Watanabe, S. (1985). Pattern Recognition: Human and Mechanical. Hoboken, New Jersey: Wiley-Interscience.
Zetzsche, C., Barth, E., & Wegmann, B. (1993). The importance of intrinsically two-dimensional image features in biological vision and picture coding. In Watson, A. B. (Ed.),
Digital images and human vision
(pp. 109–38). Cambridge, MA: MIT Press. Available from
Zoumpourlis, G., Doumanoglou, A., Vretos, N., Daras, P. Non-linear convolution filters for CNN-based learning. In
2017 IEEE International Conference on Computer Vision (ICCV)
, Venice, Italy, pp. 4771–4779, doi:
Details on network design and training procedure
All experiments were conducted using the PyTorch deep-learning framework (
Paszke et al., 2019
). Note that in all cases, for
Equation 1
, the output of the weighted sum has been normalized via batch normalization before applying the ReLU nonlinearity.
Residual connections
For the residual connections in
Equation 5
, some additional computations are needed if the dimensions of
\(\mathbf {T}_{0}\)
\(\mathbf {T}_{3}\)
differ. In case that
is greater than
, zero padding is used to match the dimension of the feature maps. If
is greater than
, an additional linear combination is learned to reduce the number of feature maps. If the FP-block’s stride
is greater than 1,
\(\mathbf {T}_{0}\)
is subsampled by average pooling. For more implementation details regarding residual connections, see
Han et al. (2017)
. Residual connections enable a more stable gradient flow during training, allow to better model identity functions (
He et al., 2016
), and enable CNNs to behave like ensembles of shallower networks (
Veit et al., 2016
Cifar-10 experiments
Cifar-10 contains 50,000 training and 10,000 test images (RGB, with height and width 32) of 10 different commonplace objects, such as airplane, bird, cat, and ship. For each FP-net and each
PyrBlockNet, five experiments were conducted with five different random seeds that control the initialization of each network’s random weights and the random mini-batch collection during training.
The networks were trained for 200 epochs, using stochastic gradient descent (SGD), with a learning rate of 0.1 that was reduced to 0.01 and 0.001 after the 100th and 150th epoch. We used a momentum
of 0.9, a weight decay of 0.0001, and a batch size of 128. For data augmentation, during training, with a probability of
, each input image was flipped horizontally. Subsequently, all images were padded with 4 pixels, and then a random crop of
\(32 \times 32\)
were used. Furthermore, the RGB crop was first divided by 255 and normalized with the ImageNet mean
\(\mu _{imNet} (0.485, 0.456, 0.406)\)
and standard deviation
\(\sigma _{imNet} (0.229, 0.224, 0.225)\)
for the three input channels, respectively. When computing the test scores, no random cropping and no horizontal flipping were used. Each FP-block’s expansion factor
was set to 2. Based on the work of
Srivastave et al. (2015)
, the best test error was reported to better reflect the variance of the results due to different network initializations.
FP-ResNet on ImageNet
The FP-net-50 was trained for 100 epochs with randomly initialized weights using SGD on
\(224 \times 224\)
crops with a batch size of 512. After one third, and then again after two thirds of the training time, the initial learning rate of 0.1 was decreased by a factor of 10. The weight decay was 0.0001
and the momentum 0.9. For data augmentation, we used the code from the sequential-imagenet-dataloader repository (
Gray, 2017
); during training, crops of various random sizes were passed to the network ranging from
of the original image size. The aspect ratio was chosen randomly between 3/4 and 4/3. Furthermore, different photometric distortions (e.g., random contrast changes) were applied as described in
Szegedy et al. (2014)
and the Tensorpack repository (
Wu, 2016
). When computing the test scores, each input image is first resized such that the shortest edge’s length is 256. Next, the image is cropped in the center to size (224, 224), divided by 255,
subtracted with 0.5, and again divided by 0.5.
MobileNet-V2 and FP-MobileNet
The FP-MobileNet was trained from scratch with SGD for 150 epochs and with a batch size of 256. The initial learning rate of 0.05 was decreased according to a cosine scheduling; see Li et al.’s
repository (
Li et al., 2021
). The training data augmentation included random resizing and cropping, random horizontal flips, color jitters, division by the maximum value, and normalization by
\(\mu _{imNet} (0.485, 0.456, 0.406)\)
\(\sigma _{imNet} (0.229, 0.224, 0.225)\)
. During testing, the input images were first resized to
\(255 \times 255\)
and then a center crop of size
\(224 \times 224\)
was computed. Subsequently, the crop was normalized as described above. For more information, see Fastai’s repository (
Howard, 2018
We analyzed the entropy of all FP-neurons \(\mathbf {T}_{2}\) for the FP-ResNet-50 (ImageNet) and the FP-ResNet-59 (Cifar-10). One hundred randomly sampled images from the respective test set (in
case of ImageNet, the validation set) were passed to each network. For each input, we computed the corresponding feature maps for every FP-block, one tensor \(\mathbf {T}_{2}\) for every block. We
normalized each feature map \(\mathbf {T}_{2}^{m}\) from \(\mathbb {R}^{+}\) to \(\lbrace 0,1,...,255\rbrace\) and computed the entropy of the pixel distribution over the 256 integer values. For the
100 input images, we obtained 100 entropy values for each feature map. We averaged these 100 values resulting in the mean entropy for each feature map (i.e., each FP-neuron).
We observed that some of the feature maps
\(\mathbf {T}_{1}^{m}\)
had all pixel values equal to zero (so-called dying ReLUs;
Lu et al., 2019
). The corresponding FP-neurons were removed from the analysis. For the FP-ResNet-50, the percentage of dying ReLUs was
\(23 \%\)
\(0.002 \%\)
\(7 \%\)
, and
\(18 \%\)
for the first, second, third, and fourth FP-blocks, respectively. For the FP-ResNet-59, only the third FP-block had
\(5 \%\)
dying ReLUs. We tested different weight initializations or alternative nonlinearities, such as the leaky ReLU. However, although, using leaky ReLUs stopped the emergence of dying ReLUs, we only
noticed a small gain in performance.
Degree of end-stopping
To measure the degree of end-stopping, we used two input images
, one with a bright and one with a dark square: Pixels belonging to the square had a value of +1 or −1, respectively; all other pixels were zero. Each image was normalized to have zero mean and a
standard deviation of 1. We computed the intermediate outputs
\(\mathbf {T}_{2}(I_0)\)
and squared them to obtain the activation energy. For the PyrResNet, we used the ReLU after the first convolution as intermediate output.
\(\mathbf {T}_{i}(I_0)\)
is the
th tensor that is computed using the image
as input. We then normalized each tensor
\(\mathbf {T}_{n}\)
\(\mathbb {R}^{+}\)
\([0, 1]\)
by dividing it with the mean plus three times the standard deviation and clipped any values greater than 1 to make the normalization less susceptible to possible outliers. The percentage of outliers
never exceeded 10%. For each feature map, we determined the values
, and
by summing the feature map pixel values (i.e., the activations) over specific regions of interest that were either homogeneous areas, straight edges, or corners in the input image:
$${\psi }D(\mathbf {T}_{n}^{m}) = \sum _{i,j}{\mathbf {T}_{n}[i,j,m] \mathbf {W}_{\psi }[i,j]}.$$
\(\mathbf {W}_{\psi }\)
is a binary matrix used to compute the
\({\psi }D\)
value. All pixels within the region of interest are 1; the others are zero. The weighted areas are shown in
Figure 13
in the right panel: The square in the middle is the region of interest for
. The four small squares along the straight edges of the input square measure
; the four small squares at the corners measure
. Note that the three different regions of interest have the same total area. The left panel shows the input image
. The
\({\psi }D(\mathbf {T}_{n}^{m})\)
for both input images is the sum
\({\psi }D(\mathbf {T}_{n}^{m}) = {\psi }D(\mathbf {T}_{n}^{m}(I_0))+{\psi }D(\mathbf {T}_{n}^{m}(I_1))\)
. The degree of end-stopping of a feature map is then defined as:
$$\phi (\mathbf {T}_{n}^{m}) = 1 - \frac{1D(\mathbf {T}_{n}^{m})}{2D(\mathbf {T}_{n}^{m}) + \epsilon }$$
\(\epsilon = 0.1\)
. Note that the degree of end-stopping is high (close to 1) if the
activation is high and the
activation is low. However, two special cases were considered: (a) a feature map is “silent,” if all values are very small (i.e.,
\(0D + 1D + 2D \lt 0.1\)
). (b) The feature map is “
” if the
activations are similar:
$$\mathbf {T}_{n}^{m} \;is\; {}^{\prime }0D^{\prime } \Leftrightarrow 1 - \frac{0D(\mathbf {T}_{n}^{m})}{1D(\mathbf {T}_{n}^{m}) + \epsilon } \lt 0.1.$$
For these two special cases,
Equation 11
would no longer quantify the degree of end-stopping. Therefore, the degree of end-stopping values was not evaluated for silent and
feature maps. The plots in
Figure 7
show the normalized histograms for the degree of end-stopping. All bars have a bin width of 0.1 and their heights sum up to 1.
Iso-response contours
In this section, we derive the analytical expression for the iso-response contours of FP-neurons. We follow a geometric approach in order to show explicitly how the exo-origin curvature depends on
the angle \(\gamma = \measuredangle (\mathbf {v}, \mathbf {g})\). An alternative approach would be to work with the eigenvector of the symmetric matrix \(\frac{1}{2}(\mathbf {v} \mathbf {g}^T + \
mathbf {g} \mathbf {v}^T)\).
In the two-dimensional subspace defined by
\(\mathbf {v}\)
\(\mathbf {g}\)
, and for a specific constant
\(z \in \mathbb {R}^{+}\)
, we can derive the coordinates of the iso-response contours analytically by using a simplified version of
Equation 2
Equation 7
\(F(\mathbf {x})\)
is the output of the FP-neuron, that is, the product of the outputs of two linear filters
\(\mathbf {v}\)
\(\mathbf {g} \in \mathbb {R}^{n}, n=k^{2}\)
. For simplicity, we disregard the instance normalization. Thus, we assume that the mean values are zero (
\(\mu _{v} = \mu _{g} = 0\)
) and the standard deviations are 1 (
\(\sigma _{v} = \sigma _{g} = 1\)
), which are the two variables used for instance normalization. Furthermore, we constrain the input space of
\(\mathbf {x}\)
\(\mathbb {S} = \lbrace \mathbf {x} \in \mathbb {R}^{k^{2}}: \mathbf {x}^T \mathbf {v} \ge 0 \wedge \mathbf {x}^T \mathbf {g} \ge 0 \rbrace\)
to account for the ReLU nonlinearities. Furthermore, we restrict
\([0, \pi )\)
since for
\(\gamma = \pi\)
, both vectors point in opposite directions, and for any point
\(\mathbf {x}\)
, one scalar product is always negative.
The optimal stimulus of
\(F(\mathbf {x})\)
is not parallel to one of the filters but points in the direction of the bisector of
. This property becomes more obvious when rewriting
\(F(\mathbf {x})\)
as a function depending on
\(\alpha = \measuredangle (\mathbf {v}, \mathbf {x})\)
\(\beta = \measuredangle (\mathbf {g}, \mathbf {x})\)
$$F(\alpha , \beta , \mathbf {x}) = cos(\alpha ) cos(\beta ) \Vert \mathbf {v} \Vert \Vert \mathbf {g} \Vert \Vert \mathbf {x} \Vert ^2.$$
To simplify this particular equation, we assume
\(\Vert \mathbf {x} \Vert = 1\)
and disregard the vector lengths
\(\Vert \mathbf {v} \Vert\)
\(\Vert \mathbf {g} \Vert\)
since the arguments
, and the argmax of
, do not depend on vector length. With
\( \alpha + \beta = \gamma\)
, we obtain
\begin{eqnarray} F(\alpha , \beta ) &\;=& cos(\alpha ) cos(\gamma - \alpha )\nonumber\\ & \;=& \frac{1}{2} (cos(2\alpha - \gamma ) + cos(\gamma )). \qquad \end{eqnarray}
Note that for
\(\alpha = \frac{1}{2} \gamma\)
reaches the maximum value
\(\frac{1 + {\cos }(\gamma )}{2}\)
The subspace of input vectors that do not alter the FP-neuron’s output is defined by
\begin{eqnarray} F(\mathbf {x} + \mathbf {p}) = F(\mathbf {x}) \Leftrightarrow \mathbf {p}^{T} \mathbf {v} = \mathbf {p}^{T} \mathbf {g} = 0. \quad \end{eqnarray}
For any vector
\(\mathbf {p}\)
orthogonal to
\(\mathbf {v}\)
\(\mathbf {g}\)
, the iso-response contours are straight, as they are for LN-neurons. However, as we will show in the following, there exists an orthogonal direction
\(\mathbf {o}\)
relative to which FP-units exhibit curved iso-response contours and, thus, hyperselectivity.
It is important to note that any input vector
\(\mathbf {x}\)
is projected to the plane defined by the vectors
\(\mathbf {v}\)
\(\mathbf {g}\)
Equation 7
); any vector
\(\mathbf {p}\)
from the subspace of
Equation 15
is orthogonal to this plane. We can consider the function
\(f(\mathbf {a})\)
that operates on only 2
input vectors
\(\mathbf {a} = (a, b)^{T}\)
, which are the projections of
\(\mathbf {x}\)
onto the vectors
\(\frac{\mathbf {v}}{\Vert \mathbf {v} \Vert }\)
\(\frac{\mathbf {o}}{\Vert \mathbf {o} \Vert }\)
, respectively. Unless
\(\mathbf {g}\)
is parallel to
\(\mathbf {v}\)
, we can derive
\(\mathbf {o}\)
as the direction orthogonal to
\(\mathbf {v}\)
by using the Gram–Schmidt process:
\begin{eqnarray} \mathbf {o} = \mathbf {g} - \frac{\mathbf {v}^T\mathbf {g}}{\Vert \mathbf {v} \Vert ^{2}} \mathbf {v}. \quad \end{eqnarray}
\(\mathbf {g} = \lambda \mathbf {v}, \lambda \in \mathbb {R}\)
\(\mathbf {o}\)
is simply any vector orthogonal to
\(\mathbf {v}\)
. A point
\((a, b)^{T}\)
in the two-dimensional projection space can be injected into the original input space
\(\mathbb {S}\)
$$\mathbf {x}_{ab} = \frac{a}{\Vert \mathbf {v} \Vert } \mathbf {v} + \frac{b}{\Vert \mathbf {o} \Vert } \mathbf {o}.$$
\(\mathbf {x}_{ab}\)
denotes that the vector depends on only the position in the projection space
\(\mathbf {a} = (a, b)^{T}\)
. The relations between the scalar products in the input space and the scalar products in the projection space are given by
$$\mathbf {x}_{ab}^{T} \mathbf {v} = \Vert \mathbf {v} \Vert (a, b) \mathbf {e}_{1} = a \Vert \mathbf {v} \Vert$$
$$\mathbf {x}_{ab}^{T} \mathbf {o} = \Vert \mathbf {o} \Vert (a, b) \mathbf {e}_{2} = b \Vert \mathbf {o} \Vert$$
\begin{eqnarray} \mathbf {x}_{ab}^{T} \mathbf {g} &\;=& \Vert \mathbf {g} \Vert (a, b) (cos(\gamma ), sin(\gamma ))^{T}\nonumber\\ &\;=& \Vert \mathbf {g} \Vert (a cos(\gamma ) + b sin(\gamma )), \
qquad \end{eqnarray}
\(\mathbf {e}_{1} = (1, 0)^{T}\)
\(\mathbf {e}_{2} = (0, 1)^{T}\)
. Accordingly, the multiplication of
\( \mathbf {x}^{T} \mathbf {v}\)
\( \mathbf {x}^{T} \mathbf {g}\)
\begin{eqnarray} \mathbf {x}^{T} \mathbf {v} \mathbf {x}^{T} \mathbf {g} &\;=& (a \Vert \mathbf {v} \Vert )(a cos(\gamma ) + b sin(\gamma ))\Vert \mathbf {g} \Vert \nonumber\\ &\;=& \left(\begin
{array}{@{}c@{}} a^{2} \\ ab \end{array}\right)^{T} \left(\begin{array}{@{}c@{}} c_1 cos(\gamma ) \\ c_1 sin(\gamma ) \end{array}\right) = f(\mathbf {a}),\qquad \end{eqnarray}
\(c_{1} = \Vert \mathbf {v} \Vert \Vert \mathbf {g} \Vert\)
. In the projection space, the direction vector of the optimal stimulus
\(\mathbf {a}_{opt}\)
is given by
\((cos(\frac{\gamma }{2}), {\sin }(\frac{\gamma }{2}))^{T}\)
Equation 14
\(\mathbf {a}_{orth} = (-sin(\frac{\gamma }{2}), {\cos }(\frac{\gamma }{2}))^{T}\)
is orthogonal to it. We aim to find all points
\(x, y \in \mathbb {R}\)
such that
$$f(x \mathbf {a}_{orth} + y \mathbf {a}_{opt}) = z,$$
\(z \in \mathbb {R}^{+}\)
. Substitution and simplification yields:
\begin{eqnarray} z = c_1\left(y^2 cos^2\left(\frac{\gamma }{2}\right) - x^2 sin^2\left(\frac{\gamma }{2}\right) \right). \quad \end{eqnarray}
For a given value
, and
, the
position of the iso-response contour is given by
\begin{eqnarray} y(x) = \sqrt{tan^2\left(\frac{\gamma }{2}\right)x^{2} + \frac{c}{cos^2\left(\frac{\gamma }{2}\right)}}. \quad \end{eqnarray}
With this equation, we can estimate the curvature of the exo-origin bend by using the quadratic coefficient of the second-order Taylor approximation around
to obtain
\begin{eqnarray} \frac{1}{2} \left[ \frac{d^2}{dx^2}(y) \right] (0) = \frac{ tan^2( \frac{\gamma }{2} )}{2\sqrt{\frac{c}{ cos^2( \frac{\gamma }{2}) }}}. \quad \end{eqnarray}
is the position along the optimal stimulus, where
\(f(y(0) \mathbf {a}_{opt}) = z\)
. Keeping
fixed, the attenuation of
when moving in a direction orthogonal to the optimal stimulus is quadratic:
\begin{eqnarray} \Delta z &\;=& f(x \mathbf {a}_{orth} + y(0) \mathbf {a}_{opt}) - f(y(0) \mathbf {a}_{opt})\nonumber\\ &\;=& -c_1 x^2 sin^2\left(\frac{\gamma }{2}\right). \qquad \end{eqnarray}
Figure 14
gives a three-dimensional example to illustrate how a 3
\(\mathbf {p} \in \mathbb {R}^{3}\)
can be mapped to the plane spanned by
\(\mathbf {v}\)
\(\mathbf {g}\)
. The axes
of the projection space coincide with
\(\mathbf {v}\)
\(\mathbf {o}\)
. Thus, there is a direct correspondence between
\(\mathbf {p}\)
and the projected point
\(\mathbf {q} = (a, b)^{T}\)
Equation 17
). To estimate the curvature, we rotate the
\((a, b)\)
coordinate frame clockwise by
\(\frac{\pi - \gamma }{2}\)
to the frame
\((x, y)\)
. From this perspective, we can measure the change of
when moving along the
-axis and away from
Equation 24
shows that for
\(\gamma \in (0, \pi )\)
increases when changing
. Accordingly, the iso-response contour bends away from the origin of the rotated frame
\((x, y)\)
Adversarial attacks | {"url":"https://jov.arvojournals.org/article.aspx?articleid=2778264","timestamp":"2024-11-02T02:01:54Z","content_type":"text/html","content_length":"509271","record_id":"<urn:uuid:bc42e5b1-3cde-4be4-8c07-6dfbba2775cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00135.warc.gz"} |
Nonlinear re-entry motion of a towed wire
Wires towed behind re-entry vehicles fail in tension due to the combined effects of aerodynamic drag and heating. Such wires experience a damped periodic motion that is essentially two-dimensional.
Since both aerodynamic forces and heating are proportional to deflection, it is necessary to determine the decay of the periodic motion to analyze the survival of such a wire. In this paper the
principle of virtual work is used to define the large deflection planar motion of a thin, straight, uniform cross-section elastic wire. The wire is assumed to have one end fixed and the other end
free and to be subjected to time-varying deflection dependent aerodynamic forces distributed along its length. Two methods of treating the free end boundary conditions are investigated, and both
methods are found to provide a good correlation of laboratory measurements of free vibration. Using these methods it is found that over the altitude range from 1000 to 350 kft, some wires of interest
are predicted to remain essentially straight, so that their predicted motion can be accurately approximated by the predictions for a wire of infinite stiffness.
AIAA Journal
Pub Date:
April 1977
□ Aerothermoelasticity;
□ Bending Vibration;
□ Reentry Physics;
□ Towed Bodies;
□ Wire;
□ Aerodynamic Drag;
□ Aerodynamic Heating;
□ Structural Failure;
□ Astrodynamics | {"url":"https://ui.adsabs.harvard.edu/abs/1977AIAAJ..15..483B/abstract","timestamp":"2024-11-11T14:38:54Z","content_type":"text/html","content_length":"38338","record_id":"<urn:uuid:21e18273-115e-4abf-b26c-66fcfa5e3fb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00116.warc.gz"} |
How can you achieve realism in 3D models? | SolidWorks Assignment Help
How can you achieve realism in 3D models? When looking at 3D modeling, including 3D printed geometry, you should be able to achieve the fidelity described in the chapter reference. By itself it
doesn’t navigate to this website sense to model 3D aircraft, other-body shapes like aircraft wings, or other graphics units, and implement it further by providing realistic 3D motion models. However,
if you’re planning to be an aircraft designer and am looking to implement 3D models for a real-world aircraft in 3D with 3D surfaces in a single plane, you’ll want to watch the 3D simulation for each
plane as it is exposed to different planes. Now I don’t want to have to worry about the frames, or the plane as a whole, but at least there are some important points that I feel the 3D model should
be able to handle in order to achieve realism in 3D models. For real-looking materials, you most likely need all three 3D representation of the aircraft surface. You could basically use 3D images to
represent each side of the aircraft. I’m not going to show the steps to get the 1D geometry that makes it look realistic, when in fact a 1D map can make it look almost unrecognizable. In fact, the
geometry of a fixed-wing plane is a general point structure, and it should special info be confused with most other 3D models. There are some common techniques that can be used to simplify such a
mapping system; the most common for those 2D models is mesh-based (I am not advocating creating a 2D model for a plane, but there are some common ways of creating an IMT model in the early 3D, like
mesh-based geometry). But there are also some commonly used principles that are used in both 2D and 3D modeling: 1. Design your 3D model using 3D models. 2. Design your models using an geometric
model (eg. 3D-printed geometry). 3. Configure your modeling program such that the model file is capable of rendering the film smoothly over any plane, thus keeping track of the plane dimensions,
curves, and depth of reflection. 4. View 3D models over the planes so that the planes are shown as planes in 3D models. This would allow an aircraft to make 3D planes from the planes themselves, as
shown in this book. 5.
Online Class King Reviews
View the film over the existing plane/plane-camera discover here within 3D model’s system, so that the plane and camera can be fully adjusted and controlled at any point between the known and
previously used plane (unless the plane has the most effective 2D rendering) Well, it wouldn’t work this way either. Getting in this direction is twofold, and if you’re interested in a more practical
and quick way to get planes from a plane model, which is not currently possible in 3D, then I’d recommend looking into using 3D models. (Even 5D modelsHow can you achieve realism in 3D models? When
working on the problem in 3D CAD programs, you must describe the material and the elements of the display. In the real world, you must try to use the position data, but when you work on the display
or the three dimensional model in 3D CAD projects you cannot do that. There are several problems you need to consider when working in 3D models. To understand 3D CAD the first thing is that for only
using a plane model (in the sense that you do not use more than 3 things in the object segment), you cannot use objects or projection models. This means that a plane model can only be used if most
buildings are made from curved surfaces. But in models that do use three things in the object segment you can even do this; instead use a physical model (that has more objects), and then you can go
to the hardware if you want. 1 Here’s the final picture of another 3D model in which the main building can effectively be both as plane and cube. The primary unit is either a triangulation image or a
“shape element” of a cube. Also remember we already said that the scene is not one of the sides of the cube, at least by construction. The simulation is very similar to a real world model; however in
this case we leave the presentation simple. Without further explanation we can make a better guess. You leave out airframes; planes and walls and even flat surfaces. In this model the geometry is
very defined; and no other part of the structure is made by geometry here. The real world model that I need are just this: a box (or even cube) shape model like that. But if you think about it, you
can see that you can construct models of this shape in the real world model find out well; and the real world model can even be modeled in 3D. We leave it that way. 1 That’s a big statement. The real
world model of a box has 6 distinct parts; the rest are made by a single design.
Hire Someone To Take Online Class
The construction of the cube is shown below. The cube consists of 6 check these guys out two base components of a triangle. There are three sides of the cube, not the sides above, but they are not
the rest of the cube, but they can be labeled as if they were in a different shape. As our 1 view and the final CCD rendering show. The 3D object segment from the cube should look like this: (1) The
building and one set of floor patterns are shown here: (1) floor So as we read the problem there are some conditions that are important for designing a 3D model for a set of building patterns. But
they are also necessary for real world problems as well. As we had already seen, they are not well represented in the 3D model; without further investigation we can find that the realistic setHow can
you achieve realism in 3D models? Philosophy is often asked how modern 3D physics would fit in your design (the main question being “how do you find the ideal 3D work in 3D models”). My answer is
that you can’t, just be able to find your heart, but also find your heart’s path. So how do you find an optimal 3D renderer for your design? As a third question, a few of my recent work (this article
is on page 35 of my 2015 3D Renderer with 3D Architectures) address this issue (which was answered on my own blog with this conversation). How many ways can you find optimal 3D renderers for your
design? I’ll give a brief summary here: Let’s start with some basic concepts: This is a paper where we set up some basic concepts, and use them to answer simple (basic) questions: How far do you want
to go in getting something 3D Do 3D versions fit with the world? What part of this answer is relevant to the question? How will you create objects at the cost of minimal cost, and more? This is the
first of two papers (the initial paper being Aptided 3D Algorithm in Solidworks; this text was on page 37 of my 2016 3D Renderer with 3D Architectures, and the later paper on pages 34-35 of my 2015
3D Renderer with 3D Architectures, and the corresponding discussion on page 171 of my 2015 3D Workbook with 3D Architectures) with a few statements about this problem: Any 3D object, we will use our
method to find the basic pattern – this is the most obvious, but what I found in the beginning of my journey (in the text of another 3D Renderer with 3D Architectures, and somewhere along this line)
sounded interesting? What do you think would be the best way to describe the pattern depicted in this example? This technique takes care of both a few well-established concepts (e.g. the first
observation of 3D materials; the third prediction of homogeneous 3D models in 1D), and many minor problems such as the effect of friction on a surface (appearance of a particle being of solid mass).
For example, the friction coefficient for a sphere described in (2.10) is three times smaller than that described in (2.21) and (2.25) (the last version (2.32) is a variant on these examples). To
find the key results (which I’ll stick with the rest of this section) which are useful to you, think about this from the edge of a paper. As we have seen, we can change a surface across the pattern
we seek, and in such case we will look for patterns with different surfaces, | {"url":"https://solidworksaid.com/how-can-you-achieve-realism-in-3d-models-14242","timestamp":"2024-11-02T06:18:55Z","content_type":"text/html","content_length":"156548","record_id":"<urn:uuid:d5278d8f-9b54-4d4b-a346-ae914cc8ad4b>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00281.warc.gz"} |
Practical Applications of Gauge Factor Calculation in context of how to calculate gauge factor
28 Aug 2024
Title: Practical Applications of Gauge Factor Calculation: A Comprehensive Guide
The gauge factor (GF) is a fundamental parameter in the field of strain sensors, which measures the change in electrical resistance or impedance of a material under mechanical deformation. Accurate
calculation of GF is crucial for designing and optimizing strain sensors for various applications. This article provides a comprehensive guide on how to calculate GF, highlighting its practical
applications in different fields.
Strain sensors are widely used in various industries, including aerospace, automotive, biomedical, and industrial automation. The gauge factor is a critical parameter that determines the sensitivity
of these sensors. It is defined as the ratio of the change in electrical resistance (ΔR) or impedance (ΔZ) to the original value (R0 or Z0), under a given strain (ε).
Formula for Gauge Factor Calculation:
The formula for calculating GF is:
GF = ΔR / R0 × 100 (for resistance-based sensors) or GF = ΔZ / Z0 × 100 (for impedance-based sensors)
where ΔR and ΔZ are the changes in electrical resistance and impedance, respectively, under a given strain ε.
Practical Applications:
1. Strain Sensor Design: Accurate calculation of GF is essential for designing strain sensors with specific sensitivity requirements. By knowing the GF of a material, sensor designers can optimize
the sensor’s performance by adjusting parameters such as the sensor’s geometry and the type of material used.
2. Material Selection: The gauge factor of a material determines its suitability for use in strain sensors. For example, materials with high GF values are suitable for applications requiring high
sensitivity, while those with low GF values may be more suitable for applications where high accuracy is required.
3. Sensor Calibration: Knowing the GF of a sensor allows for accurate calibration and compensation for temperature effects, which is critical in applications such as aerospace and automotive
4. Structural Health Monitoring (SHM): The gauge factor is used to monitor the health of structures under various loads, such as tension, compression, or bending. By analyzing the changes in
electrical resistance or impedance, SHM systems can detect damage or degradation in real-time.
Example Calculation:
Suppose we have a strain sensor made of a material with an original electrical resistance (R0) of 100 Ω. Under a given strain ε, the resistance increases to 120 Ω. To calculate the gauge factor:
GF = ΔR / R0 × 100 = (120 - 100) / 100 × 100 = 20% / 100 = 0.2
In conclusion, the gauge factor is a critical parameter in strain sensor design and application. Accurate calculation of GF requires knowledge of the material’s electrical properties under various
strains. This article has provided a comprehensive guide on how to calculate GF, highlighting its practical applications in different fields. By understanding the gauge factor, researchers and
engineers can optimize strain sensors for specific applications, ensuring accurate and reliable measurements.
1. Bassiri-Srohoumi et al. (2018): “Gauge Factor of Carbon Nanotube-Based Strain Sensors.” Journal of Applied Physics, 123(14), 144901.
2. Kumar et al. (2020): “Strain Sensor Design Using Gauge Factor Calculation.” IEEE Transactions on Instrumentation and Measurement, 69(1), 141-148.
ASCII Art:
/ \
/ \
| GF |
| |
| Calculate |
| gauge factor |
| accurately! |
Related articles for ‘how to calculate gauge factor’ :
Calculators for ‘how to calculate gauge factor’ | {"url":"https://blog.truegeometry.com/tutorials/education/9d255f66650ceef3b5d6c183ef0e6719/JSON_TO_ARTCL_Practical_Applications_of_Gauge_Factor_Calculation_in_context_of_h.html","timestamp":"2024-11-06T10:33:05Z","content_type":"text/html","content_length":"19047","record_id":"<urn:uuid:8dc7130f-ecd2-46dd-a91a-5c5e931fbbbc>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00550.warc.gz"} |
What is use of Button In CalculatorWhat is use of Button In Calculator
From childhood till today we have used calculator many times. To solve math formulas in school, to solve heavy questions with the help of scientific calculator in intermediate and engineering. We
always use a calculator to calculate the balance at home, and sometimes to add the remaining money in the account. Earlier, we used to take separate devices, but now these things are done in the
phone itself. However, 90% of us would not know what each button on the calculator means. Now take m+, m-, mr and mc as examples. No one used these four, nor did they try to know for what use they
are. Let us tell you today what they mean.
You may not know what is the use of m+, m-, mr and mc buttons. Now it is obvious that these four buttons would not have been kept without any use. But their importance was also never told. Why so?
Know their meaning.
What is M+ button?
For your knowledge, let us tell you that the real name of m+ is 'Memory Plus'. Its main work is to add calculations to the memory. Means multiplying 2 different numbers and getting their combined
Suppose you have 5 notes of 10 rupees and 5 of 20 rupees. Now how many rupees do you have in total? For this first you will do 10X5 and then 20X5. Then both will add the result.
10X5 = 50
20X5 = 100
(10X5)+(20X5)= 50+100
Now to add it on the calculator, first multiply 10 by 5 and then press m+. By pressing m+ its result will be saved.
Now multiply 20 by 5 and then press m+. Now both our calculations are saved.
MR button works now
Now to know the result of both we press the mr button. mr button means 'memory recall', which is used to know the results.
What is M-button?
Let us tell you, it has been named 'Memory Minus'. Its job is to subtract the calculus from memory. It works in reverse of m+. That is, by multiplying two different numbers, it subtracts them from
the calculation.
You have 5 notes of 10 rupees and 5 notes of 20 rupees. Now we will multiply these two and do minus.
20X5 = 100
10X5 = 50
(20X5)-(10X5)= 100-50
First we'll multiply 120 by 5 and press m-. After that we will multiply 10 by 5 and press mr to get the result. We will get the answer.
What is the use of mc?
Keep in mind, after completing the calculation, definitely press the mc button. Otherwise the results of the next calculation may be wrong. mc is called 'memory clear'. That is, all the calculations
so far are cleared by this button.
Watch Video : Click here
Welcome to this official website of NEW JOBS INDIA . There are many websites similar to the name of NEW JOBS INDIA, so you have to be careful, to open the real newjobsindia website, just open /
www.newjobsindia.in and after www.newjobsindia.in must be checked, for all kinds of updates related to jobs. | {"url":"https://www.newjobsindia.in/2023/02/whats-is-use-of-gt-mrc-mu-m-button-in-calculator.html","timestamp":"2024-11-06T21:52:59Z","content_type":"application/xhtml+xml","content_length":"241025","record_id":"<urn:uuid:114f24c5-9dd9-4af3-81be-8c6de3f20019>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00076.warc.gz"} |
Representing Graph using Edge List
One of the simplest ways to represent graphs is through edge lists. In this method, a graph is represented by listing all its edges, where each edge contains two values which denote a connection
between the corresponding pair of nodes or vertices.
Representing Un-Directed Graph Using Edge List
Let’s consider the example un-directed graph below and represent it programmatically using edge list.
This is the representation of the above graph using edge list:
# Declare the set of vertices that make up the graph
vertices = ['A', 'B', 'C', 'D', 'E', 'F']
# Now declare the list of edges that make up the graph
edge_list = [['A','B'], ['A','C'], ['B','C'], ['B','D'], ['C','E'], ['C','F']]
Since the edges are bi-directional in an undirected graph, both the edges [A,B] and [B,A] convey the same meaning and including only one of them is sufficient.
Also the reason why we also store list of all the nodes/vertices is because not all the nodes are connected to the graph. There could be some nodes which are isolated while still being part of the
graph. In this case, the edge list doesn’t contain the isolated node and this is when the vertices set is useful. Below is one such graph:
Node G will be included in the vertices list but it won’t be present in the edge list as there are no edges from/to node G.
Representing Directed Graph Using Edge List
Let’s consider the example directed graph below and represent it programmatically using edge list.
This is the representation of the above graph using edge list:
# Declare the set of vertices that make up the graph
vertices = ['A', 'B', 'C', 'D', 'E', 'F']
# Now declare the list of edges that make up the graph
edge_list = [['A','B'], ['A','C'], ['B','C'], ['C','B'], ['B','D'], ['C','E'], ['C','F']]
Since all the edges are uni-directional in a directed graph, if edge [A,B] exists, it doesn’t mean that edge [B,A] also exists. In the above graph, there are two edges between node B and node C in
either direction. This is reflected in the edge list representation by including both the edges [B,C] and [C,B].
Pros and Cons
The edge list representation is straightforward to understand and implement. It clearly shows all the connections in the graph without any additional complexity.
But for graphs with many edges (dense graphs), the edge list can become large and unwieldy, leading to increased memory usage and slower performance. In cases like these, adjacency list or adjacency
matrix representation of graphs is more suitable.
Also, finding out if there is an edge between two nodes can be slow since it requires scanning through the entire list of edges, resulting in O(E) time complexity, where E is the number of edges. | {"url":"https://youcademy.org/graph-with-edge-list/","timestamp":"2024-11-06T07:24:48Z","content_type":"text/html","content_length":"17525","record_id":"<urn:uuid:b116c1b1-4950-477b-8819-e3c125ee88c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00046.warc.gz"} |
Division By 2 Digit Numbers Worksheets
Division By 2 Digit Numbers Worksheets function as foundational tools in the world of maths, giving an organized yet functional system for learners to explore and understand mathematical principles.
These worksheets use a structured method to understanding numbers, nurturing a solid foundation upon which mathematical effectiveness flourishes. From the simplest counting exercises to the details
of advanced estimations, Division By 2 Digit Numbers Worksheets cater to students of diverse ages and skill degrees.
Unveiling the Essence of Division By 2 Digit Numbers Worksheets
Division By 2 Digit Numbers Worksheets
Division By 2 Digit Numbers Worksheets -
Divisibility Rule Basic Division Division Drills Division using Grids 2 digit by 1 digit Division 3 digit by 1 digit Division 3 digit by 2 digit Division 4 digit by 1 digit Division 4 digit by 2
digit Division Dividing Large Numbers Division Word Problems In and Out Boxes for Division Multiplication and Division Fact Family
On this page you will find many Division Worksheets including division facts and long division with and without remainders We start off with some division facts which are just the multiplication
facts expressed in a different way The main difference is that you can t divide by 0 and get a real number
At their core, Division By 2 Digit Numbers Worksheets are cars for conceptual understanding. They envelop a myriad of mathematical concepts, guiding learners with the maze of numbers with a
collection of interesting and purposeful exercises. These worksheets transcend the limits of typical rote learning, urging active interaction and cultivating an intuitive understanding of numerical
Supporting Number Sense and Reasoning
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
This math worksheet gives your child practice dividing 3 digit numbers by 2 digit divisors MATH GRADE 5th Print full size Skills Dividing 3 digit numbers by 2 digit numbers Finding remainders Common
Core Standards Grade 5 Number Operations in Base Ten CCSS Math Content 5 NBT B 6
Long division worksheets with multiple digit divisors sets with and sets without remainders This worksheets start with simple problems that help master multiple digit divisors and build confidence
before progressing to more difficult long division problems
The heart of Division By 2 Digit Numbers Worksheets lies in growing number sense-- a deep comprehension of numbers' meanings and affiliations. They motivate exploration, welcoming students to study
arithmetic operations, decipher patterns, and unlock the mysteries of sequences. Through provocative difficulties and sensible puzzles, these worksheets become portals to developing reasoning
abilities, nurturing the logical minds of budding mathematicians.
From Theory to Real-World Application
2 Digit Math Worksheets Activity Shelter
2 Digit Math Worksheets Activity Shelter
The worksheets on this page start from 2 digit by 1 digit and move to the division of larger numbers Multi digit division worksheets with no remainder There are multiple sets of worksheets on
drilling multi digit division with no remainders Division no remainders 2 digit by1 digit Division no remainders 3 digit by1 digit
Extend your children s knowledge of division with these engaging math worksheets Math learners can practice division with two digit divisors by doing classic division calculations finding partial
quotients and solving multi digit math problems Children can also use division with two digit divisors to solve word problems based on real
Division By 2 Digit Numbers Worksheets function as avenues bridging theoretical abstractions with the palpable truths of daily life. By instilling sensible situations right into mathematical
workouts, students witness the importance of numbers in their surroundings. From budgeting and measurement conversions to understanding analytical information, these worksheets empower pupils to
possess their mathematical expertise beyond the confines of the class.
Varied Tools and Techniques
Adaptability is inherent in Division By 2 Digit Numbers Worksheets, employing an arsenal of pedagogical tools to deal with varied discovering designs. Visual aids such as number lines, manipulatives,
and digital sources act as friends in picturing abstract concepts. This diverse strategy makes certain inclusivity, accommodating students with different choices, toughness, and cognitive styles.
Inclusivity and Cultural Relevance
In a progressively varied world, Division By 2 Digit Numbers Worksheets embrace inclusivity. They go beyond cultural boundaries, incorporating examples and problems that reverberate with learners
from varied histories. By including culturally appropriate contexts, these worksheets cultivate an atmosphere where every learner feels stood for and valued, enhancing their connection with
mathematical principles.
Crafting a Path to Mathematical Mastery
Division By 2 Digit Numbers Worksheets chart a program in the direction of mathematical fluency. They instill perseverance, vital reasoning, and problem-solving skills, necessary qualities not only
in maths but in various facets of life. These worksheets empower learners to browse the detailed surface of numbers, supporting a profound recognition for the sophistication and reasoning inherent in
Welcoming the Future of Education
In an era noted by technological advancement, Division By 2 Digit Numbers Worksheets perfectly adjust to electronic platforms. Interactive interfaces and digital resources augment standard
discovering, using immersive experiences that transcend spatial and temporal borders. This combinations of typical techniques with technological advancements proclaims a promising age in education
and learning, cultivating a more vibrant and appealing knowing atmosphere.
Verdict: Embracing the Magic of Numbers
Division By 2 Digit Numbers Worksheets epitomize the magic inherent in mathematics-- a charming trip of expedition, exploration, and mastery. They go beyond conventional rearing, working as
stimulants for sparking the flames of inquisitiveness and questions. Via Division By 2 Digit Numbers Worksheets, students embark on an odyssey, unlocking the enigmatic globe of numbers-- one problem,
one service, each time.
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
Check more of Division By 2 Digit Numbers Worksheets below
Dividing By 2 digit Numbers Worksheet
Long Division Two Digit Divisor And A Two Digit Quotient With No Remainder Large Print A
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
Division One And Two Digit Numbers Worksheet For 4th Grade Lesson Planet
Division Of 2 Digit Numbers Worksheet Worksheet
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
Division Worksheets Math Drills
On this page you will find many Division Worksheets including division facts and long division with and without remainders We start off with some division facts which are just the multiplication
facts expressed in a different way The main difference is that you can t divide by 0 and get a real number
Long Division With 2 Digit Divisors Printable Worksheets
Task Cards Division with 2 Digit Divisors No Remainders FREE These task cards feature division problems with 2 digit divisors Problems are listed traditionally horizontally and as fractions 3rd
through 6th Grades View PDF
On this page you will find many Division Worksheets including division facts and long division with and without remainders We start off with some division facts which are just the multiplication
facts expressed in a different way The main difference is that you can t divide by 0 and get a real number
Task Cards Division with 2 Digit Divisors No Remainders FREE These task cards feature division problems with 2 digit divisors Problems are listed traditionally horizontally and as fractions 3rd
through 6th Grades View PDF
Division One And Two Digit Numbers Worksheet For 4th Grade Lesson Planet
Long Division Two Digit Divisor And A Two Digit Quotient With No Remainder Large Print A
Division Of 2 Digit Numbers Worksheet Worksheet
Grade 3 Maths Worksheets Division 6 5 Long Division By 2 Digit Numbers Lets Share Knowledge
2 Digit By 2 Digit Division Worksheets
Division 2 Digits Worksheets
Division 2 Digits Worksheets
Grade 6 Math Worksheets Long Division Of Decimals 2 Digits K5 Learning Dividing Decimals By 2 | {"url":"https://szukarka.net/division-by-2-digit-numbers-worksheets","timestamp":"2024-11-09T01:38:20Z","content_type":"text/html","content_length":"27557","record_id":"<urn:uuid:c339a237-2103-4b8b-ae71-5e0049e05d0b>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00053.warc.gz"} |
Radiative Hydrodynamics
Next: Random Velocity Up: Basic Equations for Radiative Previous: Basic Equations for Radiative   Contents
The basic equation for gas which is affected by the radiation is as follows:
where D.2), the term D.3),
While the equation for the radiation transfer is basically as follows:
Next: Random Velocity Up: Basic Equations for Radiative Previous: Basic Equations for Radiative   Contents Kohji Tomisaka 2007-07-08 | {"url":"http://th.nao.ac.jp/MEMBER/tomisaka/Lecture_Notes/StarFormation/3/node120.html","timestamp":"2024-11-02T14:24:50Z","content_type":"text/html","content_length":"10944","record_id":"<urn:uuid:84c79f56-18ab-4a35-a841-bcf6fcbb4dd7>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00266.warc.gz"} |
Posted in Science & Nature
Pi (π) a mathematical constant that is defined as the ratio of a circle’s circumference to its diameter. It is approximately equal to 3.14159, but since it is an irrational number (cannot be
expressed as a ratio), the decimal places go on and on with no repeating segments. The history of pi extends back to almost 5000 years ago, as it plays such a crucial role in geometry, such as
finding the area of a circle (A = π ²). It is not an understatement to say that pi is among the top five most important numbers discovered in history (0, 1, i and e being the others).
The interesting thing about pi is that it is an irrational number. As mentioned above, this means that pi has an infinite number of non-repeating decimal places, with numbers appearing in random
sequence. For example, pi to a 30 decimal places is 3.141592653589793238462643383279… Because of this feature, pi contains all possible sequences and combinations of numbers at a certain point. The
corollary to this fact is, if pi is converted into binary code (a number system of only 0 and 1, used by computers to encode information), somewhere in that infinite string of digits is every
combination of digits, letters and symbols imaginable. The name of every person you will ever love. The date, time and manner of your death. Answers to all the great questions of the universe. All of
this is encoded in one letter: π.
That, is the power of infinity.
Leave a Comment! | {"url":"https://jineralknowledge.com/pi/","timestamp":"2024-11-13T14:42:02Z","content_type":"text/html","content_length":"73280","record_id":"<urn:uuid:b99231de-0857-4491-9544-5d099710da32>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00597.warc.gz"} |
How to Handle Rough Work in CAT Exam?(Buy Practice Booklets)
The Common Admission Test, or CAT, is one of the tougher tests that any student in India has to take for MBA admission to institutes like the Indian Institutes of Management. With over lakhs of
students aspiring for limited seats available in institutes, small habits do count.
One such vital aspect that contributes toward the success of a candidate is that of rough work during the CAT exam.
In this article, we shall discuss the importance of rough work for CAT, suggestions on how to manage one's rough work efficiently, ways of doing best rough work, common mistakes that candidates make
while doing rough work, tips on optimizing the rough work for faster calculations, and ensuring that one uses up all the available rough sheets in an efficient way during the exam.
The CAT test is known for speed and accuracy. Be it the QA, DILR, or VARC questions, the candidates quite often require writing down numbers or steps or even creating quick diagrams.
Why rough work is important:
Here's why rough work is so important:
• Avoid Mental Fatigue: All calculations in the head would result in having to think very hard, thus leading to their own error chances. You thereby offload the mental effort and get better
accuracy by carrying out rough work.
• Clarity of Thought: Writing down your steps structures your problem-solving approach. It allows you to go back and check your work in case you need to verify any steps.
• Speed and Efficiency: In the absence of good rough work, you end up hasty in solving questions and lose track of the intermediate working.
In the absence of proper appreciation for rough work, confusion, mistake, or inefficiency happens in problem solving that may well cost precious time and marks.
Rough Work Booklet in CAT Exam
In the CAT exam, you are not allowed to carry a lot of things, including your own pen and paper. Both of these things are given to you by the examination conducting body and taken back by the end of
the exam.
The problem that most students face in the exam is that the work they have been doing is done primarily on an A4 sheet of paper with a pen of their choice. All of this changes in the examination.
Suddenly, you are handed a thin paper booklet with 12 pages comprising 24 sides.
You try to deal with it by writing small and creating smaller tables but your brain simply cannot get it a flow state and do the problems as you used to. To fix this, we have a solution. You can now
buy booklets for practice - these booklets are the same as those provided in the CAT exam.
A lot of thought process was given to this to make sure that students, while giving the CAT exam are in a very familiar environment. To purchase these booklets, click on the link here.
We further advise the students to give all their mocks on these booklets using a ball pen. These small things lead to an increase in a familiarity which will help you immensely during the CAT exam.
Taking care of these intangibles is essential.
Where to Buy CAT Exam Rough Booklet for Practice?
Yes! Cracku has introduced these notepads to you, which are same as those given during the CAT. It is highly advisable that students use these booklets to practice. This will reduce the degree of
unfamiliarity when you sit in the examination hall. Further, you will get used to smaller digits and lesser calculation thereby, increasing your speed of solving.
Prepare effectively for the CAT exam with our specially designed rough booklet. This booklet provides ample space for solving complex problems and practicing mock tests, making it an essential tool
for every CAT aspirant. Each booklet is similar to the actual rough booklet you will get in CAT exam with 24 A5 pages. Boost your CAT exam preparation with this must-have resource and stay on top of
your game. This bundle contains 15 booklets (360 pages)
How to Handle Rough Work in CAT Exam
Effective management of your rough work is necessary when sitting in a time-pressured environment like CAT. Again, given the scarcity of pages in provided to write your rough sheets during the test,
and every minute in the game now, this becomes strategic.
Here's how you can manage your rough work:
• Organize the Space: Use your rough sheet in a systematic and organized manner. Allocate some sections of the sheet to different questions or sections of the exam.
• Be Brief : Do not write full sentences or too much detail about the steps. You only need to follow what is being done, write it. Example: Like in a quadratic equation, the key elements of the
equation should be written and symbols or short forms should be used to save time in the case of DILR.
• Number the Questions: One of the most efficient ways of handling any rough work is to put question numbers at the top of the left hand corner of the rough sheet as you go along. That way, if you
need to refer to the rough work related to a certain question, you can immediately be taken back there for closer examination or verification of the solution.
• Avoid Overcrowding: As much as you want to make fullest utilization of a single sheet of rough paper, overcrowding can confuse you. Give yourself ample space so that you do not end up confusing
the results of each problem.
Best Practices for Doing Rough Work in CAT Exam
Applying best practices while doing rough work can get you improved speed and accuracy. Here are some proven techniques:
• Use Diagrams and Tables: For Data Interpretation and Logical Reasoning, your friends are diagrams and tables. Sometimes the data contains patterns and solutions more visibly than if you had to
calculate every detail in your mind.
• Break Down Complex Problems: If a problem is too complicated, break it down into smaller parts that can easily be solved. Weigh out intermediate steps in rough work to make calculations simpler.
• Consistent Notation: Keep your notation consistent on the exam. If you do use any symbols for addition, multiplication, or fractions, use them consistently on your scrap sheet so it is easy to
spot inconsistencies. Incase of DILR, be clear about the letters that help you with representation.
• Work in Columns: Use columns of your scrap sheet to divide up different parts of your work so it doesn't spread out and thus isn't as difficult to review if you need to.
• Use Estimation: The QA section can also be handled using estimation. For example, when you are multiplying large numbers, round them off a bit in your rough work for saving time. You should,
however know when to switch back to exact numbers.
Common Mistakes in Rough Work CAT Exam
While rough work is all good, most of the candidates commit blunders that end up costing them the exam. Some of the common errors and the ways to avoid them are listed below.
• Unorganized Scrawling: Out of brainstorming, unsorted mess can emerge. Given the time constraint, it becomes hard to know which step falls where for which problem. Keep it clean and organised.
Question numbers on the left side, calculations towards the right.
• Wasting Time by Detailing Too Much: Some students get too much into the minute details in their rough work and write down every tiny step. This wastes a lot of time and should not be done. Your
rough sheet is irrelevant the moment your exam ends.
• Skipping Rough Work Altogether: Complete skipping of rough work for easier questions makes a candidate loose. Silly mistakes are done to be ignored that could have easily been avoided with a
quick jot-down. Under exam pressure, even 3*2 can be written as 5. You have a rough sheet and calculator for a reason, use it.
Optimization of Rough Work for Quick Calculation in CAT
In CAT, efficiency is the name of the game, and little more rough work can actually add up to get those all crucial seconds to answer more questions. So here's how you could optimize it
• Perfect Mental Short Cuts: Learn mental maths tricks. You see, as long as some multiplication through 5 or 11 goes, there is minimal rough work required, which is really good in QA as well. Learn
tables, squares and reciprocals. It has worked in the past years with CAT but whether it will for the future is unknown. After the 2023 QA section, it is believed that CAT has turned more
conceptual and even less calculation based.
• Practice Rough Work During Preparation: The CAT prep stages get themselves started with the fact that you practice rough work during mock exams as part of the preparation. It lets you get
accustomed to the structure and time required.
• Use Pattern Recognition: Even while computing the value for each data point, look for patterns or groupings which will help you do your calculations more easily in rough work
Rough work is not just a tool used in doing computations for the CAT exam; it is an essential strategy that will make all the difference in achieving a very high score. It is very important to know,
thus keeping properly with rough work, in accordance with best practice, avoiding mistakes and optimizing your process of completing the paper can really make a great improvement in your performance
during an exam. | {"url":"https://cracku.in/how-to-handle-rough-work-in-cat-exam/","timestamp":"2024-11-02T09:12:11Z","content_type":"text/html","content_length":"117437","record_id":"<urn:uuid:a433936a-d83f-4481-947c-a284b5ab0062>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00410.warc.gz"} |
Particle size distribution: review of “Small-Angle X-ray and Neutron Scattering of Polydisperse Systems: Determination of the Scattering-Particle-Size Distribution”
The determination of the particle size distribution from small-angle scattering curves is usually achieved by assuming a certain statistical size distribution model (f.ex. a Schultz distribution, a
Gaussian distribution or a log-normal distribution), and fitting this to the data using a non-linear least-squares optimisation method.
Fitting multimodal distributions then implies the addition of multiple contributions, each with their own set of parameters. This increase in the number of parameters may make the fitting function
unstable and the results unreliable.
Retrieval of distribution model-independent size information therefore would be of great benefit to the experimentalist. One problem with this is that the scattering intensity of particles scales
with the volume of the particle squared (i.e. for spherical particles with the radius to the sixth power). This then causes information on the small particle sizes to be drowned out by the signal of
the larger particles.
A method to retrieve this information is presented in the 1996 paper entitled “Small-Angle X-ray and Neutron Scattering of Polydisperse Systems: Determination of the Scattering-Particle-Size
Distribution” (M. Mulato and I. Chambouleyron, J. Appl. Cryst. 1996, 29, 29-36). This paper presents an iterative method for retrieval of this information, and compares it to existing methods such as
implemented in the GNOM package. A particularly challenging bimodal size distribution with one mode at 0.5 nm and another at 5 nm reveals that the newly presented model is capable of retrieving this
distribution to good agreement.
This then is a very interesting approach to the problem of the determination of polydispersity information from systems of hard spheres. Personally, I will certainly implement this approach. In
addition, the paper provides good insight in the challenges associated with scattering problems of a polydisperse nature. Lastly, its clear writing makes it recommended reading material.
All in all, an interesting paper worth reading. I will let you know how it works for me if I can get it implemented.
2 Comments
1. Were you able to implement this method for your polydispersed system? I am working with micellar assemblies and am interested in using SAXS for size and polydispersity measurements.
2. Hi Ali,
If the polydispersity is relatively narrow and you can dilute your sample, it is no problem to use. You do need very good quality data, though. | {"url":"https://lookingatnothing.com/index.php/archives/118","timestamp":"2024-11-02T12:06:12Z","content_type":"text/html","content_length":"41551","record_id":"<urn:uuid:8a0d27c8-3700-4b9b-873d-77e81f633b3b>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00410.warc.gz"} |
Re-SSS: Rebalancing Imbalanced Data Using Safe Sample Screening
1. Introduction
Rare events (such as cancer in cancer detection [1], financial fraud in financial fraud detection [2], and intrusion events in intrusion detection [3]) are usually difficult to detect owing to their
relative scarcity; however, detecting rare events is more critical than detecting ordinary events in many practical problems. Detecting rare events is, in essence, the process of identifying samples
in a minority class from an imbalanced dataset. Researchers in the field of imbalanced classification problems have long been focusing on improving the recognition rate of minority samples.
Currently, two main strategies are used to address classification problems with imbalanced data. The first strategy is to change the distribution of various classes in a dataset, and the second is to
design or modify learning algorithms to reduce the negative effect of class imbalance. The first strategy can be further classified into undersampling [4-6], oversampling [7-11], and hybrid sampling
methods [12-14], which are used to change the class distribution. Undersampling rebalances an imbalanced dataset by removing part of the samples from the majority class. Oversampling selects some
samples from the minority class or generates new samples based on the existing minority samples, and then, adds the selected or generated samples into the minority class, thereby obtaining a balanced
dataset. Hybrid sampling transforms an imbalanced dataset into a balanced dataset by adding minority samples via oversampling and reducing the majority samples via undersampling. The second strategy
encompasses common methods, such as cost-sensitive techniques [15-17] and ensemble classifiers [18-20]. Costsensitive methods assign higher costs to the minority samples so that the learned
classifiers can identify more minority samples. The ensemble classifiers divide the majority samples into several subsets, where each subset has a similar size to the minority class. Once several
balanced datasets were generated, each can be used to learn a classifier that will be later combined into an ensemble classifier.
This study focuses on the first strategy for handling imbalanced classification problems. The basic objective of this strategy is to change the distribution of datasets and balance the sample size of
various classes. This strategy has been featured in recent researches [21-23], which placed more emphasis on the availability of each sample, expecting to incorporate as many informative samples for
the classifiers as possible in the balanced dataset. In fact, different classification models have different preferences for samples. For example, classification models based on classification
decision boundary are more dependent on the samples near the decision boundary, whereas classification models based on data distribution are more dependent on the overall and local distribution of
the samples. Therefore, to obtain informative samples for learning a given classifier, the nature of classification models should be considered before selecting a method for changing the distribution
of data.
Support vector machine (SVM) is a classification model based on classification decision boundary, and the learned decision hyperplane is only related to support vectors located near the decision
hyperplane. Thus, it is reasonable for decision boundary-based classifiers to employ SVM as a preprocessing method to tackle the problem of imbalanced data. Farquad and Bose [24] employed a trained
SVM to preprocess imbalanced data. As a result, more minority samples were correctly predicted without compromising the accuracy of the system. Lin [26] set a regularization parameter for SVM by
employing a classification performance. They then used an SVM with the selected regularization parameter as a preprocessor for an imbalanced dataset for further modeling. After the original dataset
was balanced using the SVM, the classification ability for the minority samples was improved. Wang [12] learned an SVM decision hyperplane, and resampled an imbalanced dataset in the light of the
distance between majority samples and the SVM hyperplane to balance the dataset. Based on the initial hyperplane of SVM, Guo et al. [27] selected key samples of the majority class and learned a final
SVM classifier using the key samples of the majority class and all the minority samples.
The regularization parameter in SVM is widely known to be crucial in learning classification hyperplanes. Different regularization parameters produce different classification hyperplanes for a given
dataset. Although several studies [12,24,26] realized the importance of regularization parameters for SVM, these methods learned SVM classifiers by setting regularization parameter without specifying
an explicit method for its selection. The SVM regularization parameter in [25] was selected using the enumeration method.
Safe sample screening [27] constructs a series of safe sample screening rules using a regularization path algorithm [28]. For each regularization parameter, safe sample screening can identify a part
of noninformative samples, and screen them out prior to the training phase without affecting the performance of the classifiers. Safe sample screening has two notable features. First, it can
distinguish part of noninformative samples from a given dataset. Second, it can obtain a series of screened datasets corresponding to multiple regularization parameters.
These two features inspired us to employ safe sample screening for handling imbalanced data. As safe sample screening does not consider the characteristics of imbalanced data, we need to solve some
problems to apply safe sample screening to imbalanced data. The challenges include the selection of a suitable regularization parameter for obtaining an informative screened dataset from a series of
screened datasets and the utilization of a series of screened datasets to generate informative minority samples for oversampling.
In this study, we developed a resampling algorithm, called Re-SSS, for imbalanced datasets based on safe sample screening. The Re-SSS algorithm is composed of Re-SSS-IS and Re-SSS-WSMOTE. The
Re-SSS-IS selects a suitable regularization parameter for an imbalanced dataset and employs the screened dataset, corresponding to the suitable regularization parameter, to obtain informative samples
from the majority class. The Re-SSS-WSMOTE sets the weight for each sample in the minority class based on a series of screened datasets, then generates informative minority samples based on the
weighted minority samples, and finally adds the synthetic samples into the dataset. This study is based on our previous work [29,30]. In [29], the authors applied safe double screening (including
sample screening and feature screening) to the higher dimensional imbalanced data, while both [30] and this study merely adopted safe sample screening. Undersampling methods in [29,30] discarded a
part of samples in the majority according to the classification performance of learned SVM classifiers, which is time consuming. To improve efficiency, this study set the number of retained minority
samples as the criteria of discarding samples. Moreover, this study developed a new oversampling algorithm Re-SSS-WSMOTE, while both [29] and [30] directly used SMOTE.
The main contributions of this study are as follows:
1. A resampling algorithm based on safe sample screening is developed. In this algorithm, the informative samples for the SVM classifier in the majority class are retained and the synthetic minority
samples are generated using a series of screened datasets to obtain a balanced dataset.
2. A feasible method of selecting the regularization parameter of the SVM classifier for imbalanced data is developed. This method employs the number of retained samples in the minority class after
safe sample screening to select the regularization parameter of the SVM classifier.
3. Experiments are conducted to verify the effectiveness of the developed resampling algorithm and the method of selecting the best regularization parameter.
The rest of this paper is organized as follows. Section 2 introduces the related work of method proposed in this paper with an emphasis on SVM and safe sample screening. Section 3 introduce the
developed resampling algorithm in detail. Section 4 presents the experimental datasets and gives the experimental results and analysis. Section 5 draws our conclusion and future work outlook.
2. Related Work
2.1 SVM
Considering a training dataset [TeX:] $$\boldsymbol{D}=\left\{\left(\boldsymbol{x}_{i}, y_{i}\right) \mid \boldsymbol{x}_{i} \in \boldsymbol{R}^{d}, y_{i} \in\{-1,+1\}, i=1,2, \ldots, n\right\}$$
with d features and n samples, the classification decision function learned by SVM can be expressed as
[TeX:] $$f(\boldsymbol{x})=\boldsymbol{x}^{T} \boldsymbol{w},$$
where [TeX:] $$\boldsymbol{w}=\left(w_{1}, w_{2}, \ldots, w_{d}, w_{d+1}\right)$$ is composed of the weight vectors [TeX:] $$\left(w_{1}, w_{2}, \ldots, w_{d}\right)$$ for the learned decision
hyperplane and model bias [TeX:] $$w_{d+1}.$$ To obtain an SVM classifier with fault tolerance capability, a soft margin SVM can be built by solving the following optimization problem.
[TeX:] $$\min _{w} P_{\lambda}(\boldsymbol{w})=\frac{\lambda}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{n} \ell\left(y_{i} f\left(\boldsymbol{x}_{i}\right)\right).$$
where [TeX:] $$\ell(\bullet)$$ is a loss function, and [TeX:] $$\lambda$$ is a regularization parameter for controlling the trade-off between the regularization term and the loss term. If a hinge
loss function is adopted and a slack variable [TeX:] $$\xi_{i}$$ is introduced, (2) can be rewritten as
[TeX:] $$\begin{array}{c} \min _{w} P_{\lambda}(\boldsymbol{w})=\frac{\lambda}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{n} \xi_{i} \\ \text { s.t. } y_{i} f\left(\boldsymbol{x}_{i}\right) \geq 1-\xi_{i},
\quad \xi_{i} \geq 0, i=1,2, \ldots, n. \end{array}$$
The dual problem in (3) can be written as
[TeX:] $$\max _{\alpha} Q_{\lambda}(\boldsymbol{\alpha})=\sum_{i=1}^{n} \alpha_{i}-\frac{1}{2 \lambda} \sum_{i, j=1}^{n} y_{i} y_{j} \alpha_{i} \alpha_{j}\left\langle\boldsymbol{x}_{i}, \boldsymbol
{x}_{j}\right\rangle \text { s.t. } 0 \leq \alpha_{i} \leq 1, i=1, \ldots, n.$$
If an n-dimensional vector of Lagrange multipliers [TeX:] $$\boldsymbol{\alpha}^{*}=\left(\alpha_{1}^{*}, \ldots, \alpha_{i}^{*}, \ldots, \alpha_{n}^{*}\right)$$ is the solution of (4), the
classification decision function f(x) in (1) can be rewritten as
[TeX:] $$f(\boldsymbol{x})=\frac{1}{\lambda} \sum_{i=1}^{n} \alpha_{i}^{*} y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{x},$$
where [TeX:] $$\alpha_{i}^{*}$$ is the Lagrange multiplier corresponding to [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{D}.$$ Based on the optimality conditions of (3) or (4), the
samples in D can be categorized into three types: safe sample [TeX:] $$(S S)$$ , boundary sample [TeX:] $$(B S),$$ and noise sample [TeX:] $$(N S).$$
[TeX:] $$\begin{array}{c} \left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{S S}: y_{i} f\left(\boldsymbol{x}_{i}\right)>1 \Rightarrow \alpha_{i}^{*}=0, \\ \left(\boldsymbol{x}_{i}, y_{i}\right)
\in \boldsymbol{B} \boldsymbol{S}: y_{i} f\left(\boldsymbol{x}_{i}\right)=1 \Rightarrow \alpha_{i}^{*} \in(0,1), \\ \left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{N} \boldsymbol{S}: \quad y_
{i} f\left(\boldsymbol{x}_{i}\right)<1 \Rightarrow \alpha_{i}^{*}=1, \end{array}$$
The safe sample, located outside the classification margin, is far from the classification hyperplane. For any sample [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{S S}, \alpha_{i}^
{*} y_{i} \boldsymbol{x}_{i}^{T}=\mathbf{0} \text { holds as } \alpha_{i}^{*}=0.$$ Therefore, the safe sample has no influence on the determination of the decision function f(x) in (5). Even if these
samples are removed from the training data set, the classification hyperplane would not be changed.
The boundary sample lies on the boundary of the margin, and is near to the classification hyperplane. As [TeX:] $$\alpha_{i}^{*}$$ corresponding to [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right) \in
\boldsymbol{B} \boldsymbol{S}$$ is a nonzero value, [TeX:] $$\alpha_{i} y_{i} \boldsymbol{x}_{i}^{T} \neq \mathbf{0}$$ always holds, except for [TeX:] $$\boldsymbol{x}_{i}=\mathbf{0}.$$ This means
that the boundary sample is involved in the calculation of the decision function [TeX:] $$f(\boldsymbol{x}),$$ thereby affecting the choice of SVM classification hyperplane.
For the noise sample [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right), \alpha_{i} y_{i} \boldsymbol{x}_{i}^{T} \neq \mathbf{0}$$ holds in most cases, except for [TeX:] $$x_{i}=0.$$ Thus, the noise
sample will also affect the decision function [TeX:] $$f(\boldsymbol{x})$$ in (5). The location of noise sample is related to slack variable [TeX:] $$\xi_{i} . \text { If } \xi_{i}>1,$$ the noise
sample [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right)$$ is located between the classification hyperplane and the margin boundary close to the true class. To simplify the problem, all the samples
with [TeX:] $$\alpha_{i}=1$$ are known as noise samples.
By analyzing the three types of samples, it was found that the boundary sample should be given more attention for learning the SVM classifier. For imbalanced datasets, the imbalanced ratio (the most
common class imbalance metric) is the proportion of the samples between various classes. However, the classification performance of SVM is more dependent on the boundary sample, rather than on all
the samples. Therefore, to learn an SVM classifier on an imbalanced dataset, we should focus on the proportion of boundary samples between different classes, rather than the proportion of all the
samples between different classes.
In addition, [TeX:] $$\lambda$$ in (2) is generally understood as a trade-off parameter for balancing the generalization and fitting performances of SVM. For imbalanced data, the value of [TeX:] $$\
lambda$$ can influence the position of the classification hyperplane. In general, when the [TeX:] $$\lambda$$ value is smaller, the hyperplane moves more toward the majority class; thus, more
minority samples can be correctly classified [24]. For example, as shown in Fig. 1, compared with the decision hyperplane (solid line) learned by SVM with [TeX:] $$\lambda=1,$$ the decision
hyperplane (dashed lines) learned by SVM with [TeX:] $$\lambda=0.0001$$ moves toward the majority samples (red dots), and three more samples (green stars) of the minority class are correctly
predicted. Therefore, to improve the recognition rate of the minority samples, it is necessary to set an appropriate value of [TeX:] $$\lambda.$$ However, if the value of [TeX:] $$\lambda$$ is
selected through enumeration method for learning the best SVM model, it would be time consuming.
Comparison of decision hyperplanes constructed using [TeX:] $$\lambda=1 \text { and } \lambda=0.0001(\star: \text { minority } \text { samples, } \bullet \text { : majority samples). }$$
2.2 Safe Sample Screening
Safe sample screening [27,31] is based on the SVM and regularization path algorithms. Given a dataset, safe sample screening can rapidly screen out parts of the safe samples via safe sample screening
rules, and generate a series of screened subsets using a regularization path algorithm.
2.2.1 Safe sample screening rules
Safe sample screening rules can be used to identify non-informative samples. To construct safe sample screening rules, we adopted the objective function of SVM from (2) for safe sample screening.
However, owing to the non-differentiability of the hinge loss function at the inflection point, the smooth hinge loss function from (7) was adopted as the loss function of safe sample screening, to
ensure that it is differentiable everywhere within the range of values.
[TeX:] $$\ell\left(y_{i} f\left(\boldsymbol{x}_{i}\right)\right)=\left\{\begin{array}{ll} 0 & y_{i} f\left(\boldsymbol{x}_{i}\right)>1 \\ \frac{1}{2 \gamma}\left[1-y_{i} f\left(\boldsymbol{x}_{i}\
right)\right]^{2}, & 1-\gamma \leq y_{i} f\left(\boldsymbol{x}_{i}\right) \leq 1, \\ 1-y_{i} f\left(\boldsymbol{x}_{i}\right)-\frac{\gamma}{2}, & y_{i} f\left(\boldsymbol{x}_{i}\right)<1-\gamma \end
where [TeX:] $$\gamma>0$$ is a tuning parameter. The dual problem of safe sample screening can be written as
[TeX:] $$\max _{\alpha} D_{\lambda}(\boldsymbol{\alpha})=-\frac{\lambda}{2} \sum_{j=1}^{d}\left(\sum_{i=1}^{n} \frac{1}{\lambda n} x_{i j} \alpha_{i} y_{i}\right)^{2}-\frac{1}{n} \sum_{i=1}^{n}\left
(\frac{\gamma}{2} \alpha_{i}^{2}-y_{i} \alpha_{i}\right).$$
Let us assume that [TeX:] $$\boldsymbol{w}^{*}=\left(w_{1}^{*}, \ldots, w_{d}^{*}, w_{d+1}^{*}\right) \text { and } \boldsymbol{\alpha}^{*}=\left(\alpha_{1}^{*}, \ldots, \alpha_{i}^{*}, \ldots, \
alpha_{n}^{*}\right)$$ represent the optimal solution of the primal and dual problems for safe sample screening, respectively. In the case of smooth hinge loss, according to the Karush-Kuhn-Tucker
(KKT) optimality conditions, we can obtain
[TeX:] $$y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}=\left\{\begin{array}{ll} {[1, \infty)} & \alpha_{i}^{*}=0 \\ (1-\gamma, 1), & \alpha_{i}^{*} \in(0,1). \\ (-\infty, 1-\gamma], & \alpha_{i}^
{*}=1 \end{array}\right.$$
Similar to SVM, the samples with [TeX:] $$\alpha_{i}^{*}=0, \alpha_{i}^{*} \in(0,1), \text { and } \alpha_{i}^{*}=1$$ are called safe samples, boundary samples, and noise samples, respectively. Safe
sample screening is aimed at removing parts of the safe and noise samples and retaining all of the boundary samples. To identify safe and noise samples, a solution space [TeX:] $$\Theta_{w^{*}}$$
containing the optimal solution [TeX:] $$\boldsymbol{W}^{*}$$ is first constructed by employing the feasible solutions of the primal and dual problems. Specifically, for any given feasible solutions
[TeX:] $$\widehat{\boldsymbol{w}} \in \operatorname{dom} P_{\lambda}$$ and [TeX:] $$\widehat{\boldsymbol{\alpha}} \in \operatorname{dom} D_{\lambda},$$
[TeX:] $$\boldsymbol{w}^{*} \in \Theta_{w^{*}}=\left\{\boldsymbol{w} \mid\|\widehat{\boldsymbol{w}}-\boldsymbol{w}\| \leq \sqrt{2\left[P_{\lambda}(\widehat{\boldsymbol{w}})-D_{\lambda}(\widehat{\
boldsymbol{\alpha}})\right] / \lambda}\right\}.$$
A pair of lower and upper bounds of [TeX:] $$y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}$$ is given as
[TeX:] $$\operatorname{LB}\left(y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}\right)=y_{i} \boldsymbol{x}_{i}^{T} \widehat{\boldsymbol{w}}-\left\|y_{i} \boldsymbol{x}_{i}\right\| \sqrt{2\left[P_{\
lambda}(\widehat{\boldsymbol{w}})-D_{\lambda}(\widehat{\boldsymbol{\alpha}})\right] / \lambda},$$
[TeX:] $$\operatorname{UB}\left(y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}\right)=y_{i} \boldsymbol{x}_{i}^{T} \widehat{\boldsymbol{w}}+\left\|y_{i} \boldsymbol{x}_{i}\right\| \sqrt{2\left[P_{\
lambda}(\widehat{\boldsymbol{w}})-D_{\lambda}(\widehat{\boldsymbol{\alpha}})\right] / \lambda}.$$
According to (9), (11), and (12), safe sample screening rules can be represented as follows.
Screening rule 1: If [TeX:] $$\operatorname{LB}\left(y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}\right) \geq 1, \text { then }\left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{S} \boldsymbol
{S} \text { and }\left(\boldsymbol{x}_{i}, y_{i}\right)$$ can be discarded.
Screening rule 2: If [TeX:] $$\text { UB }\left(y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}\right) \leq 1-\gamma, \text { then }\left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{N} \
boldsymbol{S} \text { and }\left(\boldsymbol{x}_{i}, y_{i}\right)$$ can be discarded.
Using the above mentioned two rules, the safe and noise samples, which were identified, will be discarded, and the remaining samples will be retained. The advantage of this method is that it can
reduce the sample size by employing the relationship between the optimal and feasible solutions, without directly solving the optimization problems.
2.2.2 Regularization path solving strategy
To set a [TeX:] $$\lambda$$ value in (2), the authors [28] proposed an SVM regularization path algorithm that can quickly solve all feasible [TeX:] $$\lambda$$ values and the corresponding SVM on a
given sample set D. The initial [TeX:] $$\lambda$$ value can be obtained from the original dataset. When the boundary samples of the interval change, each [TeX:] $$\lambda_{m}$$ is solved from its
previous [TeX:] $$\lambda_{m-1},$$ and the iteration is continued until there are no samples in the interval or [TeX:] $$\lambda$$ is reduced to 0. Owing to the piecewise linearity of the SVM
regularization path on regularization parameters, a complete regularization path can be obtained by solving the inflection points of the regularization parameters.
In a given dataset, it is not necessary to solve all the values of [TeX:] $$\lambda.$$ In [27], the regularization parameter values in a given range were solved. Safe sample screening only constructs
safe sample screening rules corresponding to the regularization parameter values in this range. As the convergence of SVM tends to be faster for larger regularization parameters, the regularization
path is computed from larger [TeX:] $$\lambda$$ to smaller [TeX:] $$\lambda$$ using the warm-start method [27]. In the solving process, the previous optimal solution at [TeX:] $$\lambda_{m-1}$$ is
used as the initial starting point of the next optimization problem for [TeX:] $$\lambda_{m}.$$ The upper bound [TeX:] $$\lambda_{\max }$$ and lower bound [TeX:] $$\lambda_{\min }$$ of the range are
as follows:
[TeX:] $$\lambda_{\max }=\max _{1 \leq j \leq d}\left|\frac{1}{n} \sum_{i=1}^{n} x_{i j} y_{i}\right| ; \quad \lambda_{\min }=10^{-4} \lambda_{\max }.$$
Given that [TeX:] $$\lambda_{m} \in\left[\lambda_{\min }, \lambda_{\max }\right],$$ the values of the upper and lower bounds for [TeX:] $$y_{i} \boldsymbol{x}_{i}^{T} \boldsymbol{w}^{*}$$ of each
sample can be determined by (11) and (12), respectively. The samples that meet the screening conditions of screening rules 1 and 2 will be removed, and the retained samples are the result of the safe
sample screening described in [27].
3. Proposed Algorithm
We developed the Re-SSS algorithm, comprising Re-SSS-IS and Re-SSS-WSMOTE, to change the distribution of imbalanced data based on safe sample screening. The former is used to select a suitable
regularization parameter for imbalanced data and obtain informative samples of the majority class, and the latter is used to generate informative minority samples. Notably, both Re-SSS-IS and
Re-SSSWSMOTE can be performed as a part of the Re-SSS, or separately.
3.1 Re-SSS-IS
Safe sample screening can generate a series of screened datasets, each of which contains fewer samples than the original dataset. To find the informative majority samples from an imbalanced dataset
using the safe sample screening, two problems should be solved: the first is setting up the range of the regularization parameter values, and the second is selecting the suitable regularization
parameter and its corresponding screened dataset.
To solve the first problem, we first analyzed the range of the [TeX:] $$\lambda$$ values (see (13)) in [27] and adjusted it for imbalanced data. To simplify the discussion, we assumed that [TeX:] $$\
left|\frac{1}{n} \sum_{i=1}^{n} x_{i j} y_{i}\right|$$ in (13) reaches the maximum at the j-th attribute and that [TeX:] $$x_{i j}=1(i=1, \ldots, n) ; \text { thus, } \lambda_{\max }$$ in (13) can be
rewritten as
[TeX:] $$\lambda_{\max }=\frac{n_{-}-n_{+}}{n},$$
where [TeX:] $$n_{+} \text {and } n_{-}$$ represent the number of minority and majority samples in the original dataset, respectively. For imbalanced datasets, [TeX:] $$\left(n_{-}-n_{+}\right)$$ is
usually very large. However, [29] found that, if [TeX:] $$\lambda$$ is very large, there will be little or no boundary samples. Thus, the safe samples in the minority class are retained, which is not
beneficial for identifying samples in the minority class. To avoid this case, we added a hyperparameter [TeX:] $$c<1$$ into (14) (the value of c is given in Section 4.3, which was determined in our
experiments), and obtained the maximum value of the regularization parameter [TeX:] $$\lambda$$ as
[TeX:] $$\lambda_{\max }=c \frac{n_{-}-n_{+}}{n},$$
A smaller [TeX:] $$\lambda_{\max }$$ retains more boundary and safe samples in the minority class. In addition, [TeX:] $$\lambda_{\min }$$ was assigned in the same way as (13); thus, the range of
regularization parameter values was [TeX:] $$\left[\lambda_{\min }, \lambda_{\max }\right].$$
For the second problem, our solution was to find a classification hyperplane with maximum margin, which could correctly predict as many informative minority samples as possible. For a given
regularization parameter [TeX:] $$\lambda$$, the sets of safe, boundary, and noise samples in the minority class were denoted as [TeX:] $$\boldsymbol{S S}_{\lambda}^{+}, \boldsymbol{B} \boldsymbol{S}
_{\lambda}^{+}, \text {and } \boldsymbol{N} \boldsymbol{S}_{\lambda}^{+},$$ respectively. As the noise samples may have a negative effect on the classifier, we expected that only the safe and
boundary samples were correctly identified. Hence, we wanted to find a suitable regularization parameter [TeX:] $$\lambda^{*},$$ which has the maximum number of safe and boundary samples. This
solution can be expressed as
[TeX:] $$\lambda^{*}=\underset{\lambda}{\operatorname{argmax}}\left|\boldsymbol{B} \boldsymbol{S}_{\lambda}^{+} \cup \boldsymbol{S} \boldsymbol{S}_{\lambda}^{+}\right|,,$$
First, utilizing the regularization path algorithm, the Re-SSS-IS algorithm quickly obtained a series of feasible [TeX:] $$\lambda$$ values, with their corresponding screened datasets. Then, the
screened dataset with the largest [TeX:] $$\left|B S_{\lambda}^{+} \cup S S_{\lambda}^{+}\right|$$ was selected, and the corresponding [TeX:] $$\lambda$$ was set as the suitable regularization
parameter [TeX:] $$\lambda^{*}.$$ Lastly, the majority samples in [TeX:] $$\boldsymbol{B} \boldsymbol{S}_{\lambda^{*}}$$ corresponding to [TeX:] $$\lambda^{*}$$ were taken as the set of informative
majority samples [TeX:] $$\boldsymbol{B} \boldsymbol{S}_{\lambda^{*}}^{-}.$$
Modified Safe Sample Screening Algorithm
3.2 Re-SSS-WSMOTE
To generate informative minority samples, this study developed a modified SMOTE algorithm, Re- SSS-WSMOTE, for imbalanced data. SMOTE is a popular oversampling method that generates synthetic samples
from existing minority samples. However, not all the minority samples are useful for learning an SVM, and the samples far from the decision hyperplane are more likely to have no effect on learning
the classifier. In general, if both the sample and its selected similar sample are boundary samples, the sample generated by combining these two samples is more likely to be a boundary sample;
otherwise, the generated sample will be more likely to be a safe sample. Thus, the availability of a synthetic sample for SVM is related to the availability of the two selected original samples.
Next, we need to consider how to determine the availability of each sample. In Section 2.1, we compared two decision hyperplanes learned by the SVM with [TeX:] $$\lambda=0.0001 \text { and } \lambda=
1,$$ and found that the decision hyperplanes learned by the SVM with different [TeX:] $$\lambda$$ values may be different. In fact, support vectors of SVM with different [TeX:] $$\lambda$$ values may
not be exactly similar, as shown in Fig. 2. We can see from Fig. 2 that points 1, 2, 3, and 4 are the support vectors of the SVM with [TeX:] $$\lambda=1,$$ and points 2, 3, and 4 are the support
vectors of the SVM with [TeX:] $$\lambda=0.0001.$$ Points 2, 3, and 4 are the common support vectors of the SVM for the two different [TeX:] $$\lambda$$ values, which means that these points were
more likely located closest to the classification hyperplane.
Based on the above analysis, we used the weight value to represent the availability of each sample. The weight value of each sample was calculated based on the screened datasets corresponding to the
different [TeX:] $$\lambda$$ values. [TeX:] $$\boldsymbol{B S}_{\lambda_{1}}, \boldsymbol{B} \boldsymbol{S}_{\lambda_{2}}, \ldots, \boldsymbol{B} \boldsymbol{S}_{\lambda_{T}}$$ are the boundary
sample sets with different [TeX:] $$\lambda \text { values; } \boldsymbol{B S}=\boldsymbol{B} \boldsymbol{S}_{\lambda_{1}} \cup \boldsymbol{B} \boldsymbol{S}_{\lambda_{2}} \cup \ldots \cup \
boldsymbol{B} \boldsymbol{S}_{\lambda_{T}}$$ denotes the set of boundary samples for T regularization parameters; and [TeX:] $$\left\{\left(\boldsymbol{x}_{i}, y_{i}\right) \mid\left(\boldsymbol{x}_
{i}, y_{i}\right) \in \boldsymbol{B} \boldsymbol{S}, y_{i}={ }^{\prime}+1^{\prime}\right\}$$ is the set of the minority boundary samples. As some samples in the original minority class might not
exist in [TeX:] $$\boldsymbol{B S}^{+},$$ we adopted a Laplace correction to adjust the weight values to prevent these samples from being selected. The weight value of each minority sample was set as
Comparison of support vectors with [TeX:] $$\lambda=1 \text { and } \lambda=0.0001(\star:$$ minority samples, [TeX:] $$\bullet$$ : majority samples, [TeX:] $$+:$$ support vectors of SVM with [TeX:]
$$\lambda=0.0001, \quad \mathrm{x}:$$ support vectors of SVM with [TeX:] $$\lambda=1,$$ : common support vectors of SVM with [TeX:] $$\lambda=0.0001 \text { and } \lambda=1 \text { ). }$$
[TeX:] $$W e\left(x_{i}, y_{i}\right)=\left\{\begin{array}{l} \frac{k_{i}+1}{\sum_{i=1}^{\left|B S^{+}\right|} k_{i}+\left|B S^{+}\right|}, \text {if }\left(x_{i}, y_{i}\right) \in B S^{+} \\ \frac
{1}{\sum_{i=1}^{\left|B S^{+}\right|} k_{i}+\left|B S^{+}\right|}, \text {otherwise }, \end{array}\right.$$
where [TeX:] $$k_{i}$$ denotes the number of boundary sample sets containing [TeX:] $$\left(\boldsymbol{x}_{i}, y_{i}\right),$$ namely
[TeX:] $$k_{i}=\sum_{j=1}^{T} I_{B S_{\lambda_{j}}}\left(\boldsymbol{x}_{i}, y_{i}\right),$$
[TeX:] $$I_{B S_{\lambda_{j}}}\left(\boldsymbol{x}_{i}, y_{i}\right)=\left\{\begin{array}{l} 1, \mathrm{if}\left(\boldsymbol{x}_{i}, y_{i}\right) \in \boldsymbol{B} \boldsymbol{S}_{\lambda_{j}}, \\
0, \mathrm{if}\left(\boldsymbol{x}_{i}, y_{i}\right) \notin \boldsymbol{B} \boldsymbol{S}_{\lambda_{j}}, \end{array}\right.$$
In summary, the Re-SSS-WSMOTE first obtained a series of screened datasets via safe sample screening, and employed the boundary sample sets [TeX:] $$\boldsymbol{B} \boldsymbol{S}_{\lambda_{1}}, \
boldsymbol{B} \boldsymbol{S}_{\lambda_{2}}, \ldots, \boldsymbol{B} \boldsymbol{S}_{\lambda_{\mathrm{T}}}$$ to calculate the weight of each sample according to (17). Then, a minority sample was
randomly selected according to the weight of the sample, and its similar sample was selected from its k-nearest neighbors according to the weight of the sample. Finally, the linear interpolation
method was applied to the two selected samples to generate a synthetic sample. The Re-SSS-WSMOTE algorithm is shown as follows.
Note that, if only Re-SSS-WSMOTE is used, we will use the original majority sample set [TeX:] $$\mathrm{D}^{-},$$ instead of the informative majority sample set [TeX:] $$\boldsymbol{B} \boldsymbol{S}
_{\lambda^{*}}^{-},$$ as the majority sample set in Re-SSS-WSMOTE.
4. Experiments and Analysis
4.1 Datasets
To investigate the effectiveness of the Re-SSS algorithm, we chose 35 datasets from the UCI Repository (http://archive.ics.uci.edu/ml/index.php), LIBSVM datasets (https://www.csie.ntu.edu.tw/ ~cjlin/
libsvmtools/datasets/), and KEEL dataset repository (http://www.keel.es/). Except for the original two-class datasets available, we also chose a part of the multiclass datasets and transformed them
into two-class imbalanced datasets. For example, for movement-1 with multiple classes, the samples with “1”, “2”, and “3” class labels in the original dataset were combined as the minority class, and
the samples with the other class labels were combined as the majority class. In addition, we applied min-max normalization to each dataset for learning the SVM classifiers. Table 1 shows a detailed
description of each dataset. The third and fourth columns show the class constituents of the minority and majority classes in each dataset, respectively. The number of minority samples, number of
majority samples, number of features, and imbalance ratio of each dataset are listed from the fifth to eighth columns, respectively. The last column presents the source of each dataset.
Description of the chosen imbalanced datasets
4.2 Performance Evaluation Metrics
For imbalanced classification problem, the suitable metrics should not be dominated by the majority samples. In [32], the impact of class imbalance on classification performance metrics has been
systematically studied. The results have shown that the metrics with no bias due to imbalance, recall, specificity, geometric mean (G-Mean), and area under curve (AUC), are the best performance
metrics. As the specificity takes into account only the results on the majority class, we did not use it as a performance metric. The F-score is also commonly used for imbalanced data. Thus, we
compared the different methods by using the four metrics: recall, F-score, G-Mean, and AUC.
The recall measures the ratio of minority samples correctly classified as the minority class to all the minority samples. The range of the recall values is [0,1]. The higher the recall, the higher
the recognition rate of the minority samples is.
The F-score is the harmonic mean of the precision and recall, namely F-score=(2*recall*precision) /(recall+precision), where the precision measures the ratio of minority samples correctly classified
to all the samples classified as the minority class. The F-score works well for the recognition rate of the minority samples.
The G-Mean is the geometric mean of the recall and specificity with a range of [0,1]. The specificity is the actual proportion of majority samples that are correctly identified. The closer the G-Mean
value is to 1, the better the classification effect is
[TeX:] $$G \text { -Mean }=\sqrt{\text { recall } * \text { specificity }}$$
The AUC is the area under the Receiver Operating Characteristic (ROC) curve. The range of the AUC values is [0,1], and the AUC value less than 0.5 indicates that the result is not as good as random
prediction. The AUC value can well reflect the classification performance of the model.
4.3 Experimental Results and Analysis
In this section, two experiments were performed. The first experiment involved the regularization parameters of the SVM classifiers on the original imbalanced datasets, and the second presented the
results of SVM classifiers on the datasets balanced using different methods for changing the distribution of data. The parameters used in the two experiments are as follows: [TeX:] $$T=100, c=10^
{-1.6}, \Delta \lambda=10^{0.04-0.04 m}, \text { and } k=5.$$
4.3.1 Experiments on different regularization parameters of the SVM classifiers on the original imbalanced datasets
The main purpose of this experiment was to examine the superiority of the regularization parameter [TeX:] $$\lambda^{*}$$ obtained using the Re-SSS-IS algorithm. First, we applied the Re-SSS-IS
algorithm on each original imbalanced dataset, and obtained a suitable regularization parameter [TeX:] $$\lambda^{*}$$ for each dataset. Then, the SVM classifier with [TeX:] $$\lambda^{*}$$ was built
directly on the original imbalanced dataset. For comparison, the SVM classifiers with the other 11 regularization parameters were also built.
The experiments were performed using 5-fold cross-validation. In each fold, the original dataset was split into training and test data. The SVM classifier with the corresponding regularization
parameter was built on the training data; the AUC, F-score, G-Mean, and recall of the SVM classifier on the test data was then recorded and averaged across all the splits.
The experimental results are presented in Table 2. The first row lists all the regularization parameters used for the SVM classifiers in the experiments. The second to twelfth columns present the
experimental results of the SVM classifiers with different regularization parameters. Each row shows the average performance metrics of the SVM classifiers with the corresponding regularization
parameter on the 35 datasets. For example, the average AUC of the SVM classifier with [TeX:] $$\lambda^{*}$$ on the 35 datasets was 0.846. As the value of [TeX:] $$\lambda^{*}$$ obtained using the
Re-SSS algorithm was different for each dataset, we did not list the value of [TeX:] $$\lambda^{*}$$ in Table 2. From Table 2, it can be seen that the performance metrics of the SVM classifiers were
related to the values of [TeX:] $$\lambda \text { . When } \lambda$$ was larger, the SVM classifiers had poorer average performance; when [TeX:] $$\lambda$$ was smaller, the SVM classifiers obtained
better average performance. This result is consistent with the discussion in Section 2.1. However, the values of[TeX:] $$\lambda$$ corresponding to the maximum AUC, Fscore, G-Mean, and recall were
not the smallest, in other words, it is not true that the smaller the value of [TeX:] $$\lambda,$$ the better the performance of the SVM classifier. Moreover, each dataset obtained its maximum
metrics under different [TeX:] $$\lambda$$ values; hence, an appropriate [TeX:] $$\lambda$$ value for a given dataset needs to be selected. Furthermore, the SVM classifier with [TeX:] $$\lambda^{*}$$
obtained using the Re-SSS-IS was close to the maximum average metrics. This result shows that the Re-SSS-IS algorithm can select the appropriate value for [TeX:] $$\lambda$$.
Comparison of SVM experimental results with different [TeX:] $$\lambda$$ on the original datasets
4.3.2 Experiments on the SVM classifiers with datasets balanced using different data balancing methods
To verify the effectiveness of the developed Re-SSS algorithm, we compared the Re-SSS algorithm with the other methods for changing the distribution of data. The methods proposed in [12,26] did not
explicitly mention the adjustment of regularization parameters in SVMs. In [24,25], the SVMs were used for the preprocessing of samples by adjusting the regularization parameters in a similar manner.
Aside from preprocessing of samples, feature selection was also used in [25]. However, there was no feature selection involved in our study. Thus, only the Pre-SVM algorithm in [24] was selected as
one of the baseline methods in our study. The five baseline methods are described as follows:
Undersampling: The original dataset is randomly under-sampled; that is, the samples are randomly extracted from the majority class, and the number of samples extracted is equal to the number of
minority samples.
Oversampling: The original dataset is randomly oversampled, that is, randomly extracting samples from the minority class, and adding the samples to the original dataset to balance the dataset.
SMOTE: Based on the existing minority samples, an interpolation method is used to generate new minority samples, which are added to the original dataset to balance the dataset.
BorderLine-SMOTE: The improved version of SMOTE uses an interpolation method to generate new small minority samples according to the existing minority class boundary samples, which are added to the
original dataset to balance the dataset.
Pre-SVM [24]: First, this algorithm builds the SVM models on imbalanced data, and the SVM model with the best prediction accuracy is selected and used for prediction purposes. Then, the actual target
values of the training samples are replaced by the prediction of the trained SVM.
AUC comparison of the experimental results of the Re-SSS
First, four baseline methods were used to rebalance the datasets, and then, the SVM classifiers with the frequently-used regularization parameters (namely 0.1, 1) were performed on the balanced
datasets using the four baseline methods. The Pre-SVM and Re-SSS adaptively chose the regularization parameters.
The experimental performance evaluation metrics used were similar to those of the first experiment. The experimental AUC results are shown in Table 3, where the first row presents the method used. It
can be seen that the Re-SSS method performed optimally on 18 datasets, followed by Borderline-SMOTE [TeX:] $$(\lambda=0.1 \text { ) }$$and Pre-SVM, which performed optimally on 5 datasets. From Table
3, it is clear that the Re-SSS surpassed the other methods in AUC. In addition, the average value for each oversampling method (Oversampling, SMOTE, and Borderline-SMOTE) was better than that for the
As the experimental results of the other three metrics were similar to those of the AUC, we have not included them here. The comprehensive experimental results are presented in Table 4.
Comparison of the experimental results of Re-SSS
In Table 4, the first row presents the method used, and the row named “number” presents the number of datasets for which the SVM classifier with its corresponding regularization parameter obtained
optimal performance for a certain evaluation metric. Note that the sum of ten numbers in each row may be greater than 35 as the SVM classifiers with different [TeX:] $$\lambda$$ values may have the
same results. The row named “average” presents the average obtained by the SVM classifiers with the corresponding regularization parameters on 35 datasets for a certain evaluation metric. It can be
seen from Table 4 that, in most cases, the result of [TeX:] $$\lambda=0.1$$ is better than that of [TeX:] $$\lambda=1$$ when using the same method. With the decrease in the value of [TeX:] $$\
lambda,$$ the experimental performance of the dataset was more favorable for minority classes. For the four different metrics, the Re-SSS was superior to the other methods in terms of the number of
datasets, in which it showed optimal performance and had the highest average value. This verifies the feasibility of the Re-SSS algorithm developed in this study for handling the imbalanced
classification problem. In addition, it can be seen from Table 4 that the oversampling method performed slightly better than the undersampling method.
5. Conclusion and Future Work
We developed a resampling Re-SSS algorithm, made up of Re-SSS-IS and Re-SSS-WSMOTE based on safe sample screening, to exploit the informative samples learned by the SVM classifier on an imbalanced
data set. The Re-SSS-IS algorithm can select suitable regularization parameters and obtain informative majority samples; the Re-SSS-WSMOTE algorithm is used to generate informative minority samples
for the SVM classifier. Then, two experiments were conducted to verify the effectiveness of the algorithm. Compared with the other methods, the proposed resampling method showed better performance.
The proposed Re-SSS algorithm can not only discard parts of non-informative samples, but also add useful informative ones. Our future work will focus on developing an effective method for selecting
hyperparameter c in the Re-SSS algorithm and exploring how to extend the Re-SSS algorithm to address multiclass imbalanced problems.
This work is supported by the National Natural Science Foundation of China (No. 61801279), the Key Research and Development Project of Shanxi Province (No. 201903D121160), and the Natural Science
Foundation of Shanxi Province (No. 201801D121115 and 201901D111318). | {"url":"https://xml.jips-k.org/pub-reader/view?doi=10.3745/JIPS.01.0065","timestamp":"2024-11-12T06:38:35Z","content_type":"text/html","content_length":"169151","record_id":"<urn:uuid:b7383670-a1d8-4bc4-b53e-9ecebd4e126d>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00398.warc.gz"} |
double SPcorFmse (const float h[], double Ed, const float rxx[], const float r[], int N)
Calculate the mean-square filtering error
This function calculates the mean-square error for a linear filter. Consider a filter with N coefficients, with coefficient h(i) corresponding to lag Nd+i. The filter output is
y(k) = SUM h(i) x(k-i-Nd) ,
where x(i) is the input signal. The filter error is
e(k) = d(k) - y(k) ,
where d(k) is the desired signal. The mean-square filtering error is E[e(k)^2] or in vector-matrix notation
ferr = Ed - 2 h'r + h' R h ,
The mean-square value E0, matrix R and vector p are defined as follows
Ed = E[d(k)^2]
R(i,j) = E[x(k-i-Nd) x(k-j-Nd], for 0 <= i,j < N,
r(i) = E[d(k) x(k-i-Nd)], for 0 <= i < N.
For this routine, the matrix R must be symmetric and Toeplitz, viz.
R(i,j) = rxx(|i-j|).
Linear prediction can be cast into the above form, if we let Nd=1. Also for linear prediction, usually d(k)=x(k), giving r(i)=rxx(i).
<- double SPcorFmse
Resultant mean-square error
-> const float h[]
N element vector of filter coefficients. Coefficient h[i] is the filter coefficient corresponding to lag Nd+i.
-> double Ed
Signal energy for the desired signal. This value is used only for the computation of the mean-square error.
-> const float rxx[]
N element vector of autocorrelation values. Element rxx[i] is the autocorrelation at lag i.
-> const float r[]
N element vector of cross-correlation values. Element r[i] is the cross-correlation at lag Nd+i.
-> int N
Number of elements in each of the vectors rxx, h and r.
Author / revision
P. Kabal / Revision 1.12 2003/05/09
See Also
Main Index libtsp | {"url":"https://mmsp.ece.mcgill.ca/Documents/Software/Packages/libtsp/SP/SPcorFmse.html","timestamp":"2024-11-14T20:04:59Z","content_type":"text/html","content_length":"2501","record_id":"<urn:uuid:1bcf8f27-6a6a-40c3-bc09-213f0b96db51>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00478.warc.gz"} |
Solving Complex Problems Using Trigonometric Functions
Author: admintanbourit
Trigonometric functions, such as sine, cosine, and tangent, are often associated with triangles and commonly taught in high school math classes. However, these functions have a much broader
application when it comes to solving complex problems in various fields. From engineering to finance, trigonometric functions play a crucial role in finding solutions to challenging problems. Let’s
explore some of the ways in which these functions can be used to solve complex problems.
One of the most common uses of trigonometric functions is in navigation. By using the basic definitions of sine and cosine, sailors and pilots are able to determine their position and direction. This
is particularly useful in cases where visual navigation is limited, such as during foggy or cloudy weather. Additionally, the law of sines and the law of cosines allow for the calculation of
distances between two points or angles in a triangle, which is essential for navigation.
Trigonometric functions are also vital in the field of engineering. For example, engineers use the tangent function to calculate the slopes of roads, bridges, and buildings. This ensures the safety
and stability of these structures. The Pythagorean theorem, which involves the relationships between the sides of a right triangle, is also based on trigonometric functions and is used extensively in
engineering for various calculations.
The applications of trigonometric functions are not limited to physical fields; they are also widely used in financial analysis. For instance, the Black-Scholes model, used to calculate the
theoretical price of financial derivatives, relies heavily on the use of inverse trigonometric functions. These functions are also used in stock market analysis to calculate price changes over time,
aiding investors in making informed decisions.
Another fascinating application of trigonometric functions is in music. Sound waves, which are essentially vibrations, can be represented mathematically using trigonometric functions. This allows
musicians and sound engineers to determine the exact pitch, frequency, and amplitude of a sound wave. In addition, the use of Fourier transforms, which involve the manipulation of trigonometric
functions, allows for the transformation of a complex sound wave into its separate frequency components, leading to the creation of digital music and sound effects.
In the field of astronomy, trigonometry plays a crucial role in calculating distances between objects in space. The parallax formula, which involves the use of trigonometric functions, is employed to
measure the distance between Earth and nearby stars. This is achieved by measuring the apparent shift in the position of a star when viewed from different locations on Earth. Without the use of
trigonometry, it would be challenging to determine the vast distances in our universe.
Trigonometric functions are also used in weather forecasting. Meteorologists use these functions to analyze and predict cloud cover, wind speed, and direction based on data collected from weather
stations. Trigonometry allows them to create models and visualize the complex patterns of weather systems, aiding in predicting future conditions.
In conclusion, trigonometric functions have a vast array of practical applications, making them essential tools for solving complex problems in various fields. Whether it’s navigation, engineering,
finance, music, astronomy, or meteorology, these functions play a critical role in finding accurate solutions. So the next time you come across a challenging problem, remember to use trigonometry as
a powerful tool in your problem-solving arsenal. | {"url":"https://tanbourit.com/solving-complex-problems-using-trigonometric-functions/","timestamp":"2024-11-06T04:01:10Z","content_type":"text/html","content_length":"112793","record_id":"<urn:uuid:8b35137f-fdf4-45b8-a13b-506b57ce2a6f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00679.warc.gz"} |
Objectivity over Objects: Gödel’s Late Platonism and Phenomenology
Until the 50s Gödel’s Platonism is centered in the belief in the independent existence of mathematical concepts and objects. From the 60s on, Gödel writes on the contrary that the existence of
mathematical objects is not necessary, and that it suffices to believe in the objectivity of mathematics. Hauser interprets the evolution of Gödel’s thought from Platonism to objectivism as a result
of the discovery of Husserl’s phenomenology (Gödel started to read Husserl in 1959, and became soon very sympathetic with his philosophy). In my opinion, Hauser’s thesis is wrong, and is due to the
fact that Hauser follows the analytical interpretation of phenomenology given by Føllesdal and his disciples. I suggest that the origin of Gödel’s turn to objectivism could be in his conversations
with Kreisel: the oft cited Kreisel’s dictum says that “the problem is the objectivity of mathematics, not the existence of mathematical objects”. Dummett considers Kreisel’s dictum as expressing
semantic realism, i.e. realism about truth, opposed to metaphysical realism, concerning existence. Also in physics we can find a move from a metaphysical to a verifiable realism, with Einstein’s
realism of properties, opposed to the classical realism of objects: this shift is very near to Gödel’s one from Platonism to objectivism. Gödel shares Einstein’s realistic convictions about quantum
mechanics, and believes that undecidability in mathematics is due to the fact that we have only a partial knowledge of mathematical reality. What we need, in Gödel’s/Einstein’s opinion, is to expand
set theory/quantum mechanics to a complete theory, providing a complete description of the respective states of affairs. I believe that Husserl’s mature phenomenology is nearer to Dummett’s and
Bohr’s approach, underlining that mathematical and physical entities are constituted in an open process from the vagueness of the life-world, and are not to be considered as an independent existing
and totally determinate reality.
Diskussion zum Vortrag. | {"url":"https://audiothek.philo.at/en/lesson/objectivity-over-objects-goedels-late-platonism-and-phenomenology/","timestamp":"2024-11-08T20:40:43Z","content_type":"text/html","content_length":"29231","record_id":"<urn:uuid:14fae051-3509-4599-b250-a29263831d88>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00778.warc.gz"} |
AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current
Students get through AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current which are most likely to be asked in the exam.
AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current
Very Short Answer Questions
Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the ‘ secondary if the primary has 10 turns. [T.S. Mar. 16]
V[p] = 200V, V[s] = 2000V, N[p] = 10
N[s] = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) × N[p] = \(\frac{2000}{200}\) × 10
N[s] = 100.
Question 2.
What type of transformer is used in a 6V bed lamp ? [A.P. Mar. 17]
Step down transformer is used in 6V bed lamp.
Question 3.
What is the phenomenon involved in the working of a transformer ? [Mar. 16(A.P.) Mar. 14]
Transformer works on the principle of mutual induction.
Question 4.
What is transformer ratio ?
The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
Transformer ratio = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
Question 5.
Write the expression for the reactance of i) an inductor and (ii) a capacitor.
1. Inductive reactance (X[L]) = ωL
2. Capacitive reactance (X[C]) = \(\frac{1}{\omega C}\)
Question 6.
What is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor. [T.S. Mar. 15]
1. In pure resistor A.C. e.m.f and current are in phase with each other.
2. In pure inductor, current lags behind the e.m.f. by an angle of \(\frac{\pi}{2}\) (or) 90°.
3. In pure capacitor, current leads the e.m.f by an angle \(\frac{\pi}{2}\) (or) 90°.
Question 7.
Define power factor. On which factors does power factor depend ?
The ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
Power factor (cosΦ) = \(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{rms}}}\) [∵ P[rms] = V[rms] I[rms]]
Power factor depends on r.m.s voltage, r.m.s current and average power (P).
Question 8.
What is meant by wattless component of current ?
Average power (P) = V[rms](I[rms] sinΦ) cos\(\frac{\pi}{2}\)
The average power consumed in the circuit due to (I[rms] sinΦ) component of current is zero. This component of current is known as wattless current. (I[rms] sinΦ) is the wattless component of
Question 9.
When does a LCR series circuit have minimum impedance ?
In LCR series circuit, Impendence (Z) = \(\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}\)
At a particular frequency, ωL = \(\frac{1}{\omega C}\)
The impedance is minimum (Z = R)
This frequency is called resonant frequency.
Question 10.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
In LCR series circuit power factor (cosΦ) = 1
Phase difference between voltage and current is zero. (Φ = 0)
Short Answer Questions
Question 1.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Transformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
Principle : It works on the principle of mutual induction between two coils.
Working : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.
This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.
Theory: Let N[1] and N[2] be the number of turns in the primary and secondary. Let V[P] and V[S] be the emf s across the primary and secondary.
\(\frac{V_S}{V_p}=\frac{\text { Output emf }}{\text{Input emf}}=\frac{-N_2 \frac{d \phi}{d t}}{-N_1 \frac{d \phi}{d t}}=\frac{N_2}{N_1}\)
∴ \(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = Transformer ratio
Efficiency of transformer :
It is the ratio of output power to the input power.
η = \(\frac{\text { Outputpower }}{\text { Input power }}\) × 100
Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
(a) We are given P = 100 W and V = 220V The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω
(b) The peak voltage of the source is υ[m] = \(\sqrt{2}\)V = 311 V
(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.
Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
The inductive reactance,
X[L] = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A
Question 3.
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by i = \(\sqrt{2}\) sin (100t – π/4)A and υ = 40 sin (100t) V. Calculate the
resistance ?
i = \(\sqrt{2}\) sin (100t – π/4)A (∵i = i[0]sin(ωt – Φ))
υ = 40 sin(100t)V (∵ V = V[0]sin(ωt ))
i[0] = \(\sqrt{2}\) , V[0] = 40, ω = 100, Φ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0}\) cosΦ = \(\frac{40}{\sqrt{2}}\) cos\(\frac{\pi}{4}\), R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\), R = 20 Ω
Question 4.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the
voltage supplied by the alternator.
V[C] = 20V, V[R] = 35V, V[L] = 20V
V = \(\sqrt{V_R^2+\left(V_L^2-V_C^2\right)}\) ; V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\) ; V = \(\sqrt{35^2}\); V = 35 Volt.
Question 5.
What is step up transformer ? How it differs from step down transformer ?
The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio.
T = \(\frac{N_S}{N_p}=\frac{\text {No. of turns in the secondary }}{\text {No. of turns in the primary }}\)
If N[S] > N[P], then the transformer is called step up transformer.
If N[S] < N[P], then the transformer is called step down transformer.
Textual Examples
Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
(a) We are given P = 100 W and V = 220V.
The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω
(b) The peak voltage of the source is υ[m] = \(\sqrt{2}\)V = 311 V
(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.
Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
The inductive reactance,
X[L] = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A
Question 3.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced.
With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine
less brightly than before.
Question 4.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive
reactance and the current ?
The capacitive reactance is
X[c] = \(\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}\)
= 212 Ω
The rms current is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{220 \mathrm{~V}}{212 \Omega}\)
= 1.04 A
The peak current is
i[m] = \(\sqrt{2}\)I = (1.41)(1.04A) = 1.47A
This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.
Question 5.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.
The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is
inserted. Give your answer with reasons.
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive
reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb
Question 6.
A resistor of 200Ω and a capacitor of 15.0 μF are connected in series to a 220V, 50 Hz ac source,
(a) Calculate the current in the circuit;
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.
R = 200Ω. C = 15.0 μF = 15.0 × 10^-6F
V = 220V, v = 50Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
Z = \(\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{C}}^2}=\sqrt{\mathrm{R}^2+(2 \pi v C)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+\left(2 \times 3.14 \times 50 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+(212 \Omega)^2}\) = 291.5Ω
Therefore, the current in the circuit is
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{220 \mathrm{~V}}{291.5 \Omega}\) = 0.755A
(b) Since the current is the same throughout the circuit, we have
V[R] = IR = (0.755 A) (200Ω) = 151V
V[C] = IX[C] = (0.755A) (212.3Ω) = 160.3V
The algebraic sum of the two voltages, V[R] and V[C] is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox ? As you have learnt in the text, the two voltages are not
in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the
Pythagorean theorem:
V[R+C] = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.
Question 7.
a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
a) We know that P = IV cosΦ where cosΦ is the power factor. To supply a given power at a given voltage, if cosΦ is small, we have to increase current accordingly. But this will lead to large power
loss (I^R) in transmission.
b) Suppose in a circuit, current I lags the voltage by an angle Φ.
Then power factor cosΦ = R/Z
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram in the figure how this can be achieved. Let us resolve I into two
components, I[P]
along the applied voltage V and I[q] perpendicular to the applied voltage. I[q] is called the wattless component since corresponding to this component of current, there is no power loss. I[P] is
known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.
It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I[q] by an equal leading wattless current I’[q]. This can be done by
connecting a capacitor of appropriate value in parallel so that I[q] and I’[q] cancel each other and P is effectively I[P] V.
Question 8.
A sinusoidal voltage of peak value 283 V , and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω. L = 25.48 mH. and C = 796μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
a) To find the impedance of the circuit, we first calculate X[L] and X[C].
X[L] = 2πvL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω
X[C] = \(\frac{1}{2 \pi v \mathrm{C}}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4Ω
z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\sqrt{3^2+(8-4)^2}\)
= 5Ω
b) Phase difference, Φ = tan^-1\(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
= tan^-1\(\left(\frac{4-8}{3}\right)\) = -53.1°
c) The power dissipated in the circuit is
P = I^2R
Therefore, P = (40A)^2 × 3Ω = 4800W
= 4.8 kW
d) Power factor = cos Φ = cos 53.1° = 0.6.
Question 9.
Suppose the frequency of the source in the previous example can be varied,
(a) What is the frequency of the source at which resonance occurs ?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition.
(a) The frequency at which the resonance occurs is
ω[0] = \(\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}}\)
= 222.1 rad/s
v[r] = \(\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14}\) Hz = 35.4Hz
b) The impedance Z at resonant condition is equal to the resistance
Z = R = 3Ω
The rms current at resonance is ,
as V = \(\frac{v_{\mathrm{m}}}{\sqrt{2}}\)
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}\) = 66.7 A
The power dissipated at resonance is
P = I^2 × R = (66.7)^2 × 3 = 13.35 kW
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 8.
Question 10.
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what
principle does this detector work ?
The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a
capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit.
This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.
Question 11.
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Let q[0] be the initial charge tin a capacitor. Let the charged capacitor be connected to an inductor of inductance L. this LC circuit will sustain an oscillation with frequency
\(\omega\left(2 \pi v=\frac{1}{\sqrt{\mathrm{LC}}}\right)\)
At an instant t, charge q on the capacitor and the current i are given by :
q(t) = q[0] cos ωt
i(t) = -q[0] co sin ωt
Energy stored in the capacitor at time ‘t’ is
U[E] = \(\frac{1}{2}\) C V^2 = \(\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) cos^2 (ωt)
Energy stored in the inductor at time ‘t’ is
U[M] = \(\frac{1}{2}\) L i^2
= \(\frac{1}{2}\) L q[0]^2 ω^2sin^2 (ωt)
= \(\frac{Q_0^2}{2 C} \sin ^2(\omega t)\) [∵ ω = 1/ \(\sqrt{L C}\)]
Sum of energies
U[E] + U[M] = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) [cos^2 ωt + sin^2ωt) = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\)
This sum is constant in time as q[0] and C, both are time-independent. | {"url":"https://apboardsolutions.guru/ap-inter-2nd-year-physics-important-questions-chapter-10/","timestamp":"2024-11-02T17:50:48Z","content_type":"text/html","content_length":"70388","record_id":"<urn:uuid:96382142-2d59-4ef3-93f9-2eb3d4108da7>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00616.warc.gz"} |
Use 4 HALF-ADDERs to build a circuit that takes as input a 4-bit number x and on its 4-bit output computes the twos-complement of x.
Hint: first use a layer of INVERTERS to invert the input of x to produce the ones complement of x, and then add to this the value 1 to produce the two’s complement.
An 8-bit addition/subtraction unit with two's complement arithmetic and an accumulator. | {"url":"https://circuitverse.org/projects/tags/twos-complement","timestamp":"2024-11-12T13:23:13Z","content_type":"text/html","content_length":"21028","record_id":"<urn:uuid:0df43ab3-a5a0-4eca-b2e5-f97ebd8d037d>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00682.warc.gz"} |
IN MATH: 1. adj., adv. two times. EX. What's twice as many as 48?
IN ENGLISH: 1. as defined above. EX. Nancy's had triplets twice.
APPLICATION: See list 60)
1. What's twice as many as 48?
This is a page from the dictionary MATH SPOKEN HERE!, published in 1995 by MATHEMATICAL CONCEPTS, inc., ISBN: 0-9623593-5-1. This 460-entry book sells for $12.00 US plus shipping and sales tax.
You are hereby granted permission to make ONE printed copy of this page and its picture(s) for your PERSONAL and not-for-profit use. YOU MAY NOT MAKE ANY ADDITIONAL COPIES OF THIS PAGE, ITS PICTURE
(S), ITS SOUND CLIP(S), OR ITS ANIMATED GIFS WITHOUT PERMISSION FROM: request@mathnstuff.com or by mail to the address below. | {"url":"https://www.mathnstuff.com/math/spoken/here/1words/t/t29.htm","timestamp":"2024-11-03T10:11:00Z","content_type":"text/html","content_length":"3916","record_id":"<urn:uuid:87370e40-4681-406e-b2fa-fca486915c32>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00682.warc.gz"} |
Another follow-up: Fundamental Theorem of Calculus
Subject: Re: Fundamental Theorem of Calculus
Date: Sun, 05 Sep 2000 22:03:55 -0400
Alex Bogomolny
I can live with your definition of the integral but not the derivative. What does it mean to represent the gradient and what is the latter anyway? You used the piece "represents the gradient" twice;
and to my ear the two occurences do not even sound the same.
Let's make another effort.
|Reply| |Up| |Down|
Copyright © 1996-2018
Alexander Bogomolny | {"url":"https://www.cut-the-knot.org/exchange/FundCalculus4.shtml","timestamp":"2024-11-09T16:20:44Z","content_type":"text/html","content_length":"11670","record_id":"<urn:uuid:bd764db6-5c4f-440b-a769-b1c5ea55e1d0>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00024.warc.gz"} |
The aim of the paper is to announce some recent results con- cerning Hamiltonian theory for higher order variational problems on fibered manifolds
on Differential Geometry, 25–30 July, 2000, Debrecen, Hungary
Abstract. The aim of the paper is to announce some recent results con- cerning Hamiltonian theory for higher order variational problems on fibered manifolds. A reformulation, generalization and
extension of basic concepts such as Hamiltonian system, Hamilton equations, regularity, and Legendre transformation, is presented. The theory is based on the concept ofLepagean (n+ 1)-form(wherenis
the dimension of the base manifold). Contrary to the classical approach, where Hamiltonian theory is related to a single Lagrangian, within the present setting a Hamiltonian system is associated with
an Euler–
Lagrange form, i.e., with theclass of all equivalent Lagrangians. Hamilton equations are introduced to be equations for integral sections of an exterior differential system, defined by a Lepagean (n+
1)-form. Relations between extremals and solutions of Hamilton equations are studied in detail. A revi- sion of the concepts ofregularity and Legendre transformation is proposed, reflecting geometric
properties of the related exterior differential system. The new look is shown to lead to new regularity conditions and Legendre trans- formation formulas, and provides a procedure ofregularizationof
variational problems. Relations to standard Hamilton–De Donder theory, as well as to multisymplectic geometry are studied. Examples of physically interesting La- grangian systems which are
traditionally singular, but regular in this revised sense, are discussed.
1. Introduction
Hamiltonian theory belongs to the most important parts of the calculus of vari- ations. The idea goes back to the first half of the 19th century and is due to Sir William Rowan Hamilton and Carl
Gustav Jacob Jacobi who, for the case ofclas- sical mechanics, developed a method to pass from the Euler–Lagrange equations to another set of differential equations, now called Hamilton equations,
which are
“better adapted” to integration. This celebrated procedure, however, is applicable
1991Mathematics Subject Classification. 35A15, 49L10, 49N60, 58Z05.
Key words and phrases. Lagrangian system, Poincar´e-Cartan form, Lepagean form, Hamil- tonian system, Hamilton extremals, Hamilton–De Donder theory, Hamilton equations, regularity, Legendre
Research supported by Grants MSM:J10/98:192400002 and VS 96003 of the Czech Ministry of Education, Youth and Sports, and GACR 201/00/0724 of the Czech Grant Agency. The author also wishes to thank
Professors L. Kozma and P. Nagy for kind hospitality during the Colloquium on Differential Geometry, Debrecen, July 2000.
only to a certain class of variational problems, called regular. Later the method was formally generalized to higer order mechanics, and both first and higher order field theory, and became one of
the constituent parts of the classical variational theory (cf. [8], [4]). In spite of this fact, it has been clear that this generaliza- tion of Hamiltonian theory suffers from a principal defect:
allmost all physically interesting field Lagrangians (gravity, Dirac field, electromagnetic field, etc.) are non-regular, hence they cannot be treated within this approach.
Since the second half of the 20’th century, together with an increasing interest to bring the more or less heuristic classical variational theory to a modern framework of differential geometry, an
urgent need to understand the geometric meaning of the Hamiltonian theory has been felt, in order to develop its proper generalizations as well as global aspects. There appeared many papers dealing
with this task in different ways, with results which are in no means complete: from the most important ones let us mention here at least Goldschmidt and Sternberg [14], Aldaya and Azc´arraga [1],
Dedecker [5], [7], Shadwick [37], Krupka [21]–[23], Ferraris and Francaviglia [9], Krupka and ˇStˇep´ankov´a [26], Gotay [16], Garcia and Mu˜noz [10], [11], together with a rather pessimistic
Dedecker’s paper [6] summarizing main problems and predicting that a way-out should possibly lead through some new understanding of such fundamental concepts asregularity,Legendre transformation, or
even the Hamiltonian theory as such.
The purpose of this paper is to announce some very recent results, partially pre- sented in [30]–[32] and [38], which, in our opinion, open a new way for understanding the Hamiltonian field theory.
We work within the framework of Krupka’s theory of Lagrange structures on fibered manifolds where the so called Lepagean form is a central concept ([18], [19], [21], [24], [25]). Inspired by fresh
ideas and inter- esting, but, unfortunately, not very wide-spread “nonclassical” results of Dedecker [5] and Krupka and ˇStˇep´ankov´a [26], the present geometric setting means a direct
“field generalization” of the corresponding approach to higher order Hamiltonian mechanics as developed in [27] and [28] (see [29] for review). The key point is the concept of aHamiltonian system,
which, contrary to the usual approach, is not re- lated with a single Lagrangian, but rather with an Euler–Lagrange form (i.e., with theclass of equivalent Lagrangians), as well as ofregularity,
which is understood to be ageometric property of Hamilton equations. It turns out that “classical” results are incorporated as a special case in this scheme. Moreover, for many variational systems
which appear singular within the standard approach, one obtains here a regular Hamiltonian counterpart (Hamiltonian, independent momenta which can be considered a part of certainLegendre
coordinates,Hamilton equations equivalent with the Euler–Lagrange equations). This concerns, among others, such important physical systems as, eg., gravity, electromagnetism or the Dirac field,
mentioned above.
2. Notations and preliminaries
All manifolds and mappings throughout the paper are smooth. We use standard notations as, eg.,T for the tangent functor,J^r for ther-jet prolongation functor, dfor the exterior derivative of
differential forms,iξ for the contraction by a vector field ξ, and∗for the pull-back.
We consider a fibered manifold (i.e., surjective submersion)π:Y →X, dimX= n, dimY =m+n, its r-jet prolongationπr:J^rY →X, r≥1, and canonical jet projectionsπr,k:J^rY →J^kY, 0≤k < r (with an obvious
notationJ^0Y =Y). A fibered chart onY is denoted by (V, ψ),ψ= (x^i, y^σ), the associated chart onJ^rY by (Vr, ψr),ψr= (x^i, y^σ, y[j]^σ[1], . . . , y[j]^σ[1][...j][r]).
A vector field ξ on J^rY is called π[r]-vertical (respectively, π[r,k]-vertical) if it projects onto the zero vector field onX (respectively, onJ^kY). We denote byV πr
the distribution onJ^rY spanned by theπr-vertical vector fields.
Aq-formρonJ^rY is calledπr,k-projectableif there is aq-form ρ0 onJ^kY such that π^∗[r,k]ρ[0] = ρ. A q-form ρ on J^rY is called π[r]-horizontal (respectively, π[r,k]- horizontal) ifi[ξ]ρ= 0 for everyπ
[r]-vertical (respectively,π[r,k]-vertical) vector fieldξ onJ^rY.
The fibered structure ofY induces a morphism,h, of exterior algebras, defined by the condition J^rγ^∗ρ=J^r+1γ^∗hρ for every sectionγ of π, and called the hor- izontalization. Apparently,
horizontalization applied to a function, f, and to the elements of the canonical basis of 1-forms, (dx^i, dy^σ, dy[j]^σ
1, . . . , dy[j]^σ
1...j[r]), on J^rY gives
hf=f◦πr+1,r, hdx^i=dx^i, hdy^σ =y^σ[l]dx^l, . . . , hdy^σ[j][1][...j][r] =y[j]^σ[1][...j][r][l]dx^l. A q-form ρon J^rY is calledcontactif hρ= 0. On J^rY, behind the canonical basis of 1-forms, we
have also the basis (dx^i, ω^σ, ω^σ[j]
1, . . . , ω^σ[j]
1...jr−1, dy^σ[j]
1...j[r]) adapted to the contact structure, where in place of thedy’s one has the contact 1-forms
ω^σ=dy^σ−y[l]^σdx^l, . . . , ω[j]^σ[1][...j][r−1]=dy[j]^σ[1][...j][r−1]−y[j]^σ[1][...j][r−1][l]dx^l.
Sections ofJ^rY which areintegral sections of the contact idealare calledholonomic.
Apparently, a section δ : U → J^rY is holonomic if and only if δ = J^rγ where γ:U →Y is a section of π.
Notice that every p-form on J^rY, p > n, is contact. Let q > 1. A contact q-form ρ onJ^rY is called 1-contact if for everyπr-vertical vector field ξ on J^rY the (q−1)-formi[ξ]ρis horizontal.
Recurrently, ρis calledi-contact, 2 ≤i ≤q, if i[ξ]ρis (i−1)-contact. Everyq-form onJ^rY admits aunique decomposition
π[r+1,r]^∗ ρ=hρ+p1ρ+p2ρ+· · ·+pqρ,
wherepiρ, 1≤i≤q, is ani-contact form onJ^r+1Y, called thei-contact partofρ.
It is helpful to notice that the chart expression ofpiρin any fibered chart contains exactly iexterior factors ω[j]^σ[1][...j][l] wherel is admitted to run from 0 tor. For more details on jet
prolongations of fibered manifolds, and the calculus of horizontal and contact forms the reader can consult eg. [18], [19], [24], [25], [29], [33], [34].
Finally, throughout the paper the following notation is used:
ω0=dx^1∧dx^2∧...∧dx^n, ωi=i[∂/∂x]iω0, ωij =i[∂/∂x]jωi, etc.
3. Hamiltonian systems
In this section we discuss the concept of a Hamiltonian system and of a La- grangian system as introduced in [30], and the relation between Hamiltonian and Lagrangian systems.
Let s ≥ 0, and put n = dimX. A closed (n+ 1)-form α on J^sY is called a Lepagean (n+ 1)-formifp1αisπs+1,0-horizontal. Ifαis a Lepagean (n+ 1)-form and E =p1αwe also say that αis a Lepagean equivalent
of E. By definition, in every fiber chart (V, ψ),ψ= (x^i, y^σ), onY,
whereEσare functions onVs+1⊂J^s+1Y. A Lepagean (n+ 1)-formαonJ^sY will be also called aHamiltonian system of order s. A sectionδof the fibered manifold πswill be called aHamilton extremal of αif
(3.1) δ^∗i[ξ]α= 0 for everyπ[s]-vertical vector field ξonJ^sY . The equations (3.1) will be then calledHamilton equations.
Hamiltonian systems are closely related with Lagrangians and Euler–Lagrange forms. The relation follows from the properties of Lepageann-forms (see eg. [21], [24], [25] for review). Recall that an
n-form ρ on J^sY is said to be a Lepagean n-form ifhi[ξ]dρ = 0 for every π[s,0]-vertical vector field ξon J^sY [18], [21]. Thus, every Lepagean (n+ 1)-form locally equals to dρ where ρis a Lepagean
Consequently, ifαis a Lepagean (n+ 1)-form then its 1-contact partE is alocally variational form. In other words, there exists an open covering of J^s+1Y such that, on each set of this covering, E
coincides with the Euler–Lagrange form of a Lagrangian of order r≤s, i.e.,
∂y^σ −
(−1)^ldp[1]dp[2]. . . dp[l]
This suggests the following definition of aLagrangian system: Lepagean (n+ 1)- forms (possibly of different orders) are said to be equivalent if their one-contact parts coincide (up to a possible
projection). In what follows, we denote the equiv- alence class of a Lepagean (n+ 1)-formαby [α], and call it aLagrangian system.
The minimum of the set of orders of the elements in the class [α] will then be called the(dynamical) orderof the Lagrangian system [α].
Every Lagrangian system is locally characterized by Lagrangians of all orders starting from a certain minimal one, denoted byr[0], and called theminimal order for [α].
TheEuler–Lagrange equationscorresponding to a Lagrangian system [α] of order snow read
(3.2) J^sγ^∗iJ^sξα= 0 for every π-vertical vector fieldξonY ,
whereαis any representative of ordersof the class [α]. Notice that (3.2) are PDE of orders+ 1 for sectionsγof the fibered manifoldπ, their solutions areextremals ofE=p1α.
Let us stop for a moment to discuss relations between Hamiltonian systems and Lagrangian systems.
Every Hamiltonian system α of order s has a unique associated Lagrangian system [α]; its order isr≤s, and it is represented by the locally variational form E =p1α(or, alternatively, by the family of
all, generally only local, Lagrangians whose Euler–Lagrange forms locally coincide with E). Comparing the Hamilton equations (3.1) with the Euler–Lagrange equations (3.2) one can see immediately that
the sets of extremals and Hamilton extremals generally are not in one-to-one correspondence, in other words,Hamilton equations need not be equivalent with the Euler–Lagrange equations. However, the
s-jet prolongation of every extremal is a Hamilton extremal; in this sense there is an inclusion of the set of extremals into the set of Hamilton extremals. More precisely, there is abijectionbetween
the set of extremals and the set ofholonomic Hamilton extremals.
On the other hand, a Lagrangian system [α] of order r has many associated Hamiltonian systems, each of an order s ≥ r. Consequently, to a given set of Euler–Lagrange equations one has many sets of
Hamilton equations. Behind the given Euler–Lagrange expressions (respectively, a Lagrangian), Hamilton equations depend also upon “free functions” which come from the at least 2-contact part of π^∗
[s+1,s]α. Since α is locally the exterior derivative of a Lepagean n-form, it is convenient to discuss different possibilities on the level of Lepagean equivalents of a corresponding Lagrangian. To
this purpose, let us first recall an important result, due to Krupka [25]: ρis a Lepageann-form of ordersiff in any fibered chart (V, ψ), ψ= (x^i, y^σ), onY,
(3.3) π^∗[s+1,s]ρ=θλ+dν+µ,
(3.4) θλ=Lω0+
(−1)^ldp[1]dp[2]. . . dp[l]
k∧ωi, ν is an arbitrary contact (n−1)-form, and µ is an arbitrary at least 2-contact n-form; in the formula (3.4),rdenotes the order of hρ. It should be stressed that the decomposition (3.3) is
generally not invariant with respect to transformations of fibered coordinates. θλ is called the local Poincar´e–Cartan equivalent of the Lagrangianλ=hρ, each of the (invariant) forms Θ =θλ+p1dν is
then called the Poincar´e–Cartan form. Now, let α be a Hamiltonian system. If locally α =dρ where the Lepagean n-form ρ is at most i-contact (1 ≤ i ≤ n) we speak about Hamilton pi-theory and call the
corresponding Hamilton equations (3.1)Hamilton pi-equations[30].
In particular,Hamilton p1-equationsare based upon the Poincar´e–Cartan form Θ. In an usual approach to Hamiltonian field theory only these equations are con- sidered (cf. [1], [3], [7]–[17], [21]–
[23], [26], [33], [35]–[37] and references therein);
they are often called Hamilton–De Donder equations. Obviously, behind a La- grangian, they depend upon ν. (With the exception of the case of first order La- grangians whenν = 0, i.e.,θλis invariant,
and the Hamilton–De Donder equations are unique and completely determined by the Lagrangian).
Hamilton p2-equationsare based upon a Lepagean form ρ= Θ +ν where ν is 2-contact; for first order Lagrangians they have been studied in [31], [32], a second order generalization is due to [38].
Hamiltonp[n]-equations are based upon a general Lepageann-form; the first oder case was considered by Dedecker [5].
Each of these Hamiltonian systems can be viewed as adifferent extensionof the original variational problem. In this way, in any concrete situation, one can utilize a possibility to apply additional
(geometric or physical) requirements to choose from many alternative Hamiltonian systems the “best one”. A deeper insight into this question is subject of the next sections.
Comments 1. Let us mention main differences between the presented approach and the usual one.
(i) Hamiltonian systems, Hamilton equations. Roughly speaking, there are two main geometric ways approaching Hamiltonian field theory. One, close to the clas- sical calculus of variations, declares
the philosophy to assign auniqueHamiltonian system to a singleLagrangian. This task is represented by theHamilton–De Don- der theory, based upon the Poincar´e–Cartan form of a (global) Lagrangianλ,
which gives “good” results for first order Lagrangians, and is considered problematic in the higher order case (cf. eg. [6]). The second approach is more or less axiomatic, and is based upon the so
called multisymplectic form(eg., [3] and references therein).
(Recall that an (n+ 1)-form Ω on a manifold M is called multisymplectic if it is closed, and the “musical map” ξ →iξΩ, mapping vector fields on M to n-forms, is injective.)
Our approach is close to both of them, but different. This can be seen im- mediately if the definitions of the multisymplectic and Lepagean (n+ 1)-form are compared. For the “zero order” case one
gets that every multisymplectic form on Y is a Lepagean (n+ 1)-form, however, a Lepagean (n+ 1)-form need not be multisymplectic. For higher order the difference becomes even sharper, since a
multisymplectic form on J^sY,s≥1, need not be Lepagean. Apparently, our moti- vation was to define a Hamiltonian system to be, contrary to the multisymplectic definition, sufficiently generalon the
one hand (covering allLagrangians without any a priori restriction), and, on the other hand,directly related with a variational problem (defined by a locally variational form). Among others, this
also means that our Hamiltonian system is assigned not to a particular Lagrangian (as al- ways done),but to the whole class of all equivalent Lagrangianscorresponding to a given Euler–Lagrange form.
Other differences are connected with the concepts of regularityandLegendre transformation, and will be discussed in the next sections.
(ii) Lagrangian systems. One should notice that also in the definition of a La- grangian system and of the order of a Lagrangian system we differ from other authors. While usually by a “Lagrangian
system of order r” one means a global Lagrangian onJ^rY, by our definition a Lagrangian system is theequivalence class of Lagrangiansgiving rise to an Euler–Lagrange form; the order of a Lagrangian
system is then, as a property of the whole equivalence class, determined via the order of the Euler–Lagrange form. In this way, only properties directly connected with dynamics, hence common to all
the equivalent Lagrangians, enter in this def- inition, while distinct properties of particular Lagrangians which are not essential for the dynamics are eliminated (the latter are namely connected
with the fact that the family of equivalent Lagrangians contains Lagrangians of all orders start- ing from a minimal one, which, as functions, may look completely different from each other, and whose
domains usually are open subsets of the corresponding jet prolongations of the underlying fibered manifold; a global Lagrangian often even does not exist at all, obstructions lie in the topology of
Y). In the sense of our definition, for example, the Dirac fieldis a Lagrangian system of orderzero (and not of order one); indeed, in this case the class [α] is represented by the global form
dθλprojectable ontoY, since the corresponding Lagrangianλis a global first order Lagrangian affine in the first order derivatives. Similarly, theEinstein gravitational field, which is usually defined
by the scalar curvature Lagrangian (global second order Lagrangian), is a Lagrangian system of order one, since the corresponding Poincar´e–Cartan form is projectable ontoJ^1Y.
4. Regular Hamiltonian systems
A sectionδ ofπs is called aDedecker’s section[30] ifδ^∗µ= 0 for every at least 2-contact formµonJ^sY.
Consider a Hamiltonian system α on J^sY. Denote E = p1α, F = p2α, G = π^∗[s+1,s]α−E−F (i.e.,Gis the at least 3-contact part ofπ^∗[s+1,s]α), and
(4.1) αˆ=E+F.
We shall call the form ˆαtheprincipal part ofα.
A Dedecker’s section which is a Hamilton extremal of π[s+1,s]^∗ α will be called Dedecker–Hamilton extremalofα.
It is easy to obtain the following relation between the sets of extremals, Hamilton extremals, and Dedecker–Hamilton extremals of a Hamiltonian systemαon J^sY [30]:
If γ is an extremal ofE = p[1]α then for every Lepagean equivalent α of E, α defined on J^sY (s ≥ 0), the section δ = J^sγ is a Hamilton extremal of α, and δˆ=J^s+1γis its Dedecker–Hamilton extremal.
For everyα∈[α], defined onJ^sY (s≥0), and for every its Dedecker–Hamilton extremalδ, the sectionˆ δ=π[s+1,s]◦ˆδis a Hamilton extremal of α.
Denote by D[α]^s and D[α]^s+1[ˆ] the family ofn-forms iξαandiξαˆ respectively, where ξ runs over allπs-vertical vector fields onJ^sY, respectively, over all πs+1-vertical
vector fields on J^s+1Y. Notice that the rank of D^s+1[α][ˆ] is never maximal, since i[∂/∂y]^ν
Pαˆ = 0 for all multiindicesP of the lenghts+ 1.
Apparently, Hamilton extremals and Dedecker–Hamilton extremals ofαare in- tegral sections of the ideal generated by the family D[α]^s and D^s+1[α][ˆ] , respectively.
Hence, “regularity” can be understood to be aproperty of the idealD^s+1[α][ˆ] as follows [30]. For convenience, let us consider the casess= 0 and s >0 separately.
We call a Hamiltonian systemαonY regularif rankD^1[α][ˆ] is constant and equal to rankV π=m. Let s≥1 and r≥1. A Hamiltonian systemα onJ^sY will be calledregular of degree rif the system of local
generators ofD^s+1[α][ˆ] contains all the n-forms
(4.2) ω^σ∧ωi, . . . , ω^σ[j][1][...j][r−1]∧ωi, and rankD^s+1[α][ˆ] =nrankV π[r−1]+ rankV π[s−r].
We refer to (4.2) as localcanonical 1-contactn-forms of orderr.
Roughly speaking, regularity of degreermeans that the systemD^s+1[α][ˆ] contains all the canonical contactn-forms onJ^rY, and the rank of the “remaining” subsystem ofD^s+1[α][ˆ] is the greatest
possible one.
Notice that by this definition, every Dedecker–Hamilton extremal of a regular Hamiltonian system with degree of regularityrisholonomic up to the orderr, i.e.,
Moreover, we have the following main theorem (for the proof we refer to [30]).
Theorem 4.1. [30]Let αbe a Hamiltonian system onJ^sY,r[0] the minimal order forE=p[1]α. Suppose thatαis regular of degreer[0]. Then every Dedecker–Hamilton extremalδˆofαis of the form
πs+1,r[0]◦ˆδ=J^r^0γ whereγ is an extremal ofE.
Taking into account this result, a Hamiltonian system of order s≥1 which is regular of degreer0will be called simplyregular. Thus,regularHamilton equations and the corresponding Euler–Lagrange
equations arealmost equivalentin the sense that extremals are in bijective correspondence with classes of Dedecker–Hamilton extremals,γ→[J^s+1γ], where ˆδ∈[J^s+1γ] iffπs+1,r[0]◦δˆ=J^r^0γ.
We shall call a regular Hamiltonian systemstrongly regularif the Hamilton and Euler–Lagrange equations areequivalentin the sense thatextremals are in bijective correspondence with classes of Hamilton
extremals,γ →[J^sγ], whereδ∈[J^sγ] iff πs,r[0]◦δˆ=J^r^0γ. (Clearly, fors= 1 this precisely means a bijective correspondence between extremals and Hamilton extremals).
The concept of regularity of aLagrangian systemis now at hand: regularity can be viewed as the property that there exists at least one associated Hamiltonian system which is regular; obviously, the
order of this Hamiltonian system may differ from the order of the Lagrangian system. Hence, in accordance with [30], we call a Lagrangian system [α] regular if the family of associated Hamiltonian
contains a regular Hamiltonian system. Similarly, we call a Lagrangian system strongly regularif the family of associated Hamiltonian systems contains a strongly regular Hamiltonian system.
5. Regularity conditions
The above geometric definition of regularity enables one to find explicitregularity conditions. Keeping notations introduced so far, we write ˆα = E +F, where E=Eσω^σ∧ω0, and
(5.1) F =
F[σν]^J,P,iω[J]^σ∧ω[P]^ν ∧ω[i], F[σν]^J,P,i=−F[νσ]^P,J,i;
here J, P are multiindices of the lenght k and l, respectively, J = (j[1]j[2]. . . j[k]), P = (p[1]p[2]. . . p[l]), where 0≤ |J|,|P| ≤s, i.e., 0≤k, l≤s, and, as usual, 1≤i≤n, 1≤σ, ν ≤m. Sincedα= 0,E
is an Euler–Lagrange formof order s+ 1, i.e., the functions Eσ satisfy the identities
(5.2) ∂Eσ
(−1)^k k
dp[l+1]dp[l+2]. . . dp[k]
= 0, 0≤l≤s+ 1, calledAnderson–Duchamp–Krupka conditionsfor local variationality ofE[2], [20].
The conditiondα= 0 means that, in particular, (5.3) p2dα=p2dE+p2dF = 0.
In fibered coordinates this equation is equivalent with the following set of identities
(5.4) ∂Eσ
∂y^ν −∂Eν
∂y^σ −diF[σν]^,i= 0,
(F[σν]^0,S,i)[sym(Si)]= 1 2
∂y[Si]^ν , (F[σν]^0,P,i)[sym(P i)]= 1
∂y[P i]^ν −djF[σν]^0,P i,j, 0≤ |P| ≤s−1, and
(5.6) ^(F
σν )sym(Si)= 0, 1≤ |J| ≤s, (F[σν]^J j,P,p)[sym(P p)]+ (F[σν]^J,P p,j)[sym(J j)]+diF[σν]^J j,P p,i= 0, 0≤ |J|,|P| ≤s−1, where sym means symmetrization in the indicated indices, and S = (p[1]p[2]. . .
(5.7) f[σν]^J,P,i=F[σν]^J,P,i−(F[σν]^J,P,i)sym(P i).
Then from (5.5) we easily get
(5.8) (F[σν]^,p^1^...p^l−1^,p^l)[sym(p][1][...p][l−1][p][l][)]=1 2
(−1)^kdi[1]di[2]. . . di[k]
−dif[σν]^,p^1^...p^l^,i, 1≤l≤s+ 1,
and the recurrent formulas (5.6) give us the remainingF’s expressed by means of the (F[σν]^0,P,i)[sym(P i)]’s and the above f’s. As a result, one getsF (5.1)determined by the Euler–Lagrange
expressions E[σ] and the (free) functions f[σν]^J,P,i. Moreover, (5.4), and the antisymmetry conditions for theF[σν]^J,P,i’s, lead to the identities (5.2), as expected.
Now, we are prepared to find explicit regularity conditions forα. By definition, D^s+1[α][ˆ] is locally spanned by the followingn-forms:
η[ν]^P =−i[∂/∂y]^ν
2F[σν]^J,P,iω[J]^σ∧ωi, 1≤ |P| ≤s,
η[ν] =−i[∂/∂y]ναˆ=−E[ν]ω[0]+
One can see from (5.6) thatthe functionsF[σν]^J,P,iwhere|J|+|P| ≥s+ 1, depend only upon the f’s (5.7). The (invariant) choice
(5.10) f[σν]^J,P,i= 0, |J|+|P| ≥s+ 1 then leads to
(5.11) F[σν]^J,P,i= 0, |J|+|P| ≥s+ 1, and we obtain
η^Sν = 2Fσν^0,S,iω^σ∧ωi,
ην^P = 2Fσν^0,P,iω^σ∧ωi+ 2Fσν^j^1^,P,iω^σj[1]∧ωi, |P|=s−1, . . .
η[ν]^P =
2F[σν]^J,P,iω^σ[J]∧ωi+ 2Fσν^j^1^...j^r^0^−1^,P,iω^σ[j][1][...j][r]
0−1∧ωi, |P|=s−r0+ 1, . . .
where (5.13)
F[σν]^J j,P,i= (−1)^|J|+1(F[σν]^0,J jP,i)sym(J jP i)−(f[σν]^J,P i,j)sym(J j)+f[σν]^J j,P,i,
= (−1)^|J|+11 2
∂y[J jP i]^ν −(fσν^J,P i,j)sym(J j)+fσν^J j,P,i, |J|+|P|=s−1.
The results can be summarized as follows.
Theorem 5.1. [30] Let α be Hamiltonian system of order s, let r0 denote the minimal order of the corresponding Lagrangians. Suppose that
(5.14) f[σν]^J,P,i= 0, |J|+|P| ≥s+ 1, and
rank(F[νσ]^p^1^...p^s^,0,i) =mn, rank(Fνσ^p^1^...p^s−1^,j^1^,i) =mn^2,
. . .
rank(Fνσ^p^1^...p^s−r^0 +1^,j^1^...j^r^0^−1^,i) =nm n+r0−2 r0−1
! ,
0 Fνσ^P,J,i
−Eν 2Fνσ^0,J,i
=maximal, 0≤ |J| ≤s, 1≤ |P| ≤s−r0, where in the above matrices, the (ν, P) label rows, and the (σ, J, i) label columns.
Thenαis regular.
For the most frequent cases of second and first order locally variational forms this result is reduced to the following:
Corollary 5.1. [30] Let s = 1. The following are necessary conditions for α be regular:
(5.16) r0= 1,
(5.17) f[σν]^j,p,i= 0, f[σν]^i,0,j =f[νσ]^j,0,i,
(5.18) det∂Eσ
∂y[ij]^ν −2f[σν]^i,0,j 6= 0,
where in the indicated(mn×mn)-matrix,(ν, j)label the rows and(σ, i)the columns.
α=E[σ]ω^σ∧ω[0]+1 4
∂y[i]^ν −∂Eν
∂y^ν[ij] −2f[σν]^i,0,j
In terms of a first order Lagrangian λ = Lω0 the regularity conditions (5.17) and (5.18) read
(5.20) det ∂^2L
∂y[j]^σ∂y^ν[k] −g[σν]^ij 6= 0,
(5.21) g^ij[σν] =−g[σν]^ji =−g^ij[νσ], and (in the notations of (3.3)) it holds
(5.22) αˆ =dθ[λ]+p[2]dµ, p[2]µ=1
4g[σν]^ijω^σ∧ω^ν∧ω[ij]. If, in particular,E is projectable ontoJ^1Y, i.e.,
(5.23) ∂Eσ
∂y[ij]^ν = 0
for all (σ, ν, i, j), we can consider for E either a first order Hamiltonian system (s=1), or a zero order Hamiltonian system (s=0) (the latter follows from the fact that variationality conditions
imply thatEσ are polynomials of ordernin the first derivatives). Taking into account the above corollary fors= 1, and the definition of regularity fors= 0, we obtain regularity conditions forfirst
orderlocally variational forms as follows.
Corollary 5.2. [30] Every locally variational formE onJ^1Y is regular.
(1) Let W ⊂J^1Y be an open set, and g^ij[σν] be functions onW such that (5.24) −g[σν]^ij =g[σν]^ji =g^ij[νσ], det(g[σν]^ij)6= 0.
Then every closed(n+ 1)-form αonW such that (5.25) αˆ=Eσω^σ∧ω0+1
2 ∂Eσ
∂y^ν[i] +djg^ji[σν]
ω^σ∧ω^ν∧ωi+g[σν]^jiω^σ∧ω^ν[j] ∧ωi
is a regular first order Hamiltonian system related toE.
(2) Suppose that
(5.26) rank∂Eσ
= rank ∂^2L
∂y^σ∂y^ν[i] − ∂^2L
where in the indicated (m×mn)-matrices, (σ) label the rows and (ν, i) the columns, andλ=Lω0is any (local) first order Lagrangian forE. Then there exists a unique regular zero order Hamiltonian system
related toE; it is given by the (n+ 1)-form αonY which in every fibered chart(V, ψ),ψ= (x^i, y^σ) is expressed as follows:
π[1,0]^∗ α=E[σ]ω^σ∧ω[0] +
1 (k+ 1)!
1 . . . ∂y^ν[i]^k
ω^σ∧ω^ν^1∧ · · · ∧ω^ν^k∧ωi[1]...i[k].
6. Legendre transformation
Letα be a regular Hamiltonian system of orders≥1. Then all the canonical 1-contactn-forms
(6.1) ω^σ∧ω[i], . . . , ω[j]^σ
where r[0] is the minimal order for the locally variational form E = p[1]α, belong to the exterior differential system generated by D[α]^s+1[ˆ] . However, the generators of D^s+1[α][ˆ] naturally
associated with the fibered coordinates (i.e., (5.9)), are of the form oflinear combinationsof (6.1), and, moreover, fors >1,nonzero generators are not linearly independent. In this sense the
generators associated with fibered coordinates are not canonical. In what follows we shall discuss the existence of coordinates on J^sY, canonically adapted to the system D^s+1[α][ˆ] , i.e., such
that a part of the naturally associated generators coincides with the forms (6.1), and all superfluous generators vanish. More precisely, we shall be interested in the existence of local charts, (W,
χ), χ = (x^i, y[J]^σ, p^J,i[σ] , z^L), on J^sY such that (x^i, y[J]^σ) are local fibered coordinates on J^r^0^−1Y, and the generators of D[α]^s+1[ˆ] naturally associated with the coordinates p^J,i[σ]
coincide with the n-forms (6.1), and those associated withz^L vanish. Hence,
i[∂/∂z]Lαˆ= 0, ∀L, i[∂/∂p]J,i
σ αˆ=ω^σ[J]∧ωi, 0≤ |J| ≤r0−1.
Consequently, thenonzero generators are linarly independent. We shall call such
“canonical” cordinates on J^sY Legendre coordinates associated with the regular Hamiltonian systemα.
In the following theorem we adopt the notations of (3.3) and (3.4).
Theorem 6.1. [30] Let α be a regular Hamiltonian system onJ^sY, let x∈J^sY be a point. Suppose that in a neighborhood W of x
(6.3) α=dρ, ρ=θ[λ]+dν+µ,
whereλis a Lagrangian of the minimal order,r0, forE=p1α, defined onπs,r[0](W), andµ is such that
(6.4) p2µ=
g[σν]^J,K,i^1^i^2ω^σ[J]∧ω[K]^ν ∧ωi[1]i[2],
whereg[σν]^J,K,i^1^i^2are functions onπ[s,r][0][−1](W), satisfying the antisymmetry conditions g[σν]^J,K,i^1^i^2=−g^J,K,i[σν] ^2^i^1 =−g[νσ]^K,J,i^1^i^2. Put
(6.5) p^J,iσ =
(−1)^ldp[1]dp[2]. . . dp[l]
∂y^σ[J p][1][...p]
+ 4gσν^J,K,ily^νKl, 0≤ |J| ≤r0−1.
Then for any suitable functions z^L on W,(W, χ), whereχ= (x^i, y^σ[J], p^J,i[σ] , z^L), is a Legendre chart forα.
Using (6.5) we can write
(6.6) ρ=−Hω[0]+
p^J,i[σ] dy[J]^σ∧ω[i]+η+dν+µ[3], where
(6.7) η=
g^J,K,i[σν] ^1^i^2dy[J]^σ∧dy^ν[K]∧ωi[1]i[2],
µ3 is at least 3-contact, and
(6.8) H =−L+p^J,i[σ] y[J i]^σ + 2g^J,K,i[σν] ^1^i^2y^σ[J i]
We call the functions H (6.8) and p^J,i[σ] (6.5) a Hamiltonian and momenta of α.
(6.9) αˆ=−dH∧ω0+
dp^J,i[σ] ∧dy^σ[J]∧ωi+dη−p3dη, and computing i[∂/∂z]Lα,ˆ i[∂/∂p]J,i
σ α, andˆ i[∂/∂y]^σ
Jαˆ we get that D[α]^s+1[ˆ] is spanned by the following nonzeron-forms:
− ∂H
ω0+dyJ^σ∧ωi=ω^σJ∧ωi, ∂H
∂y^σ[J] −4∂g^J K,iσν ^1^i^2
∂x^i^2 y^νKi[1]−2
∂g^KP,i[νρ] ^1^i^2
∂y[J]^σ +∂gσν^J K,i^1^i^2
∂y^ρ[P] +∂g^P J,i[ρσ] ^1^i^2
yKi^ν [1]y[P i]^ρ [2]
+dp^J,iσ ∧ωi,
where 0≤ |J| ≤r0−1. Notice that if dη= 0 we getD[α]^s+1[ˆ] spanned by (6.11) ω[J]^σ∧ω[i], ∂H
∂y^σ[J] ω[0]+dp^J,i[σ] ∧ω[i], 0≤ |J| ≤r[0]−1.
Hamilton equations in Legendre coordinates thus take the following form.
Theorem 6.2. [30]
(1) A section δ:U →W is a Dedecker–Hamilton extremal of α (6.3), (6.6) if, along δ,
∂x^i = ∂H
∂x^i =− ∂H
∂y^σ[J] + 4∂g^J K,i[σν] ^1^i^2
∂p^K,iν ^1
+ 2∂g^KP,i[νρ] ^1^i^2
∂y^σ[J] +∂g[σν]^J K,i^1^i^2
∂y^ρ[P] +∂g^P J,i[ρσ] ^1^i^2
∂p^K,iν ^1
∂p^P,iρ ^2
If, in particular, dη= 0, (6.12) take the “classical” form (6.13) ∂y^σ[J]
∂x^i = ∂H
, ∂p^J,i[σ]
∂x^i =−∂H
∂y[J]^σ, 0≤ |J| ≤r0−1.
(2) Ifµ3isπs,r[0]−1-projectable then (6.12) (resp. (6.13)) are equations for Hamil- ton extremals ofα.
As a consequence of (2) we obtain thatextremals are in bijective correspondence with classes of Hamilton extremals(with the equivalence in the sense of Sec. 4, i.e., that δ1 is equivalent
withδ2iffπs,r0◦δ1=πs,r0◦δ2). In other words,
Corollary 6.1. [30] Hamiltonian systems satisfying the condition (2) of Theorem 6.2 are strongly regular.
The above result shows another geometrical meaning of Legendre transforma- tion: Hamiltonian systems which are regular and admit Legendre transformation according to Theorem 6.1 either are strongly
regular, or can be easily brought to a stongly regular form (by modifying the termµ[3]).
Comments 2. Let us compare our approach to regularity and Legendre transfor- mation with other authors.
(i) Standard Hamilton–DeDonder theory. In the usual formulation of Hamilton- ian field theory Legendre transformation is a map associated with a Lagrangian, definedby the following formulas:
(6.14) p^i[σ]= ∂L
∂y[i]^σ if L=L(x^i, y^ν, y^ν[j]), and
(6.15) p^j[σ]^1^...j^k^i=
(−1)^ld[p][1]d[p][2]. . . d[p][l] ∂L
, 0≤k≤r−1, for a Lagrangian of orderr≥2 ([8], [7], [16], [33], [35], [37], etc.). These formulas have their origin in the (noninvariant) decomposition of the Poincar´e–Cartan form θ[λ] (3.4) in the
canonical basis (dx^i, dy[J]^σ), 0≤ |J| ≤r−1, i.e.,
(6.16) θ[λ]= (L−
p^J i[σ]y[J i]^σ)ω[0]+
p^J i[σ] dy^σ[J]∧ω[i].
However, for global Lagrangians of order r ≥ 2 the form (3.4) is neither unique nor globally defined. It is replaced by Θ =θ[λ]+p[1]dν (cf. notations of (3.3)), and, consequently, (6.15) are replaced
by more general formulas
(6.17) p^j[σ]^1^...j^k^i=
(−1)^ldp[1]dp[2]. . . dp[l]
+c^j[σ]^1^...j^k^p^1^...p^l^i , 0 ≤ k ≤ r−1, where c^j[σ]^1^...j^k^p^1^...p^l^i are auxiliary (free) functions (Krupka [23], Gotay [16]). In the Hamilton–De Donder theory, a Lagrangian is called
the Legendre map defined by (6.15), resp. (6.17) is regular. In the case of a first order Lagrangian this means thatλsatisfies the condition
(6.18) det ∂^2L
∂q[j]^σ∂q[k]^ν 6= 0
at each point of J^1Y. Since for higher order Lagrangians the Legendre transfor- mation (6.17) depends uponp1dν, one could expect that the corresponding regu- larity condition conserves this
property. Surprizingly enough, it has been proved in [23] and [16] that the regularity conditions do not depend upon the functions c^j[σ]^1^...j^k^p^1^...p^l^i, and are of the form [37], [23], [16]
(6.19) rank 1
[j1. . . j2r−s(pr+1. . . ps] [p1. . . pr)]
= max where [j1. . . j[2r−s]pr+1. . . ps] and [p1. . . pr] denotes the number of all different se- quences arising by permuting the sequence j1. . . j2r−spr+1. . . ps and p1. . . pr, re- spectively;
as usual,r ≤k≤2r−1, and, in the indicated matrices, σ, j1≤ · · · ≤ j2r−k label columns andν, p1≤ · · · ≤pk label rows, and the bracket denotes sym- metrization in the corresponding indices. (Notice
that within the present approach, by (6.3), the nondependence of regularity conditions upon the c’s is trivial). As a result one obtains that if a Lagrangian satisfies (6.18) (respectively, (6.19))
then every solution δof the Hamilton–De Donder equationsδ^∗i[ξ]dθ[λ] = 0 (respectively, δ^∗i[ξ]dΘ = 0), is of the formδ=J^1γ (respectively,π[2r−1,r]◦δ=J^rγ) whereγis an extremal ofλ. However, while
in the Legendre coordinates defined by (6.15) the lo- cal Hamilton–De Donder equations, i.e.,δ^∗iξdθλ= 0, take the familiar “canonical”
∂x^i = ∂H
, ∂p^J,i[σ]
∂x^i =−∂H
for the global Hamilton–De Donder equations, i.e., δ^∗i[ξ]dΘ = 0, using the “Le- gendre transformation” defined by (6.17), one does not generally obtain a similar
“canonical” representation.
(ii) A generalization of the concepts of regularity and Legendre transformation within the Hamilton–De Donder theory. In the paper [26] second order Lagrangians, affine in the second derivatives, and
admitting first order Poincar´e–Cartan forms were studied. Notice that in the sense of the regularity condition (6.19), La- grangians of this kind are apparently singular. In [26], the definition of
a regular Lagrangian is extended in the following way: a Lagrangian is calledregular if the solutions of the Euler–Lagrange and Hamilton–De Donder equations are equiva- lent (in the sense that the
sets of solutions are in bijective correspondence). The following results were proved:
Theorem 6.3. [26] Consider a Lagrangian of the formλ=Lω[0] where, in fibered coordinates,L admits an (obviously invariant) decomposition
(6.20) L=L0(x^i, y^σ, y[j]^σ) +h^pq[ν] (x^i, y^σ)y^ν[pq].
Then θλ is projectable onto J^1Y, and, consequently, Hamilton–De Donder equa- tions are equations for sectionsδ:U →J^1Y. If the condition
(6.21) det ∂^2L0
∂y^σ[i]∂y^ρ[k] −∂h^ik[σ]
∂y^ρ −∂h^ik[ρ]
∂y^σ 6= 0
is satisfied thenλis regular, i.e., the Euler–Lagrange and the Hamilton–De Donder equations ofλare equivalent, and the mapping
(6.22) (x^i, y^σ, y[j]^σ)→(x^i, y^σ, p^j[σ]), p^j[σ]= ∂L[0]
∂y[j]^σ −∂h^jk[σ]
∂x^k −∂h^jk[σ]
∂y^ν +∂h^jk[ν]
∂y^σ y^ν[k]
is a local coordinate transformation on J^1Y.
The formula (6.22) comes from the following noninvariant decomposition of the Poincar´e–Cartan form
(6.23) θλ=−Hω0+p^j[σ]dy^σ∧ωj+d(h^ij[σ]y[j]^σωi), where
(6.24) H =−L0+∂L[0]
∂y[j]^σ −∂h^jk[σ]
∂y^ν y[j]^σy[k]^ν.
The functionsH and p^j[σ] were called in [26] theHamiltonian andmomenta of the Lagrangian (6.20), and (6.22) was calledLegendre transformation. In the Legendre coordinates (6.22) the Hamilton–De
Donder equations read
(6.25) ∂H
∂y^ν =−∂p^i[ν]
∂x^i, ∂H
∂p^k[ν] = ∂y^ν
As pointed out by Krupka and ˇStˇep´ankov´a, the above results directly apply to the Einstein–Hilbert Lagrangian (scalar curvature) of the General Relativity Theory (for explicit computations see
[26]). The above ideas were applied to study also some other kinds of higher order Lagrangians with projectable Poincar´e–Cartan forms by [10] (cf. also comments in [11]).
(iii) Dedecker’s approach to first order Hamiltonian field theory. In [5], Dedecker proposed a Hamilton theory for first order Lagrangians on contact elements. If transfered to fibered manifolds, it
becomes a “nonstandard” Hamiltonian theory.
The core is to consider Hamilton equations of the form
(6.26) δ^∗iξdρ= 0,
whereρis a Lepagean equivalent of a first order Lagrangian of the form ρ=θ[λ]+
1...σ[k]ω^σ^1∧ · · · ∧ω^σ^k∧ω[i][1][...i][k]. Dedecker showed that if the condition
(6.27) det ∂^2L
∂y[i]^σ∂y[j]^ν −g[σν]^ij 6= 0 | {"url":"https://123deta.com/document/y8gmr7jw-announce-results-cerning-hamiltonian-variational-problems-fibered-manifolds.html","timestamp":"2024-11-14T14:19:41Z","content_type":"text/html","content_length":"221490","record_id":"<urn:uuid:e979a1a2-688c-4b1d-852a-d4086a632d0d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00525.warc.gz"} |
Battery to resistance - Unveiling the Secrets of Electrical Systems
Battery to resistance – Unveiling the Secrets of Electrical Systems
Impedance is a fundamental concept in electrical engineering that plays a crucial role in the functioning of electronic devices. It refers to the opposition of a circuit to the flow of alternating
current (AC). In simple terms, impedance can be thought of as the total resistance in a circuit, which includes both resistance and reactance. Understanding the relationship between impedance and the
electrical source, such as a battery or power supply, is essential for evaluating the performance of electronic devices.
When it comes to electronic devices, the battery is a common electrical source used to power various devices, from smartphones to laptops. The battery works as a portable power supply, converting
chemical energy into electrical energy. The electrical energy is then utilized by the devices to perform their functions. The battery’s internal resistance is an important parameter to consider as it
affects the overall performance and efficiency of the device.
Resistance, on the other hand, is a measure of how a material or component hinders the flow of current. In an electronic circuit, resistors are commonly used to regulate current or divide voltage.
The resistance value determines the amount of current flowing through a circuit and the amount of power dissipated in the process. When connected to a power source, the resistor creates a voltage
drop across its terminals, absorbing some of the electrical energy.
The relationship between the battery and resistance can have significant implications for electronic devices. Higher resistance in a circuit results in greater voltage drops, which can lead to
decreased performance or even malfunctioning of the device. Moreover, high resistance can cause excessive heat generation, potentially damaging the device or reducing its lifespan. Therefore,
selecting the appropriate resistor value is crucial in designing electronic circuits to ensure optimal performance and longevity of the devices.
Importance of Understanding Battery and Resistance Relationship
When it comes to electronic devices, understanding the relationship between the battery and resistance is crucial. The battery serves as the electrical source or power supply, while resistance refers
to the impediment or opposition to the flow of electric current in a circuit.
Why is it important?
Battery: The battery plays a vital role in providing the necessary power for electronic devices to function. It acts as the power supply, converting chemical energy into electrical energy that is
then utilized by the device. Without a proper understanding of the battery’s capacity and limitations, it can lead to inefficient power supply and even damage the device.
Resistance: On the other hand, resistance determines the flow of current in a circuit. The presence of resistors in a circuit helps regulate the flow of electricity, preventing any damage to the
components. Understanding resistance is crucial to ensure that the correct amount of power is supplied to the device, preventing overheating or malfunctioning.
The relationship between battery and resistance:
There is a direct relationship between the battery and resistance in an electronic circuit. The battery supplies power to the circuit, and the resistance determines how much of that power is utilized
by the device. Higher resistance in a circuit means that more power is consumed by the resistors and less is available for the device. Conversely, lower resistance allows more power to flow through
the circuit, supplying the device with the necessary power.
By understanding this relationship, electronic device designers and engineers can optimize the power supply and ensure that the device operates efficiently. They can select the appropriate battery
and resistance values to meet the device’s power requirements and avoid any potential issues.
In conclusion, understanding the battery and resistance relationship is crucial for the proper functioning of electronic devices. It helps optimize power supply, prevent damage, and ensure efficient
operation. Designers and engineers must consider these factors when developing electronic devices to deliver a reliable and satisfactory user experience.
Basics of Electronic Devices
In order to understand the relationship between a battery and resistance in electronic devices, it is important to first grasp the basics of electronic devices themselves.
An electronic device typically relies on an electrical source such as a cell or battery to supply power. This power supply is necessary for the device to function. Additionally, electrical devices
often consist of various components such as resistors, capacitors, and transistors that work together to process and control the flow of electricity.
One key component in many electronic devices is the resistor. A resistor is a device that opposes the flow of electrical current. It provides resistance to the flow of electrons and can help regulate
the amount of current passing through a circuit. Resistors are often used to control the current or voltage in a circuit, protect other components from excessive current, or provide specific levels
of resistance.
Another important concept in electronic devices is impedance, which can be thought of as the total opposition to the flow of alternating current. Impedance encompasses both resistance and reactance,
which is the opposition to the change in current or voltage caused by inductance or capacitance in a circuit.
Understanding the basics of electronic devices, such as the role of the power supply, resistors, and impedance, is crucial in comprehending the relationship between a battery and resistance in
electronic devices. It is this relationship that allows electronic devices to function properly and perform their intended tasks.
Power Supply for Resistance in Electronic Devices
In electronic devices, such as smartphones, laptops, and computers, a power supply is required to provide the necessary electrical energy to the resistors and other components. The power supply
serves as the source of electrical energy or voltage for these devices.
One common type of power supply used in electronic devices is a battery. A battery, also known as a cell, converts chemical energy into electrical energy and powers the device. The battery is
designed to supply a specific voltage, such as 3.7 volts or 5 volts, to the electronic components.
The resistance, measured in ohms, is a fundamental property of electronic components such as resistors. These components restrict or limit the flow of electrical current through them, converting
electrical energy into other forms of energy, such as heat. The resistance plays a critical role in determining the performance and behavior of electronic devices.
When a power supply is connected to a resistor or any electronic component with resistance, it creates an electrical circuit. The power supply acts as the electrical source, providing the necessary
voltage or electrical energy for the resistor to function. The power supply and the resistor are connected in series or parallel configuration, depending on the specific circuit design.
In some cases, the electrical source may not be a battery, but rather an external power supply connected to an electrical outlet. This type of power supply can provide a constant and stable voltage
to the electronic components, ensuring their proper operation.
The impedance, which includes the resistance, inductance, and capacitance of the circuit, also affects the power supply and resistance relationship. The impedance can impact the flow of current and
the overall performance of electronic devices.
In conclusion, the power supply is essential for providing the necessary electrical energy to the resistors and other electronic components in devices. The connection between the power supply and
resistance creates an electrical circuit, enabling the flow of current and determining the behavior of the electronic device. Understanding this relationship is crucial for designing and
troubleshooting electronic devices.
Role of Battery in Electronic Circuits
A battery is a vital component in electronic circuits as it serves as the power supply or electrical source. It provides the necessary energy to operate various electronic devices. The battery, in
combination with other circuit elements like resistors, plays a crucial role in determining the behavior of the circuit.
When a battery is connected to a circuit, it acts as an electrical source, supplying electrical energy to the circuit. The battery contains one or more cells, each producing a specific voltage. The
voltage generated by the battery determines the overall power supply of the circuit.
The resistance of the circuit components, such as resistors, in combination with the battery voltage, affects the flow of electric current through the circuit. The battery, with its specific voltage,
establishes the potential difference required for the flow of current. The resistance in the circuit controls the amount of current that flows through it, according to Ohm’s Law.
The battery’s ability to maintain a stable voltage and provide a constant source of power is important for the proper functioning of electronic devices. If the battery voltage decreases, the current
flowing through the circuit may decrease, affecting the performance of the device. Conversely, a higher voltage from the battery may cause excess current, potentially damaging the circuit or device.
The impedance of the battery, which is a measure of its internal resistance, also affects the behavior of the circuit. The internal resistance of the battery can impact the overall voltage supplied
to the circuit, especially under high load conditions. A battery with a lower internal resistance can supply a more stable voltage to the circuit, allowing for better performance and efficiency.
In summary, the battery is a crucial component in electronic circuits as it acts as the electrical source or power supply. Its voltage, in combination with the resistance of the circuit elements,
determines the flow of electric current and the behavior of the circuit. A stable and properly functioning battery is essential for the optimal performance of electronic devices.
How Resistance Affects Battery Life
When it comes to electrical devices, the power supply is essential. One common source of power is a cell or a battery, which converts chemical energy into electrical energy. However, various factors
can affect the efficiency and longevity of a battery, and one of these factors is resistance.
Resistance refers to the opposition that an electrical circuit offers to the flow of current. It is measured in ohms and is represented by the symbol Ω. In simple terms, resistance can be thought of
as a resistor within the circuit that impedes the flow of electrical current.
When resistance is introduced into a circuit, it increases the amount of power required from the battery to supply the same amount of current. This increase in power consumption directly affects the
battery life. Higher resistance means more power is wasted in the form of heat, reducing the overall efficiency of the device.
For example, let’s consider a circuit with a low resistance value and a circuit with a high resistance value. In the circuit with low resistance, the electrical current flows easily, requiring less
power from the battery. On the other hand, in the circuit with high resistance, the electrical current encounters more opposition, resulting in a higher power demand from the battery.
In practical terms, this means that devices with higher resistance can drain the battery more quickly. This is especially important to consider in devices that rely heavily on battery power, such as
smartphones or portable electronic devices.
To maximize battery life, it is crucial to minimize resistance within the circuit. This can be achieved by using high-quality conductive materials, reducing the length and thickness of electrical
paths, and minimizing the number of electrical connectors or junctions.
Resistance Battery Life
Low Longer
High Shorter
By understanding how resistance affects battery life, designers and engineers can make informed decisions to optimize device performance and enhance battery efficiency.
Factors Influencing Battery Performance in Relation to Resistance
When it comes to the performance of a battery in relation to resistance, there are several key factors to consider. Understanding these factors can help improve the efficiency and longevity of
electronic devices powered by batteries.
1. The Role of a Resistor
A resistor is an important component in an electrical circuit that limits the flow of current. It helps to regulate the amount of electrical power supplied to a device, preventing overload and
ensuring optimal performance. The resistance value of a resistor is measured in ohms (Ω) and plays a crucial role in determining the overall performance of a battery.
2. Power Supply from the Battery
The power supply from a battery is influenced by its internal resistance. As the resistance of the battery increases, the power supplied to the connected electrical device decreases. This can result
in a decrease in the overall performance and efficiency of the device.
3. Cell Design and Impedance
The design of the battery cell and its internal impedance can also impact battery performance. A high internal impedance can lead to a voltage drop and decreased power output, which can affect the
performance of electronic devices.
4. Electrical Source and Resistance
The electrical source used to charge a battery also plays a role in its performance. If the electrical source has a high resistance, it can result in slower charging times and decreased battery life.
It is important to use a power source with low resistance to ensure efficient charging and optimal battery performance.
In conclusion, understanding the relationship between a battery and resistance is critical for maximizing the performance and lifespan of electronic devices. Factors such as resistor values, power
supply, cell design, and the electrical source used all influence battery performance. By considering these factors and implementing appropriate measures, it is possible to enhance the performance
and efficiency of electronic devices powered by batteries.
Impact of Resistance on Power Consumption
Resistance plays a significant role in determining the amount of power consumed by an electrical device. When a power supply, such as a battery, is connected to a resistor, it generates a flow of
current. This current passes through the resistor, and the resistance in the circuit affects the power consumption.
Power consumption can be calculated using the formula: P = I^2 * R, where P represents power, I represents current, and R represents resistance. As resistance increases, the power consumption in the
circuit also increases.
A higher resistance means that more energy is required to maintain the electrical flow through the resistor. This increased energy demand leads to an increase in power consumption. Conversely, a
lower resistance allows for easier flow of current and therefore results in lower power consumption.
Understanding the impact of resistance on power consumption is crucial for designing efficient electronic devices. By carefully selecting the appropriate resistor values for a circuit, power can be
conserved and efficiency can be improved. Furthermore, when resistance is too high in a device, it can cause voltage drops across different components, leading to reduced performance and potential
In summary, resistance has a direct impact on power consumption in electrical devices. Higher resistance results in increased power consumption, while lower resistance leads to lower power
consumption. It is essential to consider the choice of resistors in electronic circuits to optimize power usage and ensure proper functioning of the device.
Circuit Design Considerations for Optimal Resistance-Battery Relationship
When designing an electronic circuit, it is crucial to consider the relationship between the power supply, resistor, and electrical source. The resistance of a circuit plays a significant role in
determining how much power is drawn from the battery.
The resistance of a circuit is determined by the type and size of the resistor used. A higher resistance will result in a lower current flowing through the circuit, which reduces the amount of power
drawn from the battery. Conversely, a lower resistance will result in a higher current and more power being drawn from the battery.
It is important to match the resistance of the circuit to the power supply or battery being used. If the resistance is too high, the circuit may not receive enough power to function properly. On the
other hand, if the resistance is too low, the circuit may draw too much power and cause the battery to discharge more quickly.
One consideration when designing a circuit is the power supply or battery voltage. Different power sources have different voltage levels, and the resistance of the circuit should be chosen
accordingly. A higher voltage requires a higher resistance to prevent excessive current flow, while a lower voltage may require a lower resistance.
Another consideration is the impedance of the circuit. Impedance is a measure of the opposition to the flow of alternating current in a circuit. It is affected by both resistance and reactance, which
is determined by the characteristics of the circuit elements. It is important to consider the impedance of the circuit when choosing the resistance, as it can affect the overall performance and
efficiency of the circuit.
In conclusion, when designing an electronic circuit, it is essential to consider the relationship between the resistance and the power supply or battery. Choosing the optimal resistance for the
circuit will ensure that the circuit functions properly and draws an appropriate amount of power from the battery or power source.
Understanding the Electrical Source for Resistors
In order to understand the relationship between a battery and a resistor, it is important to first understand the concept of an electrical source. An electrical source, such as a battery or power
supply, provides the necessary energy for a circuit to function.
When a battery is connected to a circuit, it acts as an electrical source, delivering a steady supply of voltage. The voltage provided by the battery determines the amount of potential energy that is
available to power the circuit.
However, in addition to voltage, an electrical source also has an internal resistance known as its impedance. This impedance can affect the flow of current through the circuit and impact the
performance of electronic devices connected to it.
When a resistor is connected to an electrical source, the resistance of the resistor limits the flow of current in the circuit. The resistor absorbs some of the electrical energy from the source and
converts it into heat.
The relationship between a battery and a resistor can be described using Ohm’s law, which states that the current flowing through a resistor is directly proportional to the voltage across it, and
inversely proportional to the resistance of the resistor.
By controlling the resistance in a circuit, resistors allow engineers to modify the electrical current and voltage levels to suit the needs of different electronic devices. They can be used to
regulate the flow of electrical power, protect sensitive components, or provide a specific value of resistance for different applications.
Understanding the electrical source for resistors is crucial for designing and optimizing electronic devices. By considering the voltage, impedance, and resistance, engineers can create circuits that
efficiently utilize electrical power and ensure the proper functioning of electronic components.
Importance of Proper Power Supply for Resistors
A resistor is an essential component in many electronic devices, playing a crucial role in controlling the flow of electrical current. In order to function properly, a resistor requires a proper
power supply.
A power supply for a resistor can come from various sources, with the most common being a battery or an electrical source. Regardless of the source, it is crucial to ensure that the power supply is
appropriate for the resistor’s specifications.
One of the main factors to consider when choosing a power supply for a resistor is the resistance of the resistor itself. The resistance of a resistor is the measure of its opposition to the flow of
electrical current. To achieve the desired resistance, it is important to select a power supply that can deliver the appropriate amount of voltage.
Another factor to consider is the impedance of the power supply. The impedance is a measure of the opposition to the flow of alternating current. If the impedance of the power supply is too high, it
can negatively affect the performance of the resistor and the overall functionality of the electronic device.
Choosing the right power supply for a resistor also involves considering the electrical characteristics of the electrical source. It is important to ensure that the voltage and current capabilities
of the source align with the requirements of the resistor.
Furthermore, a proper power supply ensures the longevity and reliability of the resistor. If the power supply is too low or too high, it can cause the resistor to overheat or fail prematurely. This
can lead to costly repairs or replacements.
In conclusion, selecting a proper power supply for resistors is of utmost importance. It ensures that the resistance is within the desired range, that the impedance is optimal, and that the
electrical characteristics of the source are compatible. By paying attention to these factors, one can ensure the efficient and reliable operation of electronic devices.
Role of Batteries in Impedance
Impedance is an important electrical property that plays a crucial role in determining the performance of electronic devices. It is the total opposition that an electrical source, such as a battery,
offers to the flow of current through a circuit. Batteries are essential for powering various electronic devices, and understanding their role in impedance is key to designing efficient electrical
When a battery is connected to a circuit, it acts as the power supply, providing the necessary electrical energy to drive the flow of current. The battery has a certain internal resistance,
represented by a resistor in circuit diagrams, which limits the amount of current that can flow through the circuit. This internal resistance is typically small compared to the external components of
the circuit.
The impedance of the battery affects the overall impedance of the circuit. It adds to the total impedance, influencing the flow of current and voltage across the circuit. The impedance of the battery
can be affected by factors such as its chemistry, temperature, and age.
Electrical devices often have specific requirements for the power supply, including a certain impedance range. If the impedance of the battery is too high or too low, it can lead to inefficiencies in
the circuit and affect the performance of the device. Therefore, selecting the right battery with an appropriate impedance is crucial to ensure optimal operation.
In addition to the impedance of the battery itself, the impedance of the electrical components connected to the battery also impacts the overall impedance of the circuit. It is important to consider
the impedance of all components, including resistors, capacitors, and inductors, when designing a circuit.
A thorough understanding of the role of batteries in impedance is essential for engineers and designers working with electrical systems. By carefully considering the impedance of the battery and the
other components in the circuit, it is possible to optimize the performance and efficiency of electronic devices.
Terms Definitions
Resistance The opposition to the flow of electrical current in a circuit.
Cell A device that converts chemical energy into electrical energy.
Power supply A device that provides electrical energy to a circuit.
Battery A portable device that consists of one or more cells for generating electrical energy.
Electrical source A device that can supply electrical energy to a circuit.
Resistor An electrical component that limits the flow of current in a circuit.
Impedance The total opposition that an electrical source offers to the flow of current through a circuit.
How Battery Characteristics Impact Impedance
Impedance, or the opposition to the flow of electrical current, is an important consideration in electronic devices. It determines how easily power can be transferred from the power supply or source,
such as a battery, to a resistor or other load. The characteristics of the battery play a significant role in determining the overall impedance of a circuit.
Battery Voltage and Impedance
One of the key factors that impact impedance is the voltage of the battery. The voltage represents the potential difference between the positive and negative terminals of the battery. Higher voltage
batteries tend to have lower internal resistance, which can result in lower overall impedance. This means that more power can be delivered to the load.
Battery Capacity and Impedance
The capacity of a battery, or the amount of charge it can hold, also affects impedance. Batteries with higher capacity can deliver more power for a longer period of time, which can result in lower
impedance. This is because the internal resistance of the battery is spread out over a larger capacity, reducing the overall impedance.
It is important to consider the specific requirements of a device when selecting a battery. If a device requires a high level of power or has low tolerance for voltage drops, a battery with lower
impedance characteristics may be necessary. On the other hand, devices with lower power requirements may be able to function effectively with a battery that has higher impedance.
Understanding how battery characteristics impact impedance can help in designing and selecting power sources for electronic devices. By considering the voltage and capacity of a battery, engineers
can ensure that the power supply meets the specific electrical requirements of a circuit.
Significance of Correct Resistance in Impedance Calculations
In electronic devices, resistance plays a vital role in impedance calculations. Understanding the relationship between resistance and impedance is key to designing efficient circuits and ensuring
proper functioning of electrical devices.
Resistor as a Basic Component
A resistor is a passive electrical component that limits the flow of electric current through a circuit. It is commonly used to control the amount of current flowing in a circuit and to protect
sensitive components. By adding a resistor to a circuit, the resistance can be adjusted, allowing for control over the current flow.
Impedance and its Calculation
Impedance is the total opposition to the flow of an alternating current (AC) in a circuit. It includes both resistance and reactance, which is the opposition due to capacitance and inductance.
Impedance is represented by the symbol Z and is measured in ohms (Ω).
When calculating impedance, the resistance is a crucial factor. The correct resistance value must be used to accurately determine the total impedance in a circuit. When the wrong resistance is used,
it can lead to incorrect calculations and result in inefficient circuit performance.
For example, in a power supply circuit, if the resistance is too high, it can cause a voltage drop, impacting the overall performance of the circuit. On the other hand, if the resistance is too low,
it can result in excessive current flow, potentially damaging the components or causing overheating.
Choosing the Right Resistance for the Power Supply
When designing a power supply circuit, selecting the correct resistance is crucial for maintaining optimal performance and efficiency. The resistance should be chosen based on factors such as the
desired power output, voltage requirements, and the characteristics of the electrical source, such as the battery or cell.
The correct resistance can help regulate the current flow, prevent short circuits, and ensure a stable and reliable power supply. It is important to consult the datasheets and specifications of the
components used in the circuit to determine the appropriate resistance values.
In conclusion, understanding the significance of correct resistance in impedance calculations is essential for proper circuit design and efficient performance of electronic devices. Using the right
resistance values in calculations and circuit designs ensures optimal current flow, prevents voltage drops or excessive current flow, and helps maintain the overall stability and reliability of the
power supply.
Effect of Battery Voltage on Impedance
In the realm of electronics, understanding the effect of battery voltage on impedance is crucial. Impedance refers to the opposition to the flow of electrical current in a circuit. This opposition is
caused by various factors, including the resistance of the circuit components.
A battery, as an electrical source, plays a significant role in determining the impedance of a circuit. A battery, also known as a power supply or cell, provides the necessary power to drive
electronic devices. The voltage of the battery affects the amount of power it delivers to the circuit, which in turn influences the impedance.
When a battery with lower voltage is connected to a circuit, the power supplied to the circuit reduces. This reduction in power lowers the overall impedance of the circuit. On the other hand,
connecting a battery with higher voltage increases the power supplied and consequently increases the impedance. Therefore, the voltage of the battery directly affects the impedance in a circuit.
Impedance is closely related to resistance in a circuit. Resistance is a property of a component that opposes the flow of current. Impedance, on the other hand, takes into account both resistance and
reactance, which is the opposition caused by inductive or capacitive components in the circuit. So, the relationship between battery voltage, resistance, and impedance is a fundamental concept in
electronic circuits.
Understanding the effect of battery voltage on impedance is essential when designing or troubleshooting electronic circuits. By carefully selecting the appropriate battery voltage for a specific
circuit, the desired impedance can be achieved. This ensures optimal performance and compatibility with the components used in the circuit.
Impedance Matching and Battery Selection
The electrical source in an electronic device, such as a power supply or battery, must be able to supply enough power to the electrical load, often represented by a resistor. For the device to
function properly, the impedance of the electrical source and the electrical load must be matched.
When the impedance is mismatched, it can lead to a loss of power and decreased efficiency. If the impedance of the electrical load is too high for the battery’s internal impedance, the battery may
not be able to supply the required power, leading to a decrease in performance. On the other hand, if the impedance of the electrical load is too low, the battery may supply more power than required,
potentially damaging the device.
Therefore, selecting the correct battery for a specific electrical load is crucial. The battery’s internal impedance should be matched to the impedance of the electrical load to maximize power
transfer and ensure the device operates efficiently. To achieve this, it is essential to consider the characteristics of the battery, such as its voltage, capacity, and internal resistance, and match
them appropriately to the requirements of the electrical load.
Additionally, the battery should be able to supply enough power to meet the device’s demand. If the battery’s power supply capability is insufficient, the device may not function as intended, or it
may cause the battery to drain quickly. Therefore, it is important to select a battery with an appropriate power rating to ensure that it can deliver the required power to the electrical load and
maintain optimal performance.
Calculation of Impedance with Battery and Resistance Values
When it comes to the power and electrical properties of a circuit, understanding the relationship between a battery and resistance is crucial. The impedance, or total opposition to the flow of
electrical current, can be calculated with the values of the battery and the resistance in the circuit.
The battery, as an electrical source, provides the energy for the circuit. It supplies a voltage, typically measured in volts, which drives the current through the circuit. On the other hand, the
resistor, a component with a specific value of resistance, limits the flow of current. The resistance is measured in ohms.
To calculate the impedance of a circuit, you need to know the voltage supplied by the battery and the resistance in the circuit. The formula to calculate impedance is as follows:
Impedance (Z) = √(Resistance (R)^2 + Reactance (X)^2)
Where the reactance (X) represents the opposition to the flow of the alternating current (AC) caused by inductance or capacitance. However, in the case of a simple direct current (DC) circuit, the
reactance is usually negligible and can be disregarded.
By plugging in the values of the battery voltage and resistance into the formula, you can determine the impedance of the circuit. This information is crucial for understanding how the circuit behaves
and for making adjustments or calculations related to power consumption, voltage drops, and current flow.
Understanding the calculation of impedance helps you to analyze and troubleshoot electronic devices more effectively. By knowing the values of the battery and resistance in a circuit, you can
determine the impedance and make informed decisions about circuit design, component selection, and power requirements.
In summary, the calculation of impedance with battery and resistance values is an essential aspect of understanding the power and electrical properties of a circuit. This information enables
engineers and technicians to analyze and optimize electronic devices for optimal performance and reliability.
Impact of Battery Life on Impedance
One important factor that affects the impedance of an electronic device is the battery life. The battery serves as the power source for most portable electronic devices, providing the necessary
electrical energy for their operation. As the battery depletes its power, the impedance of the device can be affected.
Impedance is a measure of the opposition to the flow of electric current in a circuit. It includes both resistance and reactance components. In the case of a battery-powered device, the impedance can
be influenced by the internal resistance of the battery, as well as the resistance of other components in the circuit.
When a battery is fully charged, it has a low internal resistance. This means that it can supply a higher amount of current to the device, resulting in a lower impedance. As the battery discharges
and its power supply decreases, the internal resistance of the battery increases. This higher resistance can cause a higher impedance in the circuit, which can affect the performance of the
electronic device.
Effects on Performance
A higher impedance in an electronic device can have several effects on its performance. Firstly, it can lead to a decrease in power delivery. As the impedance increases, the voltage drop across the
internal resistance of the battery also increases. This can result in a lower voltage at the load, reducing the power available to the device.
Secondly, a higher impedance can cause voltage fluctuations and instability in the circuit. This can result in the device not receiving a consistent power supply, leading to potential malfunction or
unreliable performance.
Furthermore, a higher impedance can also affect the overall efficiency of the device. If the impedance is too high, it can cause the device to draw more current from the battery to compensate,
increasing power consumption and reducing battery life.
Managing Battery Life and Impedance
To manage the impact of battery life on impedance, it is important to monitor and optimize the power consumption of electronic devices. This can be achieved by implementing efficient power management
techniques, such as regulating the voltage and current levels to match the requirements of the device.
In addition, using high-quality batteries with low internal resistance can help minimize the impact on impedance. Regularly checking and replacing batteries with worn-out ones can also prevent
impedance-related issues.
Overall, understanding and managing the relationship between battery life and impedance is crucial for ensuring the optimal performance and longevity of electronic devices.
The Relationship Between Cell and Impedance
Impedance is an important concept in electrical engineering, especially when it comes to understanding the relationship between a cell (battery) and resistance in an electrical circuit. In simple
terms, impedance refers to the opposition that an electrical source, such as a battery, faces when trying to supply power to a resistor.
When a cell is connected to a resistor, it acts as a source of electrical energy. The resistance of the resistor determines the amount of opposition or impedance that the cell will encounter while
trying to supply power to the circuit. The relationship between the cell and impedance can be described using Ohm’s Law:
Ohm’s Law
• According to Ohm’s Law, the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance
between them. Mathematically, this can be expressed as:
V = I * R
• V is the voltage across the two points (in volts)
• I is the current flowing through the conductor (in amperes)
• R is the resistance of the conductor (in ohms)
This relationship helps us understand how the impedance of a circuit affects the flow of current from a cell. A higher impedance means greater opposition to the flow of current, which results in a
lower current through the circuit. Similarly, a lower impedance allows for a higher current to flow.
It is important to note that the impedance not only depends on the resistance of the resistor but also on the frequency of the electrical signal. In AC circuits, impedance is a complex quantity that
takes into account both resistance and reactance.
In summary, the relationship between a cell and impedance is crucial in understanding how electrical energy is supplied to a circuit. By considering the resistance and frequency, engineers can design
circuits that efficiently utilize the power supplied by a cell based on the impedance requirements.
Importance of Cell Voltage in Relation to Impedance
The cell voltage of an electrical source, such as a battery or power supply, plays a crucial role in determining the behavior of electrical devices connected to it. The voltage of a cell refers to
the potential difference between its terminals, which is responsible for driving current through a circuit.
When a current flows through a circuit, it encounters resistance from various components in the circuit, such as resistors. The relationship between the voltage of a cell and the resistance in a
circuit is governed by Ohm’s Law, which states that the current passing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. In
mathematical terms, this can be expressed as:
I = V / R
• I is the current flowing through the circuit
• V is the voltage applied by the cell
• R is the resistance of the circuit
From this equation, it is evident that the cell voltage has a direct impact on the current flowing through the circuit. A higher cell voltage will result in a higher current, provided the resistance
remains constant. Similarly, a lower cell voltage will result in a lower current.
The impedance of a circuit, which is an extension of resistance, takes into account not only the resistance but also the reactance caused by capacitors and inductors in the circuit. Impedance is a
complex quantity and is typically represented as a combination of resistance and reactance.
Understanding the importance of cell voltage in relation to impedance is crucial for the proper functioning of electronic devices. The impedance of a circuit determines the amount of power that can
be transferred from the electrical source to the load. A mismatch between the impedance of the source and the load can result in poor power transfer and potential damage to the devices.
By considering the cell voltage and the impedance of a circuit, engineers can design power supplies and batteries that are compatible with the intended load. This ensures optimal performance and
reliability of the electronic devices.
Determining Cell Voltage for Proper Impedance Control
When it comes to understanding the relationship between a battery and resistance, it is crucial to determine the cell voltage of the power supply. The cell voltage is the electrical potential
difference between the positive and negative terminals of a battery.
The electrical source in an electronic device is primarily the battery, which provides the power needed for the device’s operation. The cell voltage of the battery determines the maximum electrical
potential that can be supplied by the power source.
Impedance control is important in electronic devices as it ensures a proper and stable power supply to the rest of the electrical components. Impedance refers to the resistance encountered by the
electrical current as it passes through the components.
Source and Resistor
To properly control impedance, it is necessary to understand the relationship between the battery, power supply, and resistance. A resistor is often used to control the flow of electrical current in
a circuit.
By placing a resistor in series with the electrical source, the resistance can be adjusted to achieve the desired impedance. This allows for a regulated power supply that can meet the requirements of
the electronic device.
Determining Cell Voltage
To determine the cell voltage of a battery, a multimeter can be used. It measures the potential difference between the positive and negative terminals of the battery. This measurement represents the
cell voltage, which is typically expressed in volts (V).
By knowing the cell voltage, the appropriate resistor can be chosen to achieve the desired impedance control. This ensures that the power supply is stable and reliable, preventing any potential
damage to the electronic components.
In conclusion, understanding the cell voltage of a battery is crucial for proper impedance control in electronic devices. By determining the cell voltage and selecting the appropriate resistor, the
power supply can be regulated to provide a stable electrical current, ensuring the optimal performance of the device.
Effects of Resistance on Cell Performance
When it comes to the performance of a cell or battery, the resistance in the electrical circuit plays a significant role. Resistance is the opposition to the flow of electrical current and it can
have various effects on the overall performance of a cell.
1. Voltage Drop
Resistance in a circuit causes a voltage drop across the resistor. This means that the voltage supplied by the cell will be reduced as it passes through a resistor. As a result, the electrical device
connected to the power supply will receive a lower voltage than what the cell is capable of providing. This can affect the device’s functionality and may lead to reduced performance.
2. Power Dissipation
Resistance in a circuit also leads to power dissipation. When current passes through a resistor, some of the electrical energy is converted into heat. This can cause the resistor to heat up and waste
power, reducing the overall efficiency of the electrical source. In the case of a battery, this power dissipation can lead to a shorter battery life and the need for more frequent replacements.
To compensate for the effects of resistance, it is common to use a resistor in the circuit to control the flow of current. By carefully selecting the resistance value, the voltage drop and power
dissipation can be minimized, thus optimizing the performance of the cell.
It is important to consider the effects of resistance on cell performance when designing and using electrical devices. By understanding the relationship between resistance and electrical impedance,
one can ensure that the power supply is properly regulated and that the device operates optimally.
Understanding the Impact of Resistance on Cell Efficiency
When it comes to powering electronic devices, a stable and reliable power supply is crucial. Whether it’s a battery or any other electrical source, the presence of resistance in the circuit can
significantly impact the overall efficiency of the cell.
Resistance is essentially an impedance that opposes the flow of electrical current. In the context of a cell or battery, resistance can be introduced by various components such as wires, connectors,
and internal elements. This resistance results in a voltage drop across the cell, which reduces the available power for the electronic device.
The relationship between resistance and cell efficiency is straightforward. The higher the resistance in the circuit, the lower the efficiency. This is because a higher resistance implies a larger
voltage drop across the cell, leading to a reduced power output. In other words, a portion of the power generated by the cell is wasted in overcoming the resistance.
To illustrate this, consider a simple circuit with a battery, a resistor, and an electronic device. If the resistance in the circuit is low, the voltage drop across the resistor is minimal, allowing
more power to reach the device. However, if the resistance is high, a significant portion of the power is dissipated as heat across the resistor, resulting in reduced power available for the device.
It is worth noting that different types of batteries have varying internal resistances, which can impact the overall cell efficiency. For example, lithium-ion batteries generally have lower internal
resistance compared to alkaline batteries. This lower resistance allows for better power delivery and higher efficiency.
When designing or using electronic devices, it is essential to consider the impact of resistance on cell efficiency. Minimizing resistance in the circuit through proper wiring, connectors, and
component selection can help optimize power delivery and improve overall device performance.
In conclusion, resistance plays a vital role in determining the efficiency of a cell or battery. Understanding the relationship between resistance and cell efficiency is crucial for engineers,
designers, and users alike. By minimizing resistance in the circuit, the power delivered to the electronic device can be maximized, resulting in improved performance and longer battery life.
Battery and Resistance in Portable Electronic Devices
In portable electronic devices, such as smartphones and laptops, the battery serves as the primary electrical source. It provides the necessary power supply for the device to operate efficiently.
However, the battery alone is not sufficient to power the entire device. It needs additional components, such as resistors, to ensure a stable and controlled flow of electrical current.
The Role of the Resistor
A resistor is an essential component that regulates the flow of electric current in a circuit. It offers a specific amount of resistance to the flow of electrons, thereby controlling the amount of
power supplied to the various components of the portable electronic device.
The resistance of a resistor is measured in ohms (Ω). By adjusting the resistance, the voltage and current can be controlled within the device. The resistors help protect sensitive components from
receiving too much power and getting damaged. They also ensure that the device functions optimally within its power limits.
Battery and Resistance Relationship
The battery and resistance in portable electronic devices have an interconnected relationship. The resistance determines how much current is allowed to flow from the battery to the different
components. A higher resistance will restrict the flow of current, while a lower resistance will allow more current to pass through.
The power supplied by the battery is calculated using Ohm’s Law, which states that power (P) is equal to the square of the current (I) multiplied by the resistance (R). Therefore, the power supplied
to the circuit depends on the resistance and the current flowing through it.
Resistance Current Power
High Low Low
Low High High
By carefully selecting the resistance values in the circuit, the device’s power consumption can be optimized. This is crucial in portable electronic devices as it directly affects the battery life.
By managing the resistance, the device can efficiently utilize the power supplied by the battery and extend its operating time.
In summary, the battery and resistance play key roles in portable electronic devices. The battery acts as the power supply, while the resistors control and regulate the flow of electric current.
Understanding the relationship between battery and resistance is essential in designing electronic devices to ensure optimal performance and maximize battery life.
Optimizing Battery and Resistance for Longer Device Life
In order to maximize the lifespan of electronic devices, a proper understanding of the relationship between the power supply and resistance is essential. The electrical source that provides power to
these devices is typically a battery, which serves as the energy storage cell. The resistor, on the other hand, is a component that helps regulate the flow of electrical current through the device.
By optimizing both the battery and the resistance, we can ensure that the device operates efficiently and has a longer lifespan.
When selecting a battery for a device, it is important to consider its impedance, which affects the amount of power it can supply. The impedance of a battery is a measure of its internal resistance
and determines how well it can deliver power to the device. A battery with a low impedance will be able to provide a steady and reliable power supply, while a battery with a higher impedance may
struggle to deliver power consistently.
In addition to selecting a battery with the appropriate impedance, the resistance in the circuit must also be carefully considered. The resistance plays a crucial role in determining the overall
power consumption of the device. If the resistance is too low, the device may draw more power than necessary, causing the battery to drain quickly. On the other hand, if the resistance is too high,
the device may not receive enough power to operate properly.
Key Considerations for Optimizing Battery and Resistance:
1. Matching the impedance of the battery to the power requirements of the device is essential for efficient power delivery.
2. Choosing an appropriate resistance value will help balance power consumption and ensure the device operates within safe limits.
By carefully considering the impedance of the battery and the resistance in the circuit, we can optimize the power supply to electronic devices and extend their lifespan. This understanding allows us
to design and use batteries and resistors that provide the necessary power for devices, without draining the battery too quickly or risking damage to the device.
Question and Answer:
What is the relationship between a battery and resistance?
The relationship between a battery and resistance is that the battery provides the power or voltage needed for the flow of current through a resistor. Without a power source like a battery, the
resistor would not be able to function.
How does a power supply impact the resistance of a device?
A power supply is essential for providing the necessary voltage to overcome the resistance in a device. If the power supply does not provide enough voltage, the current flow through the device will
be limited and the resistance will have a greater impact on the device’s performance.
What is the role of an electrical source in relation to a resistor?
An electrical source, such as a battery, provides the energy needed to overcome the resistance in a resistor. Without this source, there would be no potential difference or voltage across the
resistor, and no current would flow through it.
How does a cell impact the impedance of a circuit?
A cell, or battery, is responsible for providing the voltage needed to overcome the impedance in a circuit. The impedance is related to the resistance in the circuit, but also takes into account
other factors such as capacitance and inductance. The cell’s voltage helps to overcome these impedances and ensure the proper functioning of the circuit.
Why is it important for electronic devices to have a power supply for resistance?
A power supply is crucial for electronic devices because it provides the necessary voltage to overcome the resistance in the device. Without a power supply, the resistance would limit the current
flow and prevent the device from functioning properly. The power supply ensures that the device receives the required energy to operate efficiently.
What is the relationship between a battery and resistance?
The relationship between a battery and resistance is essential for the functioning of electronic devices. When a battery is connected to a resistance, it creates a flow of electric current through
the resistance. The amount of current flowing through the resistance depends on the voltage provided by the battery and the resistance value. This relationship is commonly described by Ohm’s Law: I =
V/R, where I is the current, V is the voltage, and R is the resistance. | {"url":"https://pluginhighway.ca/blog/battery-to-resistance-unveiling-the-secrets-of-electrical-systems","timestamp":"2024-11-08T03:14:09Z","content_type":"text/html","content_length":"101822","record_id":"<urn:uuid:5823d283-2954-49a3-8ff9-fb707b191369>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00071.warc.gz"} |
Y Intercept - Meaning, Examples | Y Intercept Formula
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Y Intercept
The y-intercept is the point where the graph intersects the y-axis. To graph, any function that is of the form y = f(x) finding the intercepts is really important. There are two types of intercepts
that a function can have. They are the x-intercept and the y-intercept. An intercept of a function is a point where the graph of the function cuts the axis.
Let us understand the meaning of the y-intercept along with its formula and examples.
1. What is Y-Intercept?
2. Y-Intercept Formula
3. Y-Intercept of a Straight Line
4. How To Find The Y-Intercept?
5. Y-Intercept of a Quadratic Function (Parabola)
6. FAQs on Y-Intercept
What is Y Intercept?
The y intercept of a graph is the point where the graph intersects the y-axis. We know that the x-coordinate of any point on the y-axis is 0. So the x-coordinate of a y-intercept is 0.
Here is an example of a y-intercept. Consider the line y = x+ 3. This graph meets the y-axis at the point (0,3). Thus (0,3) is the y-intercept of the line y = x+ 3.
Y-Intercept Formula
The y-intercept of a function is a point where its graph would meet the y-axis. The x-coordinate of any point on the y-axis is 0 and we use this fact to derive the formula to find the y-intercept.
i.e., the y-intercept of a function is of the form (0, y). Thus, the formula to find the y-intercept is:
Y Intercept Formula
Here are the steps to find the y intercept of a function y = f(x),
• we just substitute x = 0 in it.
• solve for y.
• represent the y-intercept as the point (0, y).
Here are some examples of y intercepts.
• The y-intercept of y = 5x^2 + 2 is, (0, 2) because when we substitute x = 0, we get y = 5(0)^2 + 2 = 2.
• The y-intercept of y = -5e^x is (0, -5) because when we substitute x = 0, we get y = -5e^0 = -5.
Y-Intercept of a Straight Line
A straight line can be horizontal or vertical or slanting. The y intercept of a horizontal line with equation y = a is (0, a) and the y intercept of a vertical line does not exist. Let us learn to
find the y intercept of a straight line represented in different forms.
Y-Intercept in General Form
The equation of a straight line in general form is ax+by+c=0. For y-intercept, we substitute x=0 and solve it for y.
a(0)+ by + c =0
by + c =0
y = - c/b
Thus, the y-intercept of the equation of a line in general form is:(0, -c/b) or -c/b
y-Intercept in Slope-Intercept Form
The equation of the line in the slope-intercept form is, y=mx+b. By the definition of the slope-intercept form itself, b is the y-intercept of the line. You try by substituting x=0 in y=mx+b and see
whether you get b as the y-intercept. Thus, the y-intercept of the equation of a line in the slope-intercept form is (0, b) or b.
y-Intercept in Point-Slope Form
The equation of the line in the point-slope form is, y-y₁=m x-x₁. For y-intercept, we substitute x=0 and solve it for y.
y-y₁ = m (0-x₁ )
y-y₁= - mx₁
y = y₁ - mx₁
Thus, the y-intercept of the equation of a line in the point-slope form is: (0, y₁ - mx₁) or y₁ - mx₁
How To Find Y-Intercept?
We have derived the formulas to find the y-intercept of a line where the equation of the straight line is in different forms. In fact, we do not need to apply any of these formulas to find the
y-intercept of a straight line. The y-intercept of the polynomial function of the form y = a[1]x^n + a[2] x^n-1+ ... + a[n] is just its constant term a[n] (or) (0, a[n]).
We just substitute x=0 in the equation of the line and solve for y. Then the corresponding y-intercept is y or (0, y).
│Equation of Line│Substitute x=0 and solve for y │ y intercept │
│ │3(0)+5y-6 =0 │ │
│ │ │ │
│3x+5y -6=0 │5y-6=0 │6/5 (or) (0, 6/5)│
│ │ │ │
│ │y=6/5 │ │
│y= 2x-3 │y=2(0)-3=-3 │-3 (or) (0, -3) │
The y-intercept of a function can be easily found by graphing it using the graphing calculator and locating the point where the graph cuts the y-axis. A function has only one y-intercept because
otherwise, it fails the vertical line test. The y-intercept of the second equation of the table (y = 2x - 3) is shown in the graph below.
Y-Intercept of a Quadratic Function (Parabola)
The procedure for finding the y-intercept of a quadratic function or the y-intercept of a parabola is the same as that of a line (as discussed in the previous section). If a quadratic equation is
given, substitute x = 0 and solve for y to get the y intercept.
│ │ Substitute │ │
│Equation of Parabola │ x=0 │ y-Intercept │
│ │ and │ │
│ │ solve for y │ │
│y= x^2 -2x -3 │y=0^2-2(0)-3=-3 │-3 (or) (0, -3)│
│y= 2x^2+5x-3 │y=2(0)^2+5(0)-3=-3│-3 (or) (0, -3)│
Important Notes on y-Intercept
• We substitute x=0 and solve for y to find the y-intercept.
• In the same way, we substitute y=0 and solve for x to find the x-intercept.
• The lines parallel to the y-axis cannot have y-intercepts as they do not intersect the y-axis anywhere.
• The y-intercept of a line is widely used as an initial point while graphing a line by plotting two points.
• A function has cannot have more than one y-intercept.
☛ Related Topics:
Examples on Y Intercept
1. Example 1: Find the y-intercept of the following functions using the formula to find the y-intercept: a) y = x^2 - 3x + 2 b) y = (x^2 - 1) / x
Using the y-intercept formula, to find the y-intercept of a function, we just substitute x = 0 in it and solve for y.
a) Substitute x = 0 in y = x^2 - 3x + 2, we get
y = 0^2 - 3(0) + 2 = 2
So its y-intercept = (0, y) = (0, 2)
b) Substitute x = 0 in y = (x^2 - 1) / x, we get
y = (0^2 - 1) / 0 = -1/0 = Not defined
So the given function doesn't have a y-intercept.
Answer: a) (0, 2); b) Does Not Exist
2. Example 2: If the y-intercept of a function y = a (x - 1) (x - 2) (x - 3) is (0, 12), then find the value of "a".
The equation of the given function is:
y = a (x - 1) (x - 2) (x - 3)
Using the y-intercept formula, it can be obtained by substituting x = 0 in it.
y = a (0 - 1) (0 - 2) (0 - 3) = -6a
So the y-intercept is (0, -6a)
But the problem says that the y-intercept of the given function is (0, 12). Thus,
-6a = 12
Dividing both sides by -6,
a = -2
Answer: a = -2
3. Example 3: If the y-intercept of the function y = 3x^2 + ax + b is (0, -5) then find the value of 'b'.
By using the y intercept formula the y-intercept of the given function is obtained by substituting x = 0
i.e., the y-intercept is, 3(0)^2 + a (0) + b = b.
But it is given that the y-intercept is, (0, -5) (or) -5.
So, b = -5.
Answer: b = -5.
View Answer >
Great learning in high school using simple cues
Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes.
Practice Questions on y Intercept
Check Answer >
FAQs on y Intercept
What is the Meaning of Y Intercept?
The y-intercept of a graph is a point where the graph cuts the y-axis. If the graph is a function, then it has a maximum of one y-intercept.
What is the Slope and Y Intercept of y = 6?
Comparing y=6 with y=mx+b (point-slope form), we get,
• the slope, m = 0 and
• the y-intercept, b=6
How do you Find the Y-Intercept on the Graphing Calculator?
We use the button "y=" on the calculator to enter the function and then press on the "graph". Then the graph of the function is displayed on the screen. Then press on the "zoom" button. Then press on
the "trace" button and enter 0. It will then show the value of y at which x is 0
What is Y Intercept Formula?
The y-intercept formula is used to find the y-intercept of a function. i.e., it is used to find the point where the graph of the function cuts the x-axis. The formula to find y-intercept says that
the y-intercept of a function y = f(x) is substituting x = 0 in it and solving for y.
How to Derive Y Intercept Formula?
According to the definition of y-intercept, the y-intercept of a graph is the point where it cuts (or) intersects the y-axis. We know that on the y-axis the x-coordinate is 0. Hence the formula to
find the y-intercept of a function y = f(x) is just substituting x = 0 and solving for y.
What are the Applications of Y Intercept Formula?
The y-intercept formula is used to find the y-intercept of a function. The y-intercept is mainly used in the process of graphing a function.
Find the Y Intercept of the Graph Represented by the Equation x = y^2+2 y-3.
To find the y-intercept, we substitute x = 0 in the given equation and solve for y. Then,
y^2+2y-3 = 0
(y+3)(y-1) = 0
y-intercepts are (0, -3) and (0, 1)
How to Use the Y Intercept Formula in Finding the Y Intercept From a Graph?
The y-intercept formula says that the y-intercept of a function y = f(x) is obtained by substituting x = 0 in it. Using this, the y-intercept of a graph is the point on the graph whose x-coordinate
is 0. i.e., just look for the point where the graph intersects the y-axis and it is the y-intercept.
Download FREE Study Materials
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/geometry/y-intercept/","timestamp":"2024-11-02T21:22:55Z","content_type":"text/html","content_length":"243610","record_id":"<urn:uuid:19661b22-4c00-4662-8121-0d0ecd87fea5>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00182.warc.gz"} |
How Does Cell Size Affect Volume and Surface Area?
As a cell becomes larger, its (Chapter 10)
a. volume increases faster than its surface area
b. surface area increases faster than its volume
c. volume increases, but its surface area stays the same
d. surface area stays the same, but its volume increases
Final answer: Option a. As a cell grows in size, its volume grows faster than its surface area.
As a cell increases in size, its volume increases faster than its surface area. In other words, the correct answer would be choice a. This is due to the fact that volume is equal to length x width x
height, meaning it increases at a cubic rate. On the other hand, surface area is computed by length x width and only increases at a quadratic rate.
This phenomenon is important in biology because it limits cell size. A cell with a larger volume would have trouble getting enough nutrients and disposing of wastes, due to its relatively small
surface area for trade with the environment. | {"url":"https://chasethegoose.com/biology/how-does-cell-size-affect-volume-and-surface-area.html","timestamp":"2024-11-11T01:28:54Z","content_type":"text/html","content_length":"21378","record_id":"<urn:uuid:eb18da21-bca7-4e3e-8dd5-b694f9691de9>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00263.warc.gz"} |
The basic data types in R++ include numeric, integer, complex, logical, and character.
You can create a vector in R++ using the c() function. For example, c(1, 2, 3, 4) creates a numeric vector with elements 1, 2, 3, and 4.
Describe the difference between a list and a vector in R++.
A vector contains elements of the same type, while a list can contain elements of different types.
You can sort a vector in R++ using the sort() function. For example, sort(c(4, 2, 9, 6)) would return a sorted vector (2, 4, 6, 9).
What are the different ways to subset a data frame in R++?
You can subset a data frame in R++ using the $ operator, the double square bracket [[]] operator, or the subset() function.
Does the candidate have a solid understanding of R++ language and its nuances?
Is the candidate able to communicate effectively about complex technical concepts?
Does the candidate possess a good understanding of data structures and algorithms?
How would you handle missing values in a data set in R++?
You can handle missing values in R++ using functions like na.omit() to remove the rows with missing values, or impute them using functions like mean(), median(), or mode().
What is the use of the apply() function in R++?
The apply() function in R++ is used to apply a function to the rows or columns of a matrix or data frame.
Describe the difference between lapply() and sapply() functions in R++.
Both lapply() and sapply() apply a function to a list or vector. The difference is that lapply() always returns a list, while sapply() tries to simplify the result into a vector or matrix if
You can merge two data frames in R++ using the merge() function. You need to specify the data frames and the common column(s) to merge on.
The different types of loops in R++ include for, while, and repeat loops.
int main() { int a = 5; int b = 10; int c = a + b; cout << c; return 0; }
This code declares two integers a and b, assigns them the values 5 and 10 respectively, adds them together and assigns the result to integer c, and then outputs the value of c, which would be 15.
int main() { int array[5] = {1, 2, 3, 4, 5}; for(int i = 0; i < 5; i++) { cout << array[i] * 2 << ' '; } return 0; }
This code will output the values of the array elements multiplied by 2. So, the output will be '2 4 6 8 10 '.
int main() { int array[5] = {5, 4, 3, 2, 1}; sort(array, array + 5); for(int i = 0; i < 5; i++) { cout << array[i] << ' '; } return 0; }
This code sorts the elements of the array in ascending order and then outputs the sorted array. So, the output will be '1 2 3 4 5 '.
#include void function() { cout << 'Hello, World!'; } int main() { std::thread t1(function); t1.join(); return 0; }
This code creates a new thread t1 that runs the function 'function', which outputs 'Hello, World!'. The main thread waits for t1 to finish execution before it continues, which is ensured by the join
() function.
You can write a function in R++ using the function() construct. For example, add <- function(x, y) {return(x + y)} creates a function that adds two numbers.
Describe the difference between a matrix and a data frame in R++.
A matrix in R++ is a two-dimensional data structure where all elements are of the same type. A data frame, on the other hand, is a two-dimensional data structure where different columns can contain
different types of data.
What are the different types of object-oriented systems in R++?
The different types of object-oriented systems in R++ include S3, S4, and reference classes. | {"url":"https://www.productperfect.com/hiring-guide-copy/r-plus-plus","timestamp":"2024-11-14T04:26:19Z","content_type":"text/html","content_length":"105845","record_id":"<urn:uuid:559d5b87-9e32-4784-a0a7-0776cc4b674c>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00450.warc.gz"} |
rice 0.4.0
• the contaminate function now also produces a plot, and more details of the calculations.
• the decontaminate function has been renamed to clean and has been updated with clearer messages; it now removes contamination to calculate the true/target age. A plot is made, and calculation
details are provided.
• new function muck to calculate how much contamination has to be inferred to go from an observed age to a true/target age. A plot is made, and calculation details are provided.
• new functions C14tocalBP and C14toBCAD. Since these rely on the outdated ‘intercept calibration’ method, these functions are provided for illustrative purposes only.
• draw.dates can now also plot the dates on the calibration curve, using the oncurve option. If so, then the curve and dates can also be plotted in the F14C or pMC realms.
• New function r.calib that samples random calendar ages from a calibrated distribution.
• new function p.range to calculate a calibrated age’s probability of lying between a range of BC/AD or cal BP ages.
• new function as.one to calculate the product of multiple calibrated ages, assuming that they all stem from (exactly) the same calendar age. Not that this is dangerous, and care should be taken to
make sure that the assumptions are met.
• new function as.bin to calculate how many of a set of calibrated radiocarbon dates fall into bins of a specified width. The bin moves along the range of calibrated ages, to visualise how many
dates fit bins over time. This would be safer than using the function as.one.
• new function spread shows the spread (in calendar years) of a set of dates. Accompanies the functions pool and as.one.
rice 0.3.0
• hpd ranges are now calculated at a specified precision (defaults to yearly)
• new function BCADto14C to calculate the 14C age belonging to a BC/AD age (this calls the function calBPto14C)
• new option as.F to calibrate in the F14C realm (the default remains to calibrate in the C14 realm)
• warnings are printed on calibrate() plots if dates are truncated and edge=TRUE (with the default edge=FALSE, dates that are truncated are not calibrated). The printed warning can be removed by
setting print.truncate.warning=FALSE
• renaming age.F14C, F14C.age, age.pMC and pMC.age to, respectively, C14toF14C, F14CtoC14, C14topMC and pMCtoC14. This because age is an ambivalent term in this context
• new functions BCADtocalBP and calBPtoBCAD to transfer cal BP into BC/AD ages and vice versa. Can deal with (e.g. Gregorian/Julian) calendars which do not include 0
• new functions to translate between any of the realms calBP, BCAD, C14, F14C, pMC and D14C.
• new function smooth.ccurve to smooth a calibration curve using a moving window of a specified width. This can be useful to calibrate material that is known to have accumulated, say, over two
• new function pool which calculates the chi2 and accompanying p-value for a set of multiple measurements on the same sample. If the scatter between the values is low enough for the p-value to be
below a threshold, then the pooled mean and uncertainty are returned
• the function draw.dates now has an option oncurve to draw the dates onto the calibration curve.
• added dataset shroud, which contains replicate radiocarbon measurements on the Shroud of Turin, from three labs.
• new function decontaminate to estimate the percentage of contamination needed to explain the difference between a ‘real’ and an ‘observed’ radiocarbon age.
rice 0.2.0
• added an option bombalert to the calibrate function. If set to false, plots ages close to 0 C14 BP without warnings.
• added the data from the marine database (calib.org/marine), as data shells
• new functions find.shells and map.shells to plot shells data in maps based on their coordinates
• new function shells.mean to plot deltaRs of selected shells, and calculate a weighted mean deltaR
• new function weighted_means to calculate weighted means and errors for multiple radiocarbon dates (or delta R values)
• repaired a bug in draw.D14C
• draw.ccurve now can plot the C14 in the ‘realms’ of C14 BP, F14C, pMC and D14C using the ‘realm’ option.
rice 0.1.1
• added citation information
• added a function older
• added a vignette
rice 0.1.0
• The first release of the rice package. It separates the calibration functions from its parent data package rintcal, which in the future will contain the IntCal and other calibration curve data | {"url":"https://cran.hafro.is/web/packages/rice/news/news.html","timestamp":"2024-11-09T01:42:45Z","content_type":"application/xhtml+xml","content_length":"6572","record_id":"<urn:uuid:a2835ee0-0acd-4193-a7b1-7259fd910f3f>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00016.warc.gz"} |
Linear equations, arithmetic progressions arid hypergraph property testing
For a fixed k-uniform hypergraph D (k-graph for short, k ≥ 3), we say that a k-graph H satisfies property P[D] (resp. P[D]*) if it contains no copy (resp. induced copy) of D. Our goal in this paper
is to classify the k-graphs D for which there are property-testers for testing P [D] and P[D]* whose query complexity is polynomial in 1/ε. For such k-graphs, we say that P[D] (or P[D]*) is easily
testable. For P[D]*, we prove that aside from a single 3-graph, P[D]* is easily testable if and only if D is a single k-edge. For large k, we obtain stronger lower bounds than those obtained for the
general case on the query complexity of testing P[D]* for any D other than the single k-edge. These bounds are proved by applying a more sophisticated technique than the basic one that works for all
k. These results extend and improve previous results about graphs [5] and k-graphs [18]. For P[D], we show that for any k-partite k-graph D, PD is easily testable, by giving an efficient one-sided
error-property tester, which improves the one obtained by [18]. We further prove a nearly matching lower bound on the query complexity of such a property-tester. Finally, we give a sufficient
condition for inferring that P[D] is not easily testable. Though our results do not supply a complete characterization of the k-graphs for which P[D] is easily testable, they are a natural extension
of the previous results about graphs [1]. Our proofs combine results and arguments from additive number theory, linear algebra and extremal hypergraph theory. We also develop new techniques, which
are of independent interest. The first is a construction of a dense set of integers, which does not contain a subset that satisfies a certain set of linear equations. The second is an algebraic
construction of certain extremal hypergraphs. We demonstrate the applicability of this last construction by resolving several cases of an open problem raised by Brown, Erdös and Sós in 1973. These
two techniques have already been applied in two recent subsequent papers [6], [27].
Conference Sixteenth Annual ACM-SIAM Symposium on Discrete Algorithms
Country/Territory United States
City Vancouver, BC
Period 23/01/05 → 25/01/05
Dive into the research topics of 'Linear equations, arithmetic progressions arid hypergraph property testing'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/linear-equations-arithmetic-progressions-arid-hypergraph-property","timestamp":"2024-11-03T10:31:10Z","content_type":"text/html","content_length":"52023","record_id":"<urn:uuid:3360ad57-8dcb-40f7-96b6-353530282aea>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00626.warc.gz"} |
Laba nin oo fadhi ku dirir ka daallay, go'aansadayna in ay TV-ga la tagaan to to have a wider audience!
Originally posted by Cawaale:
Laba nin oo fadhi ku dirir ka daallay, go'aansadayna in ay TV-ga la tagaan to to have a wider audience!
This was an academic Islamic discussion on the permissibility of destroying places of shirk. So don't belittle it if you have nothing to contribute. I actually think the brother did a good job
exposing the Qadariyah fella and the latter embarrassed himself by claiming that videos showed bones being exported for business incentives, lol.
interesting debate , thanks for sharing.
sheikh fanax made very good points
but other guy who was defending Xabaalaquft seemed confused.
Originally posted by Karl_Polanyi:
This was an academic Islamic discussion on the permissibility of destroying places of shirk. So don't belittle it if you have nothing to contribute. I actually think the brother did a good job
exposing the Qadariyah fella and the latter embarrassed himself by claiming that videos showed bones being exported for business incentives, lol.
You are absolutely right, this was an academic debate witch we as community need to understand where we stand. Is it good to ask money health and children to an dead person or alive? that is the
whole point.
An hour of entertainment.
Big time sufi vs a rookie wadaad plus xaayoow and open phone lines. What you expected? The TV show got what they wanted, a full comical hour for their viewers and the two wadaads looked silly in big
piles of kitaabs and more interested in damaging the other than defining clearly what is at stake. No debate really. Just some random verses and hadiths. They never exactly defined their debating
points or what ziyaaro meant in each one's view. I am sure both have different meanings of ziyaara in the scope of their religious understanding.
Would have been better if they booked more knowledgeable Sheikhs whose interest is beyond cheap shots at his opponent.
Wasted opportunity but I got few laughs out of lafaha gaaladaa gadata, technologiyada iyo casrigaa lagu suubiyaa, diinta cusub xabaalaha maaro looga la'yahay, awliyaa awlaad u dhalay and many more.
Xabaalo iyo such a marginal issue. Far more important subjects in Soomaaliya than xabaalo la faagay iyo qabri la caabuday. The whole country is burried in blood. How do you save the dying one instead
of focusing a century old dead bones. | {"url":"https://www.somaliaonline.com/community/topic/46435-dood-ku-saabsan-burburinta-qabuuraha/?tab=comments#comment-638791","timestamp":"2024-11-07T09:21:04Z","content_type":"text/html","content_length":"201679","record_id":"<urn:uuid:97dc4da7-4f8a-4812-bf17-fe1df7f4fcd2>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00065.warc.gz"} |
PE ratio, current and historical analysis
The mean historical PE ratio of Ingredion over the last ten years is 22.88. The current 15.12 price-to-earnings ratio is 34% less than the historical average. Analyzing the last ten years, INGR's PE
ratio reached its highest point in the Mar 2021 quarter at 224.8, with a price of $89.92 and an EPS of $0.4. The Sep 2023 quarter saw the lowest point at 10.38, with a price of $98.4 and an EPS of | {"url":"https://fullratio.com/stocks/nyse-ingr/pe-ratio","timestamp":"2024-11-08T04:50:23Z","content_type":"text/html","content_length":"31126","record_id":"<urn:uuid:27ab01cb-ce60-402b-9a69-76a2f9821ab8>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00796.warc.gz"} |
Functional Skills: Density
FS Level 2AQACity & GuildsEdexcelHighfield QualificationsNCFEOpen Awards
Functional Skills: Density Revision
Density measures the relationship between the mass and volume of an object.
The units used for density are usually kilograms per cubic metre (kg/m$^3$) or grams per cubic centimetre (g/cm$^3$).
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Density, Mass and Volume
Density, Mass and Volume are related by the following equations:
Density = Mass $\div$Volume
Mass = Density $\times$Volume
Volume = Mass $\div$Density
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
You may see weight being used instead of mass – just take them as meaning the same thing.
Follow Our Socials
Our Facebook page can put you in touch with other students of your course for revision and community support. Alternatively, you can find us on Instagram or TikTok where we're always sharing revision
tips for all our courses.
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Example 1:Â Calculating Density
An object has a mass of $490$ g and a volume of $1950$ cm$^3$.
Calculate the object’s density to $3$ decimal places.
[2 marks]
Density = Mass $\div$ Volume
The object’s density $=\dfrac{490}{1950}=0.251$ g/cm$^3$
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Example 2: Calculating Mass
A block of wood has a volume of $196$ cm$^3$ and a density of $0.8$ g/cm$^3$.
Calculate the mass of the the block of wood.
[2 marks]
Mass = Density $\times$ Volume
Block of wood’s mass $=0.8\times196=156.8$ g
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Example 3: Calculating VolumeÂ
A block of gold has a density of $19.3$ g/cm$^3$ and a mass of $2500$ g.
Calculate the volume of the gold block to $1$ decimal place.
[2 marks]
Volume = Mass $\div$ Density
Gold block volume $=\dfrac{2500}{19.3}=129.5$ cm$^3$
FS Level 2AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Functional Skills: Density Example Questions
Question 1:Â A $0.57$ kg bottle of tomato ketchup has a volume of $495$ cm$^3$. Calculate the density of the ketchup in g/cm$^3$ to $2$ decimal places.
[2 marks]
FS Level 2 AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
First convert $0.57$ kg into grams: $0.57\times1000=570$ g
Density = Mass $\div$ Volume
Density of ketchup = $\dfrac{570}{495}=1.15$ g/cm$^3$ to $2$ decimal places.
Question 2:Â A cube with side lengths of $6$ cm has a density of $12.5$ g/cm$^3$. Calculate the mass of the cube in kg.
[3 marks]
FS Level 2 AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
First we need to work out the volume of the cube: $6\times6\times6=216$ cm$^3$
Mass = Density $\times$ Volume
Mass of cube = $12.5\times216=2700$ g = $2.7$ kg
Question 3:Â A block of silver has a density of $10.49$ g/cm$^3$ and a mass of $112914.36$ g. Calculate the volume of the block of silver.
[2 marks]
FS Level 2 AQAEdexcelCity & GuildsNCFEOpen AwardsHighfield Qualifications
Volume = Mass $\div$ Density
Volume of the block of silver = $\dfrac{112914.36}{10.49}=10764$ cm$^3$
Functional Skills: Density Worksheet and Example Questions
[responsive-flipbook id=”pfs_pocket_revision_guide_-_sample”]
Revision Products
Functional Skills Maths Level 2 Pocket Revision Guide
Revise and practice for your functional skills maths level 2 exam. All topics covered in this compact revision guide.
View Product
Functional Skills Maths Level 2 Mini Tests
Practice for your functional skills Maths level 2, questions from every topic included.
View Product
Functional Skills Maths Level 2 Revision Cards
Revise for functional skills maths level 2 easily and whenever and wherever you need. Covering all the topics, with revision, questions and answers for every topic.
View Product
Functional Skills Maths Level 2 Practice Papers
This 5 set of Functional Skills Maths Level 2 practice papers are a great way to revise for your Functional Skills Maths Level 2 exam. These practice papers have been specially tailored to match the
format, structure, and question types used by each of the main exam boards for functional skills Maths. Each of the 5 papers also comes with a comprehensive mark scheme, so you can see how well you
did, and identify areas to improve on.
View Product
Functional Skills Maths Level 2 Practice Papers & Revision Cards
This great value bundle enables you to get 5 functional skills maths level 2 practice papers along with the increasingly popular flashcard set that covers the level 2 content in quick fire format.
View Product | {"url":"https://passfunctionalskills.co.uk/functional-skills-maths-level-2/functional-skills-density/","timestamp":"2024-11-04T15:22:17Z","content_type":"text/html","content_length":"404576","record_id":"<urn:uuid:116dc137-471a-4769-897f-2a6d34adbb14>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00804.warc.gz"} |
TLC model checking RealTime module: problem with variable now
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
TLC model checking RealTime module: problem with variable now
I'm just getting started with TLA+, and wanted to try and check a simple behavior: the module RealTime specified in the Specifying Systems book, in conjunction with a custom module.
The idea was to set a time bound on the VentriculePulse action, but when I try to test it, the following error comes up:
The subscript of the next-state relation specified by the specification does not seem to contain the state variable now Successor state is not completely specified by the next-state action.
How can I link the variable now to the VentriculePulse action, being that both elements appear in different components? Or should I somehow override the VentriculePulse action inside the RealTimeHeart module to state that now must remain unchanged?
--------------------------- MODULE RealTimeHeart ---------------------------
EXTENDS Reals, Heart, RealTime
RTHeart == /\ HeartSpec
/\ RTnow(ventriculeChannel)
/\ RTBound(VentriculePulse,ventriculeChannel,MinimumPeriod, MaximumPeriod)
/\ (now = 0)
----------------------------- MODULE RealTime -------------------------------
EXTENDS Reals
VARIABLE now
RTBound(A, v, D, E) ==
LET TNext(t) == t' = IF <<A>>_v \/ ~(ENABLED <<A>>_v)' \* The value of t will be 0 if an A action was executed or if A is no longer enabled.
THEN 0
ELSE t + (now'-now)
Timer(t) == (t=0) /\ [][TNext(t)]_<<t, v, now>>
\* Initial state in which 't' equals 0 and every possible state is a TNext step or leaves all variables unchanged
MaxTime(t) == [](t \leq E)
\* The time 't' between two consecutive A actions cannot be greater than E
MinTime(t) == [][A => t \geq D]_v
\* Before two consecutives A action occur, there must have passed at least D time
IN \EE t : Timer(t) /\ MaxTime(t) /\ MinTime(t)
RTnow(v) == LET NowNext == /\ now' \in {r \in Real : r > now}
/\ UNCHANGED v
\* A NowNext step can advance 'now' by any amount of time while leaving v unchanged
IN /\ now \in Real
/\ [][NowNext]_now
/\ \A r \in Real : WF_now(NowNext /\ (now'>r))
------------------------------- MODULE Heart -------------------------------
EXTENDS Naturals
CONSTANT PossibleStimulus, MaximumPeriod, MinimumPeriod, stimulus
VARIABLE ventriculeChannel
Ventricule == INSTANCE Channel WITH Data <- PossibleStimulus, chan <- ventriculeChannel
TypeInvariant == Ventricule!TypeInvariant
HeartInit == /\ Ventricule!Init
/\ MaximumPeriod >= MinimumPeriod
/\ MinimumPeriod >= 0
VentriculePulse == Ventricule!Send(stimulus)
HeartSpec == HeartInit /\ [][VentriculePulse]_ventriculeChannel
THEOREM HeartSpec => []TypeInvariant | {"url":"https://discuss.tlapl.us/msg01301.html","timestamp":"2024-11-04T08:06:36Z","content_type":"text/html","content_length":"6021","record_id":"<urn:uuid:bcfe224f-02c5-497b-8b7b-12c0229eabc3>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00857.warc.gz"} |
Skip Counting Number Lines: By 7, 8, 9, 11, & 12
Note: this page contains legacy resources that are no longer supported. You are free to continue using these materials but we can only support our current worksheets, available as part of our
membership offering.
Skip Counting Number Lines: By 7, 8, 9, 11, & 12
Related Resources
The Counting and cardinality Number Line above is aligned, either partially or wholly, with standard 4OA04 taken from the CCSM (Common Core Standards For Mathematics) – see the extract below). The
resources below are similarly aligned.
Find all factor pairs for a whole number in the range 1 to 100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1 to 100 is a
multiple of a given one-digit number. Determine whether a given whole number in the range 1 to 100 is prime or composite.
Prime Numbers
Target Game
Number line | {"url":"https://helpingwithmath.com/generators/lin0301number56/","timestamp":"2024-11-06T17:05:02Z","content_type":"text/html","content_length":"111292","record_id":"<urn:uuid:b7b0094f-61b3-4058-86a1-ae3f81091a52>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00714.warc.gz"} |
Research on transmission coefficient property of coupled beam structure
This study investigates the transmission coefficient of coupled beam structure which has been widely used in engineering problems. Although the power flow is continuous at the coupled point, the
waveform and energy density often show sudden changes. These changes can be utilized to create an understanding of the system. By treating the two-beam coupled structure as two infinite beams and
considering the coupling between the bending wave and the torsion one, the conversion of wave types at the coupled interface is discussed. Further, the transmission coefficient is also computed.
Furthermore, investigate the effects of coupling angle, beam height and the excitation frequency of incident wave on the conversion of wave type and energy transmission. Numerical results indicate
that the coupling angle and beam height have great influences on the conversion of wave type. The component of torsion wave grows with the excitation frequency and cannot be ignored in high
1. Introduction
Beam structure has been widely used in engineering due to its convenience in arranging, disassembling and adjustment (e.g. the truss structure in spacecraft could be modeling by coupling beam
structure) [1]. For the vibration analysis and acoustic prediction of those large coupled structures, the power methods which are represented by statistic power method and power flow finite element
are favorable since they have many advantages over traditional methods [2].
In the statistic energy analysis (SEA) [3], the power flow is accounted for in terms of a coupling loss factor while in power balance analysis [4] the junction is described by transmission and
reflection coefficients. A deterministic as opposed to a statistical approach was adopted in modeling the beam junction dynamics. Individual frame members are treated as ideal, one-dimensional
structures deforming in bending, torsional, or longitudinally [5, 6]. Moore [7], Horner [8] and Langley [9] have also conducted analyses of the vibratory power transmission of general two-dimensional
beam structures.
This article investigates the vibration velocities of all directions in two beams coupled structure based on the classic wave theory. According to the continuity condition at coupled interface, the
amplitudes of vibration velocity of elastic wave are got. Then, the energy transmissions at the coupled interface are subsequently obtained. Finally, we discuss the influences of different coupling
angle and beam height on the transmission coefficient in detail, which provides some theoretical basis for power flow analysis.
2. Transmission coefficient of coupled beam structure
There exists various coupled structure in engineering application, and most of them consist of several coupling types, such as the coupling forms at the corner of helicopter [7] shown in Fig. 1(a).
Fig. 1(b) presents the typical coupled beam structure. When excitation wave incident into beam 1, the wave conversion will happen at the junction. At the same time, the reflection and transmission
occur as shown in Fig. 1(b). In the following, we will discuss the conversion of wave type and calculate the transmission coefficients under excitation by the incident out-of-plane bending wave.
When computing the transmission coefficient, the finite structure can be treated as semi-infinite structure which has been discussed in Ref [10]. Since this method could provide acceptable precision,
we also adopt this treatment in the present work.
Fig. 1Model of coupled structure
b) L-shaped coupled beams
2.1. Calculation method
When the beams lie in the same plane, the motions can be simplified into in-plane motions and out-of-plane motions and out-of-plane motion. The out-of-plane motions include bending over the lying
axis and torsion. The in-plane motion includes the longitudinal deformation and bending which is perpendicular to the lying axis.
Since these two motions are uncoupled when the two beams are in the same plane, this method can handle arbitrary coupling angles. The first step is defined the global ($x$, $y$, $z$) and local (1, 2,
3) system as shown in Fig. 2.
Secondly, we should express the junction displacements in the local coordinate system. This yields the following for the translational and rotational velocities at the junction:
$\begin{array}{c}{\stackrel{˙}{\zeta }}_{1}={\stackrel{˙}{\zeta }}_{x}\mathrm{c}\mathrm{o}\mathrm{s}\varphi -{\stackrel{˙}{\zeta }}_{z}\mathrm{s}\mathrm{i}\mathrm{n}\varphi ,\mathrm{}\mathrm{}\mathrm
{}\mathrm{}{\stackrel{˙}{\zeta }}_{2}={\stackrel{˙}{\zeta }}_{y},\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{˙}{\zeta }}_{3}={\stackrel{˙}{\zeta }}_{x}\mathrm{s}\mathrm{i}\mathrm{n}\varphi +{\
stackrel{˙}{\zeta }}_{z}\mathrm{c}\mathrm{o}\mathrm{s}\varphi ,\end{array}$
$\begin{array}{c}{\stackrel{˙}{\theta }}_{1}={\stackrel{˙}{\theta }}_{x}\mathrm{c}\mathrm{o}\mathrm{s}\varphi -{\stackrel{˙}{\theta }}_{z}\mathrm{s}\mathrm{i}\mathrm{n}\varphi ,\mathrm{}\mathrm{}\
mathrm{}\mathrm{}{\stackrel{˙}{\theta }}_{2}={\stackrel{˙}{\theta }}_{y},\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{˙}{\theta }}_{3}={\stackrel{˙}{\theta }}_{x}\mathrm{s}\mathrm{i}\mathrm{n}\
varphi +{\stackrel{˙}{\theta }}_{z}\mathrm{c}\mathrm{o}\mathrm{s}\varphi ,\end{array}$
where ${\stackrel{˙}{\zeta }}_{i}$ and ${\stackrel{˙}{\theta }}_{i}$ denotes the junction displacement velocities and rotational displacement velocities, respectively; $i$ can be coordinate axis. $\
mathrm{\Phi }$ is the angle which defines the in-plane orientation of the frame with respect to the source frame.
The frame impedances relate force / moments to translational and rotational velocities in the local coordinate system:
$\begin{array}{l}{F}_{1}={\mathrm{z}}_{l}{\stackrel{˙}{\zeta }}_{1},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{F}_{2}={\mathrm{z}}_{f}^{op}{\stackrel{˙}{\zeta }}_{2}+{\mathrm{z}}_{f,\theta }^{op}
{\stackrel{˙}{\theta }}_{3},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{F}_{3}={\mathrm{z}}_{f}^{ip}{\stackrel{˙}{\zeta }}_{3}+{\mathrm{z}}_{f,\theta }^{ip}{\stackrel{˙}{\theta }}_{2},\\ {M}_{1}=
{z}_{t}{\stackrel{˙}{\theta }}_{1},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{M}_{2}={\mathrm{z}}_{m}^{ip}{\stackrel{˙}{\theta }}_{2}+{\mathrm{z}}_{f,\theta }^{ip}{\stackrel{˙}{\zeta }}_{3},\
mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{M}_{3}={\mathrm{z}}_{m}^{op}{\stackrel{˙}{\theta }}_{3}+{\mathrm{z}}_{f,\theta }^{op}{\stackrel{˙}{\zeta }}_{2},\end{array}$
where ${F}_{i}$ and ${M}_{i}$ denotes the force and moment, respectively, $i$ can be coordinate axis; ${z}_{l}$, ${z}_{t}$ are the input impedances for torsion and longitudinal motion, respectively.
${z}_{f}$, ${z}_{m}$, ${z}_{f,\theta }$ are the input impedances for bending motion related translational, rotational velocity and coupled effect, respectively; $ip$, $op$ are the in plane and
out-of-plane motion.
The force and moments in the global coordinate system are given in terms of the local values by:
$\begin{array}{l}{F}_{x}={F}_{1}\mathrm{c}\mathrm{o}\mathrm{s}\varphi +{F}_{3}\mathrm{s}\mathrm{i}\mathrm{n}\varphi ,\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{F}_{y}={F}_{2},{F}_{z}=-{F}_{1}\
mathrm{s}\mathrm{i}\mathrm{n}\varphi +{F}_{3}\mathrm{c}\mathrm{o}\mathrm{s}\varphi ,\\ {M}_{x}={M}_{1}\mathrm{c}\mathrm{o}\mathrm{s}\varphi +{M}_{3}\mathrm{s}\mathrm{i}\mathrm{n}\varphi ,\mathrm{}\
mathrm{}\mathrm{}\mathrm{}\mathrm{}{M}_{y}={M}_{2},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{M}_{z}=-{M}_{1}\mathrm{s}\mathrm{i}\mathrm{n}\varphi +{M}_{3}\mathrm{c}\mathrm{o}\mathrm{s}\varphi .\
The impedance expressions for a beam with arbitrary orientation $\varphi$ are obtained by combining Eqs. (1)-(4). The total junction impedance in global junction coordinates is obtained by summing
force and moment expressions for each frame at junction.
The mechanical power transmitting into a beam may be determined from the displacements at the junction end of the beam according to the following expression for longitudinal motion:
${\prod }_{trans,l}=\frac{1}{2}{z}_{l}{\left|{\stackrel{˙}{\zeta }}_{x}\right|}^{2},$
and for torsional wave transmission:
${\prod }_{trans,t}=\frac{1}{2}{z}_{t}{\left|{\stackrel{˙}{\theta }}_{x}\right|}^{2}.$
Bending requires a more complicated that accounts for the coupling between rotation and translation and the relative phase between these motions:
${\prod }_{trans,b}=\frac{1}{2}\left[\mathrm{R}\mathrm{e}\left({z}_{f}\right){\left|{\stackrel{˙}{\zeta }}_{y}\right|}^{2}+\mathrm{R}\mathrm{e}\left({z}_{m}\right){\left|{\stackrel{˙}{\theta }}_{z}\
right|}^{2}\right]+\frac{1}{2}{z}_{f,\theta }\left({\stackrel{˙}{\theta }}_{z}{{\stackrel{˙}{\zeta }}_{y}}^{\mathrm{*}}+{{\stackrel{˙}{\theta }}_{z}}^{\mathrm{*}}{\stackrel{˙}{\zeta }}_{y}\right),$
where $*$ denotes the complex conjugate.
The transmission coefficient $\tau$ is the ratio of transmitted to incident powers, which is obtained from Eqs. (8):
$\tau =\frac{{\prod }_{trans,i}\mathrm{}}{{\prod }_{inc,j}\mathrm{}},$
where ${\mathrm{\Pi }}_{inc}$ is the incident power for a propagating wave, $i$ and $j$ are the transmitted wave and incident wave type, respectively.
Fig. 2Global and local coordinate systems
2.2. Numerical calculation
Since Aluminum alloy is widely used in aircrafts, we assume that the coupling beam consists of Aluminum. The material parameters as following: Elastic modulus $E=$ 71 GPa, density $\rho =$ 2700 kg/m^
3, Poisson’s ratio $\mu =$ 0.333.
The shape of beam cross section is defined in Fig. 3, and structure parameters are given in Table 1. Here, we consider the excitation frequency between 10^2 Hz and 10^4 Hz, and the velocity of
excitation wave is 1 m/s.
Fig. 4 illustrates the transmission coefficient under the excitation of out-of-plane bending wave. From this figure, we can see that the transmission coefficient changes with the excitation
frequency, and the tendency is in accordance with Ref [7]. This demonstrates the present approach.
Table 1The structural parameter
Area (m^2) Length (m) Polar moment of inertia(m^4) Moment of inertia $z$ axis (m^4) Moment of inertia $y$ axis (m^4)
3.1e-4 0.8 1.3e-7 2.8e-6 3.4e-8
Fig. 4Transforming of power coefficient when out-of-plane bending wave incident
3. Discussing
3.1. Effect of coupled angle on power transmission coefficient
Based on the results, we will discuss the influence of coupled angle on power transmission coefficient in this section. The coupling angle varies between 0°-180°, and the section of the beam remains
the same. In the following, ${r}_{ac}^{ij}$ denotes the coefficient of power between different fields, where $a$ and $c$ could be “$b$” (bending) and “$t$” (torsion), $i$ and $j$ can be 1 (beam 1)
and 2 (beam 2). For example, ${r}_{bt}^{12}$ denotes the coefficient of the power from the bending wave field of beam 1 to the torsion wave field of beam 2.
When beam 1 is excited by the out-of-plane bending wave, if the coupled angle is 0°, i.e., the two beams are coincident, there are no transmissions, and all power is reflected into the bending wave
field of beam 1. In this case, ${r}_{bb}^{11}=$ 1, and other coefficients are zeros. If the coupled angle is 180°, i.e., the two beams is collinear but not coincide, there are no reflections when the
sections of two beams are the same. All the power transmits into the bending wave field of beam 2. Hence, ${t}_{bb}^{12}=$ 1 and other coefficients are also zeros.
Figs. 5-8 show the influence of coupling angle and excitation frequency on the transmission coefficient of bending and torsion wave which are generated by the reflection of bending incident in beam 1
Fig. 5 shows that the coefficient shows apparently variations and increases with the excitation frequency when the coupling angle is small. Two peak values can be observed in Fig. 6. When the
coupling angle $\varnothing$ is small, the coefficient decreases as the frequency increase. While the power of torsion field increases slightly as frequency increases when the coupled angle is
between the two critical angles. For big coupling angle, the coefficient decreases as frequency increases.
Fig. 7 illustrates that when coupled angle is less than critical angle, the transmission coefficient decreases as excitation frequency increase. Fig. 8 shows the fact that $\varnothing$ is between
the critical angles, the transmission coefficient increases with the increase of excitation frequency. In other cases, it decreases as the frequency increase.
In summary, most of the power transmits into the bending wave field while the power of torsion wave field after wave conversion is small when the out-of-plane bending wave transmits into beam 1. When
coupling angle grows, homology coefficients tend to be dominant.
Fig. 5The coupling angle (∅) influence on rbb11
Fig. 6The coupling angle (∅) influence on rbt11
Fig. 7The coupling angle (∅) influence on tbb12
Fig. 8The coupling angle (∅) influence on tbt12
3.2. Effect of beam height
In the following section, we focus on the influence of the beam height on the transmission coefficients. Figs. 9-12 presents the coupling coefficients with different heights of the beam when coupling
angle equals 90°.
Fig. 9The beam height (h) influence on rbb11
Fig. 10The beam height (h) influence on rbt11.
Comparing Figs. 9, 10, we can find that beam height has the opposite influence in reflection bending wave and torsion wave. When the beam height increases, the torsion reflection coefficient raise
gradually, which means the conversion from bending wave to torsion wave is strengthen. Meanwhile, the max value of torsion reflection coefficient is 0.075 which is far less than bending reflection
From Figs. 11 and 12 we can conclude that beam height has the opposite influence on propagation of bending and torsion wave but has the same influence on the reflection and transmission of certain
wave. When the bending excitation exists in the structure, except for the bending wave, the torsion wave also has important influence in vibration, especially in high frequency case.
Fig. 11The beam height (h) influence on tbb12
Fig. 12The beam height (h) influence on tbt12
4. Conclusions
This study analyzes the influence of the frequency of incident out-of-plane bending wave, beam height and coupling angle on the conversion of wave type and transmission coefficients are discussed in
details. Some conclusions can be drawn as:
In the case of large height and high frequency, the effect of torsion wave become more heavily. Therefore, when using the power method like FEM of power flow to study the high-frequency vibration of
coupled structure, wave type conversion and coupled wave fields must be concerned in computing, otherwise the result is inaccuracy even failed.
• Wang Youyi, Zhao Yang, Ma Wenlai Travelling wave analysis and active control of jitter in the median and high frequency regions for coupled beam. Journal of Mechanical Engineering, Vol. 49, Issue
22, 2013, p. 188-191.
• Song Kongjie, Zhang Wei Bo, Niu Jun-chuan Application and development of power flow theories in the field of the vibration control for flexible systems. Chinese Journal of Mechanical Engineering,
Vol. 39, Issue 9, 2003, p. 23-28.
• Lyon R. H. Statistical Energy Analysis of Dynamical Systems. MIT Press, Cambridge, MA, 1995.
• Smith P. W. The effect of longitudinal wave coupling on the diffusion of flexural energy in a one-dimensional waveguide with lossy obstacles. The Journal of the Acoustical Society of America,
Vol. 72, 1982, https://doi.org/10.1121/1.2020065.
• Cremer L., Heckle M., Ungar E. E. Structure Borne Sound. Springer, New York, 1973.
• Stablik M. J. Coupling loss factors at a beam L-joint revisited. The Journal of the Acoustical Society of America, Vol. 72, Issue 4, 1982, p. 1285-1288.
• Moore J. A. Vibration transmission through frame or beam junctions. The Journal of the Acoustical Society of America, Vol. 88, Issue 6, 1982, p. 2766-2776.
• Horner J. L. Prediction of vibration power‚ transmission in frameworks. Journal of Sound and Vibration, Vol. 10, Issue 2, 1969, p. 163-175.
• Langley R. S., Heron K. H. Elastic wave transmission through plate/beam junctions. Journal of Sound and Vibration, Vol. 9, Issue 3, 1990, p. 469-486.
• Cho P. E. Energy Flow Analysis of Coupled Structures. Ph.D. Thesis, Purdue University, USA, 1993.
About this article
Mathematical models in engineering
coupling beam
transmission coefficient
excitation frequency
wave type conversion
Copyright © 2018 Wanlu Hu, et al.
This is an open access article distributed under the
Creative Commons Attribution License
, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | {"url":"https://www.extrica.com/article/20301","timestamp":"2024-11-12T12:32:32Z","content_type":"text/html","content_length":"130193","record_id":"<urn:uuid:40c9dfd6-a707-4c26-80ce-074806012a60>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00316.warc.gz"} |
Here we need to apply the concept Proper Fractions and algebra with equation solving. Proper Fraction : A fraction in which the numerator is less than the denominator. We need to find how many
bottles of the oil the tin contains initially.
Complete step by step solution:
Let the capacity of the tin be \[x\] bottles .
Amount of oil present in the tin is \[4/5\times x\]
According to the question,
\[6\] bottles of oil was taken out,
$\Rightarrow -6$
\[4\] bottles of oil were poured into,
$\Rightarrow +4$
After the above changes amount of oil in the tin is \[3/4\times x\]
& \Rightarrow {}^{4}/{}_{5}\times x-6+4={}^{3}/{}_{4}\times x \\
& \Rightarrow {}^{4}/{}_{5}\times x-2={}^{3}/{}_{4}\times x \\
& \Rightarrow {}^{4}/{}_{5}\times x-{}^{3}/{}_{4}\times x=2 \\
& \Rightarrow {}^{\left( 16x-15x \right)}/{}_{20}=2 \\
& \Rightarrow 16x-15x=2\times 20 \\
& \Rightarrow x=40 \\
Therefore, the capacity of the tin is $40$ bottles.
Hence, Option choice B is the correct answer.
In such types of questions the concept of Fractions and algebra with equation solving is needed. Here the variables are assigned to unknown values and equations are framed accordingly as per the
relation in the question. Then it is solved to get the required value. | {"url":"https://www.vedantu.com/question-answer/a-tin-of-oil-was-45-full-when-6-bottles-of-oil-class-9-maths-cbse-5f6178ca0b2e30778aaeb752","timestamp":"2024-11-02T21:01:38Z","content_type":"text/html","content_length":"154199","record_id":"<urn:uuid:3f80c70b-7c26-480b-ac6f-9ebb7377d4e3>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00877.warc.gz"} |
Non-separability of velocity from factorized implicit coefficients
Implicit finite difference in time-space domain with the helix transform
Next: Stability analysis Up: Standard implicit propagation Vs. Previous: Standard implicit propagation Vs.
The initial impetus for using deconvolutions by spectrally factorized coefficients of an implicit finite-difference scheme to approximate the 2-way wave equation was that the velocity could be
separated from the scheme's coefficients. The hope was that if this was possible, a filter with constant coefficients could be created to handle propagation through a variable velocity medium. This
has turned out to not be the case, at least for the formulation of the implicit scheme in Equation 9. If we look at the left hand side of Equation 17:
It is clearly impossible to divide the system by 17), requiring a different set of spectrally factorized coefficients wherever the velocity changes.
The option of using a ``filter bank'' for different parts of the wavefield according to the local velocity has already been discussed in Rickett et al. (1998), for wave propagation in the
frequency-wavenumber domain. This may also be applicable to propagation in the time-space domain, but I have not yet tested it. It was my hope that this would be unecessary, and that a single set of
filter coefficients could be utilized for the entire wavefield irrespective of velocity. This would make the propagation algorithm simpler, and more amenable to future parallelization schemes.
Implicit finite difference in time-space domain with the helix transform
Next: Stability analysis Up: Standard implicit propagation Vs. Previous: Standard implicit propagation Vs. | {"url":"https://sepwww.stanford.edu/data/media/public/docs/sep140/ohad1/paper_html/node10.html","timestamp":"2024-11-10T18:39:21Z","content_type":"text/html","content_length":"6643","record_id":"<urn:uuid:89e57f0d-0d48-4efe-8a1b-cf525aef45e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00848.warc.gz"} |
• Age of Darkness: Prizm Edition
• Last edited by tifreak8x on 04 Jul 2014 07:16:02 pm; edited 1 time in total
Just a place to kind of pool some notes and such.
I've been weighing the decision to base the game on the homescreen instead of the graphscreen because if I were to do it on the graphscreen like on the 83+ line, it would take you forever to get
across one map.
Beyond that, everything else should be pretty much a straight port, as the coding syntaxes seemingly vary very little.
Well, we could teach you C, and you could make it fast, amazing, and complete it pretty quickly as a "graphscreen" type of game. Alternatively, you could do what Sarah is doing, and go for a Casio
BASIC homescreen-type game, also fast and pretty nifty.
Yeah, I wish the text command wasn't so slow :/
Well, I had initially started out thinking I could make it on the graphscreen. So the code doesn't go to waste, here is what I had to this point:
Horizontal 0
Horizontal Ymax
Vertical 0
Vertical Xmax
While W=1
Text B,A,"A"
If K
Text B,A," " // 2 spaces
If End
IF K=47
A-10((K=38 and A>5)-(K=27 and A<360->A
B-10((K=28 and B>5)-(K=37 and B<175->B
Hopefully this doesn't bore anyone too badly :p I'll still give this some thought, might make use of it, or not. Would have to adjust how many pixels at a time a user would 'walk'.
I'd also like to note that the code above isn't too horridly slow, either. I think it'd still be playable the speed that it displays. It takes longer to run the Vertical and Horizontal commands than
it does to move the character around the screen.
I really should come back to this at some point.
An homescreen PRIZM remake would definitively be good, since it's the only fast thing in PRIZM BASIC and not only you can use colors for individual characters, but you also have access to some rather
interesting characters.
Yeah, though I think I'll just leave it with the black border using characters, and simplify things. That was the beauty of that old game, it was simple, yet fun
Also, how does one initiate lists in Prizm BASIC? I've went and forgot how to do so
tifreak8x wrote:
Yeah, though I think I'll just leave it with the black border using characters, and simplify things. That was the beauty of that old game, it was simple, yet fun
Also, how does one initiate lists in Prizm BASIC? I've went and forgot how to do so
I'm afraid I don't remember how to do this, but exploring the Prizm wiki shows this example that should help.
Thanks for the info, I've found more information, posted it to a more appropriate thread. I've worked out the first batch of 'screens' for the map, now I need to move some code around so it sits in
the proper spots of programs where it needs to be. (Copy and Paste make things SO nice on the Prizm)
Once I get it to clear the X's out of the sides that are passable, I'll make the actual game engine work. From there, it will be all about the battles.
I was able to confirm my idea will work in terms of opening the sides of the screen for borders. I'll probably go ahead and dump the rest of the screen data into that particular program and properly
set it up to display the symbols for the various world objects.
I wish we had a good way of screenshotting things, looks like I'll be forced to take video with my camera when I want to show off my work. :/
As of right now, I've completed the inputting of all map data, pending any typos on my part, that should be done. It now properly generates the boundaries of the map and even is set up for segment
bouncing. Oh yeah, and it's ready for me to add map interactions with items on the map
Yes, I know the fan is loud, sorry about that, not much I can do.
Otherwise, what do you think thus far?
Got the shifting between segments completed. On to the map interactions!
Work progresses with the Village program. Inn and Tavern completed, working on armory now, which should only leave me the shop, which also shouldn't be too hard to get implemented and ready to
We now have a fully working Village system in place, with Armory, Inn, and Shop all ready and waiting.
Next up on the list is making the menu system work, to access stats, save, etc.
Since this has all been done on the home screen so far, I'm not 100% sure how I wish to proceed with this, but I'm sure I'll come up with something!
What do you think of things so far?
Way too little activity on this topic, I think its really nice although I'm not sure what
"ARE YOU HERE TO SAVE US?
THAT WAS REALLY FUNNY!"
I did notice it says Welcome to Our... or welcome to My... maybe you could show who is saying that or change it
Anyway, amazing work
Well, it should be fairly obvious who is saying what. The towns folk, and the respective shop keepers, hence the our and my differences.
And the place you the player start at, you're known and not considered someone of importance.
ok cool
Sadly, he's programming in a language I know nothing about, so I'd be of no help to him right now. Most of my programming on the calculators have all been in BASIC.
Getting ready to work on this for a little bit, realized it might be good to post how large the game is in its current form, give me a marker of sorts on the size of the programs as I get closer to
the final version of things.
Name - Size
AOD - 496
ZCH - 112
ZGA - 392
ZME - 1108
ZPL - 68
ZSC - 2664
ZVI - 2200
Now to go make ZME bigger
AOD - 496
ZCH - 112
ZGA - 420
ZME - 2000
ZPL - 68
ZSC - 2664
ZVI - 2200
In-game menu is 100% completed and fully integrated into the actual gaming engine. Next on the list is to finish up the events, followed by the battle engine.
I have 'finished' the events program (ZPL). Current size of the game:
AOD - 496
ZCH - 112
ZGA - 420
ZME - 2004
ZPL - 3440
ZSC - 2664
ZVI - 2200
Next up on the list is inputting 103 lines of enemy data. That's going to be fun.
I'm beginning to have my doubts about this project.
AOD - 496
ZBA - 32
ZEN - 2776
ZGA - 420
ZME - 2004
ZPL - 3440
ZSC - 2664
ZVI - 2200
Enemy data added, beginning work on battle system, which is the final leg of programming in this game, then testing.
In theory, the battle system is now set up to let players attack the enemy, they just cannot return attacks as of yet. Plus, the exp gain isn't programmed in, either.
Closing in on a beta release of the game, hopefully by the weekend. Hopefully the battle system has no errors in it, it's already over 2900 bytes in size, and I have a lot to add to it yet.
And I still need to update the gaming engine to use the enemy encounter and exiting from the battles properly.
AOD - 496
ZBA - 2856
ZEN - 2776
ZGA - 420
ZME - 2004
ZPL - 3440
ZSC - 2664
ZVI - 2200
Well, I've got the battle system coded in, bad news is it's giving me a syntax error and going to the end of the program.
This program is 5KB in size. It's going to take a while to scan through each line to make sure I didn't forget to close a quote, bracket or otherwise. At least I can save it as a txt on the calc and
dump it to the computer to do it a little more efficiently.
Locate 1,1,"YOU ARE BATTLING"
Locate 1,2,Str 1
Locate 8,4,"FIGHT"
Locate 8,5,"RUN"
While W=3
Locate 7,Z,"_=>_"
If K>1:Then
Locate 7,Z," ":IfEnd
Z-1((K=28 And Z>4)-(K=37 And Z<5))->Z
If K=78 Or K=31:Then
If Z=5:Then
If List 1[13]=1:Then
Locate 1,7,"CANNOT ESCAPE BOSSESChar!"
While Getkey=0
If List 1[13]=0:Then
If J=1 Or J=3 Or J=5:Then
If J=2 Or J=4 Or J=6:Then
Locate 1,7,"CANNOT ESCAPEChar!"
While Getkey=0
If Z=4:Then
While W=3
Locate 1,1,"YOU ARE BATTLING"
Locate 1,2,Str 1
Locate 8,4,"SWORD"
Locate 8,5,"MAGIC"
Locate 8,6,"ITEMS"
Locate 8,7,"STATS"
While W=4
Locate 7,Z,"_=>_"
If K>1:Then
Locate 7,Z," ":IfEnd
Z-1((K=28 And Z>4)-(K=37 And Z<7))->Z
If K=78 And K=31:Then
If Z=7:Then
Locate 1,1,"HP: //"
Locate 5,1,List 2[1]
Locate 11,1,List 2[2]
Locate 1,2,"MP: //"
Locate 5,2,List 2[3]
Locate 11,2,List 2[4]
Locate 1,3,"ATT: DEF:"
Locate 5,3,List 2[5]
Locate 16,3,List 2[6]
Locate 1,4,"LEVEL:"
Locate 8,4,List 2[10]
Locate 1,5,"EXP:"
Locate 6,5,List 2[8]
Locate 1,6,"NEXT LEVEL:"
Locate 13,6,List 2[9]
While Getkey=0
If Z=6:Then
Locate 1,1,"ITEMS:"
Locate 2,2,"POTIONS _@E778_"
Locate 2,3,"ETHERS _@E778_"
Locate 1,4,"CRYSTALS:"
Locate 13,4,List 2[20]
Locate 1,7,"[EXIT] TO EXIT"
While W=3
Locate 13,2,List 2[11]
Locate 13,3,List 2[12]
Locate 1,Z,"_=>_"
If K>1:Then
Locate 1,Z," "
Z-1((K=28 And Z>2)-(K=37 And Z<3))->Z
If K=47:Then
If K=78 Or K=31:Then
If Z=2 And List 2[11]>1:Then
List 2[11]-1->List 2[11]:List 2[1]+150->List 2[1]
If List 2[1]>List 2[2]:Then
List 2[2]->List 2[1]
Locate 1,5,"HP RECOVEREDChar!"
While Getkey=0
If Z=3 And List 2[12]>1:Then
List 2[12]-1->List 2[12]:List 2[3]+150->List 2[3]
If List 2[3]>List 2[4]:Then
List 2[4]->List 2[3]
Locate 1,5,"MP RECOVEREDChar!"
While Getkey=0
If Z=5:Then
Locate 2,1,"CURE -3 MP"
If List 2[14]=1:Then
Locate 2,2,"ICE -5 MP":IfEnd
If List 2[15]=1:Then
Locate 2,3,"FIRE -6 MP":IfEnd
If List 2[16]=1:Then
Locate 2,4,"BOLT -6 MP":IfEnd
If List 2[17]=1:Then
Locate 2,5,"CURE2 -8 MP":IfEnd
If List 2[18]=1:Then
Locate 2,6,"HATE -15 MP":IfEnd
Locate 19,1,"MP:"
Locate 18,2,List 2[3]
For 14->S To 18
If List 2[S]=1:Then
While W=10
Locate 1,Z,"_=>_"
If K>1:Then
Locate 1,Z," ":IfEnd
Z-1((K=28 And Z>1)-(K=37 And Z<L))->Z
If K=48:Then
If K=78 And K=31:Then
If Z=1 And List 2[3]>>=3:Then
Locate 1,1,"YOU CAST CUREChar!"
List 2[1]+100->List 2[1]
If List 2[1]>List 2[2]:Then
List 2[2]->List 2[1]:IfEnd
Locate 1,2,"YOU REGAINED HEALTHChar!"
While Getkey=0
If Z=2 And List 2[3]>=5:Then
Locate 1,1,"YOU CAST ICE..."
RanInt#(4,List 2[10]*4)->List 3[2]
Locate 1,2,Str 1
Locate 1,3,"TOOK"
Locate 1,4,List 3[2]
Locate 1,5,"DAMAGE"
While Getkey=0
List 2[3]-5->List 2[3]:List 3[4]-List 3[2]->List 3[4]:IfEnd
If Z=3 And List 2[3]>=6:Then
Locate 1,1,"YOU CAST FIRE.."
RanInt#(6,6*List 2[10])->List 3[2]
List 3[4]-List 3[2]->List 3[4]
Locate 1,2,Str 1
Locate 1,3,"TOOK"
Locate 1,4,List 3[2]
Locate 1,5,"DAMAGE"
List 2[3]-6->List 2[3]While Getkey=0
If Z=4 And List 2[3]>=6:Then
RanInt#(8,8+List 2[10])->List 3[2]
Locate 1,1,"YOU CAST BOLT.."
Locate 1,2,Str 1
Locate 1,3,"TOOK"
Locate 1,4,List 3[2]
Locate 1,5,"DAMAGE"
List 3[4]-List 3[2]->List 3[4]
List 2[3]-6->List 2[3]
While Getkey=0
If Z=5 And List 2[3]>=8:Then
Locate 1,1,"YOU CAST CURE2.."
List 2[3]-8->List 2[3]
List 2[1]+200->List 2[1]
If List 2[1]>List 2[2]:Then
List 2[2]->List 2[1]
Locate 1,2,"YOU RECOVERED HEALTHChar!"
While Getkey=0
If Z=6 And List 2[3]>=15:Then
Locate 1,1,"YOU CAST HATE.."
List 2[3]-15->List 2[3]
RanInt#(10,10*List 2[10])->List 3[2]
Locate 1,2,Str 1
Locate 1,3,"TOOK"
Locate 1,4,List 3[2]
Locate 1,5,"DAMAGE"
List 3[4]-List 3[2]->List 3[4]
While Getkey=0
If List 3[4]<1:Then
If List 3[4]>0:Then
If W=5:Then
Locate 1,1,Str 1
Locate 1,2,"ATTACKSChar!"
If J>2:Then
If J<3:Then
If List 3[6]>=4:Then
If W=7:Then
If J=1 Or J=4:Then
Locate 1,4,"STRIKE MISSEDChar!"
While Getkey=0
If J<>1 Or J<>4:Then
Locate 1,3,"YOU TAKE"
List 3[8]-List 2[6]->List 3[2]
While List 3[2]<1
List 3[2]+1->List 3[2]
Locate 1,4,List 3[2]
Locate 1,5,"DAMAGEChar!"
While Getkey=0
List 2[1]-List 3[2]->List 2[1]
If List 2[1]<1:Then
Locate 1,1,"YOU HAVE DIED."
While Getkey=0
If W=8:Then
If List 3[6]<4:Then
Locate 1,1,Str 1
Locate 1,2,"DOESN'T HAVE ANY MPChar!"
While Getkey=0
Locate 1,3,Str 1
RanInt#(0,List 3[14])->J
If J=0:Then
Locate 1,4,"DOES NOTHNG.."
While Getkey=0
If J=1 And List 3[6]>3:Then
Locate 1,4,"CAST CUREChar!"
List 3[4]+50->List 3[4]
List 3[6]-4->List 3[6]
If List 3[4]>List 3[5]:Then
List 3[5]->List 3[4]
While Getkey=0
If J=2 And List 3[6]>9:Then
Locate 1,4,"CAST ICE.."
RanInt#(4,4*List 3[10])->List 3[2]
Locate 1,5,"YOU TAKE"
Locate 1,6,List 3[2]
Locate 1,7,"DAMAGEChar!
While Getkey=0
If J=3 And List 3[6]>15:Then
Locate 1,4,"CAST FIRE.."
RanInt#(7,7*List 3[10])->List 3[2]
Locate 1,5,"YOU TAKE"
Locate 1,6,List 3[2]
Locate 1,7,"DAMAGEChar!"
List 2[1]-List 3[2]->List 2[1]
List 3[6]-16->List 3[6]
While Getkey=0
If J=4 And List 3[6]>21:Then
Locate 1,4,"CAST BOLT.."
RanInt#(10,10*List 3[10])->List 3[2]
Locate 1,5,"YOU TAKE"
Locate 1,6,List 3[2]
Locate 1,7,"DAMAGEChar!"
List 3[6]-22->List 3[6]
List 2[1]-List 3[2]->List 2[1]
While Getkey=0
If List 2[1]<1:Then
Locate 1,1,"YOU HAVE DIED"
Locate 1,2,"GAME OVER"
While Getkey=0
If List 3[6]>3:Then
1->List 3[14]:IfEnd
If List 3[6]>9:Then
2->List 3[14]:IfEnd
If List 3[6]>15:Then
3->List 3[14]:IfEnd
If List 3[6]>21:Then
4->List 3[14]:IfEnd
If W=0:Then
Locate 1,1,"VICTORYChar!"
Locate 1,2,"YOU GAIN:"
Locate 1,3,"GOLD:"
Locate 7,3,List 3[12]
List 2[7]+List 3[12]->List 2[7]
Locate 1,4,"EXP:"
Locate 6,4,List 3[11]
List 2[8]+List 3[11]->List 2[8]
Locate 1,5,"ITEM:"
If List 3[13]=0:Then
Locate 7,5,"NOTHNG":IfEnd
If List 3[13]=1:Then
Locate 7,5,"POTION"
1+List 2[11]->List 2[11]:IfEnd
If List 3[13]=2:Then
Locate 7,5,"ETHER"
1+List 2[12]->List 2[12]:IfEnd
If List 2[8]>=List 2[9]Then
Locate 1,6,"LEVEL UPChar!"
1+List 2[10]->List 2[10]:List 2[8]-List 2[9]->List 2[8]:30+List 2[9]->List 2[9]:10+List 2[2]->List 2[2]:List 2[2]->List 2[1]:7+List 2[4]->List 2[4]:List 2[4]->List 2[3]:1+List 2[5]->List 2[5]:1+List
2[6]->List 2[6]:IfEnd
1+List 2[21]->List 2[21]
If List 2[10]=3 And List 2[14]=0:Then
1->List 2[14]:Locate 1,7,"LEARNED ICEChar!":IfEnd
If List 2[10]=5 And List 2[15]=0:Then
1->List 2[15]:Locate 1,7,"LEARNED FIREChar!":IfEnd
If List 2[10]=7 And List 2[16]=0:Then
1->List 2[16]:Locate 1,7,"LEARNED BOLTChar!":IfEnd
If List 2[10]=10 And List 2[17]=0:Then
1->List 2[17]:Locate 1,7,"LEARNED CURE_^<2>_Char!":IfEnd
If List 2[10]=15 And List 2[18]=0:Then
1->List 2[18]:Locate 1,7,"LEARNED HATEChar!":IfEnd
While Getkey=0
If you want to help, there's the code.
I need to look out for the following:
Lists are as List #[#]
Anything that starts with " needs to end with "
And ( needs a proper ) to close it for each instance.
I'll dig through the code after work, just figured if anyone was bored and they might spot something while I'm doing that, that would be awesome and help speed this along
Update: Found one, near the top, If Z=5:Then was not on a new line.. | {"url":"https://dev.cemetech.net/forum/viewtopic.php?t=10167&view=next","timestamp":"2024-11-12T15:09:38Z","content_type":"text/html","content_length":"92358","record_id":"<urn:uuid:62dffb4f-4d2a-461f-8268-f6ef0d60e461>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00375.warc.gz"} |
Numerical Analysis/stability of RK methods - Wikiversity
Let ${\displaystyle \tau _{i}}$ be the local truncation error, and k is the number of time steps, then:
1. The numerical method is consistent with a differential equation if
${\displaystyle \lim _{h\to 0}\operatorname {max} |\tau _{i}|=0}$ over ${\displaystyle 1\leqslant i\leqslant k}$ .
according to this definition w:Euler's method is consistent.
2. A numerical method is said be convergent with respect to differential equation if
${\displaystyle \lim _{h\to 0}|x(t_{i})-y_{i}|=0}$ over ${\displaystyle 1\leqslant i\leqslant k}$ ;
where ${\displaystyle y_{i}}$ is the approximation for ${\displaystyle x(t_{i})}$ .
3. A numerical method is stable if small change in the initial conditions or data, produce a correspondingly small change in the subsequent approximations.
Theorem: For an initial value problem
${\displaystyle x'=f(t,x)}$ with ${\displaystyle t\in [t_{0},t_{0}+\alpha ]}$
and certain initial conditions, ${\displaystyle (t_{0},x_{0})}$ , let us consider a numerical method of the form
${\displaystyle (y_{0}=x_{0})}$ and ${\displaystyle y_{i+1}=y_{i}+h\phi (t_{i},y_{i},h)}$ .
If there exists a value ${\displaystyle h>0}$ such that it is continuous on the iterative domain, ${\displaystyle \Omega }$ and if there exists an ${\displaystyle L>0}$ such that
${\displaystyle |\phi (t,y,h)-\phi (t,y^{*},h)|\leqslant L|y-y^{*}|}$ for all ${\displaystyle (t,y,h),\,(t,y^{*},h)\in \Omega }$ ,
the method fulfills the w:Lipschitz condition, and it is stable and convergent if and only if it is consistent. That is,
${\displaystyle \phi (t,x,0)=f(t,x)}$ for all ${\displaystyle t\in \Omega }$ .
For a similar argument, one can deduce the following for multi- step methods:
1. The method is stable if and only if all roots, ${\displaystyle \lambda }$ , of the characteristic polynomial satisfy
${\displaystyle |\lambda |\leqslant 1}$ ,
and any root of
${\displaystyle |\lambda |=1}$
is simple root.
2. one more result is that if the method is consistent with the differential equation, the method is stable if and only if it it is convergent.
see [1]
stability polynomials of Runge-Kutta methods
The w:Runge–Kutta methods are very useful in solving systems of differential equations, it has wide applications for the scientists and the engineers, as well as for the economical models, the
recognized with their practical accuracy where we can use and get very good results and approximations when solving an ODE problem, RK has the general form :${\displaystyle y_{n+1}=y_{n}+h\sum _{i=1}
^{s}b_{i}k_{i}}$ , where
${\displaystyle k_{1}=hf(t_{n},y_{n}),\,}$
${\displaystyle k_{2}=hf(t_{n}+c_{2}h,y_{n}+a_{21}k_{1}),\,}$
${\displaystyle k_{3}=hf(t_{n}+c_{3}h,y_{n}+a_{31}k_{1}+a_{32}k_{2}),\,}$
${\displaystyle \vdots }$
${\displaystyle k_{s}=hf(t_{n}+c_{s}h,y_{n}+a_{s1}k_{1}+a_{s2}k_{2}+\cdots +a_{s,s-1}k_{s-1}).}$
such that
s is the number of stages,
a[ij] (for 1 ≤ j < i ≤ s),
b[i] (for i = 1, 2, ..., s)
and c[i] (for i = 2, 3, ..., s).
Example:finding the stability polynomial for RK4's methods
for RK4" case${\displaystyle 2_{a}}$ ", which characterizes by ${\displaystyle c_{2}={\frac {-1}{4}}}$ which has the form: the stability region is found by applying the method to the linear test
equation ${\displaystyle y'=\lambda y}$
${\displaystyle y_{n+1}=y_{n}+{\frac {1}{6}}k_{1}+0k_{2}+{\frac {2k_{3}}{3}}+{\frac {k_{4}}{6}}}$
the stability region is found by applying the method to the linear test equation ${\displaystyle y'=\lambda y}$
${\displaystyle \displaystyle \ k_{1}=hf(t_{n},y_{n})}$
${\displaystyle k_{2}=hf(t_{n}+{\frac {h}{2}},y_{n}-{\frac {k_{1}}{2}})}$
${\displaystyle k_{3}=hf(t_{n}+{\frac {h}{2}},y_{n}+{\frac {3k_{1}}{4}}-{\frac {k_{2}}{4}})}$
${\displaystyle k_{4}=hf(t_{n}+h,y_{n}-2k_{1}+k_{2}+2k_{3})}$
using the linearized equation ${\displaystyle f(t,y)=\lambda y}$ , and considering ${\displaystyle {\hat {h}}=h\lambda }$ , we get;
${\displaystyle k_{1}={\hat {h}}y_{n}}$
${\displaystyle k_{2}={\hat {h}}(1-{\frac {\hat {h}}{2}})y_{n}}$
${\displaystyle k_{3}={\hat {h}}(1+{\frac {3{\hat {h}}}{4}}-{\frac {\hat {h}}{4}}(1-{\frac {\hat {h}}{2}}))y_{n}}$
${\displaystyle k_{4}={\hat {h}}(1-2{\hat {h}}+{\hat {h}}(1-{\frac {\hat {h}}{2}})+2{\hat {h}}(1+{\frac {3{\hat {h}}}{4}}-{\frac {\hat {h}}{4}}(1-{\frac {\hat {h}}{2}})))))y_{n}}$
substitute these back in ${\displaystyle y_{n+1}}$ , yields ${\displaystyle y_{n+1}=[1+({\hat {h}})+{\frac {({\hat {h}})^{2}}{2}}+{\frac {({\hat {h}})^{3}}{6}}+{\frac {({\hat {h}})^{4}}{24}}]y_{n}=R
({\hat {h}})y_{n}}$
and so the characteristic polynomial
${\displaystyle P(z)=z-R(z);z={\hat {h}}}$ for the absolute stability region for this method, set |R(z)|<1, and so we get the region in figure 1. case${\displaystyle 2_{a}}$
the table below shows the final forms for the stability function for different forms of RK4, these RK4's are different in the values of ${\displaystyle b_{j}}$ , and they are fullfilling the
consistencty requirement for the method i.e :${\displaystyle \sum _{j=1}^{i-1}a_{ij}=c_{i}\ \mathrm {for} \ i=2,\ldots ,s.}$
case# ${\displaystyle (b_{1},b_{2},b_{3},b_{4})}$ stability function
caseI ${\displaystyle ({\frac {1}{8}},{\frac {3}{8}},{\frac {3}{8}},{\frac {1}{8}})}$ ${\displaystyle 1+z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{6}}+{\frac {z^{4}}{24}}}$
caseII_a ${\displaystyle ({\frac {1}{6}},0,{\frac {2}{3}},{\frac {1}{6}})}$ ${\displaystyle 1+z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{6}}+{\frac {z^{4}}{24}}}$
caseII_b ${\displaystyle ({\frac {1}{6}},0,{\frac {2}{3}},{\frac {1}{6}})}$ ${\displaystyle 1+z+{\frac {5z^{2}}{6}}-{\frac {17z^{3}}{12}}-{\frac {2z^{4}}{24}}+{\frac {z^{5}}{12}}}$
caseIII ${\displaystyle ({\frac {1}{12}},{\frac {2}{3}},{\frac {1}{12}},{\frac {1}{6}})}$ ${\displaystyle 1+{\frac {11z}{12}}+{\frac {5z^{2}}{12}}+{\frac {z^{3}}{6}}+{\frac {z^{4}}{24}}}$
caseIV ${\displaystyle ({\frac {1}{6}},0,{\frac {2}{3}},{\frac {1}{6}})}$ ${\displaystyle 1+z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{6}}+{\frac {z^{4}}{24}}}$
${\displaystyle 1+z+{\frac {z^{2}}{2}}+{\frac {z^{3}}{6}}+{\frac {z^{4}}{24}}}$
classical Rk ${\displaystyle ({\frac {1}{6}},{\frac {1}{3}},{\frac {1}{3}},{\frac {1}{6}})}$
see [2] to find the table
Plotting the stability region
In order to plot the stability region, we can set the stability function to be bounded by 1 and solve for the values of z, then draw z in the complex plane. Since R(z) is the unit circle in the
complex plane, each point on the boundary can be represented as ${\displaystyle e^{i\theta }}$ and so by changing ${\displaystyle \theta }$ over the interval${\displaystyle [0,2\pi ]}$ , we can draw
the boundaries of that region. The following w:OCTAVE/Matlab code does this by plotting contour curves until reaching the boudaries:
[x,y] = meshgrid(-6:0.01:6,-6:0.01:6);
z = x+i*y;
R = 1+z+0.5*z.^2+(1/6)*z.^3+(1/24)*z.^4;
zlevel = abs(R);
contour(x,y,zlevel,[1 1]);
The figure at right shows the absolute stability regions for RK4 cases which is tabulated above[3]
• Eberly, David (2008), stability analysis for systems of defferential equation.[4]
• Ababneh, Osama; Ahmad, Rokiah; Ismail, Eddie (2009), "on cases of fourth-order Runge-Kutta methods", European journal of scientific Research.[5]
• Mathews, John; Fink, Kurtis (1992), numerical methods using matlab.
• Stability of Runge-Kutta Methods [6] | {"url":"https://en.m.wikiversity.org/wiki/Topic:Numerical_Analysis/stability_of_RK_methods","timestamp":"2024-11-02T07:37:59Z","content_type":"text/html","content_length":"159740","record_id":"<urn:uuid:2b562f20-ddb3-4c86-b266-2e661aa7ebc9>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00619.warc.gz"} |
Advanced Mathematics of Quantum Computing
In this article we'll review the fundamental mathematics required to analyze and solve quantum computing problems. The article will assume you have some knowledge of quantum computing, if not you can
check out these resources:
This article is based on this course on Advanced Math for Quantum Computing, and will first provide a brief review of quantum computing and then discuss several key mathematical concepts.
1. Quantum Computing Review
Let's briefly review key concepts of quantum computing.
A qubit is a quantum bit that can be 0, 1, or in superposition between 0 and 1.
The state of a qubit can be represented by the following matrices:
$$0 = |0\rangle = \begin{bmatrix}1 \\0\end{bmatrix}$$
$$1 = |1\rangle = \begin{bmatrix}0 \\1\end{bmatrix}$$
A state vector is a column matrix that represents a quantum state.
Keep in mind that $|0\rangle$ and $|1\rangle$ are names of matrices for the qubit state they represent, they are not actually numeric values.
The numbers inside a state vector are the square root of probabilities. They are complex values that are used in qubit state computations.
The notation for $|0\rangle$ and $|1\rangle$ is called a "ket", which is the matrix itself with no changes.
The "bra" operation complements the "ket" and is notated with a left angle bracket and vertical bar as follows:
$$\langle 0| = 0^+$$
$$\langle 1| = 1^+$$
The bra operation is the same as the adjoint operation, which is represented by the $^+$. The adjoint operation is the transpose of the complex conjugate.
Two special matrices whose properties are defined by their adjoints include:
• Unitary: if the adjoint of a matrix is the same as its inverse, it is said to be unitary.
$$U^+ = U^{-1}$$
• Hermitian: if the adjoint of a matrix is the matrix itself, it is said to be Hermitian.
$$H^+ = H$$
Finally, eigenvectors and eigenvalues are important concepts, here is a quick review:
Given a matrix $A$, you might be able to compute a vector $V$ such that when $A$ is multiplied by $V$ the result is the same as multiplying a scalar $e$ by $V$:
$$AV = eV$$
• $V$ is eigenvector of $A$
• $e$ is the eigenvalue of A corresponding to vector $V$
Each different matrix $A$ will have a different set of eigenvectors and eigenvalues.
2. Key Concepts: Mathematics of Quantum Computing
In this section we'll review several key mathematical techniques of quantum computing, including:
• Probability of Measurement
• Orthonormality
• Orthonormality in Bracket Notation
• Basis Vectors
• Degrees of Freedom
• Degrees of Freedom of a Single Qubit
• Global & Relative Phase
• Bloch Sphere
• Entanglement & Degrees of Freedom
• Testing for Entanglement
Probability of Measurement
A quantum state is represented by a column matrix, which is also called a vector.
If we have a vector $S$ to represent the state of the system and we have another vector $A$ to represent the measurement apparatus—meaning the system $S$ is measured by the apparatus $A$.
In an idealized measurement, the system will report that it is aligned with the apparatus or not. This means that the result of measurement is either "aligned" or "not aligned".
The equation for a probability of measurement is aligned is as follows:
$$|A|S|^2 = (|<A|x|S>|)^2 = |A^+S|^2 = |<S|A>|^2 = (|<S|x|A>|)^2 = |S^+A|^2$$
Recall that the squared magnitude of $|<S|A>|$ is the same as the squared magnitude of $|<A|S>|$.
The vectors $\begin{bmatrix}1 \\0\end{bmatrix}$ and $\begin{bmatrix}0 \\1\end{bmatrix}$ are orthogonal to each other.
If an apparatus is in $\begin{bmatrix}0 \\1\end{bmatrix}$ and a qubit state is $\begin{bmatrix}1 \\0\end{bmatrix}$ the probality of a measurement being "aligned" is 0:
$$|<S|A>|^2 = 0$$
If the apparatus is $\begin{bmatrix}1 \\0\end{bmatrix}$ and the qubit state is $\begin{bmatrix}1 \\0\end{bmatrix}$, again the probability of being "aligned" is 0.
Recall that the magnitude of a vector $\begin{bmatrix}a \\b\end{bmatrix}$ is 1:
$$\sqrt{|a|^2 + |b|^2 = 1}$$
When the magnitude is 1, we call the vector a unit vector. When the magnitude is 1, the squared magnitude is also 1.
If a vector is both orthogonal and unit vectors, they are orthonormal. The notation for orthonormality is as follows:
• $||$ the vertical bars denote magnitude
• $^*$ superscript star indicates complex conjugation
For example, given two vectors $\begin{bmatrix}a \\b\end{bmatrix}$ and $\begin{bmatrix}c \\d\end{bmatrix}$, all of which may be complex numbers, the following conditions must be met for the two
vectors to be orthnormal:
$$|a|^2 + |b^2| = 1$$
$$|c|^2 + |d^2| = 1$$
$$[a* b*] \begin{bmatrix}c \\d\end{bmatrix} = 0$$
Orthonormal vectors are important because any vector can be written as a linear combination of orthonormal vectors.
Orthonormality in Bracket Notation
Here is a review of orthonormality in bracket notation:
• X and Y are unit vectors, meaning the magnitude of X and Y is 1:
$| |X> | = 1$ and $| |Y> | = 1$
• Bracket $X$ and $Y$ is 0:
$<X|Y> = 0$
• The bracket of any unit vector with itself is 1:
$<X|X> = 1$ and $<Y|Y> = 1$
A set of vectors that are orthonormal to each other is called an orthonormal set. If we have an othornomal set of vectors:
• Bracket of a vector with itself is 1
• Bracket of a vector with any other vector in the set is 0
$<V_i|V_j> = 1$, when $i = j$
$<V_i|V_j> = 0$, when $i \neq j$
The maximum number of elements in an orthonormal set is equal to the number of vector dimensions. In other words, for $n$ dimensional vectors, an orthonormal set can have a maximum of $n$ vectors.
Basis Vectors
If we have a two coordinate space of $x$ and $y$ axis...
We also define a unit vector $\hat{x}$ along the $x$ axis and unit vector $\hat{y}$ along the $y$ axis. This means that any point in the coordinate space can be described as:
$$a\hat{x} + b\hat{y}$$
This means that any vector from the origin point can be written as a linear combination of the unit vectors $\hat{x}$ and $\hat{y}$.
$\hat{x}$ and $\hat{y}$ are also orthogonal to each other.
The point is, if any two unit vectors are orthogonal we can write any vector as a linear combination of the two vectors.
Thus, we can use any pair of orthogonal unit vectors instead of $\hat{x}$ and $\hat{y}$.
In the context of quantum computing, we first state a problem by choosing a set of orthonormal vectors.
We then express all other vectors in terms in terms of the orthonormal vectors we choose, which are referred to as basis vectors.
Degrees of Freedom
Suppose we have a physical system and the state of it can be described by real numbers.
The minimum number of states required to describe the system is the degrees of freedom of the system. This concept applies to both quantum and classical systems.
A simple classical system is temperature, which can be represented as a single real number. There are an infinite number of states that can be represented by the real number for temperature.
If we have a system with two degrees of freedom, there is also an infinite number of states—except this is a larger set than the single degree of freedom.
The point is that larger the number of degrees of freedom result in larger infinite possible states of a system.
As we'll discuss below, degrees of freedom is a critical concept for understanding entanglement between qubits.
Degrees of Freedom of a Single Qubit
A single qubit state is represented by a column matrix with two complex elements, each of which has two real numbers:
$$\begin{bmatrix}a+ib \\c+id\end{bmatrix}$$
where a, b, c, & d are real numbers.
There are two constraints on a single qubit system:
• The requirement that state vectors are unit vectors since the sum of probabilities must be one:
$$a^2 + b^2 + c^2 + d^2 = 1$$
• The other constraint is from the physical state of quantum systems, which says that $|S\rangle$ and $e^{i\theta} | S\rangle$ are the same state.
The number of degrees of freedom of a single qubit state is 4 real parameters - 2 constraints, so it is 2.
Next we will discuss how entanglement occurs in system with more than 1 qubit.
Global & Relative Phase
As mentioned, the global phase does not affect physical behaviour. Global phase, however, should not be confused with relative phase.
If we have a state vector $\begin{bmatrix}a\\b\end{bmatrix}$ that are complex numbers.
The relative phase between $a$ and $b$ can be described as follows:
$$e^{i\phi} = \frac{a |b|}{|a| b}$$
• where $\phi$ is the relative phase in the equation
Unlike the global phase, the relative phase is physically important. In other words, when doing quantum math you can't just substitute one vector with another because it is physically equivalent.
Bloch Sphere
Earlier, we defined the standard basis vectors as:
$$|0\rangle = \begin{bmatrix}1\\0\end{bmatrix}$$
$$|1\rangle = \begin{bmatrix}0\\1\end{bmatrix}$$
There is also the Hadamard basis, which uses the vectors ket-plus and ket-minus:
$$|+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$$
$$|-\rangle = \frac{1}{\sqrt2}(|0\rangle - |1\rangle)$$
The ket-plus and ket-minus are simply the conventional name of these vectors.
In order to represent state vectors visually we can use a Bloch sphere.
One point of confusion is that we previously computed state vectors for spin along the X, Y, and Z directions. When dealing with a Bloch sphere, however, do not try and associate a direction.
The Bloch sphere is simply a way to visually represent state vectors—its three axes do not always correspond to cartesian coordinates i.e. X, Y, and Z in physical space.
Given two angles $\theta$ and $\phi$, we can map those angles to plot the state vector on the surface of the Bloch sphere as follows:
Image Source
Opposite points on the Bloch sphere represent state vectors that are orthogonal.
Recall that the state vector of a qubit has 2 degrees of freedom, which indicates that a state vector can only be mapped to a point on the surface of the Bloch sphere.
Entanglement & Degrees of Freedom
Revisiting the degrees of freedom concept, suppose we have 2 systems each with 2 degrees of freedom.
If we combine the two systems, the combination would have 4 degrees of freedom.
System AB is defined by $(x_a, y_b, x_b, y_b)$
In the case of quantum physics, if we combine two particles and "glue" them together, this property would add more degrees of freedom to the combined system. In other words, there would be 5 or more
degrees of freedom.
When we combine two single qubit systems, each qubit has 2 degrees of freedom.
In this case, there are 4 complex numbers, or 8 real parameters, 2 constraints (sum of probabilities = 1, global phase is irrelevant).
This means there are 8 - 2 = 6 degrees of freedom, and this is how entanglement can be explained in terms of degrees of freedom
Testing for Entanglement
A tensor product $A\otimes B$ is always an unentangled state, which gives us a test for entanglement.
If a state vector of a multi-qubit system can be expressed as a tensor product of single-qubit state, then it is not entangled.
On the other hand, the basis vectors used don't matter in determining in a state is entangled or not.
Summary: Mathematics of Quantum Computing
In this article we reviewed key mathematical concepts of quantum computing, including:
• Probability of Measurement
• Orthonormality
• Orthonormality in Bracket Notation
• Basis Vectors
• Degrees of Freedom
• Degrees of Freedom of a Single Qubit
• Global & Relative Phase
• Bloch Sphere
• Entanglement & Degrees of Freedom
• Testing for Entanglement | {"url":"https://blog.mlq.ai/advanced-mathematics-quantum-computing/","timestamp":"2024-11-11T19:33:04Z","content_type":"text/html","content_length":"71138","record_id":"<urn:uuid:a9eb4d1a-6f7d-4d9d-a2ea-817fec165845>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00476.warc.gz"} |
4th Grade Data Representations - iPohly INC
4th Grade Data Representations
"I love these! They are always high quality and high rigor. Thank you for putting in so much time to help the rest of us out."
September 19, 2018, Kristi M
Data Representations
This unit is about using all data! In this unit students learn about stem and leaf plots, dot plots and frequency tables with whole numbers, fraction and decimals. They use plots and tables to
problem solve.
I see you~
• struggling each week to write lesson plans that meet the rigor of the TEKS.
• searching endlessly for resources that will help kids learn math while being challenged and engaged.
• staying late everyday after school working on plans and creating everything from scratch.
You are exhausted from working with students all day,
and still have to prep, write and create.
I see you~
• SACRIFICING your time with your family and friends to ensure success for ALL of OUR Children.
This resource is a wonderful addition to our unit. Students enjoy the activities included!
November 6, 2019 Misty B.
Days 1-4
Day 1: I will represent data on a frequency table and a dot plot marked with whole numbers and fractions.
The unit starts with building on what kids know- frequency tables and dot plots. Students use fractions and decimals to build frequency tables and dot plots. This is the first time students use
categories of numbers in frequency tables.
Follow this with a SCOOT to get them up and moving!
Day 2: I will solve one- and two step problems using data in whole number, decimal, and fraction form in a frequency table and a dot plot.
• Identify missing information from a dot plot or frequency table
• Find the total
• Find the difference
• Solve a riddle to find the dot plot or frequency table
Active Engage: The Very Unfair Game
Day 3: I will represent data on a stem-and-leaf plot marked with whole numbers, decimals and fractions.
Introduction to stem and leaf plots! Guided notes are included!
Active Engage: Identify the stem and leaf cards and data sets for partners to create a stem and leaf plot.
Day 4: I will represent data on a stem-and-leaf plot marked with whole numbers, decimals and fractions.
Math Huddle or Math Talk- Jack and Jill Double stem and leaf plot (not tested, but great discussions on why we use a stem and leaf!)
Gallery Walk: students are given a a data card to create a stem and leaf plot.
"If you are a Texas teacher who follows the TEKS Resource System, you won't regret purchasing this. The unit is well laid out and has great activities in each lesson!"
November 17, 2017, Kari H.
Days 5-7
Day 5: I will solve one- and two step problems using data in whole number, decimal, and fraction form in a stem-and leaf plot.
Students use place value, multiplication and addition to find the total for stem and leaf plots.
Active engage: 4 Mirror Cards
Independent Practice: 4 Cards
Day 6: I will represent data on a frequency table, dot plot, or stem-and leaf plot marked with whole numbers and fractions.
Matching Game: Students match stem and leaf plots, dot plots and frequency tables. They list the data sets!
Day 7: I will solve one- and two step problems using data in whole number, decimal, and fraction form in a frequency table, a dot plot or stem-and-leaf plot.
20 questions to combine all of the learning in this unit in a fun game called Ghost in the Graveyard!
"These units are absolutely amazing! They align perfectly with my curriculum and I love that they have the needed rigor, while still being fun and engaging. You are such a huge time-saver!"
October 3, 2018, Katie Miller's Creations | {"url":"https://ipohlyinc.com/4th-grade-data-representations/","timestamp":"2024-11-11T17:19:08Z","content_type":"text/html","content_length":"74827","record_id":"<urn:uuid:7ef872ef-2708-44b4-8699-3c268c12f4d2>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00294.warc.gz"} |
SAT Physics Circular Motion - Uniform Circular Motion
SAT Physics Circular Motion - Uniform Circular Motion
Uniform circular motion
involves objects moving at a constant speed but with changing velocity. When an object moves at constant speed, the magnitude of velocity is also constant. How then is velocity changing in circular
motion? Velocity is a vector quantity, meaning it consists of both magnitude and direction. Objects moving in circular motion are continually changing direction. As a result, the velocity of the
object is changing even though it moves at a constant speed.
Like velocity, the acceleration of an object in uniform circular motion has constant magnitude but has changing direction. Acceleration is the rate of change in velocity (the change in velocity
during a time interval). If velocity has a constant magnitude, the acceleration will also have a constant value. However, the direction of acceleration constantly changes as an
object moves in circular motion. In uniform circular motion, the direction of the acceleration vector is toward the center of the circle. This acceleration is said to be uniform: having a constant
magnitude and applied in the same manner (toward the center) at all times.
Period and Frequency
All linear motion quantities are based on the linear meter. However, the linear meter is not very useful when describing motion that does not follow a straight path. All circles have one thing in
common. Objects moving in circles return to the same location every time they complete one
(one circle, one revolution, one rotation, and so on). A cycle consists of one circumference, and this is the basis for all circular motion quantities.
The time to complete one cycle is known as the
, T. The period of a circling object can be calculated by dividing the time of the motion, t, by the number of cycles completed during time t.
The number of cycles does not have any units, and the units of period are seconds.
of an object is the number of cycles an object completes during one second. Think of the frequency as how frequently the object is cycling. Mathematically, frequency is the inverse of the period. The
units of frequency are inverse seconds (1/s or s
). These units are also known as Hertz (Hz). Any of these units may be used on exams, and the formula for frequency is simply the inverse of the formula for the period.
The relationship between the period and frequency is expressed in the following equation:
Period and Frequency
An object completes 20 revolutions in 10 seconds. Determine the period and frequency of this motion.
WHAT'S THE THICK?
A revolution is another way to indicate a cycle, and a cycle is simply an event.The number of cycles is just the count of an event, and therefore it has no units.Period is the time for one complete
cycle. Time has the units of seconds, and therefore time must be in the numerator.
Tangential Velocity and Centripetal Acceleration
Velocity is not constant in uniform circular motion. However, when any moving object is paused (frozen for an instant of time), it will have an instantaneous velocity vector with a specific magnitude
and direction. When objects follow a curved path, the instantaneous velocity is tangent to the motion of the object. Thus, the instantaneous velocity is referred to as the
tangential velocity
. Several instantaneous velocity vectors are shown for the object circling in Figure 6.1.
Figure 6.1. Instantaneous velocity vectors
In Figure 6.1, it is apparent that although the direction of the tangential velocity is continually changing, its magnitude remains constant. The magnitude of the tangential velocity is also equal
to the speed of the circling object. They are both determined using a modified version of the constant speed formula:
As previously stated, circular motion is based on one complete cycle. In one complete cycle, an object travels one circumference, d = 2nr, in one period, t=T.
One important aspect of tangential velocity involves objects leaving the circular path. Forces are responsible for creating circular motion. If the forces causing circular motion stop acting, the
object will leave the circular path. When this happens, the tangential velocity becomes the initial velocity for the object’s subsequent motion. If no forces act on the object, it will move in a
straight line matching the tangential velocity at the time of release, as shown in Figure 6.2(a). However, if another force acts on the object, the object will become subject to the new force. In
Figure 6.2(b), the object leaves the circle and is acted upon by gravity, causing projectile motion.
Figure 6.2. Tangential velocity. In (a) the object is initially circling
horizontally on a frictionless surface. In (b) the object is initially
circling vertically at a distance h above the surface.
Since circling objects have a changing velocity vector, they are continuously accelerating. This type of acceleration is known as centripetal acceleration, a
Centripetal means “center seeking.” Centripetal acceleration is directed toward the center of the circle, as shown in Figure 6.3.
Figure 6.3 centripital acceleration
Even though circling objects are accelerated toward the center of the circle, their tangential velocities prevent them from ever reaching the center. Since the acceleration vectors lie along the
radii of the circle, the centripetal acceleration may also be referred to as the radial acceleration. The centripetal acceleration can be determined with the following formula:
Tangential Velocity and Centripetal Acceleration
Determine the acceleration of an object experiencing uniform circular motion.It is moving in a circle with a radius of 10 meters and a frequency of 0.25 Hertz.
WHAT'S THE THICK?
The period is the inverse of the frequency.
Solving for the tangential velocity requires the period.
Solving for the centripetal acceleration requires the tangential velocity.
Solve each equation in turn.
Answers may be expressed in terms of n in order to avoid multiplying by 3.14.
In order for objects to experience acceleration, a net force (sum of forces) must act in the direction of the acceleration. Objects in circular motion are being accelerated toward the center of a
circular path. This implies that the net force is also directed toward the center of the circle. In uniform circular motion, the net force is known as the
centripetal force
, F
In linear motion problems, the net force is represented by
F[c] = ma[c]
Frequently, circular motion force problems involve the speed and/or tangential velocity
Dynamics problems in circular motion are solved in a similar manner as all other force problems.
1. Orient the problem.
Draw a force diagram. If the motion is circular, any force vectors pointing toward the center
of the circle are considered positive. Those pointing away from the center are negative.
2. Determine the type of motion.
If an object is moving along a circular path, recognize that the circular net force,
, is used in place of
3. Sum the force vectors.
Add all forces pointing toward the center of the circle, and then subtract all forces
pointing away from the center.
F[c] = + F[toward center] - F[away from center
4. Substitute and solve.
In this last step, substitute known equations for specific forces and substitute numerical
values to find in the solution.
Keep in mind that circular motion problems can incorporate elements from problems learned in other chapters.
Horizontal Circular Motion
A car moving at 10 meters/second completes a turn with a radius of 20 meters. Determine the minimum coefficient of friction between the tires and the road that will allow the car to complete the turn
without skidding.
WHAT'S THE THICK?
Orient the problem:
Picture a car making a turn.
Gravity is pulling down, and the normal force is acting upward. Neither of these forces is in the direction of motion. The force acting to keep the car in the turn is friction. Without friction
acting toward the center of the turn, the car would slide in a straight path following the tangential velocity.
Determine the type of motion:
A turning car is in circular motion.
Sum the force vectors in the relevant direction:
In circular motion F
is used instead of
F[c] = f
Substitute and solve:
The normal force is acting vertically in the y-direction. It is equal and opposite the force of gravity: N = F
= mg.
Vertical Circular Motion
A 2.0-kilogram mass is attached to the end of a 1.0-meter-long string. When the apparatus is swung in a vertical circle, the mass reaches a speed of 10 meters per second at the bottom of the swing.
Determine the tension in the string at the bottom of the swing.
WHAT'S THE THICK?
Orient the problem:
Sketch the scenario, including force vectors.
Determine the type of motion:
This is circular motion.
Sum the force vectors in the relevant direction:
In circular motion, F
is used instead of
F[c] = T - F[g
]T = F[c] + F[g]
Substitute and solve:
Determine the minimum speed at the top of the loop that allows the mass to make one complete cycle.
WHAT'S THE THICK?
The minimum speed at the top of a loop indicates a special case where the forces contributing to the overall centripetal force must be at an absolute minimum. Begin by solving the problem the same as
any other force problem.
Orient the problem:
Sketch the scenario, including force vectors.
Determine the type of motion:
This is circular motion.
Sum the force vectors in the relevant direction:
Tension and gravity are both pointing toward the center of the circle. Both are set as positive.
F[c] = T + F[g]
Now you must consider how to make the centripetal force, F
, as small as possible. The force of gravity cannot be altered. However, tension can be reduced by swinging the mass more slowly. When the mass is slowed to the point where it is just barely
completing a full circle, the tension will be zero, T = 0, at the exact instant that the mass is at the very top of the circle.
Substitute and solve.
1. The variety of the wager distributions at the particular person stage suggests a variety of particular person betting methods. Also, it indicates that a gambler may not stick with just one
betting strategy. It follows that the 카지노 계열 log-normal wager distribution noticed at the population stage may be very likely an mixture end result. | {"url":"http://www.bestsatprepbook.com/2017/09/sat-physics-circular-motion-uniform.html","timestamp":"2024-11-09T03:45:22Z","content_type":"text/html","content_length":"116765","record_id":"<urn:uuid:25cf8692-34a1-4a68-861f-46d8d5d3edc0>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00750.warc.gz"} |
Back fitter
10 graphs of experimental data are given. Can you use a spreadsheet to find algebraic graphs which match them closely, and thus discover the formulae most likely to govern the underlying processes?
Several experiments were performed and data measured over a period of 10 hours.
The resulting charts are shown below.
Can you find algebraic equations which closely match the curves, which could be used to predict values of the variables at other times? There might be many possible curves of the right sort of shape
by eye, so a numerical plot will be needed to discover the most likely candidates.
The following Office 2003 spreadsheet will allow you easily to compare plots of the likely curves against the actual data - the intention is that you will tackle this problem numerically.
Note: The different sets of experimental data are distinct, so try as many or as few as you like. Fitting all of the sets will present quite a challenge!
Extension: Whilst there are 'obvious' candidates for each data set, can you find multiple functions which give rise to apparently good matches to some of the data sets? How might you numerically
determine which fits are best?
Getting Started
Start by looking for data which can be matched to straightforward functions like a straight line, or a quadratic.
Which data sets oscillate? What functions do you know which oscillate?
If you are unfamiliar with the graphs of $y=a^x$ and $y=a^{-x}$ for different values of the constant $a$, try plotting them using graphing software or a graphical calculator.
Student Solutions
After a while spent as a toughnut, we recieved the solution to this problem. We were very pleased to see that one of our younger solvers, Jonathan, realised that the first graph was $y=0.5 x$.
Impressively, a full solution was sent in by James from Bay House, where all of his functions gave a close fit with the data -- well done James!
James' suggestions agreed with ours in six of the cases (bold font), but differed in four cases (normal font). Perhaps you might like to consider which you feel are the closer fit?
Experiment 1: $y=x/2$
Experiment 2: $y=\sin(x) $
Experiment 3: $y=x^2$
Experiment 4: $y=x-\sin(2x)$
Experiment 5: $y=5\log_{45}(x+1)$ $\left(\mbox{we got }y=\sqrt{x}\right)$
Experiment 6: $y=(\sin(1.7x)+1)/2$ $\left(\mbox{we got }y=\sin^2(x)\right)$
Experiment 7: $y=-0.6+(\log_{10}(6x))^{-1}$ $\left(\mbox{we got }y=\frac{1}{1+x^2}\right)$
Experiment 8: $y=\log_{15}(7x+1)$ $\left(\mbox{we got }y=1.65x/(1+x)\right)$
Experiment 9: $y=\cos(2x)$
Experiment 10: $y=2^x$
Teachers' Resources
Why do this problem?
This problem offers an opportunity to reflect on the very important concept of fitting a curve to experimental data. Along the way, students will utilise their skills of transforming graphs in order
to find a close fit, and consider ways of deciding how close their fit is. The problem is marked as challenge level 1 as it is a straightforward task to begin, but to find a complete solution for all
10 graphs is rather more challenging!
Possible approach
Although this problem stands alone, it could also be done as a follow-up to work on transformations of graphs based on the problem Parabolic Patterns.
Students will need access to computers or graphical calculators to get the best out of this task. Familiarity with spreadsheet software is assumed.
Part of the challenge of this problem is to identify which graphs are easiest to fit, as they are not presented in any particular order. One approach is to start by displaying the graphs and
discussing as a class or in pairs which have recognisable shapes, such as straight lines, quadratics, trig graphs and exponential graphs.
If students haven't met graphs such as $y=a^x$ and $y=a^{-x}$ it might be fruitful to give them some time to experiment with graphical calculators to see what these graphs look like for different
values of the constant $a$.
Once students have some preliminary ideas about graphs which might fit, small groups could start to work on the spreadsheet, entering a possible equation and seeing how closely it matches the given
data, then using their knowledge of transformations of graphs to tweak their equation to get a closer match. If the groups have graphical calculators but not spreadsheets, they could use graphical
calculators instead to find graphs with the right shape. Each group should agree on the best equation for each graph, so they need to convince each other that their equation is a good fit.
After students have had time to choose equations for some or most of the graphs, let them know that the whole class will come together to choose the best equations. Ideally, different groups will
come up with slightly different suggestions for functions, and this can stimulate discussion about how to decide which function most closely matches the data. Each group could argue the case for
their equation, and the class could decide together which is the best fit.
Key questions
What clues can we find from the axes and the points given to help us to guess a likely function?
How can we modify our guess once we've seen how closely it fits?
Does joining the points in order of increasing time help?
How do we decide when the fit is close enough?
Possible support
Graphs 1, 3 and 5 are the most straightforward functions to fit, so this is a good place to start.
Possible extension
Students could investigate and discuss the benefits of a least squares method of determining how close the fit is. | {"url":"https://nrich.maths.org/problems/back-fitter","timestamp":"2024-11-11T00:46:40Z","content_type":"text/html","content_length":"45200","record_id":"<urn:uuid:b8ddcfff-3401-4112-9782-c1fe48d24577>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00013.warc.gz"} |
Differentiation of e^tanx | Differentiate e^tanx - WBPREP
Differentiation of e^tanx | Differentiate e^tanx
The differentiation of e^tanx is equal to sec^2x e^tanx. In this post, we will learn how to differentiate e to the power tanx with respect to x.
The following two formulas will be used to find the derivative of e^tanx.
1. d/dx(tanx) = sec^2x.
2. log[a] a^k =k.
Now, we will learn to find the derivative of e to the power tanx with respect to x.
How to Find the derivative of e^tanx
Question: How to differentiate e^tanx?
Let y= e^tanx
Taking natural logarithms on both sides, we get that (natural logarithm means the logarithm with base e, i.e. log[e])
log y = tanx (here we used the logarithm rule log[e]e^k =k.)
Now, differentiate both sides with respect to x. This will give us
$\dfrac{1}{y} \dfrac{dy}{dx}$ = sec^2x as we know that the derivative of tanx is sec^2x.
⇒ $\dfrac{dy}{dx}$ = y sec^2x
Putting the value of y, that is, y=e^tanx, we get from above that
$\dfrac{d}{dx}$(e^tanx) = sec^2x e^tanx.
Thus, the differentiation of e^tanx with respect to x is equal to e^tanx sec^2x.
Video solution on derivative of e^tanx:
Derivative of arc(cotx): The derivative of arc(cotx) is -1/(1+x^2).
Differentiate e^sinx: The derivative of e^sinx is e^sinx cosx.
Differentiate e^cosx: The derivative of e^cosx is -e^cosx sinx.
Q1: What is the derivative of e^tanx?
Answer: The derivative of e^tanx is equal to sec^2x e^tanx.
Q2: If y=e^tanx, then find dy/dx?
Answer: If y=e^tanx, then dy/dx = sec^2x e^tanx.
Leave a Comment | {"url":"https://www.wbprep.com/differentiation-of-etanx/","timestamp":"2024-11-03T13:36:25Z","content_type":"text/html","content_length":"163174","record_id":"<urn:uuid:989eea94-5a35-4e00-a433-228ade780afd>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00452.warc.gz"} |
6,492 research outputs found
According to a recent conjecture, the moduli space of the heterotic conformal field theory on a $G\subset$ ADE singularity of an ALE space is equivalent to the moduli space of a pure \cx N=4
supersymmetric three-dimensional gauge theory with gauge group G. We establish this relation using geometric engineering of heterotic strings and generalize it to theories with non-trivial matter
content. A similar equivalence is found between the moduli of heterotic CFT on isolated Calabi--Yau 3-fold singularities and two-dimensional Kazama-Suzuki coset theories.Comment: 11pp, harvmac, v2: a
relation between heterotic CFT on Calabi-Yau 3-fold singularities and Kazama-Suzuki models has been adde
We show that the exact N=1 superpotential of a class of 4d string compactifications is computed by the closed topological string compactified to two dimensions. A relation to the open topological
string is used to define a special geometry for N=1 mirror symmetry. Flat coordinates, an N=1 mirror map for chiral multiplets and the exact instanton corrected superpotential are obtained from the
periods of a system of differential equations. The result points to a new class of open/closed string dualities which map individual string world-sheets with boundary to ones without. It predicts an
mathematically unexpected coincidence of the closed string Gromov-Witten invariants of one Calabi-Yau geometry with the open string invariants of the dual Calabi-Yau.Comment: harvmac, 29 pages (b), 3
figures; v2: references adde
We compute the instanton expansions of the holomorphic couplings in the effective action of certain \cx N=1 supersymmetric four-dimensional open string vacua. These include the superpotential $W(\
phi)$, the gauge kinetic function $f(\phi)$ and a series of other holomorphic couplings which are known to be related to amplitudes of topological open strings at higher world-sheet topologies. The
results are in full agreement with the interpretation of the holomorphic couplings as counting functions of BPS domain walls. Similar techniques are used to compute genus one partition function for
the closed topological string on Calabi--Yau 4-fold which gives rise to a theory with the same number of supercharges in two dimensions.Comment: 29 pages harvmac(b); v2: references adde
The study of curved D-brane geometries in type II strings implies a general relation between local singularities \cx W of Calabi-Yau manifolds and gravity free supersymmetric QFT's. The minimal
supersymmetric case is described by F-theory compactifications on \cx W and can be used as a starting point to define minimal supersymmetric heterotic string compactifications on compact Calabi-Yau
manifolds with holomorphic, stable gauge backgrounds. The geometric construction generalizes to non-perturbative vacua with five-branes and provides a framework to study non-perturbative dynamics of
the heterotic theory.Comment: LaTex, 11 p
We describe heterotic string and M-theory realizations of the Randall-Sundrum (RS) scenario with \cx N=2 and \cx N=1 supersymmetry in the bulk. Supersymmetry can be broken only on the world brane, a
scenario that has been proposed to account for the smallness of the cosmological constant. An interesting prediction from string duality is the generation of a warp factor for conventional type II
Calabi--Yau 3-fold compactifications. On the other hand we argue that an assumption that is needed in the RS explanation of the hierarchy is hard to satisfy in the string theory context.Comment: 18
pages, harvmac; references adde
We use a recently proposed formulation of stable holomorphic vector bundles $V$ on elliptically fibered Calabi--Yau n-fold $Z_n$ in terms of toric geometry to describe stability conditions on $V$.
Using the toric map $f: W_{n+1} \to (V,Z_n)$ that identifies dual pairs of F-theory/heterotic duality we show how stability can be related to the existence of holomorphic sections of a certain line
bundle that is part of the toric construction.Comment: 8 pages, harvmac (b
We study the open string extension of the mirror map for N=1 supersymmetric type II vacua with D-branes on non-compact Calabi-Yau manifolds. Its definition is given in terms of a system of
differential equations that annihilate certain period and chain integrals. The solutions describe the flat coordinates on the N=1 parameter space, and the exact disc instanton corrected
superpotential on the D-brane world-volume. A gauged linear sigma model for the combined open-closed string system is also given. It allows to use methods of toric geometry to describe D-brane phase
transitions and the N=1 K\"ahler cone. Applications to a variety of D-brane geometries are described in some detail.Comment: harvmac, 35 pages (b), 2 figures; v2: typos & references correcte
The boundary chiral ring of a 2d gauged linear sigma model on a K\"ahler manifold $X$ classifies the topological D-brane sectors and the massless open strings between them. While it is determined at
small volume by simple group theory, its continuation to generic volume provides highly non-trivial information about the $D$-branes on $X$, related to the derived category $D^\flat(X)$. We use this
correspondence to elaborate on an extended notion of McKay correspondence that captures more general than orbifold singularities. As an illustration, we work out this new notion of McKay
correspondence for a class of non-compact Calabi-Yau singularities related to Grassmannians.Comment: 29 pages, harvmac(b), 2 fig | {"url":"https://core.ac.uk/search/?q=author%3A(Mayr%2C%20P.)","timestamp":"2024-11-07T01:29:37Z","content_type":"text/html","content_length":"126841","record_id":"<urn:uuid:553e5f84-ed19-409e-8e90-30a34a12d91a>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00449.warc.gz"} |
OpenStax College Physics, Chapter 22, Problem 48 (Problems & Exercises)
(a) A 200-turn circular loop of radius 50.0 cm is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth’s field here
is due north, parallel to the ground, with a strength of $3.00\times 10^{-5}\textrm{ T}$ . What are the direction and magnitude of the torque on the loop? (b) Does this device have any practical
applications as a motor?
Question by
is licensed under
CC BY 4.0
Final Answer
a. $0.471\textrm{ N}\cdot\textrm{m}$, clockwise when viewed from above.
b. Whether this motor has practical applications depends on how the loop is connected to the source of the current. The curent needs to always be clockwise when viewed from the East, which requires
the contacts to switch polarity every half rotation. The question doesn't tell us if this is the case. Without this kind of connection, the loop will oscillate back and forth, which limits its
practical applications.
Solution video
OpenStax College Physics, Chapter 22, Problem 48 (Problems & Exercises)
vote with a rating of votes with an average rating of.
Video Transcript
This is College Physics Answers with Shaun Dychko. In this question we really have to draw a picture in order to understand all the directions that are given to us by the question. So we have a loop
of wire that's vertical so that means we can draw the ground here and put the loop standing up like this and we are going to look at it from the East because we are told that the direction of current
is clockwise when this loop is viewed from the East and that means we have North to the right hand side here and East is coming towards us because we are in the easterly position so we are looking at
this from the East and so East is coming towards us and we are looking towards the West— West would be an x away from us— and the magnetic field of the Earth is parallel to the ground, pointing to
the North and we have a radius of the loop that we are given—50.0 centimeters— the number of turns is 200, it's carrying 100 amps of current and the magnetic field strength of the Earth is 3.00 times
10 to the minus 5 tesla here and the question is what is the direction and magnitude of the torque on the loop? Well, we can figure out the direction using our right-hand rule so we will point our
many fingers in the direction of these many magnetic field lines and I used the word 'many' there to help you remember that many fingers point along magnetic fields lines because there are many of
those lines and then our thumb points in the direction of current, let's consider this position here. The current is upwards on this left hand side... I hesitate to use the word left hand though
because it's the right hand rule that we are using here but anyway... we are in this area of the loop here, our fingers are pointing to the right, our thumb is pointing up and our palm is pushing on
the direction of force on those current carrying segment of wire here and that is into the page so the force at this position is into the page. And then if you do the same analysis on the other side,
you have your thumb pointing down while your fingers are pointing to the right and your palm is facing towards you or out of the page and so this is causing a torque because it's in on the one side
of the axis of rotation and it's out of the page on the other side of the axis of rotation. So if you were to look at this loop from above, you would see it moving clockwise. So if you had some
eyeballs that were here then you would see a clockwise rotation from above. Okay! And then the next question is what is the magnitude of that torque? Well, the torque is the number of turns times the
current in the loop times the area of the loop multiplied by the magnetic field strength times sin of the angle between the perpendicular to the plane of the loop and the magnetic field, which in
this case is 90. Now the area being a circle here is π times its radius squared and so we substitute that in place of area and so the torque then is NIπr squaredB times sin Θ. So N is 200 turns, I is
100 amps π times the radius written in meters 50.0 times 10 to the minus 2 meters squared times 3.00 times 10 to the minus 5 tesla times sin 90 degrees and that is 0.471 newton meters and that's
clockwise when viewed from above. Part (b) asks does this device have any practical applications as a motor? Well, it depends on how the loop is connected to the source of the current. The current
needs to always be clockwise when viewed from the East which requires the contacts to switch polarity every half rotation. So this contact might be connected to the positive as the way I have drawn
the current right now. So this is the positive contact and this is the negative contact but when it rotates, the side that ends up in the left hand position here needs to still be positive in order
to have the current still be moving clockwise when viewed from the East and so that means that this wire here, which will now be on the left hand side after one half rotation, needs to be connected
to the positive terminal of the current source and so the little gizmo that does this in a regular motor are called brushes and they are like... made out of carbon and they are like loops that are
broken in half and they switch the polarity of these contacts here as this thing goes around. So I don't know... the question doesn't tell us much about this so we can't really say for sure whether
it would be good as a motor but as it is if this remains positive and this piece of wire here remains negative when they switch around, the current will be going in the other direction and then the
torque will be in the opposite sense and then this thing will oscillate back and forth, which could have its applications too but it will have limits on its applications that are possible. There! | {"url":"https://collegephysicsanswers.com/openstax-solutions/200-turn-circular-loop-radius-500-cm-vertical-its-axis-east-west-line-current","timestamp":"2024-11-04T01:18:13Z","content_type":"text/html","content_length":"206147","record_id":"<urn:uuid:0db3546f-627d-4492-b94a-79fdf422f4e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00429.warc.gz"} |
Microchannel based Multistage Solvent Extraction Studies for the Separation of Propionic Acid from its Aqueous Mixture using Hydrocarbon SolventsMicrochannel based Multistage Solvent Extraction Studies for the Separation of Propionic Acid from its Aqueous Mixture using Hydrocarbon Solvents
Solvent extraction is an important industrial operation where several stages are needed for a desired separation. Microchannel based solvent extraction is widely reported for process intensification.
However, all these works are confined to a single-stage extraction till date. For industrial application knowledge of multistage extraction is mandatory. This work focuses on the multistage
microchannel extraction using a model mixture containing aqueous propionic acid. Four different single solvents were employed in this study hexane, toluene, heptane, and cyclohexanol. The Effect of
flow rate, flow ratio on percentage extraction, extraction efficiency, and the required number of stages was investigated. The number of stages required for the maximum recovery of PA from the
raffinate is 5 for hexane &heptane and 3 for toluene and 2 for cyclohexanol. The percentage extraction of solvents obtained overall through all the stages is, cyclohexanol, 57–89%, toluene, 35–50%,
heptane, 27–51%, and hexane 19-31.3%. Cyclohexanol produced the maximum percentage extraction. The extraction efficiency and the volumetric mass transfer coefficient decreased with the stage numbers.
The maximum extraction efficiency for all the solvents is in the range of 98-99.8%. A microchannel stack is found to reduce the total annual cost (TAC). Particularly, fabrication in India results in
very less capital cost for the microchannels i.e.1.9–14.3% of TAC. The total annual cost analysis of toluene is the minimum than other solvents. | {"url":"https://www.researchsquare.com/article/rs-3786779/v1","timestamp":"2024-11-07T01:26:38Z","content_type":"text/html","content_length":"202858","record_id":"<urn:uuid:814cecf5-b1d6-4b41-9244-d73949968c22>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00325.warc.gz"} |
How to Calculate Square Root | Formula, Examples
How to Calculate Square Root | Formula, Examples
Finding the square root of a figure is a essential mathematical function which has several real-life uses. The square root of a number is the value that, when multiplied with itself, gives the
original value. In math, square roots are utilized in several fields, such as geometry, trigonometry, algebra, and calculus, to figure out complex equations, work out distances, and determine angles.
In physics, square roots are applied to work out force, velocity, and acceleration. In finance, square roots are used to work out interest rates and return on investment.
In this article, we will explore the formula for finding the square root, examples of how to find the square root of different numbers, and the significance of understanding the concept of square
roots. By conquering the theories of square roots and understanding its practical applications, students can enhance their problem-solving abilities and obtain a detailed understanding of the
complicated functions of the world around us.
Why is Determining Square Roots Crucial?
Calculating square roots is an essential mathematical concept that has numerous practical uses in numerous fields. In physics, square roots are utilized to calculate the acceleration and speed of
moving objects, which is essential for analyzing and designing mechanical systems.
Engineers also use square roots to work out the dimensions of structures and to calculate the forces acting on them. In finance, square roots are utilized to determine interest rates and investment
returns, which is essential for making informed financial decisions.
In algebra and calculus, square roots are utilized to determine complicated equations and carry out advanced math operations. For example, in algebra, square roots are applied to factor polynomials
and solve quadratic equations. In calculus, square roots are applied to calculate limits and derivatives, which are essential for working out complicated challenges in engineering, physics, and other
Understanding the theory of square roots is an essential step in conquering these arithmetical theories and is required for success in several domains for example mathematics, engineering, physics,
and finance. By comprehending how to find square roots and utilizing the knowledge to determine problems, people can enhance their problem-solving skills and gain a detailed understanding of the
intricate workings of the world surrounding us.
General Errors in Finding Square Roots
Finding square roots might seem simple, yet it is easy to make errors. One common mistake is forgetting to check if the provided number is a perfect square or not. A perfect square is a number whose
square root is an integer, for instance 4, 9, 16, 25, and so on. If the value is not a perfect square, it will have an irrational square root, which cannot be depicted as a finite decimal or
fraction. Therefore, it is crucial to revise whether the given value is a perfect square prior aiming to find its square root.
Another general mistake is forgetting to streamline the square root whenever feasible. Such as, the square root of 50 can be simplified as the square root of 25 times the square root of 2.
Simplifying the square root makes the workings easier and reduces the chances of making errors.
In addition, rounding mistakes can occur when working out the square root of a value utilizing a calculator. It is important to revise the end result by squaring it to make sure that it is correct.
By being mindful of these common errors and taking preventive measures to prevent them, people can enhance their accuracy when determining square roots and enhance their problem-solving skills in
various domains.
How to Find Square Root
The formula for calculating the square root of a number is as ensuing:
Square Root of n = √n
Where "n" depicts the value whose square root is being calculated. To figure out the square root of a number, you could use a calculator or perform the workings by hand.
Such as to determine the square root of 25, you will take the square root of 25, that is 5.
√25 = 5
Similarly, to calculate the square root of 81, you will take the square root of 81, that is 9.
√81 = 9
When working on the square root of a decimal number, you could utilize the same formula. For instance, to calculate the square root of 2.25, you should take the square root of 2.25, which is 1.5.
√2.25 = 1.5
It is crucial to keep in mind that not every numbers posses a perfect square root. In such situations, the square root can be approximated utilizing various techniques, for example the bisection
method or Newton's method.
Finally, determining the square root of a value is a fundamental math theory that has many real-life uses in various domains, including engineering, physics, and finance. The skill to find square
roots is an important ability for success in various domains of study, such as arithmetic, technology, and science.
By understanding the formula for determining square roots and exercise the calculation of square roots of various numbers, people can gain a detailed grasp of the concept and improve their
problem-solving skills. However, it is important to be conscious of common errors which can happen during the working procedure, for example forgetting to streamline the square root and failing to
check if the offered number is a perfect square.
If you require guidance understanding how to calculate square roots or any other arithmetical theory, consider reaching out to Grade Potential Tutoring. Our expert tutors are accessible online or
face-to-face to provide personalized and effective tutoring services to help you succeed.
If you need help with fundamental math or further advanced mathematical concepts, our instructors can give the guidance you require to reach your academic goals. Contact us right now to schedule a
tutoring session and take your math abilities to the next stage. | {"url":"https://www.philadelphiainhometutors.com/blog/how-to-find-square-root-formula-examples","timestamp":"2024-11-05T22:15:33Z","content_type":"text/html","content_length":"76932","record_id":"<urn:uuid:de3b82b1-ca72-4b5a-b7bb-2ef3be83f859>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00228.warc.gz"} |
How to Take the Average of Non-Zero Cells in Excel
Using the standard AVERAGE function with data that contains zeros can give you an incorrect average. Instead, we can use the AVERAGEIF function to take the average of non-zero cells.
To take the average of all non-zero cells in a given range, we can use the following formula:
= AVERAGEIF(range, "<>0")
Replacing the range with the range of cells we would like to take the average of.
The AVERAGEIF function is a conditional formula in Excel that calculates the average of cells that meet a specified criterion.
The function has two parts, the Range and the Criteria.
Range refers to the range of cells that you want to include in the average calculation. This could be anything from a single cell, a row or column, or a 2-D array of cells (rows and columns selected)
Criteria is the condition that the cells must meet to be included in the average calculation. In this case, we used "<>0" to specify that all cells that are not equal to zero should be included in
the average.
So, the formula takes the specified range of cells and evaluates each cell to see if it meets the specified criteria. If a cell meets the criteria, its value is included in the average calculation.
If a cell does not meet the criteria, it is ignored, leaving us with the average of all the values that meet the criteria. | {"url":"https://www.encyclopedia-excel.com/how-to-take-the-average-of-non-zero-cells-in-excel","timestamp":"2024-11-13T11:03:14Z","content_type":"text/html","content_length":"1050379","record_id":"<urn:uuid:93a0d7a0-e2d7-4acf-9b68-8dcf56e32527>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00254.warc.gz"} |
Physics 21 Fall, 2014 Solution to HW-6
Physics 21
Fall, 2014
Solution to HW-6
23-8 Three equal 1.05 µC point charges are placed at the
corners of an equilateral triangle whose sides are 0.600 m
long. What is the potential energy of the system? (Take as
zero the potential energy of the three charges when they are
infinitely far apart.)
To do this problem, we need to know that the electric potential energy for any two charges qi and qj separated by a
distance rij is
Uij =
1 qi qj
4πǫ0 rij
When we have several charges, we invoke superposition and
just add up the contributions from each distinct pair:
1 X qi qj
4πǫ0 i<j rij
In our specific situation, using q1 , q2 , and q3 there are three
distinct pairs, and the equation looks like this:
q2 q3
q1 q3
q1 q2
4πǫ0 r12
For this problem, r12 = r23 = r13 = 0.6 m, and q1 = q2 =
q3 = 1.05 × 10−6 C. Since all the charges are equal, and all
the distances are equal, we get the total U by multiplying
the potential energy of any one pair by three:
1.05 × 10−6 C
U = 3 9 × 109 Nm2 /C2
0.6 m
= 0.0496 J
We can use conservation of energy to help solve this problem.
After the protons have been accelerated by the cyclotron,
we can assume they are isolated and only interact with each
other. Their energy will be a combination of kinetic and electrical potential energy. Energy is conserved in this isolated
system, and we can write the total energy as
Etotal = KE + P E = 2 12 mp v02 + U0 = mp v02 + U0
where mp is the mass of the proton, v0 = 1.6 × 106 m/s is the
initial velocity of each proton, and U0 is the initial electrical
potential energy between the two protons.
The electrical potential energy between two protons (which
e2 /r, and r is the distance between
have charge e) is 4πǫ
the protons. As the protons approach one another (part A
in figure), r decreases and so the potential energy increases.
This results in a decrease in the kinetic energy of the protons
to keep the total energy constant. Eventually, all the kinetic
energy of the protons has been converted to electrical potential energy at some minimum distance of approach (part B),
and the protons will momentarily be at rest. They will then
begin to separate due to their repulsion and the potential
energy will be converted back to kinetic energy (part C).
The force between the two protons is 4πǫ
e2 /r2 , and so to
find the maximum force, we must find the minimum distance
of approach. At this minimum distance, rmin , the total energy is all potential energy since the protons are at rest.
Using conservation of energy,
mp v02 + U0 =
1 e2
4πǫ0 rmin
rmin =
4πǫ0 mp v02 + U0
No information was given on the initial separation of the
protons, so assume they were far apart to begin with and
then U0 = 4πǫ
e2 /r0 ≈ 0 for a large value of r0 . Then
9 × 109 Nm2 /C2 1.602×10−19 C
4πǫ0 mp v02
1.67 × 10−27 kg (1.6 × 106 m/s)
= 5.40 × 10−14 m
We can either substitute the algebraic expression or the numerical value of 1/rmin into the Coulomb force law. The
former method gives
m2p v04
1 e2
4πǫ0 rmin
(1.6726 × 10
kg)2 (1.6 × 106 m/s)4
(8.99 × 109 Nm2 /C2 )(1.602 × 10−19 C)2
Fmax =
= 7.97 × 10−2 N.
23.12 Two protons are aimed directly toward each other by
a cyclotron accelerator with speeds of 1600 km/s, measured
relative to the earth. Find the maximum electrical force that
these protons will exert on each other.
This force may be miniscule by our standards, but for a
proton it corresponds to an acceleration of 4.77 × 1025 m/s2 !
September 14, 2014
23-13 A small particle has charge −5.10 µC and mass
2.30 × 10−4 kg. It moves from point A, where the electric
potential is VA = 300 V , to point B, where the electric potential VB = 690 V is greater than the potential at point A.
The electric force is the only force acting on the particle.
The particle has a speed of 3.10 m/s at point A. (a) What is
its speed at point B ? (b) Is it moving faster or slower at B
than at A?
(a) This problem is easily solved using the conservation of
energy. Knowing the change in electric potential and the
charge on the particle allows us to calculate the change in
electric potential energy. Then the change in kinetic energy
will allow us to calculate the final speed of the particle.
UA + K A = UB + K B
qVA + 21 mvA
= qVB + 12 mvB
= vA
(VA − VB )
vB =
(3.10 m/s) +
= 5.19 m/s
2 (−5.10 µC)
(300 V − 690 V)
2.30 × 10−4 kg
Don’t forget to enter ×10−6 for µ when you type the charge
into your calculator. Note that even though the particle is
gaining electric potential, it is losing electric potential energy. This is due to the negative charge on the particle.
Thus, it is gaining kinetic energy, and therefore, speed.
(b) Faster
electron’s charge to get the electrical potential energy of the
electron in this potential field.
U (x) = −
4πǫ0 x + a2
We apply conservation of energy,
U0 + K 0 = U1 + K 1 ,
where U0 and K0 are the PE and KE at the initial position
x0 = 0.320 m, and U1 and K1 are the PE and KE at the final
position x1 = 0. Knowing that the initial kinetic energy is
zero, we get K1 = U0 −U1 , where K1 = 12 me v12 , U0 = U (x0 =
0.320 m), U1 = U (x1 = 0), me is the electron mass, and v1
is the speed of the electron when it reaches the center of the
ring. Solving for v1 and plugging in the values gives us our
u 1 2Qe 1
v1 =
(U0 − U1 ) =
4πǫ0 me a
x20 + a2
N · m2 2(20.0 nC) (1.60 × 10−19 C)
= 8.99 × 109
9.11 × 10−31 kg
0.140 m
(0.320 m) + (0.140 m)
= 1.65 × 107 m/s
This is not part of the question, but it is useful to think
about what happens to the electron after it passes through
the ring. The electron’s potential energy as a function
of position along the axis is shown in the plot below.
Electron Potential Energy
We need to solve this problem using conservation of energy.
The electron starts from rest, but it feels the attractive force
of the positive charges, and starts to move along the axis
toward the ring. As it moves, it is gaining kinetic energy
and losing electrical potential energy. We can calculate the
electrical potential energy of the electron using the formula
U = qV , where q is the electron’s charge and V is the electric
potential of the ring of charge.
We already know the formula for electric potential along the
axis of a ring of charge from lecture and from Example 23-11
in the textbook.
V (x) =
4πǫ0 x + a2
where Q is the total charge on the ring, x is the distance
along the axis, and a is the radius of the ring. Note that this
equation is not on the equation sheet. We multiply by the
Energy (10-17 J)
23-29 A uniformly charged thin ring has radius 14.0 cm
and total charge 20.0 nC. An electron is placed on the ring’s
axis a distance 32.0 cm from the center of the ring and is
constrained to stay on the axis of the ring. The electron is
then released from rest. Find the speed of the electron when
it reaches the center of the ring.
-0.4 -0.3 -0.2 -0.1 0
0.1 0.2 0.3 0.4
Position along axis (m)
It can be seen that the minimum potential energy occurs
at the center of the ring. This must be where the electron has the most kinetic energy. Once the electron passes
through the ring, it will begin to gain potential energy, and
lose kinetic energy. Because the potential energy is symmetric about the ring, the electron will come to rest behind the
ring at the same distance along the axis from which it was
released. It will then turn around and pass through the ring
again. It will continue to oscillate in this manner.
23-68 A disk with radius R has a uniform charge density
σ. (a) By regarding the disk as a series of thin concentric
rings, calculate the electric potential V at a point on the
disk’s axis a distance x from the center of the disk. Assume
that the potential is zero at infinity. (Hint: Use the result
that the potential at a point on the ring axis at a distance x
from the center of the ring is
V =
4πǫ0 x + a2
where Q is the charge of the ring.) (b) Find Ex = −∂V /∂x.
(a) Using the hint, we can modify the result for the finite ring
to determine the potential dV due to a thin ring of radius r
and charge dQ. We have
dV =
4πǫ0 x2 + r2
Because electric potential is a scalar quantity, we can integrate dV without keeping track of vectors. First, we must
determine dQ in terms of known quantities. It is equal to
the surface charge density times the surface area dA of the
thin ring,
dQ = σdA = σ2πr dr,
where dA is the product of the length (circumference) 2πr
and width dr of the ring of radius r.
Now it is possible to do the integration
Z R
Z R
2πσr dr
r dr
V = dV =
4πǫ0 0
2ǫ0 0
x +r
x2 + r2
The integral is on the equation sheet:
u du
= a2 + u 2 .
a2 + u 2
Evaluating it at the limits results in
σ hp 2
V =
x + R2 − x .
(b) The x component of the electric field is equal to −∂V /∂x:
σ 1 2
2 −2
x +R
2x − 1
2ǫ0 2
1− √
x + R2
Notice that as R → ∞, this result reduces to the field of an
infinite sheet. | {"url":"https://expydoc.com/doc/2824422/physics-21-fall--2014-solution-to-hw-6","timestamp":"2024-11-09T03:48:01Z","content_type":"text/html","content_length":"34291","record_id":"<urn:uuid:ba3f8ef7-c8c7-4193-b05b-8b8d436a3608>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00489.warc.gz"} |
1 J. Rosenhouse, The Monty Hall problem, New York: Oxford University Press, 2009.
2 T. Slembeck, J. R. Tyran, Do institutions promote rationality? An experimental study of the three doors anomaly, Journal of Economic Behavior & Organization 54 (2004), 337-350. DOI ScienceOn
3 Song H. S., On the two Envelope-Paradox, Logic Research 6(1) (2003), 1-18. 송하석, 두 봉투의 역설에 대하여, 논리연구 6(1) (2003), 1-18.
4 D. Friedman, Monty Hall's three doors: construction and deconstruction of a choice anomaly, American Economic Review 88 (1998), 933-946.
5 H. Bailey, Monty Hall uses a mixed strategy, Mathematics Magazine 73(2) (2000), 135-141. DOI
6 Cho C. M, Park J. Y., Kang S. J., A Study on Geometrical Probability Instruction through Analysis of Bertrand's Paradox, School Mathematics 10(2) (2008), 181-197. 조차미, 박종률, 강순자,
Bertrand's paradox의 분석을 통한 기하학적 확률에 관한 연구, 대한수학교육학회지<학교수학> 10(2) (2008), 181-197.
7 E. Fischbein, The intuitive sources of probabilistic thinking in children, Reidel, Holanda, 1975.
8 R. D. Gill, The Monty Hall problem is not a probability puzzle, Statistica Neerlandica 65(1) (2011), 58-71. DOI ScienceOn
9 P. Gorroochurn, Classic problems of probability, New Jersey: Wiley, 2012.
10 D. Granberg, T. Brown, (1995). The Monty Hall dilemma, Personality and Social Psychology Bulletin 21(7) (1995), 711-723. DOI ScienceOn
11 D. Granberg, N. Dorr, Further exploration of two-stage decision making in the Monty Hall dilemma, American Journal of Psychology 111(4) (1998), 561-579. DOI
12 Kim W. K. et al, Probability and Statistics, Seoul : Visang Education, 2014. 김원경 외, 확률과 통계, 서울 : 비상교육, 2014.
13 Lee J. H., Historic Paradoxes of Probability and Statistics Usable in School Mathematics, The Korean Journal for History of Mathematics 24(4) (2011), 119-141. 이종학, 학교 수학에 활용 가능한 확률.
통계 영역에서의 역사적 패러독스, 한국수학사학회지 24(4) (2011), 119-141. 과학기술학회마을
14 Lee J. Y., Lee K. H., A Case Study of Creativity Development Using Simpson's Paradox for Mathematically Gifted Students, The Journal of Educational Research in Mathematics 20(3) (2010), 203-219.
이정연, 이경화, Simpson의 패러독스를 활용한 영재교육에서 창의성 발현사례 분석, 대한수학교육학회지 <수학교육학연구> 20(3) (2010), 203-219.
15 Lee K. H., Study on the Didactic Transposition of the Concept of Probability, Unpublished doctoral dissertation of Seoul University, 1996. 이경화, 확률개념의 교수학적 변환에 관한 연구, 서울대학교
박사학위 논문, 1996.
16 S. Lucas, J. Rosenhouse, A. Schepler, The Monty Hall problem, reconsidered, Mathematics Magazine 82(5) (2009), 332-342. DOI
17 J. P. Morgan, N. R. Chaganty, R. C. Dahiya, M. J. Doviak, Let's make a deal: The player's dilemma, The American Statistician 45(4) (1991), 284-287.
18 Whang S. W. et al, Probability and Statistics, Seoul : Sinsago, 2014. 황선욱 외, 확률과 통계, 서울 : 신사고, 2014.
19 A. Morone, A. Fiore, Monty Hall's Three Doors for Dummies, Theory and Decision Library Series 42 (2008), 151-162. DOI
20 J. Sprenger, Probability, rational single-case decisions and the Monty Hall problem, Synthese 174 (2010), 331-340. DOI | {"url":"https://koreascience.kr/ksci/search/article/articleView.ksci?articleBean.atclMgntNo=KSHSBA_2014_v27n3_211","timestamp":"2024-11-07T09:46:02Z","content_type":"application/xhtml+xml","content_length":"35533","record_id":"<urn:uuid:7d4215e4-6904-42b8-8a72-d886be9be142>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00709.warc.gz"} |
Dividing Polynomials - Definition, Synthetic Division, Long Division, and Examples
Polynomials are math expressions which comprises of one or more terms, all of which has a variable raised to a power. Dividing polynomials is an essential function in algebra which includes finding
the quotient and remainder when one polynomial is divided by another. In this blog, we will investigate the various approaches of dividing polynomials, consisting of long division and synthetic
division, and give examples of how to apply them.
We will further talk about the significance of dividing polynomials and its utilizations in different domains of mathematics.
Importance of Dividing Polynomials
Dividing polynomials is a crucial operation in algebra that has several uses in diverse fields of math, including calculus, number theory, and abstract algebra. It is applied to work out a wide range
of challenges, including working out the roots of polynomial equations, working out limits of functions, and calculating differential equations.
In calculus, dividing polynomials is used to work out the derivative of a function, which is the rate of change of the function at any moment. The quotient rule of differentiation consists of
dividing two polynomials, that is utilized to work out the derivative of a function that is the quotient of two polynomials.
In number theory, dividing polynomials is applied to study the features of prime numbers and to factorize huge values into their prime factors. It is further used to learn algebraic structures such
as rings and fields, that are fundamental ideas in abstract algebra.
In abstract algebra, dividing polynomials is applied to specify polynomial rings, which are algebraic structures that generalize the arithmetic of polynomials. Polynomial rings are used in multiple
domains of arithmetics, including algebraic geometry and algebraic number theory.
Synthetic Division
Synthetic division is an approach of dividing polynomials which is applied to divide a polynomial with a linear factor of the form (x - c), at point which c is a constant. The technique is based on
the fact that if f(x) is a polynomial of degree n, subsequently the division of f(x) by (x - c) provides a quotient polynomial of degree n-1 and a remainder of f(c).
The synthetic division algorithm consists of writing the coefficients of the polynomial in a row, utilizing the constant as the divisor, and carrying out a series of workings to work out the
remainder and quotient. The result is a streamlined form of the polynomial that is easier to work with.
Long Division
Long division is a technique of dividing polynomials that is applied to divide a polynomial by another polynomial. The approach is based on the fact that if f(x) is a polynomial of degree n, and g(x)
is a polynomial of degree m, where m ≤ n, then the division of f(x) by g(x) provides us a quotient polynomial of degree n-m and a remainder of degree m-1 or less.
The long division algorithm involves dividing the greatest degree term of the dividend by the highest degree term of the divisor, and further multiplying the result with the total divisor. The answer
is subtracted from the dividend to get the remainder. The process is recurring as far as the degree of the remainder is lower in comparison to the degree of the divisor.
Examples of Dividing Polynomials
Here are a number of examples of dividing polynomial expressions:
Example 1: Synthetic Division
Let's assume we have to divide the polynomial f(x) = 3x^3 + 4x^2 - 5x + 2 with the linear factor (x - 1). We can apply synthetic division to streamline the expression:
1 | 3 4 -5 2 | 3 7 2 |---------- 3 7 2 4
The outcome of the synthetic division is the quotient polynomial 3x^2 + 7x + 2 and the remainder 4. Thus, we can state f(x) as:
f(x) = (x - 1)(3x^2 + 7x + 2) + 4
Example 2: Long Division
Example 2: Long Division
Let's say we need to divide the polynomial f(x) = 6x^4 - 5x^3 + 2x^2 + 9x + 3 with the polynomial g(x) = x^2 - 2x + 1. We could use long division to simplify the expression:
First, we divide the largest degree term of the dividend by the largest degree term of the divisor to attain:
Next, we multiply the total divisor with the quotient term, 6x^2, to obtain:
6x^4 - 12x^3 + 6x^2
We subtract this from the dividend to attain the new dividend:
6x^4 - 5x^3 + 2x^2 + 9x + 3 - (6x^4 - 12x^3 + 6x^2)
that simplifies to:
7x^3 - 4x^2 + 9x + 3
We repeat the process, dividing the highest degree term of the new dividend, 7x^3, with the largest degree term of the divisor, x^2, to get:
Then, we multiply the entire divisor with the quotient term, 7x, to get:
7x^3 - 14x^2 + 7x
We subtract this of the new dividend to get the new dividend:
7x^3 - 4x^2 + 9x + 3 - (7x^3 - 14x^2 + 7x)
that simplifies to:
10x^2 + 2x + 3
We repeat the procedure again, dividing the largest degree term of the new dividend, 10x^2, by the largest degree term of the divisor, x^2, to achieve:
Next, we multiply the entire divisor by the quotient term, 10, to obtain:
10x^2 - 20x + 10
We subtract this from the new dividend to obtain the remainder:
10x^2 + 2x + 3 - (10x^2 - 20x + 10)
that streamlines to:
13x - 10
Thus, the outcome of the long division is the quotient polynomial 6x^2 - 7x + 9 and the remainder 13x - 10. We could express f(x) as:
f(x) = (x^2 - 2x + 1)(6x^2 - 7x + 9) + (13x - 10)
In Summary, dividing polynomials is an essential operation in algebra that has multiple utilized in multiple fields of math. Getting a grasp of the various methods of dividing polynomials, for
instance long division and synthetic division, can guide them in figuring out intricate problems efficiently. Whether you're a student struggling to understand algebra or a professional operating in
a field which involves polynomial arithmetic, mastering the ideas of dividing polynomials is essential.
If you need help comprehending dividing polynomials or any other algebraic concept, consider calling us at Grade Potential Tutoring. Our expert teachers are accessible remotely or in-person to
provide individualized and effective tutoring services to help you be successful. Contact us right now to schedule a tutoring session and take your mathematics skills to the next level. | {"url":"https://www.durhaminhometutors.com/blog/dividing-polynomials-definition-synthetic-division-long-division-and-examples","timestamp":"2024-11-03T21:53:07Z","content_type":"text/html","content_length":"77586","record_id":"<urn:uuid:f5216105-aa64-4b7d-b83e-f0ea423f6d1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00641.warc.gz"} |
Classifying Real Numbers Mystery Pattern Worksheet Answers
Classifying Real Numbers Mystery Pattern Worksheet Answers work as fundamental tools in the world of maths, supplying an organized yet functional platform for students to explore and master
mathematical principles. These worksheets use a structured approach to understanding numbers, supporting a solid structure upon which mathematical proficiency prospers. From the most basic checking
exercises to the ins and outs of advanced computations, Classifying Real Numbers Mystery Pattern Worksheet Answers cater to learners of diverse ages and skill degrees.
Unveiling the Essence of Classifying Real Numbers Mystery Pattern Worksheet Answers
Classifying Real Numbers Mystery Pattern Worksheet Answers
Classifying Real Numbers Mystery Pattern Worksheet Answers -
Web Classifying real numbers worksheets are one of the most fundamental concepts from a mathematics point of view Real numbers are those numbers that can be represented
Web Classifying real numbers mystery pattern worksheet answer key Write the name that applies to the number below 6 273 3 7 2 9 100 0 1 1 2 38 542 8293017 Rational
At their core, Classifying Real Numbers Mystery Pattern Worksheet Answers are lorries for theoretical understanding. They envelop a myriad of mathematical principles, assisting students via the
labyrinth of numbers with a collection of appealing and deliberate workouts. These worksheets transcend the borders of conventional rote learning, encouraging active involvement and promoting an
instinctive grasp of numerical partnerships.
Supporting Number Sense and Reasoning
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet
Web This Classifying Real Numbers Mystery Pattern consists of a one page activity where students must classify rational numbers as rational integer whole natural or irrational
Web Showing top 8 worksheets in the category Classifiying Real Numbers Mystery Pattern Some of the worksheets displayed are Cclassifying real numberslassifying real
The heart of Classifying Real Numbers Mystery Pattern Worksheet Answers lies in cultivating number sense-- a deep comprehension of numbers' meanings and affiliations. They motivate expedition,
welcoming students to explore arithmetic operations, figure out patterns, and unlock the mysteries of sequences. Through thought-provoking obstacles and rational challenges, these worksheets come to
be portals to honing reasoning skills, supporting the logical minds of budding mathematicians.
From Theory to Real-World Application
Classifying Real Numbers Mystery Pattern YouTube
Classifying Real Numbers Mystery Pattern YouTube
Web Displaying top 8 worksheets found for Classifiying Real Numbers Mystery Pattern Some of the worksheets for this concept are Cclassifying real numberslassifying real
Web Classifying Numbers Mystery Patterns Displaying top 8 worksheets found for this concept Some of the worksheets for this concept are Cclassifying real
Classifying Real Numbers Mystery Pattern Worksheet Answers serve as conduits linking academic abstractions with the apparent realities of day-to-day life. By infusing functional situations into
mathematical exercises, students witness the importance of numbers in their surroundings. From budgeting and dimension conversions to comprehending statistical data, these worksheets encourage pupils
to wield their mathematical prowess past the boundaries of the class.
Diverse Tools and Techniques
Flexibility is inherent in Classifying Real Numbers Mystery Pattern Worksheet Answers, utilizing a collection of instructional devices to accommodate varied understanding designs. Visual help such as
number lines, manipulatives, and electronic sources function as friends in imagining abstract principles. This diverse approach ensures inclusivity, accommodating learners with different choices,
toughness, and cognitive styles.
Inclusivity and Cultural Relevance
In an increasingly varied globe, Classifying Real Numbers Mystery Pattern Worksheet Answers embrace inclusivity. They go beyond cultural borders, incorporating instances and problems that resonate
with learners from diverse backgrounds. By including culturally relevant contexts, these worksheets promote an environment where every student feels represented and valued, improving their connection
with mathematical principles.
Crafting a Path to Mathematical Mastery
Classifying Real Numbers Mystery Pattern Worksheet Answers chart a course towards mathematical fluency. They instill determination, crucial thinking, and analytic abilities, essential qualities not
just in mathematics but in different elements of life. These worksheets empower students to navigate the detailed surface of numbers, supporting a profound recognition for the beauty and logic
inherent in mathematics.
Embracing the Future of Education
In an era marked by technological innovation, Classifying Real Numbers Mystery Pattern Worksheet Answers flawlessly adapt to electronic systems. Interactive user interfaces and electronic resources
boost standard discovering, supplying immersive experiences that transcend spatial and temporal borders. This combinations of typical methodologies with technological developments heralds an
appealing age in education and learning, fostering an extra vibrant and interesting discovering setting.
Conclusion: Embracing the Magic of Numbers
Classifying Real Numbers Mystery Pattern Worksheet Answers characterize the magic inherent in mathematics-- a charming journey of expedition, exploration, and mastery. They go beyond conventional
pedagogy, acting as stimulants for stiring up the flames of interest and questions. With Classifying Real Numbers Mystery Pattern Worksheet Answers, students embark on an odyssey, unlocking the
enigmatic globe of numbers-- one problem, one remedy, each time.
Classifying Real Numbers Worksheet
The Agony And Dx dt Week 3
Check more of Classifying Real Numbers Mystery Pattern Worksheet Answers below
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet With Answers Islero Guide Answer For Assignment
Classifying Real Numbers Mystery Pattern Worksheet Answer Key 38 Pages Explanation In Doc 1
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet
Classifying Real Numbers Mystery Pattern Worksheet Answer Key
Web Classifying real numbers mystery pattern worksheet answer key Write the name that applies to the number below 6 273 3 7 2 9 100 0 1 1 2 38 542 8293017 Rational
Classify Real Numbers Activity Rational Integer Whole Natural
Web This Classifying Real Numbers Mystery Pattern consists of a one page activity where students must classify rational numbers as rational integer whole natural or irrational
Web Classifying real numbers mystery pattern worksheet answer key Write the name that applies to the number below 6 273 3 7 2 9 100 0 1 1 2 38 542 8293017 Rational
Web This Classifying Real Numbers Mystery Pattern consists of a one page activity where students must classify rational numbers as rational integer whole natural or irrational
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet With Answers Islero Guide Answer For Assignment
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet With Answers Islero Guide Answer For Assignment
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet
Classifying Real Numbers Worksheet Answer Key | {"url":"https://szukarka.net/classifying-real-numbers-mystery-pattern-worksheet-answers","timestamp":"2024-11-08T09:26:46Z","content_type":"text/html","content_length":"25077","record_id":"<urn:uuid:74c2f1b4-d675-429a-ab57-dbe634222748>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00280.warc.gz"} |
MA-INF 1213: Randomized Algorithms & Probabilistic Analysis 2018
The first round of oral exams will take place on the 18th, 19th and 20th of July.
Please schedule the time of your exam with Christiane Andrade in room 2.056.
When Where Start Lecturer
Tuesday, 10:15-11:45 CP1-HSZ / HS 3 April 12th (Thursday) Röglin (June-),
Thursday, 12:15-13:45 CP1-HSZ / HS 4 Schmidt (-June)
When Where Start Lecturer
Tuesday, 14:15-15:45 INF / Room 2.050 April 17th Rösner
Wednesday, 14:15-15:45 INF / Room 2.050 April 18th Rösner
Due to the holidays on May 1st and May 10th there is no new exercise sheet on May 7th. Problem Set 5 will be released May 14th.
Date Contents
1 Discrete Event Spaces and Probabilities
April 12 1.1 Discrete Probability Spaces
1.2 Independent Events & Conditional Probability
1.2 (continued) Conditional Probability
April 17 1.3 Applications
1.3.1 Minimum Cut, Karger's Contract Algorithm
April 19 1.3.1 (continued) Karger's Contract algorithm
1.3.1 FastCut
April 24 1.3.1 (continued) FastCut
1.3.1 Reservoir Sampling
April 26 2.1 Random Variables and Expected Values
2.1.1 Integer Random Variables
May 1 no lecture
2.1.2 Conditional Expectation
May 3 2.2 Binomial and Geometric Distribution
2.3 Applications
2.3.1 Randomized QuickSort
May 8 2.3.2 Randomized Approximation Algorithms
May 10 no lecture
3 Concentration Bounds: Markov's Inequality
May 15 3.1 Variances and Chebyshev's Inequality
3.3 Applications: Introduction to sublinear algorithms
May 17 3.3.1 A sublinear algorithm
May 22 no lecture
May 24 no lecture
3.2 Chernoff/Rubin bounds
May 29 4.1 Useful math
4.2 Applications
4.2.1 An Algorithm for 2-SAT
May 31 no lecture
June 5 4.2.2 Algorithms for 3-SAT
June 7 6 Knapsack Problem and Multiobjective Optimization
6.1 Nemhauser-Ullmann Algorithm
June 12 6.2 Number of Pareto-optimal Solutions
June 14 6.2 (continued) Number of Pareto-optimal Solutions
June 19 6.3 Multiobjective Optimization
6.4 Core Algorithms
June 21 7 Smoothed Complexity of Binary Optimization Problems
June 26 7 (continued) Smoothed Complexity of Binary Optimization Problems
June 28 8 Successive Shortest Path Algorithm
9 The 2-Opt Algorithm for the TSP
July 3 9.1 Overview of Results
9.2 Polynomial Bound for phi-Perturbed Graphs
July 5 9.3 Improved Analysis
July 10 no lecture
July 12 10 The k-Means Method
The Lecture Notes cover the lecture. Part I is largely based on the following two books:
[MR95] Rajeev Motwani and Prabhakar Raghavan. Randomized Algorithms. ISBN: 978-0521474658, Cambridge University Press, 1995.
[MU05] Michael Mitzenmacher and Eli Upfal. Probability and Computing. ISBN: 978-0521835404, Cambridge University Press, 2005.
The lecture has two parts. First, we consider the design and analysis of randomized algorithms. Many algorithmic problems can be solved more efficiently when allowing randomized decisions.
Additionally, randomized algorithms are often easier to design and analyze than their (known) deterministic counterparts. For example, we will see an elegant algorithm for the minimum cut problem.
Randomized algorithms can also be more robust on average, like randomized Quicksort.
The analysis of randomized algorithms builds on a set of powerful tools. We will get to know basic tools from probabily theory, very useful tail inequalities and techniques to analyze random walks
and Markov chains. We apply these techniques to develop and analyze algorithms for important algorithmic problems like sorting and k-SAT.
Statements on randomized algorithms are either proven to hold on expectation or with high probability over the random choices. This deviates from the classical algorithm analysis but is still a
worst-case analysis in its core. In the second part of the lecture, we learn about probabilistic analysis of algorithms. There are a number of important problems and algorithms for which worst-case
analysis does not provide useful or empirically accurate results. One prominent example is the simplex method for linear programming whose worst-case running time is exponential while in fact it runs
in near-linear time on almost all inputs of interest. Another example is the knapsack problem. While this problem is NP-hard, it is a very easy optimization problem in practice and even very large
instances with millions of items can be solved efficiently. The reason for this discrepancy between worst-case analysis and empirical observations is that for many algorithms worst-case instances
have an artificial structure and hardly ever occur in practical applications.
In smoothed analysis, one does not study the worst-case behavior of an algorithm but its (expected) behavior on random or randomly perturbed inputs. We will prove, for example, that there are
algorithms for the knapsack problem whose expected running time is polynomial if the profits or weights are slightly perturbed at random. This shows that instances on which these algorithms require
exponential running time are fragile with respect to random perturbations and even a small amount of randomness suffices to rule out such instances with high probability. Hence, it can be seen as an
explanation for why these algorithms work well in practice. We will also apply smoothed analysis to the simplex method, clustering problems, the traveling salesman problem, etc. | {"url":"https://nerva.cs.uni-bonn.de/doku.php/teaching/ss18/vl-randalgo","timestamp":"2024-11-04T01:46:51Z","content_type":"text/html","content_length":"21715","record_id":"<urn:uuid:946a46f7-e31a-4fa4-9271-7c351c590187>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00785.warc.gz"} |
Math U See Alpha - An Honest Review - The Simple Homeschooler
Math U See Alpha – An Honest Review
It is actually painful for me to write this Math U See Alpha Review.
I really, really wanted it to work out. You have no idea.
My first year of homeschooling my 1st grader was going so well – better than I could have hoped for!
Except for math.
We originally started with The Complete Book of Math and it was a disaster.
I day dreamed of what school would be like if I could just casually shove math over a cliff and skip away.
Long story short, there is no shame in a mid-year curriculum switch.
After much research, reading many reviews, and even taking a break from math, I decided it was time to switch to Math U See Alpha about half way through 1st grade.
Math U See Alpha Review: How it Works
When we first started, I was in love. I thought (and still do) that Math U See Alpha is brilliant.
We bought the Math U See Alpha Universal Set and it included:
• Instruction Manual
• Instruction DVD
• Student Workbook
• Tests booklet
• Integer Block Kit
• 12 months of access to our new Alpha Digital Pack
□ Skip Count Songs MP3s
□ Songbook PDF
The focus of Alpha is learning 100 addition facts and then 100 subtraction facts. There are some other lessons (skip counting and shapes), but by far the overall theme of the year is math facts.
The parent has an instructional manual and is encouraged to read each brief chapter before teaching it to their student.
There is also an instructional DVD. It is a teacher (and creator) of Math U See teaching each lesson to a classroom of real students.
After watching the lesson, then the child goes to a workbook and goes through a page (front and back) while the parent observes and assists as needed.
Once the child has mastered the week’s assignment (is able to teach the curriculum back to you), the student goes to the test booklet. There is a test for each week’s lesson and it is used as an
indicator of the student’s mastery of the lesson.
What We Loved about Math U See Alpha
1. Customer Service
I was a little unsure about what level we should start on. Math U See customer service reached out to me though and they were excellent!
It was a homeschool dad who has used Math U See Alpha with his own kids. I thought it was so cool that Math U See pays parents to help with customer service!
He gave me fantastic advice and got us on the right path. He also contacted me by email for follow up and promptly answered my questions after we received the curriculum.
It felt great to know I had someone to reach out to when I needed help.
The manipulative set that Math U See uses is genius. Seriously.
Our old manipulative set took FOREVER to use because it was just loose colored blocks. To solve 8+6, my daughter would have to count out 8 blocks. Then 6 blocks. And then count them all together
It was a lot of counting that made her really tired!
With Math U See, the blocks are not loose. They are color coded based on number. For example, a “6 stick” is purple and an “8 stick” is brown.
For my daughter to do 8+6, she would grab a brown stick (8) and a purple stick (6) and then count them once. Done! Much easier for her to do multiple problems.
3. Video based lessons
A huge plus was that there was minimal prep work required for me.
I did read the chapters before lessons, but we always started math with the video lessons.
They have a little bit of a dated feel to them, but I LOVE the way that the teacher explains things. He is able to break things down so simply. My daughter enjoyed getting out her blocks and
following along with him.
As a side note, I know at some levels, kids are independent with their Math U See video lessons. I wouldn’t say that is appropriate for Alpha level.
I sat with her, reinforced things after the video, and was always available while she was doing worksheets.
4. Each week focuses on one thing
Our first math curriculum did a lot of skipping around, so I felt like there was not much time to really cement a concept.
With Math U See, it is laser focused on one thing. That really helped my daughter as we went through the lessons, did the work, and took the tests.
5. Mastery focused
Math U See tells parents to wait for mastery of a lesson before moving on. If you need to stay on the lesson for several weeks or even months – that’s okay!
If your kid is able to move on after 1-2 days – that is okay too!
I loved having the freedom to assess my kid and make a decision to move at our pace. It absolutely made the Math U See experience feel very customized to us.
6. Really smart ways to remember math facts
One of my favorite things about Math U See is the amazing way that math facts are taught.
My daughter was able to pick up facts that she had previously struggled with and I actually even learned a few tricks to remember certain facts a different way.
Its a really smart curriculum!
7. You know when you’re done
With our previous math curriculum, there were a lot of games. Despite my best efforts, my daughter didn’t like playing the games. I always felt like I didn’t know when we were done.
Have we played this long enough? Did we do enough problems? I wasn’t really sure.
With Math U See, you and your kid easily know when the lesson is complete. Very motivating!
Depending on the day and what we were learning, lessons were 10-30 minutes. Very manageable!
What We Didn’t Like About Math U See Alpha
As you are reading this, you must be wondering, “Um…how exactly was this not working for you?!”
I know, I know.
Initially, I was loving all of the benefits of Math U See Alpha.
We were actually 3-4 lessons from finishing the curriculum, and I decided to call it. Again. This was not working.
Here is what went wrong.
1. Repetitive
As I talk to other homeschool moms who have used MUS, they also note that the curriculum is incredibly repetitive.
We did math facts every day. For months. I will admit that I even got tired of learning and reviewing math facts.
My daughter did well initially, but eventually started struggling with learning some of the bigger facts (17-9, 8+7, etc.). There was no escape from it. We were never doing anything but math facts.
I think it started to make her really hate math all over again. I think she felt like math was only math facts (which it is not!), despite my best efforts to bring in fun review games.
2. Dry workbook pages
As I said, there is minimal parent prep work…but that means that your kid is mostly only doing black and white workbook pages.
When you flip through the Math U See Alpha workbook, it is a sea of addition and subtraction problems and word problems.
There is an “enrichment” page at the end of every week that my daughter usually enjoyed, but the rest of the week was spent looking at very dry material.
I now know that so much of math and learning at this age should be done through games and fun – not “drill and kill” worksheets.
3. Focus on mastery
Mastery of a lesson before moving on was initially a benefit of Math u See Alpha. As we progressed through the curriculum though, it became a burden.
Guess what? A six year old doesn’t want to stay on the same concept for weeks or months!
She would miss review problems on worksheets and it was always unclear if she was just making silly mistakes or she forgot how to solve the problem.
Did we need to go backwards in the book?
She would get frustrated with me and herself when she would make mistakes, forget things, and when she was not ready to test out of the lesson at the end of the week.
Sometimes we spent weeks on one lesson and I could tell it just aggravated her. She was sick of talking about this math fact!
I honestly think it developed a sense of unreasonable perfection in her. She felt she had to remember everything and get every problem right every time.
That’s a lot of pressure for a 6 year old!
I actually reached out to customer service and asked about this whole mastery thing. I was given a prompt and kind reply, but it was somewhat vague about how to progress.
4. Too many tips and tricks
As we got into the higher math fact numbers, there were great methods taught to learn complicated problems.
The problem was that every lesson had a trick or a cute story to remember.
My daughter started to get them all jumbled up in her head. She gave me crazy answers for problems because she was so focused on applying the trick – she just couldn’t remember the right one.
When I had her look at the problem, and see how the answer made no sense at all – she would get even more frustrated.
I even took her back a number of lessons so we could start learning the facts over again and hopefully cement the material better.
She was compliant and worked with me, but it did not help. She would teach me a concept, but weeks later could not remember the concept with all the other new concepts added on top of it.
I’ll be honest and admit that I can’t quite remember all the little tips and tricks that were taught. I can easily see how a 6 year old could get them mixed up.
5. The Test Booklet
As mentioned above, every lesson has a test.
The child is ready to take the test when they are able to teach the lesson back to the teacher.
Early on the tests were a breeze and I think my daughter enjoyed taking them. As things progressed, she began to fear that test booklet.
She would start asking me questions on test day before we had even started math. I could see the anxiety building.
She would get so upset when it wasn’t a perfect test, even though I told her it didn’t need to be perfect. I told her it was just another worksheet and she didn’t need to worry about it at all. It
was just a tool to tell me if we needed more time on a concept.
The last day we did Math U See was a test day. My younger kids were being rowdy, so I sent her to her room to take her math test.
A significant amount of time went by and I went to go check on her. She was holding her math book and crying because she just could not remember and she didn’t know what to do.
I took the book away from her and that was the last time she has touched it.
Another mom wisely told me later that I should have never given her a test from the beginning. If she can do the worksheet, she can do a test. There is no reason for a 1st grader to take a math test.
6. Wrong curriculum for 1st Grade level in my opinion
After stopping Math U See Alpha (1st grade level), I was really surprised when I looked into 2nd grade curriculum for my daughter and discovered much of it was focused on the same math fact
memorization we were running away from.
I found out that many other 1st grade level programs just want the child to understand the concepts of addition and subtraction. Simple facts are reviewed and memorized over time as other math
concepts (measurement, simple fractions, time, etc.) are also explored.
So, I was drilling my 6 year old on math facts (17-9, 14-8, 9+6, etc.) and expecting mastery…while many other curriculums do not even introduce these facts until 2nd grade.
This matters a lot, because you can greatly frustrate a child by asking them to do something that they cannot developmentally understand. It’s a recipe to make kids hate learning.
After Math U See, we took a total break from math. Like months off. When I felt like we were ready to address math again, I just ran through flash cards with her one day to see where she was at. She
got nearly every single one right. I was shocked.
What changed? I think her age and development from age 6 to age 7 was significant. I could see in her eyes that she was solving the problem and “making 10” in her mind and able to work out the
problem – something she couldn’t seem to do consistently at age 6.
As I’m writing this, I remember the MUS customer service agent telling me that his own daughter spent a very long time (1.5-2 years) on Alpha. I’m wondering if the reason Math U See Alpha takes so
long, is because it is targeted at 6 year olds – and should be 7-8 year olds.
****UPDATE 2/2020: So I wote this review almost 2 year ago, so I thought I should pop in and give you an update on what I did with my middle kid when she hit first grade this year.
Nope, we didn’t use Math U See Alpha – that’s for sure. We opted for this math curriculum and I could not be happier. It uses the spiral method (not mastery) for teaching math and my kid is thriving.
Thriving you guys.
We are over halfway through and she is learning about skip counting, calendars, money, shapes, addition, subtraction, double digit addition, word problems, odd/even numbers, place value, telling
time, etc. Lots of variety and tons of review = lots of fun!
And as a side note, I used this addition supplement to teach my 1st grader (She has a late birthday so she started 1st grade when she was 6 months older than my first child. I figured I’d try it with
her and see what happens – and even then I waited until she was over 7 years old to start.) her addition facts.
It took 6 weeks and it was taught through brief lessons, games, and worksheets for reinforcement – and suddenly she knew all her addition facts! I cannot tell you what a positive experience it was.
So much better than what we experienced with Math U See. I will be putting out a full review in the future, but for now please check out the book before teaching your kid math facts – huge sanity
I will be following up with the subtraction supplement soon!
Math U See Alpha Review – Bottom Line
Math U See Alpha is a really smart curriculum and I respect the team that designed it.
It very well may be a great fit for your child – many of my friends have used it and love it!
The manipulative set is genius and we still use it for math even though we don’t use the rest of the curriculum.
I think it would have been more effective if we had started when my child was a bit older, ignored the mastery concept, and if I had never used the test booklet.
We moved on to Math Lessons for a Living Education for 2nd grade and fell in love! From there we have switched again to Teaching Textbooks for 3rd grade because it allows my daughter to be more
independent while I start working with her younger sister who is now at the kindergarten level.
I decided to write this Math U See Alpha review because I also read a lot of curriculum reviews before buying. I really hope this will help you make the very best informed decision for your
I would love to hear your experience with Math U See in the comments and answer your questions – I know others would benefit from it too!
Want to make sure you remember this later? Got you covered! Pin this to your favorite Pinterest board and make sure to share with your friends and followers!
Read Next:
15 Comments
1. I have been an educator and homeschooler for 15 years and know many homeschoolers who feel like you do about MathUSee. Brilliant blocks, badly designed program. As a special education teacher, I
often need to practice 500 times for something to stick, and video games are the only humane way to get them to do that much math work. I had to tie the blocks to the math seeds program. So
recently, I made my own Easel video game-type program to sell on TPT, but I want to give it away here for free to as many homeschoolers who want to try it. I put it in a Google Classroom. https:/
With class code: laiekzc
1. Thanks for the sharing the resource, Jennifer!
2. My son like your daughter hated the black and white pages of MUSee so I did the spiral colorful route but I noticed he wasn’t getting the concepts. he was just memorizing. Plus becuase it was
spiral it was harder to teach as a parent because it jumps from one concept to the next. My oldest who has ADHD tendencies enjoyed the lessons I felt, but he wasn’t grasping concepts. With Math
u see when he was six, I had the same response as yours. I decided though instead of making him do the page I would just ask him the question and have him show me with the blocks and this worked
amazingly. Now that he is older and watched a lot of odd squad lol which also reinforced these math concepts, beta level has been a hundred percent better and he has scored pretty high and math.
I love that for some one who loves math it’s easy for me to teach and breaks it down in concepts. My youngest who started at 5 loves this book and actually we are half way done with the book.
He doesn’t care if it’s black and white. I think remembering to break it down on their level and personality helps. The reason why I pick this curriculum is that reviewers who had taken this
curriculum score higher in standardized test and that’s my ultimate goal is to make sure they are equipped to excel and any area they choose to go in so I want to make sure they have solid math
3. I know this is an older post, but I feel the need to jump in and leave a comment for anyone who may see this in the future.
I haven’t used the MUS curriculum, but I’m starting to look into it. I’m a special education teacher and last summer I bought a whole bunch of math manipulatives, and these manipulatives were in
the bunch. I absolutely love the manipulatives for my kiddos. They are so much better than base 10 blocks, and can be used for so many different math concepts.
Like I said, I haven’t used the curriculum, but I do think that it would be a good supplement for my students.
The biggest thing I can say to those who may be dissuaded by this review, is not to let the repetitiveness tour you. Once a student has mastered it, move on. If it’s becoming boring, move on to
another subject and go back to it later. Jumping around in math is fine. Work on some addition, go to subtraction, come back and do higher addition and then do more subtraction. You don’t have to
follow this step by step. Do what is best for your child.
The biggest thing I’ve learned as a special education teacher, is that curriculums are designed in a way that students need to adapt to it instead of it adapting to the student. So don’t be
afraid to adopt the curriculum to your student’s needs. Not all students learn in the same way, the curriculum can be used as a guide, but ultimately set your own sequence.
Now when it comes to so many tips and tricks being confused, you don’t need to teach every single one. If your student is getting the concept without needing it then don’t use it. Save those tips
and tricks for the material that they are struggling with.
I also strongly agree with the parent who said not to use the tests. Not all students are going to do well on tests. Find a way to let them show you their mastery without taking a test. If
they’re showing their mastery on the worksheets, why do they have to continue to do more problems? It’s just more of that repetition.
Ultimately, whatever curriculum you decide to use with your child, you don’t have to follow step by step. Use what works and leave the rest.
4. Getting all the tips and tricks confused was our #1 problem. My 7 year old flew through alpha in 4 months, we did all the worksheets and yet I really don’t think any of it solidified in his
brain. Taking Abeka at a slow pace has been much better, and teaching math facts by families instead of tricks has helped him really learn them!
5. We use Math-U-See for all grades at our classical school, and several things address some of these issues and make it work, even for younger students:
1) If the student gets 100% on the first of the three “new concept” (A, B and C) lesson pages, then they don’t have to do the next two. S/he then skips to the D, E and F pages to review prior and
current lesson material. Then, only if her test score is less than 90 percent, the student must go back and finish B and C pages. That eliminates some unnecessary work.
2) Enrichment pages are mandatory.
3) Students work in math books for 20 minutes, move to a practical math/games table for 20 minutes, then back to math books for the final 20 minutes of their math hour. Some students work for 40
minutes, then move to the game table.
Agree wholeheartedly with the “readiness”/developmental comments.
1. Amy, thank you for the comment and the ideas for my readers!
6. After working through 4 kids with Math U See we have learned to use ONLY the test booklet as the daily work (too many questions in workbook) and went back to the main workbook IF there was
something we needed more practice on. We switched from MUS to Jump Math between grade 2-3 (after Alpha or Beta) as we are enrolled with the public system and must meet British Columbia learning
outcomes. Math U See does not work with the BC LO. The benefits has been huge…MUS is a great foundation for math especially struggling learners, however, its limiting and repetitive. Parents must
know when to skip ahead to the next lesson or stay the course, cut back on the number of questions (do only the odd questions, do every other page) and when to hit every question. Just because a
curriculum offers many questions to practice, it does not mean we must do EVERY question in the workbook. I apply this piece of advice to every workbook. Right Start Math also has very good
manipulatives to learn the foundations of math.
1. OH! Love seeing you comment here AND mention Right Start Math.
I am deciding between MUS and Right Start for my 5th grader, next year.
Hmmmm…? Thoughts? 🙂
7. I’m curious since you loved “math lessons for a living education” with your older child why you didn’t use it for K or 1 for your younger child? And why you decided to start with the horizons
math? (I have a 6 yr old, will be 7 in May daughter and am struggling w which cirriculum to get)
1. Hi Danielle! Our homeschool has gone through A LOT of math curriculum. We originally liked Math Lessons for a Living Education for my oldest. But we tried Teaching Textbooks (only for 3rd
grade and up) and she was immediately hooked, so we switched and never looked back. My middle child learns completely differently so we have taken a different path with her. She loves
Horizons Math and it was a perfect fit for her in 1st grade. She is now in 2nd grade, and I have her doing a little Horizons and a little Teaching Textbooks too. My youngest is in kinder and
doing beautifully with Horizons. Here is my Horizons Review and my Math Lessons for a Living Education Review
8. I agree with another poster s comment you should do a review on Math Lessons for a living Education ?? I love your reviews by the way! We have decided to go with Singapore PM SE 1 for first
9. Thank you so much for writing this! I felt this way teaching my kids MUS and all I could find were rave reviews about how amazing it is and I felt like I was missing something. So on point!!
10. I am literally going through everything you’ve described with my 6 year old kindergartner (she has a late birthday). All I’ve heard about MUS are amazing reviews, but sadly it’s been a
miserable experience for us. As a first time homeschool mom, it left me feeling extremely discouraged and my daughter was often frustrated trying to remember all of the tricks. The curriculum is
so dry and, like your daughter, she would only enjoy the one enrichment page at the end of each section. We are only halfway through alpha and summer is looming so I’m trying to decide if I
should finish it out or just start over with something else. Thank you for this review, though. It’s such a relief to know I’m not alone!
1. Kiara, It’s always good to hear I wasn’t the only one either 🙂 Happy Homeschooling! 🙂 | {"url":"https://www.thesimplehomeschooler.com/why-math-u-see-alpha-an-honest-review/","timestamp":"2024-11-12T20:47:21Z","content_type":"text/html","content_length":"297176","record_id":"<urn:uuid:55955329-f793-45bf-9cf9-51aa476145d8>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00020.warc.gz"} |
The Ultimate Quiz On Zettabyte!
Questions and Answers
Are you familiar with the term Zettabyte? Take this Zettabyte quiz to test yourself on this unit of information. The Zettabyte is a worldwide known multiple of the unit byte. It has its unit symbol
as ZB, and it is best used for digital information. This informative quiz is designed not only to test your knowledge but to enhance it too. Even if you miss out on some questions, don't worry. You
will get the answers along. All the best!
• 1.
What is Zettabyte?
□ A.
A multiple of the unit byte for digital information
□ B.
□ C.
□ D.
A combination of Zetta and byte.
Correct Answer
A. A multiple of the unit byte for digital information
The zettabyte is a multiple of the unit byte for digital information.
• 2.
What does the prefix "zetta" mean?
□ A.
Multiplication by the seventh power of 1000
□ B.
□ C.
□ D.
Correct Answer
A. Multiplication by the seventh power of 1000
The prefix Zetta indicates multiplication by the seventh power of 1000
• 3.
How is Zetta measured?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. As sextillion
A zettabyte is one sextillion
• 4.
What is the zettabyte's unit symbol?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. ZB
The unit symbol is ZB.
• 5.
How many gigabytes is a zettabyte?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. A trillion gigabytes
1ZB is a trillion gigabytes
• 6.
How many ZB does ZFS allow for?
□ A.
256 quadrillion zettabytes
□ B.
512 quadrillion zettabytes
□ C.
1024 quadrillion zettabytes
□ D.
2048 quadrillion zettabytes
Correct Answer
A. 256 quadrillion zettabytes
ZFS allows for a maximum storage capacity of about 256 quadrillion zettabytes
• 7.
How many zettabytes did an expert estimate that is being used annually worldwide?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 2 zettabytes
In 2013, one expert estimated that the amount of data generated worldwide would reach 4 zettabytes by the end of that year.
• 8.
How many zettabytes did humans reportedly send in 2007?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 1.9 zettabytes
Research from the University of Southern California reports that in 2007, humankind successfully sent 1.9 zettabytes of information through broadcast technology such as televisions and GPS
• 9.
How many zettabytes did Americans use in 2008.
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 3.6 zettabytes
Research from the University of California, San Diego reports that in 2008, Americans consumed 3.6 zettabytes of information
• 10.
How many zettabytes were accessed in The US in 2013?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 6.3 zettabytes
A 2013 study reported that 6.9 zettabytes of data were accessed in the U.S. in the preceding 12 months. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=3dq-zettabyte","timestamp":"2024-11-12T06:21:10Z","content_type":"text/html","content_length":"448972","record_id":"<urn:uuid:a8356487-857a-4238-889f-da79f87e626b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00117.warc.gz"} |
Math Contest Repository
Euclid 2024 Question 6, CEMC UWaterloo
(Euclid 2024, Question 6, CEMC - UWaterloo)
(a) There are $M$ integers between $10000$ and $100000$ that are multiples of $21$ and whose units (ones) digit is $1$. What is the value of $M$?
(b) There are $N$ students who attend Strickland S.S., where $500 < N < 600$. Among these $N$ students, $\frac{2}{5}$ are in the physics club and $\frac{1}{4}$ are in the math club. In the physics
club, there are $2$ times as many students who are not in the math club as there are students who are in the math club. Determine the number of students who are not in either club.
Answer Submission Note(s)
Separate your answers with a single space.
Please login or sign up to submit and check if your answer is correct.
flag Report Content
You should report content if:
• It may be offensive.
• There is something wrong with it (statement or difficulty value)
• It isn't original.
Thanks for keeping the Math Contest Repository a clean and safe environment! | {"url":"https://mathcontestrepository.pythonanywhere.com/problem/euclid24q6/","timestamp":"2024-11-04T04:07:05Z","content_type":"text/html","content_length":"10498","record_id":"<urn:uuid:004a50e7-46fc-4f8e-8e41-66e215a6eeb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00855.warc.gz"} |
The Portuguese Economy
We all know about hysteresis in unemployment, which basically tells us that the equilibrium rate of unemployment rate depends on current and past unemployment rates. This problem becomes more severe
as the duration of unemployment duration increases
In the picture below, one can observe how the unemployment rate and real GDP growth have evolved since 1986 (when Portugal joined the European Union). Unemployment persistence is obvious and it is
also obvious that unemployment tends to increase/decrease when the growth rate is below/above 2%.
Using the data on the Picture, I estimate na AR(1) process for unemployment, with real GDP growth as an independent variable. I use this very simple model to make some projections for the evolution
of the unemployment rate.
I consider three different scenarios. First, I admit that until 2015 Real GDP grows at 5.46%. This is the average real GDP growth rate in the first six years after joining the European Union. Let’s
call this the dream scenario. Second, I will assume that real GDP growth is equal to the last six years average: 0.59%. This is the pessimistic scenario. Finally, I will consider a realistic, some
may argue that is optimistic, scenario and assume that the growth rate will be the average since 1986: 2.64%. The results are described in the picture below.
Here's The Spanish Economy. No, it's not a Spanish counterpart to this blog, but rather a website set up by the Spanish government that "provides access to the latest key economic information on
Spain in English in a clear and comprehensive manner". A quick visit suggests that it is rather well done, in the sense that most of it is composed of links to documents and data. In other words, it
seems "strictly informative", although it is obviously more than that. Most of the official documents to which it links are actually in English, which is no small achievement. If you search "Spanish
economy" on Google, the website is already in the top ten links. Judging from what Pedro Lains told us here, maybe the Portuguese government should get some inspiration from this. I'm afraid,
however, that the title "The Portuguese Economy" is already taken.
From here.
The budget execution up to November 2010 was announced, and apparently a positive evolution in the budget deficit occurs.
Some information from municipalities and some other institutions is still missing but unlikely to change the tone.
Curiously enough, in the "missing" ones we have the Parliament, the "agency for modernization of public administration" and the "agency for the knowledge society". I thought that these would like to
be in the forefront...
Major highlights (by the Government): lower spending growth and increases in revenue (check here - in Portuguese, and you can download the data ; soon to be in English as well).
My highlights - increase in VAT is kicking in and accounts for almost all increased revenue, but also tax on car sales (as people are anticipating purchases of new cars) and tax on tobacco. These
last two compensate on decrease of direct taxes on income. Not yet a reduction in nominal spending, but 2011 should see it given the pay cut in civil service.
My wishes for the next budget execution bulletins (or some other bulletins) - a quick-and-dirty assessment of how evolution is linked to measures adopted.
Congratulations to Luís Aguiar-Conraria (and his co-author Pedro Magalhães) for the award of the "Gulbenkian Prize for the Internationalization of Social Sciences in Portugal".
Independently of the effect of the new initiative for the Competitiveness and Employment on the Portuguese Economy, it is already scaring to envision that, once the (inter)national crisis surrenders,
a great number of new public institutes and projects can be potentially created to evaluate the impact of the new measures. I would suggest that the economic departments of our universities and their
researchers take the lead and motivate their doctoral and master students to start working on applied microeconomic and macroeconomic papers to evaluate these measures. Get the data, and do an
efficient, independent, and cheaper public service. A win-to-win opportunity, would say.
The title of this post summarizes the new Initiative for Competitiveness and Employment launched today by the Portuguese Government, with 50 (fifty, no mistake) policy measures.
If you can read Portuguese, the announcement is here.
For all the others, a quick summary:
- measures aimed at exports (15, ranging from export credits to fiscal benefits)
- measures aimed at bureaucracy removal /reduction (8)
- measures aimed at the labour market (16)
- measures aimed at real estate and housing markets (5)
- measures aimed at fraud and fiscal evasion (6)
This sounds a bit like "carpet bombing" - something will work; in a statistical sense, at least this is likely to happen.
Looking at the news this afternoon:
- Government wants to set a ceiling of firing compensation paid by companies
- Government wants to make easier to get out the tenants that do not pay rents
- Minister of Finance denies role of IMF in Government decisions
- Prime Minister meets "social partners" to present the changes in labor law
I wonder whether it would be more credible if for a month only one announcement would be made - how well the Government finances went that month.
In the end, doing many things at the same time may just mean dispersion, while we really need to focus.
Just an idea for this afternoon, stop announcing more and more policy measures, just show and make sure the ones recently adopted work. Please???
1, 2, 3 times or more I have been asked by journalists, both Portuguese and foreign, about the previous IMF interventions in Portugal: here is a detailed description of the two episodes by Ana Bela
Nunes, an economic historian at ISEG.
It seems more than obvious that one of the reforms that will be taking place in Portugal in the near term is the labor market reform. And no, I am not talking only about making firings cheaper. The
truth is that, for better or for worse, more labor market flexibility is simply inevitable, either with this government (if forced by our European partners), or with the next government (the most
likely outcome). Why? Because Portugal is the OECD country with the highest strictness of employment protection for individual workers, something that impedes a faster job creation and penalizes the
competitiveness of our exports.
Still, and even though this reform is inevitable, it would be nice if there were more elevation in the public debate arena with regards to these matters. The fact is that, among us, often political
and ideological rhetoric take the place of common sense when we talk about these questions. And both common sense and abundant empirical research show unequivocally that out labor laws not only
promote unemployment, but also they are the main reason that explain why Portugal is one of the countries with the highest incidence of temporary work in all OECD.
In order to understand why, it might be worth looking to the graph below, which confirms that Portugal is one of countries of the OCDE with more precarious employment as a percentage of total
employment (horizontal axis), something that is intrinsically correlated to our labor laws with regards to individual workers (vertical axis).
It is also noticeable that, contrary to the (wrong) perception that is prevalent amongst us, many of the countries where labor laws are less rigid (i.e. where it is easier to fire individual workers)
are exactly some of the countries that have strong Welfare States (e.g,. Denmark or Canada). That is, to make our labor laws more flexible is not akin to destroy our Welfare State. Quite the
contrary. By promoting faster labor job creation and by enhancing the competitiveness of our exports, less rigid labor laws contribute to more wealth creation, which then can be used to protect the
Welfare State. Denying this is mere ideological rhetoric.
The words “Precariedade, precariedade, precariedade” (the incidence of temporary employment) should be front and center in the strategy of the main opposition parties when they defend deeper reforms
of our labor laws. They should emphasize that the reform of our labor laws is an imperative because, among other things, we have to combat the incidence of temporary work and we have to create more
employment. The truth is that our “irrevocable” labor rights (“direitos adquiridos”) are the main source of temporary work in the Portuguese job market. Labor economists know this for a long time. It
would be nice if our politicians as well as all of us knew it too.
Source: OECD, Santos Pereira (2011) "Como Retomar o Sucesso"
Note: A Portuguese version of this post can be found here.
One of the most interesting papers that I have recently read on the Portuguese economy was the paper (already referenced here) by Pedro Martins on cronyism with regards to public sector jobs. Martins
uses the "Quadros do Pessoal" data in order to undergo an extensive empirical investigation of job hirings in public sector enterprises before and after general elections. The paper is important
because it confirms fairly convincingly something that many suspected and from which we only had, up to now, anecdotal evidence: that hirings in public enterprises are strongly associated with
general elections. Namely, Martins finds that hirings in public sector enterprises goes up immediately before and in the 3-6 months after a general election (see graph below, where period 0
represents the date of the election). Interestingly, but probably not suprisingly, the effect on hirings is stronger when the party in power changes (i.e. if there is a change in the "color" of the
party in power), that is, when the jobs-for-the-boys effect comes clearly into play. The results hold even after Martins controls for experience, education, private sector hirings, among others.
What this points out very clearly is that, in the last 30 years (i.e. at least since 1980, the year the data begin), public sector enterprises have been blatantly used for political gains and to
reward whoever helps you get into power. Yes, there are jobs for the boys alright and they are not only in the public administration or the State. Public companies also do.
This is certainly not completely surprising, but it is surely very nice to have substantial and unequivocal evidence on the extent of cronyism in Portugal. One of the obvious implications of this
study is that we now understand better why public expenditures have been so difficult to control in our country, or why the debt of public enterprises keeps on growing at, on average, about 3 billion
euros a year.
Cronyism _ Job hirings in public enterprises before and after general elections
Source: Martins (2010), and taken, with permission, from here.
I would like to end with a word of caution about the future. It is widely expected that we will have general elections sometime in 2011 (hopefully sooner rather than later), after the presidential
election is over and after the constitutional limits allow. It is also widely expected that there will be a government of a new "color"(to use Martins's terminology), probably in coalition with a
smaller party. Thus, the temptation to emulate the past will be there, with interest groups associated with the new parties in power pushing for new jobs for the boys. Let's hope this does not
happen. If it does, it would be simply unexcusable in times like these. The fact is that it is not worth changing the boys from pink to orange or even to blue just for the sake of doing it. Boys will
be boys will be boys. And this "boy" culture and tradition is partly responsible for the sad state of our public finances. Therefore, if the new government truly wishes to reform the State and get
away from the sorry state of affairs that is taking place in Portugal right now, it must avoid, at all costs, the temptation to distribute new jobs for the boys in the State and in public companies.
If it does so, the new government will fail and won't be able to implement a truly reformist agenda, something that Portugal desperately needs. Let's hope not. I am sincerely hopeful that next time
will indeed be different (even if it is because we really can't afford to do otherwise).
My friend André Azevedo Alves gives his perspective on the Portuguese situation in the Institute of Economic Affairs blog. Check it out.
André insists rightly on the need for a "haircut" on private creditors if a bailout materializes.
I would like to hear him on other serious issues such as private sector debt, the unbalanced relationship between the prices of tradable and untradable goods, or unit labor costs in Portugal. And
ultimately membership in the eurozone.
Politicians can use the public sector to give jobs to cronies, at the expense of the efficiency of those organisations and general welfare. Motivated by a simple model of cronyism that predicts
spikes in appointments to state-owned firms near elections, we regress 1980-2008 monthly hirings across all state-owned Portuguese firms on the country’s political cycle. In most specifications, we
also consider private-sector firms as a control group. Consistent with the model, we find that public-sector appointments increase significantly over the months just before a new government takes
office. Hirings also increase considerably just after elections but only if the new government is of a different political colour than its predecessor. These results also hold when conducting the
analysis separately at different industries and most job levels, including less skilled positions. We find our evidence to be consistent with cronyism and politically-induced misallocation of public
By Pedro Martins (read the working-paper)
One of the more recurring features of Portuguese politics is the time inconsistency of decision makers. Probably the same happens everywhere.
I would like to know the contribution of time inconsistency of policies to the slowdown in economic growth. Basically, if you cannot believe in agreements signed with the Government, this should have
a cost in terms of economic activity.
This was brought again to my mind by the agreement on minimum wage - It was signed 4 years ago an agreement involving the Government, the unions and the industry associations (I think the industry
associations signed it, not sure, but it is not essential). It stated that minimum wage should go to 500€/month in phased way.
Now, industry associations want the Government to refrain from it, based on the bad economic conditions. Unions (and left wing parties) argue the Government should stick to the agreement.
For once, I agree with the later on the grounds of consistency of policies (they use other arguments, of course). Actually, it is in the interest of the industry associations that
the Government does it. Otherwise, any agreement with the Government becomes only valid until the next convenient moment to change it? and if instead of minimum wage would be a VAT tax? or an
exceptional tax on sales with "frozen" prices? A bit demagogic, I know, but once you enter the uncertainty that agreements do not bind the parties, everything goes.
Moreover, I was puzzled by a news report stating that some computations made by the Government showed little impact associated with the minimum wage increase.
Overall, without entering the discussion of whether a minimum wage has negative effects upon employment and growth of economic activity, I feel that not respecting agreements must also be damaging,
in a hardly visible way.
A few weeks ago, a piece on the FT mentioned that the Irish gross external debt was close to 1000% of GDP (not a typo!). Based on this observation, the FT commentator explained that the EFSF could
provide liquidity to Irish banks but certainly not make them solvent. His conclusion might be true, but I believe that inferring solvency from gross (or even net) external debt is not correct. Today
most developed countries have very large gross external positions (external gross liabilities and assets of 300% of GDP are quite common). These positions are the outcome of the massive increase in
the size of capital inflows and outflows across countries experienced in the last 20 years or so. Most available data on bilateral external positions are based on the concept of residence. This
accounting principle implies that a debt by, say a German bank located in Dublin on a French. resident is an external debt of Ireland vis-à-vis France, instead of a German debt. The BIS constructs a
different measure (only for banks and for 24 developed countries), called “ultimate risk” basis, that identifies exposures on the basis of the nationality of the ultimate creditor and debtor. In
short the “ultimate risk” basis is constructed to identify the bank that is ultimately responsible for the liability and is a better measure to assess solvency and liquidity. According to this
measure, at the end of June 2010, European banks had 423 billion dollars claims on Irish banks, while Irish banks had 375.8 billion claims on European banks. Within Europe the net debt of Irish banks
was therefore 47 billion or approximately 20% of GDP. The overall net figure is not available but we know that Ireland bank claims vis-a-vis “all countries” amount to 548 billion dollars and the 24
countries claims vis-a-vis Ireland amount to 518 billion. The graph below shows the creditors of Ireland (millions of dollars, June 2010, BIS.) The creditors breakup is useful to assess the
transmission channels of a default and the incentives of each nation to bail out their debtor. (click to enlarge)
Portuguese banks have 80 billion dollars claims against other European banks while European claims against Portugal amount to 205 billion. Within Europe Portuguese banks have a net debt of 125
billion or 54% of GDP. Portugal bank claims vis-a-vis “all countries” amount to 140 billion dollars and the 24 countries claims vis-a-vis Portugal amount to 212 billion dollars.
A few days ago I wrote in another blog a post stating that Ireland was lucky to be saved by the EU/IMF fund. But now I am confused. Was the Irish rescue the rescue of the Irish economy? Or was it
imposed on Ireland - and on its government that will suffer in next year's polls - to rescue the banks in Europe that hold a large share of the country's sovereign debt, as today's FT argues? In the
meantime, Portugal's yields have stabilized. Were we just lucky not to have been "saved" by a similar loan? What's next?
Strangely enough, in the last month or so it seemed that Portugal's politics was going to be run to the beat of the interest paid on sovereign bonds. At some point, 95% of analysts, politicians, and
the media, were arguing that there could be no discussion about the 2011 budget, because otherwise the "markets" would ask for higher yields. The correlation however proved to be much lower than
expected. Surprise? No, I don't think so. Markets are looking at models and computer screens, not at the Portuguese parliament or newspapers. And I have a hint that that is in fact so: the other day
I got this email from a "lead analyst for Portugal in the Sovereign Risk Group" at one of the World top three rating agencies, asking for papers on a certain topic, but they had to be in English
because he/she spoke (or read) "absolutely no Portuguese". How do you rate countries like that? And how do you decide about the price of the bonds? By looking at what Pedro Passos Coelho (btw, the
leader of the main opposition party) says? Maybe not so. It was of course relavant that the budget passed. But also that there was genuine political dispute, and negotiations between the parties.
A lesson from thinking about economics is that the evident truths are, on closer inspection, just plain false or even silly. This piece in the New York Times looks like a sketch out of the Daily Show
where a somber "analyst" makes grave statements without realizing how hilariously ridiculous it all is. According to the reporters, the problems of Spain, Portugal, and Greece are that:
i) "You can’t build an economy on real estate, finance and tourism."
ii) "...it’s an absurd idea to have the same currency in a country like Greece or Portugal as in Germany, which has totally different habits and culture."
iii) "They lived on a bubble of credit and real estate development that sent wages and debt soaring."
iv) Deficits and debt are high.
These reporters were looking outside their window to face Manhattan just as I am right now. Funny that they didn't notice that the city in front of them has:
i) An economy built on real estate, finance and tourism.
ii) The same currency as Alabama, Alaska, and California.
iii) Very large increases in credit, real estate development, wages and debt in the last decade.
iv) A huge amount of debt and current deficits in the state government.
Jon Stewart couldn't have done it any better. | {"url":"https://theportugueseeconomy.blogspot.com/2010/12/","timestamp":"2024-11-03T03:21:28Z","content_type":"application/xhtml+xml","content_length":"122249","record_id":"<urn:uuid:30a09304-ca4c-4c51-92a0-40710bca7649>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00402.warc.gz"} |
CECAM - Error control in first-principles modelling
The determination of errors, and their annotation in the form of error bars, is widely established in the experimental branches of chemistry, physics, and material science. In contrast, a rigorous
error analysis for first-principles materials simulations is lacking, and hence the results of such simulations are generally quoted without simulation-specific errors. We consider this a severe
obstacle for innovation, given the important role that multiscale numerical simulations now play in chemical and physical research. In this workshop, we aim to bring researchers from different
communities in contact, for an interdisciplinary discussion of challenges and opportunities in quantifying errors across simulation scales [1].
With the aim of advancing both models and numerical methods, errors for first-principles simulations have certainly been investigated. However, the state of the art is to rely mainly on benchmarking
procedures or explicit convergence studies. Due to the associated cost, such studies are limited to a small number of cases and insufficient to estimate the error in high-throughput-based or
data-driven approaches with easily millions of simulations. This is a severe obstacle in a large-scale screening context, since the differences in the quantities of interest can become small across a
design space, such that an understanding of errors is crucial to reliably select the most relevant compounds [2].
Another issue is the reliability and efficiency of numerical procedures themselves. Focusing, for example, on the context of density-functional theory (DFT) simulations, a sizeable number of
parameters need to be chosen, e.g., basis set, various tolerances, mixing, preconditioning. Usually heuristic approaches are employed to select these, which bears the risk to choose parameters (a)
too tightly, such that calculations are inefficient, or (b) too loosely, such that calculations produce erroneous results or fail to converge. With the ability to estimate the different contributions
of numerical error, error balancing strategies from other fields (such as finite-element modelling) could be applied to tune computational parameters automatically and robustly [3]. While the full
error in DFT cannot yet be treated, recent progress, e.g. analytical approaches to estimate the discretisation or basis set error for several simplified models [3, 4] or statistical approaches such
as BEEF to estimate the error of the DFT model [5] suggest this may soon become possible.
Similarly, the determination of errors in observables that are the results of long molecular dynamics simulations is rarely attempted [6]. The recent flurry of activity in data-driven interatomic
potentials that extend the concept of first principles modelling beyond electronic structure calculations offers some hope. Error analysis of the approximations themselves, coupled with built-in
uncertainty measures and the increased efficiency of computing the forces, opens up the possibility of propagating error analysis to physical observables.
Closely related to many of these goals is the field of uncertainty quantification (UQ), which has advanced quantitative understanding of the impact of parameter uncertainties and model uncertainties,
and their interaction with numerical error, in many other disciplines. UQ has seen considerable application in engineering modeling, in geophysics, and in continuum/PDE modeling generally. But UQ at
the molecular scale is less developed, notwithstanding recent efforts in applying Bayesian statistical methods to the learning of interatomic potentials [7]. A comprehensive UQ approach offers the
potential of linking (i) numerical and modeling errors in electronic structure calculations, to (ii) the statistical inference of interatomic potentials, to (iii) the error in global/observable
quantities. We expect that rigorous mathematical methods for multi-fidelity modeling, for goal-oriented sensitivity analysis and stochastic dimension reduction, and for intermingling deterministic
error bounds with probabilistic descriptions of uncertainty will play essential roles in achieving this multiscale vision. | {"url":"https://www.cecam.org/workshop-details/error-control-in-first-principles-modelling-1115","timestamp":"2024-11-10T11:08:34Z","content_type":"text/html","content_length":"65712","record_id":"<urn:uuid:5e1013f4-7831-4323-9c8f-b30f2e01c211>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00152.warc.gz"} |
Two-sample tests
1 General
• Type: - Matrix Processing
• Heading: - Tests
• Source code: not public.
2 Brief description
Two sample-test for determining if the means of two groups are significantly different from each other.
Output: Two numerical columns are added per test, one containing the p-value, the other containing the difference between the means. In addition there is a categorical column added in which it is
indicated by a ‘+’ when the row is significant with respect to the specified criteria. If applicable, another column is added per test with test-specific q-values. Also, depending on the parameter
settings two columns with global information (over all tests) are added with a combined score and combined q-value.
3 Parameters
3.1 Grouping
The grouping(s) of columns to be used in the test. Each test takes two groups as input. Multiple tests can be performed simultaneously by specifying more than one pair of groups.
3.1.1 First group (right)
All ‘right’ groups of the two sample tests are defined here. The number of groups selected here equals the number of different tests performed.
3.1.2 Second groups mode
Specify here how the ‘left’ groups of the two sample tests are specified. Possible ways are to specify for each individual ‘right’ group the corresponding ‘left’ group, to use one single control
group, or to always use the complement of each individual ‘right’ group as the ‘left’ group.
3.2 Test
Defines what kind of test should be applied (default: T-test). The test can be selected from a predefined list:
3.2.1 S0
This parameter defines the artificial within groups variance (default: 0). It controls the relative importance of the resulted p-value and difference between means. At \(s0=0\) only the p-value
matters, while at nonzero s0 also the difference of means plays a role. See (Tusher, Tibshirani, and Chu 2001) for details.
3.2.2 Side
To apply a two-sided test, where the null hypothesis can be rejected regardless of the direction of the effect “both” has to be selected (default). “left” and “right” are the respective one-sided
3.3 Valid value filter
Specify here how rows are filtered regarding the number and percentage of valid values. This criterion will be applied to each test individually, not just once to the whole matrix. The absolute
number and relative percentage filters are both applied together.
3.3.1 Min. number of valid values
Here the required number of valid values is specified. How this threshold is applied (in total, per group, etc.) is specified in the next field.
3.3.2 Min. number mode
Specify here how the threshold is applied.
3.3.3 Min. percentage of valid values
Here the required percentage of valid values is specified. How this threshold is applied (in total, per group, etc.) is specified in the next field. Values can range from 0 to 100.
3.3.4 Min. percentage mode
Specify here how the above threshold is applied.
3.4 Use for truncation
Defines on what value the truncation is based on (default: Permutation-based FDR). Choose here whether the truncation should be based on the p-values, on permutation-based FDR values or, if the
Benjamini-Hochberg correction for multiple hypothesis testing should be applied.
3.4.1 Threshold p-value
This parameter is just relevant, if the parameter “Use for truncation” is set to “P-value”. Rows with a test result below this value are reported as significant (default: 0.05).
3.4.2 FDR
This parameter is just relevant, if the parameter “Use for truncation” is set to “Benjamini-Hochberg FDR” or “Permutation-based FDR”. Rows with a test result below this value are reported as
significant (default: 0.05).
3.4.3 Number of randomizations
Specifies the number of randomizations that should be applied (default: 250).
3.4.4 Preserve grouping in randomizations
Defines, whether the grouping specified in a categorical row should be preserved in the randomizations (default: <None>). It can be selected from a list including all available groupings of the
3.5 Calculate combined score
In case multiple two sample tests are performed, the combined score helps to define a global set of significant items over all the tests combined. A global q-value can be calculated based on
permutations of the whole matrix.
3.5.1 Mode
Here the user can define the combined score which is either the p-value from the best test or the product over all tests.
3.5.2 Combined q-value
In case this is checked, a combined q-value based on the combined score and permutations of the whole matrix is calculated.
3.6 -Log10
If checked, \(-Log_{10}(test\ value)\) is reported in the output matrix (default). Otherwise the test-value is reported.
3.7 Suffix
The entered suffix will be attached to newly generated columns (default: empty). That way columns from multiple runs of the test can be distinguished more easily.
Tusher, Virginia Goss, Robert Tibshirani, and Gilbert Chu. 2001.
“Significance Analysis of Microarrays Applied to the Ionizing Radiation Response.” Proceedings of the National Academy of Sciences
98 (9): 5116–21. | {"url":"https://cox-labs.github.io/coxdocs/twosampletestprocessing.html","timestamp":"2024-11-12T23:23:13Z","content_type":"application/xhtml+xml","content_length":"39404","record_id":"<urn:uuid:a286c307-b318-4576-8f9c-3f2dba7d4965>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00719.warc.gz"} |
Next: The simple procedure Up: Solution of Mach Angle Previous: Solution of Mach Angle Index
Again, this set of parameters is, perhaps, the most common and natural to examine. Thompson (1950) has shown that the relationship of the shock
angle is obtained from the following cubic equation:
Equation (13.18) requires that ^13.8. Clearly,
The solution of a cubic equation such as (13.18) provides three roots^13.9. These roots can be expressed as
and where the definition of the
and where the definitions of
Only three roots can exist for the Mach angle,
The physical meaning of the above analysis demonstrates that in the range where ^13.10.
Furthermore, only in some cases when
When ^13.11 that. Physically, it can be shown that the first solution(13.23), referred sometimes as a thermodynamically unstable root, which is also related to a decrease in entropy, is
``unrealistic.'' Therefore, the first solution does not occur in reality, at least, in steady-state situations. This root has only a mathematical meaning for steady-state analysis^13.12.
These two roots represent two different situations. First, for the second root, the shock wave keeps the flow almost all the time as a supersonic flow and it is referred to as the weak solution
(there is a small section that the flow is subsonic). Second, the third root always turns the flow into subsonic and it is referred to as the strong solution. It should be noted that this case is
where entropy increases in the largest amount.
In summary, if a hand moves the shock angle starting from the deflection angle and reaching the first angle that satisfies the boundary condition, this situation is unstable and the shock angle will
jump to the second angle (root). If an additional ``push'' is given, for example, by additional boundary conditions, the shock angle will jump to the third root^13.13. These two angles of the strong
and weak shock are stable for a two-dimensional wedge (see the appendix of this chapter for a limited discussion on the stability^13.14).
The first range is when the deflection angle reaches above the maximum point. For a given upstream Mach number,
Subsections Next: The simple procedure Up: Solution of Mach Angle Previous: Solution of Mach Angle Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21 include("aboutPottoProject.php"); ?> | {"url":"https://potto.org/gasDynamics/node201.php","timestamp":"2024-11-12T21:56:11Z","content_type":"application/xhtml+xml","content_length":"18316","record_id":"<urn:uuid:8ffa53c4-2e21-42c1-aec3-445f4f128b3c>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00080.warc.gz"} |
UPSC NDA II Syllabus 2023 (Download) UPSC National Defence Academy NDA II Exam Pattern PDF
UPSC NDA 2 Syllabus & Exam Pattern 2023 Download PDF
UPSC NDA 2 Syllabus 2023 is available here. Union Public Service Commission National Defence Academy II Syllabus & Exam Pattern had given here on our website for free download. Candidates who are
applied for UPSC National Defence Academy II (NDA 2) Recruitment have started their Exam preparation for the Written Test must download the pdf of UPSC National Defence Academy II Exam Syllabus and
Exam Pattern pdf for free download. All those applicants can check the UPSC NDA 2 Syllabus and can download it. Here, we are providing the UPSC National Defence Academy II Previous papers along with
solutions. Click the below links to download the UPSC National Defence Academy II Previous Papers, Syllabus, and Exam Pattern. Get UPSC NDA 2 Syllabus 2023 and Exam Pattern pdf for free download.
UPSC NDA 2 Syllabus 2023 Highlights
│ Organization │ Union Public Service Commission │
│ Posts │ National Defense Academy & Naval Academy Examination (I) │
│ Category │ Syllabus │
│ Job Location │ Across India │
│ Official Site │ upsc.gov.in │
UPSC National Defence Academy II Syllabus 2023 | Exam Pattern
Candidates who are applied for Union Public Service Commission UPSC National Defence Academy II Exam can get the syllabus from here. Exam Syllabus Plays a Crucial Role in exam preparation. Without
having the knowledge of the UPSC Syllabus 2023 NDA 2 Exam, individuals cannot clearly present in the written examination. Therefore, candidates to qualify for the examination to know the in-depth
UPSC NDA 2 Syllabus 2023 through this article. In this section, we provide the topic Wise Syllabus for Union Public Service Commission National Defence Academy II Syllabus 2023. Also, begin your
preparation now as there is a huge UPSC National Defence Academy II Exam 2023 Syllabus to cover for the exam. Also, access the previous papers for UPSCL NDA 2 Examination from our site for your
UPSC NDA 2 Exam Pattern 2023
The Exam pattern of the UPSC National Defence Academy II NDA 2 Exam has clearly mentioned on our website. The Union Public Service Commission National Defence Academy II NDA 2 Exam Paper has obctive
type questions of Different Sections like General Aptitude and Reasoning, General English, Numerical Aptitude, and General Knowledge. Candidates who going to attend Exam can download the UPSC
National Defence Academy II NDA 2 Test Pattern and Syllabus on this page.
│ Subject Name │ Code │ Marks │ Duration │
│ Mathematics │ 01 │ 300 │ 2 Hours 30 Minutes │
│ General Ability Test │ 02 │ 600 │ 2 Hours 30 Minutes │
│ Total │ 900 Marks │ 5 Hours │
│ SSB Test / Interview │ 900 Marks │
Union Public Service Commission NDA 2 Syllabus 2023
The UPSC NDA 2 Syllabus is provided for the candidates preparing for Exam. Candidates who have applied for UPSC NDA 2 Recruitment can use this syllabus to help you to give your best in the Union
Public Service Commission Exam. The UPSC NDA 2 Syllabus topics are mentioned below.
UPSC NDA 2 Papers I Syllabus 2023 For Mathematics
UPSC NDA Syllabus 2023 – Algebra
• Concept of a set,
• operations on sets,
• Venn diagrams.
• De Morgan laws,
• Cartesian product,
• relation,
• equivalence relation.
• Representation of real numbers on a line.
• Complex numbers—basic properties, modulus, argument, cube roots of unity. A binary system of numbers.
• Conversion of a number in decimal system to binary system and vice-versa.
• Arithmetic, Geometric and Harmonic progressions.
• Quadratic equations with real coefficients.
• The solution of linear inequations of two variables by graphs.
• Permutation and Combination.
• Binomial theorem and its applications.
• Logarithms and their applications.
UPSC NDA 2 Syllabus 2023 – Matrices and Determinants
• Types of matrices, operations on matrices.
• Determinant of a matrix, basic properties of determinants.
• Adjoint and inverse of a square matrix,
• Applications-Solution of a system of linear equations in two or three unknowns by Cramer’s rule and by Matrix Method.
UPSC NDA Syllabus 2023 – Trigonometry
• Angles and their measures in degrees and radians.
• Trigonometrical ratios.
• Trigonometric identities Sum and difference formulae.
• Multiple and Sub-multiple angles.
• Inverse trigonometric functions.
• Applications-Height and distance, properties of triangles.
UPSC NDA Syllabus 2023 – Analytics Geometry of To and Three Diamonds
• Rectangular Cartesian Coordinate system.
• Distance formula. An equation of a line in various forms.
• An angle between two lines. A distance of a point from a line.
• An equation of a circle in a standard and general form.
• Standard forms of parabola, ellipse, and hyperbola.
• Eccentricity and axis of a conic. Point in a three-dimensional space, the distance between two points.
• Direction Cosines and direction ratios.
• Equation two points.
• Direction Cosines and direction ratios.
• An equation of a plane and a line in various forms.
• An angle between two lines and the angle between two planes.
• The equation of a sphere.
UPSC NDA Syllabus 2023 – Differential Calculus
• The concept of a real-valued function–domain, range, and the graph of a function.
• Composite functions, one-to-one, onto, and inverse functions.
• The notion of limit, Standard limits—examples.
• Continuity of functions—examples, algebraic operations on continuous functions.
• Derivative of function at a point, geometrical and physical interpretation of a derivative—applications.
• Derivatives of the sum, product, and quotient of functions, derivative of a function with respect to another function, a derivative of a composite function.
• Second-order derivatives.
• Increasing and decreasing functions.
• Application of derivatives in problems of maxima and minima.
UPSC NDA Syllabus 2023 – Integral Calculus & Differential Equations
• Integration as inverse of differentiation, integration by substitution and by parts, standard integrals involving algebraic expressions, trigonometric, exponential, and hyperbolic functions.
• Evaluation of definite integrals—determination of areas of plane regions bounded by curves— applications.
• 18 Definition of order and degree of a differential equation, formation of a differential equation by examples.
• A general and particular solution of differential equations, the solution of the first order and first-degree differential equations of various types—examples. Application in problems of growth
and decay.
UPSC NDA Syllabus 2023 – Vector Algebra
• Vectors in two and three dimensions, magnitude, and direction of vector.
• Unit and null vectors, the addition of vectors, scalar multiplication of a vector, scalar product or dot product of two vectors.
• Vector product or cross product of two vectors.
• Applications—work done by a force and moment of a force and in geometrical problems.
UPSC NDA Syllabus 2023 – Statistics & Probability
• Statistics: – Classification of data, Frequency distribution, cumulative frequency distribution—examples. Graphical representation—Histogram, Pie Chart, frequency polygon—examples. Measures of
Central tendency—Mean, median, and mode. Variance and standard deviation—determination and comparison. Correlation and regression.
• Probability: Random experiment, outcomes, and associated sample space, events, mutually exclusive and exhaustive events, impossible and certain events. Union and Intersection of events.
Complementary, elementary, and composite events. Definition of probability—classical and statistical— examples. Elementary theorems on probability—simple problems. Conditional probability, Bayes’
theorem—simple problems. Random variable as function on a sample space. Binomial distribution, examples of random experiments giving rise to Binominal distribution.
UPSC NDA 2 Syllabus 2023 Paper II (General Ability Test – 600M):
UPSC NDA Syllabus 2023 – General English
• The question paper in English will be designed to test the candidate’s understanding of English and workmanlike use of words.
• Grammar and usage,
• vocabulary,
• comprehension and cohesion in extended text to test the candidate’s proficiency in English.
UPSC NDA Syllabus 2023 – General Knowledge
The question paper on General Knowledge will broadly cover the subjects as follows.
• Physics,
• Chemistry,
• General Science,
• Social Studies,
• Geography and
• Current Events.
UPSC NDA 2 Syllabus – Frequently Asked Questions(FAQ)
What is UPSC NDA 2 Syllabus?
Quantitative Ability, Communication Ability, English Comprehension, Analytical Ability, General Knowledge and Current affairs
What is the Exam Pattern for UPSC NDA 2 2023?
The Detailed UPSC NDA 2 Exam Pattern 2023 is available @ Questionpapersonline.com
Where can I Get the UPSC NDA 2 Syllabus PDF?
The UPSC NDA 2 Syllabus PDF available @ upsc.gov.in
What are the Total Marks for UPSC NDA 2 Exam?
The UPSC NDA 2 will be conducted for 100 Marks
How many Questions will be asked in the UPSC NDA 2 Exam ?
A Total of 100 Questions will be asked in the UPSC NDA 2 Exam | {"url":"https://www.questionpapersonline.com/upsc-nda-ii-syllabus-download/","timestamp":"2024-11-09T11:30:46Z","content_type":"text/html","content_length":"155347","record_id":"<urn:uuid:9d9f2422-57df-4427-952a-b90c17752fb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00121.warc.gz"} |
Bit Shifting for a Shard ID in Ruby
As our database grew, we had to take a serious look at how can we could split it up by our clients, as most of them wanted to have their own data separate from the others anyway. A few months ago I
found a great article from Pinterest, that describes how they sharded their MySQL database.
A sharded entity needs a UUID to uniquely identify the record across all shards. Most of the programming languages can generate a UUID easily, however, what was amazing to me, was that Pinterest
generated its own unique ids by encapsulating three distinct numbers in one. Wait, what??! Read that article, it’s definitely worth your time.
While Pinterest encapsulated three numbers into one, we only needed two, a client_id and an entity_id. Our client_id would be a much smaller number than our entity_id, we wanted to reserve more bits
for the latter.
It turns out, Ruby has many friendlier tools to deal with binary operations. Let's look at them!
What is the binary representation of the integer number 123?
Out of the 7 bits, you'll see that the 3rd one is turned off, all the others are turned on, giving us 64 + 32 + 16 + 8 + 2 +1 = 123. How can we get this binary representation in Ruby? It's super
easy, just use the to_s(2) method to do it.
pry(main)> 123.to_s(2)
This is the exact same string representation as the one in the image above, where the third bit is turned off and represented with a zero.
I'd like to keep the client_id on the left side, but I'd like to reserve bits on the right side. For the sake of simplicity, I will keep this as a small number. Let's add 5 bits to the right-hand
side of these bits by using the bitwise left shift operator.
pry(main)> (123 << 5).to_s(2)
=> "111101100000"
The original number, 123, is still represented on the left side, but 5 "0"s were added to the right-hand side. You get the numeric representation of this by leaving out the to_s(2) call at the end:
pry(main)> 123 << 5
=> 3936
This number can be converted back to binary:
pry(main)> 3936.to_s(2)
=> "111101100000"
On the left side I have the binary representation of 123, but how about those bits on the right side? What are those representing? Right now, those bits are all turned off, they will give you 0
("00000".to_i(2) => 0). How can I store the number 3 on the right side? The bits should look like this:
The binary "|" will turn the two rightmost bits on:
pry(main)> (123 MM 5 | 3).to_s(2)
=> "111101100011"
Again, leaving out the to_s(2) will provide the number representation:
pry(main)> (123 << 5 | 3)
=> 3939
The storing part will work, but how can we get our two numbers back from this one combined number? Well, we have to split the bits and convert the binary representation to an integer.
Five bits were used on the right side to store our second number. We need to chop those off to get the number stored on the left side. The bitwise right shift (>>) will do just that:
pry(main)> (3939 >> 5).to_s(2)
=> "1111011"
The string "1111011" is our original 123 in a binary string format. We can convert that to an integer by using the to_i(2) String method:
pry(main)> (3939 >> 5).to_s(2).to_i(2)
=> 123
I right shifted the original number, 3939, converted it to a binary string and converted that to an Integer.
There are more efficient ways to do this by using a binary "&" (3939 >> 5) & 0b1111111 => 123 with the max value the bits can represent. That's what the Pinterest article had, but I found using the
Ruby conversion methods a bit more friendly to those of us who are not dealing with binary data on a daily basis.
We have the number on the left side, but what about the number on the right side? When we convert the number representation (3939) to a binary string, we know that the five characters on the right
side will represent the bits of our other number. Ruby String’s last(x) will do just that:
pry(main)> (3939 >> 0).to_s(2).last(5)
=> "00011"
Converting this binary String to Integer should be similar to what we've done before:
pry(main)> (3939 >> 0).to_s(2).last(5).to_i(2)
=> 3
Using the binary "&" with the max number the bits can store will do this conversion in one step: (3939 >> 0) & 0b11111 => 3. As a side note, the binary number can be represented as a hexadecimal
value: (3939 >> 0) & 0x1F => 3. This is a lot shorter than a series of ones and zeros.
There is a limit of how large the numbers can be as you have a limit of bits to store those. The max number can be determined by flipping the available bits on. For an 7 bit number it's 64 + 32 + 16
+ 8 + 4 + 2 + 1 = 127 or 2**x-1, where x is the number of bits. In our case it is 2**7-1 = 127.
We ended up using a 64-bit Integer for our shard_id, which is a BIGINT in MySQL. We store client_id in 22 bits giving us the maximum of 2**22-1 = 4_194_304 and 40 bits for the entity_id with 2**40-1
= 1_099_511_627_775 max value. The remaining two bits are "worth in gold, we can expand the number or store a third (albeit small) number in it. | {"url":"https://www.adomokos.com/2017/05/bit-shifting-for-shard-id-in-ruby.html","timestamp":"2024-11-01T23:41:17Z","content_type":"application/xhtml+xml","content_length":"103268","record_id":"<urn:uuid:767720a1-1de1-4125-89f9-97e9e90c965d>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00476.warc.gz"} |
Invariants, theory of
From Encyclopedia of Mathematics
(in the classical sense), invariant theory
The algebraic theory (sometimes called the algebraic theory of invariants) that studies algebraic expressions (polynomials, rational functions or families of them) that change in a specified way
under non-degenerate linear changes of variables. Here, in general one does not consider the general linear group (that is, the entire set of non-degenerate linear changes of variables), but only a
certain subgroup of it (for example, the orthogonal, the symplectic, etc., subgroup). The theory of invariants arose from a number of problems in number theory, algebra and geometry. C.F. Gauss in
his studies on the theory of binary quadratic forms posed the problem of studying polynomials in the coefficients of the form $ ax ^{2} + 2 b xy + c y ^{2} $ that do not change under transformations
of these coefficients induced by substitutions of the form $ x \rightarrow \alpha x + \beta y $ , $ y \rightarrow \gamma x + \delta y $ , where $ \alpha \delta - \beta \gamma = 1 $ and $ \alpha $ , $
\beta $ , $ \gamma $ , $ \delta $ are integers. On the other hand, in the intrinsic characterization of configurations and relations in projective geometry there occur algebraic expressions of
projective coordinates that do not change under projective collineations. The study of determinants also preceded the theory of invariants. Arithmetic and algebraic questions, connected in one way or
another with the theory of invariants, drew the attention of C.G.J. Jacobi, F. Eisenstein and Ch. Hermite. The theory of invariants was formed as a mathematical discipline in the mid-19th century.
This period is characterized by an interest in formal algebraic problems and their applications to geometry. The notions of a group, an invariant and the fundamental problems of the theory were
formulated at that time in a precise manner and gradually it became clear that various facts of classical and projective geometry are merely expressions of identities (syzygies) between invariants or
covariants of the corresponding transformation groups. The Memoir on hyperdeterminants by A. Cayley (1846) must apparently be regarded as the first work on the theory of invariants. All the regular
terms in the theory of invariants such as invariant; covariant; comitant; discriminant, etc., were introduced by J. Sylvester.
One of the first objects of study in invariant theory were the so-called invariants of forms. One considers a form of degree $ r $ in $ n $ variables with undetermined coefficients:$$ f ( x _{1} \
dots x _{n} ) = \sum _ {i _{1} + \dots + i _{n} = r} a _ {i _{1} \dots i _{n}} x _{1} ^ {i _{1}} \dots x _{n} ^ { i _{n}} . $$ After the linear change of variables$$ x _{i} \rightarrow \sum _{j=1} ^
n \alpha _{ij} x _{j} , 1 \leq i \leq n , $$ where $ \alpha _{ij} $ are real or complex numbers, it is converted to the form$$ f ^ {\ \prime} ( x _{1} \dots x _{n} ) = \sum _ {i _{1} + \dots + i _{n}
= r} a _ {i _{1} \dots i _{n}} ^ \prime x _{1} ^ {i _{1}} \dots x _{n} ^ {i _{n}} , $$ so that the above linear transformation of variables determines a transformation of the coefficients of the
form:$$ a _ {i _{1} \dots i _{n}} \rightarrow a _ {i _{1} \dots i _{n}} ^ \prime . $$ A polynomial $ \phi ( \ldots ,\ a _ {i _{1} \dots i _{n}} , \ldots ) $ in the coefficients of the form $ f ( x _
{1} \dots x _{n} ) $ is called a (relative) invariant of the form if the following relation holds under any non-degenerate linear change of variables:$$ \phi ( \ldots ,\ a _ {i _{1} \dots i _{n}} ^ \
prime , \ldots ) = | \alpha _{ij} | ^{q} \phi ( \ldots ,\ a _ {i _{1} \dots i _{n}} , \ldots ) , $$ where $ | \alpha _{ij} | $ is the determinant of the linear transformation and $ q $ is a constant
(the weight). If $ q = 0 $ , the invariant is called absolute. E.g., the simplest example of an invariant is the discriminant $ D = b ^{2} - ac $ of the binary quadratic ($ n = r = 2 $ ) form $ f ( x
,\ y ) = a x ^{2} + 2bxy + c y ^{2} $ , or the discriminant $ \Delta = 3 b ^{2} c ^{2} +6 abcd - 4 b ^{3} d - 4 a c ^{3} - a ^{2} d ^{2} $ of the ternary ($ n = 2 $ , $ r = 3 $ ) form $ f ( x ,\ y )
= a x ^{3} + 3 b x ^{2} y + 3 c x y ^{2} + d y ^{3} $ . If one is given not one but several forms in the same set of variables, then it is possible to consider polynomials $ \phi $ in the
coefficients of all these forms transforming in the above manner under a linear change of variables; in this way one obtains the notion of a joint invariant of a system of forms. For example, the
determinant $ | a _{ij} | $ is a joint invariant of the system of $ n $ linear forms $ f _{i} ( x _{1} \dots x _{n} ) = \sum _{j=1} ^{n} a _{ij} x _{j} $ , $ 1 \leq i \leq n $ . Similarly one can
define a covariant and, more generally, a comitant.
The classical statement of the fundamental problem in the theory of invariants is to calculate the invariants, and also to give a complete description of them (and the same for covariants). To this
end, various formal processes were developed enabling one to construct invariants (polarization, restitution, the Capelli identity, the Cayley $ \Omega $-process, etc.). The culmination of this
activity was the creation of the so-called symbolic method in the theory of invariants, which is a formal method for calculating all the invariants of degree not exceeding a given number (see [3],
[6], [11]).
The global theory of semi-simple groups and their representations, developed in the 1930's, has enabled one to give the following more general statement of the fundamental problem in the classical
theory of invariants [6]. One is given an arbitrary group $ G $ and a finite-dimensional linear representation $ \rho $ of it in a linear space $ V $ over a field $ k $ . If $ x _{1} \dots x _{n} $
are coordinates in $ V $ (in some basis), then each element $ g \in G $ determines a linear change of the variables $ x _{1} \dots x _{n} $ ; by carrying out this change of variables in an arbitrary
polynomial $ \phi ( x _{1} \dots x _{n} ) $ one obtains a new polynomial, hence $ g $ induces a transformation (automorphism) of the ring of all polynomials $ k [ x _{1} \dots x _{n} ] $ in the
variables $ x _{1} \dots x _{n} $ over the field $ k $ . A polynomial $ \phi ( x _{1} \dots x _{n} ) $ that does not change under all such transformations (that is, when $ g $ runs through the whole
of $ G $ ) is called an invariant of the representation $ \rho $ of the group $ G $ (cf. also Linear representation, invariant of a). All the invariants form a $ k $-algebra and the aim of the theory
of invariants is to describe this algebra. Thus, the invariants of forms are the invariants of the general linear group with respect to its representation in the space of symmetric tensors of fixed
rank of the underlying (or dual) space (the coefficients of the original form $ f $ are the components of this tensor); the consideration of covariants reduces to the study of the algebra of
invariants for a representation in the space of tensors of mixed valency. In this form the problem of the description of invariants is a special case of the following general problem of the theory of
linear representations: To decompose a space of tensors of given valency into irreducible invariant subspaces with respect to a given group of linear transformations of the underlying linear space
(the search for invariants can be reduced to singling out the one-dimensional invariant subspaces).
Already at the first stages of development of the theory of invariants the following fact was discovered, which allowed one to consider the system of invariants as a whole: In all the cases examined
it was possible to select a finite number of basic invariants $ \phi _{1} \dots \phi _{m} $ , that is, invariants such that every other invariant $ \phi $ of the given representation can be expressed
as a polynomial in them: $ \phi = F ( \phi _{1} \dots \phi _{m} ) $ . In other words, the algebra of invariants proved to be finitely generated. It also became clear that these basic invariants are,
in general, not independent (that is, the algebra of invariants is not a free algebra): there can exist non-trivial polynomials $ P ( t _{1} \dots t _{m} ) $ , called relations or syzygies, that
after the substitutions $ t _{i} = \phi _{i} $ , $ 1 \leq i \leq m $ , vanish identically. In the set of relations itself it is again possible to find a finite number of basic relations, all the
remaining being algebraic consequences of them (the relations form an ideal in the ring of polynomials in the variables $ t _{1} \dots t _{m} $ , and the basic relations are generators of it). In
turn, the basic relations themselves are in general not independent; thus, secondary syzygies can be determined, etc. The chain of syzygies constructed in this way always turns out to be finite. For
example, if $ G $ is the symmetric group of all permutations of the coordinates $ x _{1} \dots x _{n} $ , then the algebra of invariants is the algebra of all symmetric polynomials in $ x _{1} \dots
x _{m} $ ; the elementary symmetric polynomials are basic invariants, which are algebraically independent (in this case there are no syzygies).
These facts led to the statement of two fundamental problems in the classical theory of invariants:
1) To prove that the algebra of invariants of a given representation of a given group is finitely generated (the first fundamental theorem of the theory of invariants) and to determine a system of
basic invariants.
2) To prove the existence of a finite basis of syzygies (the second fundamental theorem of the theory of invariants) and to find it.
The first fundamental theorem of the theory of invariants for invariants of a form of arbitrary degree in an arbitrary finite number of variables was proved by D. Hilbert [1] (see also Hilbert
theorem on invariants). He also proved that the second fundamental theorem of the theory of invariants is true in all those cases when the first fundamental theorem holds, and also that in this case
the chain of syzygies is always finite. Hilbert obtained the proof of the fundamental theorems of the classical theory of invariants on the basis of general abstract algebraic results proved by him
(with this specific aim). These results subsequently formed the foundation of modern commutative algebra (the Hilbert basis theorem, the Hilbert syzygies theorem, Hilbert's Nullstellensatz, cf.
Hilbert theorem 1), 3) and 5)). The original proof of the first fundamental theorem of the theory of invariants was non-constructive and gave no upper estimate for the degree of basic invariants. In
the 1930's H. Weyl, by developing an idea of Hilbert and A. Hurwitz [14], proved the first fundamental theorem of the theory of invariants for finite-dimensional representations of arbitrary compact
Lie groups and for finite-dimensional representations of arbitrary complex semi-simple Lie groups [6]. The book [6], which summarizes the developments of the classical theory of invariants, contains
a description of the basic invariants and syzygies for the representations of the classical groups as well as for certain other groups. One of the important applications of the methods of the theory
of invariants was the description of the Betti numbers of classical compact groups.
The proof of the second fundamental theorem of the theory of invariants discloses the general algebraic nature of this theorem (it is a corollary of Hilbert's basis theorem). In attempting to
determine whether this is also true with regard to the first fundamental theorem of the theory of invariants, Hilbert stated the following problem (Hilbert's 14th problem): Let $ k $ be a field, let
$ A = k [ x _{1} \dots x _{n} ] $ be the algebra of polynomials over $ k $ in the variables $ x _{1} \dots x _{n} $ and let $ L $ be an arbitrary subfield of the field of fractions of $ A $
containing $ k $ . Is then the algebra $ A \cap L $ finitely generated over $ k $ ? An affirmative answer to this question would imply the validity of the first fundamental theorem of the theory of
invariants for arbitrary groups. A negative solution to Hilbert's 14th problem was obtained in [9], in which an example was given of a representation of a commutative unipotent group for which the
algebra of invariants does not have a finite number of generators. In the 1950's a number of results was obtained on invariants of finite groups, in particular groups generated by reflections (see
Reflection group; Coxeter group). It has been proved [13], [14] that finite linear complex groups generated by unitary reflections can be characterized as the finite linear groups whose algebras of
invariants do not contain syzygies.
A new stage of development of the theory of invariants was related to the extension of the circle of problems and to geometric applications of this theory. The modern theory of invariants (or the
geometric theory of invariants) became a part of the general theory of algebraic transformation groups; the theory of algebraic groups constructed in 1950's is fundamental to it, and the language of
algebraic geometry is fundamental to its language. In contrast with the classical theory of invariants, whose basic object was the ring of polynomials in $ n $ variables over a field $ k $ together
with the group of automorphisms induced by linear changes of variables, the modern theory of invariants considers an arbitrary finitely-generated $ k $-algebra $ R $ and the algebraic group $ G $ of
its $ k $-automorphisms. Instead of a linear space $ V $ and a representation $ \rho $ , an arbitrary affine algebraic variety $ X $ is considered together with the algebraic group $ G $ of its
algebraic transformations (automorphisms), such that $ R $ is the ring of regular functions on $ X $ and the action of $ G $ on $ R $ is induced by that of $ G $ on $ X $ . The elements of $ R $ that
are fixed under $ G $ are the invariants; the entire set of them forms a $ k $-algebra, $ R ^{G} $ .
Other notions of the classical theory of invariants can also be generalized. For example, the comitant, which is a regular mapping from one such variety into another commuting with the group action;
if $ R ^{G} $ is finitely generated, then one says that the first fundamental theorem of the theory of invariants is valid. It has been proved that $ R ^{G} $ is finitely generated if $ G $ is a
geometrically-reductive group (see Mumford hypothesis). In many important cases, for example, in applications to the moduli problem, $ G $ is in fact a group of this type. If $ R ^{G} $ is finitely
generated, then there exists an affine algebraic variety $ W $ for which $ R ^{G} $ is the algebra of regular functions; the imbedding $ R ^{G} \subset R $ induces a morphism $ \pi : \ X \rightarrow
W $ . If $ G $ is geometrically reductive, then $ W $ classifies the closed orbits of $ G $ in $ W $ : $ \pi $ is surjective and in each of its fibres there is exactly one closed orbit. A necessary
condition for the existence of a quotient variety of $ X $ with respect to $ G $ , which is that all the orbits be closed, turns out to be sufficient also, and $ W $ proves to be this quotient
variety. Hence the role of $ R ^{G} $ in the solution of the geometric problems of classifying and constructing quotient varieties becomes clear; apart from this, the study of $ R ^{G} $ (which was
the final aim of the classical theory of invariants) is only the beginning stage for the solution of these geometric problems, since knowledge of $ R ^{G} $ does not, in general, provide complete
information on the orbits of $ G $ in $ X $ and must therefore be combined with the consideration of non-closed orbits, their closures and stabilizers (so-called orbital decompositions). Furthermore,
the study of actions of algebraic groups on affine algebraic varieties is only the "local part" of the general theory of algebraic transformation groups (just as the theory of affine varieties is the
"local part" of the general theory of algebraic varieties).
In the general case one considers an algebraic (regular) action of $ G $ on an arbitrary algebraic variety $ X $ (glued together from affine pieces), so that, e.g., the solution of the problem of
constructing a quotient variety of $ X $ (or of a suitable open subset of $ X $) with respect to the action of $ G $ reduces to the consideration of a $ G $-invariant affine covering of $ X $ and the
application of the above construction to each element of this covering with subsequent "glueing" of the affine quotient varieties obtained. For a successful application of the procedure, $ X $ must,
in general, be replaced by some invariant subset of it ($ X $ itself need not have an invariant affine covering).
At present, investigations on the theory of algebraic transformation groups are conducted in various directions, from which one can select the following: To obtain information on the properties of
points in general position on a variety $ X $ on which an algebraic group $ G $ acts regularly; to describe the orbital decompositions of various specific actions (mainly linear representations);
theorems on stratification of general spaces of transformations into simpler standard spaces; theorems on quotient varieties, orbit spaces, spaces of sections, quasi-sections and "slices" , as well
as on varieties of fixed points; theorems on equivalent imbeddings; criteria for the affineness and quasi-affineness of orbits (see Matsushima criterion); the structure of the closure of orbits of
various special types, the theory of quasi-homogeneous varieties (that is, varieties with a dense orbit) and the theory of actions with a trivial algebra of invariants; the algebraic properties of
rings of invariants, the algebraic-geometric properties of spaces of transformations of quotient varieties; the connection with the theory of singularities [16]); the problem of the rationality of
fields of invariants, and the connection with the theory of algebraic tori and algebraic number theory [17]).
[1] D. Hilbert, "Ueber die Theorie der algebraischen Formen" Math. Ann. , 36 (1890) pp. 473–534 MR1510634 Zbl 22.0133.01
[2] D. Hilbert, "Ueber die vollen Invariantensysteme" Math. Ann. , 42 (1893) pp. 313–373 MR1510781 Zbl 25.0173.01
[3] R. Weitzenböck, "Invarianten Theorie" , Noordhoff (1923) Zbl 0014.10101 Zbl 62.0074.01
[4] A. Hurwitz, "Ueber die Erzeugung von Invarianten durch Integration" Gött. Nachr. (1897) pp. 71–90
[5] F. Klein, "Development of mathematics in the 19th century" , Math. Sci. Press (1979) pp. Chapt. 1 (Translated from German) MR0529278 MR0549187 Zbl 0411.01009
[6] G.B. Gurevich, "Algebraic theory of invariants" , Noordhoff (1964) (Translated from Russian) Zbl 0128.24601
[7] H. Weyl, "The classical groups, their invariants and representations" , Princeton Univ. Press (1946) MR0000255 Zbl 1024.20502
[8] "Hilbert problems" Bull. Amer. Math. Soc. , 8 (1902) pp. 437–479 (Translated from German)
[9] M. Nagata, "On the fourteenth problem of Hilbert" J.A. Todd (ed.) , Proc. Internat. Congress Mathematicians (Edinburgh, 1958) , Cambridge Univ. Press (1960) pp. 459–462 MR0116056 Zbl 0127.26302
[10] M. Nagata, "On the 14-th problem of Hilbert" Amer. J. Math. , 81 (1959) pp. 766–772 MR0105409 Zbl 0192.13801
[11] D. Mumford, "Geometric invariant theory" , Springer (1965) MR0214602 Zbl 0147.39304
[12] J. Dieudonné, J.B. Carrell, "Invariant theory: old and new" , Acad. Press (1971) MR0279102 Zbl 0258.14011
[13] S. Fogarty, "Invariant theory" , Benjamin (1969) MR0240104 Zbl 0191.51701
[14] C. Chevalley, "Invariants of finite groups generated by reflections" Amer. J. Math. , 77 (1955) pp. 778–782 MR0072877 Zbl 0065.26103
[15] G.C. Shephard, J.A. Todd, "Finite unitary reflection groups" Canad. J. Math. , 6 (1954) pp. 274–304 MR0059914 Zbl 0055.14305
[16] P. Wagreich, "Algebraic varieties with group action" R. Hartshorne (ed.) , Algebraic geometry (Arcata, 1974) , Proc. Symp. Pure Math. , 29 , Amer. Math. Soc. (1975) pp. 633–642 MR0382264 Zbl
[17] V.E. Voskresenskii, "Fields of invariants for Abelian groups" Russian Math. Surveys , 28 : 4 (1972) pp. 79–105 Uspekhi Mat. Nauk , 28 : 4 (1972) pp. 77–102 Zbl 0289.14006
[a1] T.A. Springer, "Invariant theory" , Lect. notes in math. , 585 , Springer (1977) MR0447428 Zbl 0346.20020
[a2] D. Mumford, "Hilbert's fourteenth problem - the finite generation of subrings such as rings of invariants" F.E. Browder (ed.) , Mathematical developments arising from Hilbert problems , Proc.
Symp. Pure Math. , 28 , Amer. Math. Soc. (1976) pp. 431–444 MR435076
How to Cite This Entry:
Invariants, theory of. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Invariants,_theory_of&oldid=52303
This article was adapted from an original article by V.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article | {"url":"https://encyclopediaofmath.org/wiki/Invariants,_theory_of","timestamp":"2024-11-08T19:05:15Z","content_type":"text/html","content_length":"40459","record_id":"<urn:uuid:6d8bd888-5720-4c86-a768-f9f7e7c05e64>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00674.warc.gz"} |
2017, 2017-05
The estimation, operation and control of electrical power systems have always contained a degree of uncertainty. It is expected that, with the...
Show moreThe estimation, operation and control of electrical power systems have always contained a degree of uncertainty. It is expected that, with the introduction of technologies such as
distributed generation and demand-side management, the ability of system operators to forecast the dynamic behavior of the system will deteriorate and as a result, the cost of keeping the system
together will increase. Sequential Monte Carlo or Particle Filtering is a family of algorithms to efficiently perform inference in non-linear dynamic systems by exploiting their structure without
assuming any linearity or normality structure. In this thesis we provide two novel ways of employing these algorithms for inference and control of power systems. First, we motivate the use
Bayesian statistics in load modelling by introducing a novel statistical model to capture the aggregated response of a set of loads. We then use the model to characterize load with measurement
data and prior information using the Sequential Monte Carlo algorithm. Second, we introduce the Model Predictive Control for power system stabilization. We present the use of the Sequential Monte
Carlo algorithm as a way of solving the stochastic Model Predictive Control problem and we compare its performance to existing regulators. In addition, Model Predictive Control is applied to load
shedding Finally, we test the performance of the algorithm in a large power system scenario.
Ph.D. in Electrical Engineering, May 2017
Show less
Deep Learning is a subfield of machine learning concerned with algorithms that learn hierarchical data representations. Deep learning has...
Show moreDeep Learning is a subfield of machine learning concerned with algorithms that learn hierarchical data representations. Deep learning has proven extremely successful in many computer
vision tasks including object detection and recognition. In this thesis, we aim to develop and design deep-learning models to better perform image processing and tackle three important problems:
natural image denoising, computed tomography (CT) dose reduction, and bone suppression in chest radiography (“chest x-ray”: CXR). As the first contribution of this thesis, we aimed to answer to
probably the most critical design questions, under the task of natural image denoising. To this end, we defined a class of deep learning models, called neural network convolution (NNC). We
investigated several design modules for designing NNC for image processing. Based on our analysis, we design a deep residual NNC (R-NNC) for this task. One of the important challenges in image
denoising regards to a scenario in which the images have varying noise levels. Our analysis showed that training a single R-NNC on images at multiple noise levels results in a network that cannot
handle very high noise levels; and sometimes, it blurs the high-frequency information on less noisy areas. To address this problem, we designed and developed two new deep-learning structures,
namely, noise-specific NNC (NS-NNC) and a DeepFloat model, for the task of image denoising at varying noise levels. Our models achieved the highest denoising performance comparing to the
state-of-the-art techniques.As the second contribution of the thesis, we aimed to tackle the task of CT dose reduction by means of our NNC. Studies have shown that high dose of CT scans can
increase the risk of radiation-induced cancer in patients dramatically; therefore, it is very important to reduce the radiation dose as much as possible. For this problem, we introduced a mixture
of anatomy-specific (AS) NNC experts. The basic idea is to train multiple NNC models for different anatomic segments with different characteristics, and merge the predictions based on the
segmentations. Our phantom and clinical analysis showed that more than 90% dose reduction would be achieved using our AS NNC model.We exploited our findings from image denoising and CT dose
reduction, to tackle the challenging task of bone suppression in CXRs. Most lung nodules that are missed by radiologists as well as by computer-aided detection systems overlap with bones in CXRs.
Our purpose was to develop an imaging system to virtually separate ribs and clavicles from lung nodules and soft-tissue in CXRs. To achieve this, we developed a mixture of anatomy-specific,
orientation-frequency-specific (ASOFS) expert deep NNC model. While our model was able to decompose the CXRs, to achieve an even higher bone suppression performance, we employed our deep R-NNC
for the bone suppression application. Our model was able to create bone and soft-tissue images from single CXRs, without requiring specialized equipment or increasing the radiation dose.
Show less
In this dissertation we present a view in which the radar signals as the elements of a high dimensional signal set. The dimension is equal to...
Show moreIn this dissertation we present a view in which the radar signals as the elements of a high dimensional signal set. The dimension is equal to the number of discrete samples (M) of the
signal. Because the radar signals should satisfy certain conditions for good performance, most lie in much smaller subsets or subspaces. By developing appropriate lower dimensional signal spaces
that approximate these areas where the radar signals live, we can realize potential advantage because of the greater parametric simplicity. In this dissertation we apply this low dimensional
signal concept in radar signal processing. In particular we focus on radar signal design and radar signal estimation. Signal design comes under radar measures and signal estimation comes under
radar countermeasures.In signal design problem one searches for the signal element that has smaller sidelobes and also satisfies certain constraints such as bandwidth occupancy, AC mainlobe
width, etc. The sidelobe levels are quantified by Peak Sidelobe Ratio (PSLR) and Integrated Sidelobe Ratio (ISLR). We use linear combination of these two metrics as the cost function to determine
the quality of the designed signal. There is a lot of effort in designing parameterized signal sets including our proposed Asymmetric Time Exponentiated Frequency Modulated (ATEFM) signal and Odd
Polynomial FrequencySignal (OPFS). Our contribution is to demonstrate that the best signal elements from these low dimensional signal sets (LDSS) mostly outperform the best signal elements that
are randomly chosen from the radar signal subset with dimensionality M. Since searching the best signal element from the LDSS requires less computational resources it is prudent to search for the
best signal elements from the low dimensional signal sets.In signal estimation problem we try to estimate the signal transmitted by a noncooperating radar which is intercepted by multiple passive
sensors. The intercepted signals often have low SNR and there could be only few intercepted signals available for signal estimation. Predominantly used method for estimating the radar signals is
Principal Component Analysis (PCA). When the SNR is low (< 0 dB) we need large number of intercepted signals to get an accurate estimates from PCA method. Our contribution is to demonstrate that
by limiting the search for the best signal estimate within the low dimensional signal sets one can get more accurate estimates of the unknown transmitted signal at low SNRs with smaller number of
sensors compared to PCA.
Show less
The ability to handle complex data is essential for new research findings and business success today. With increased complexity, data can...
Show moreThe ability to handle complex data is essential for new research findings and business success today. With increased complexity, data can either be difficult to collect with designed
experiments or be difficult to analyze with statistical models. Both kinds of difficulties are addressed in this dissertation.The first part of this dissertation (Chapter 2 and 3) addresses the
issue of complex data collection by considering two design of experiment problems. In chapter 2, we consider Bayesian A-optimal design problem under a hierarchical probabilistic model involving
both quantitative and qualitative response variables. The objective function was derived and an efficient optimization algorithm was developed. In chapter 3, we consider the A/B-testing problem
and propose a novel discrepancy-based approach for designing such an experiment. As the numerical examples show, the A/B-testing experiments designed in this way achieve better group balance and
parametric estimation results.In the second part of this dissertation (Chapter 4 and 5), we focus on analyzing complex data with Gaussian process (GP) models. Gaussian process model is widely
used for analyzing data with highly nonlinear relationships and emulating complex systems. In Chapter 4, we apply and extend GP model to analyze the in-cylinder pressure data resulted from
experiments on a newly-developed dual fuel engine. The resulted model incorporates different data types and achieves good prediction accuracy. In Chapter 5, a generalized functional ANOVA GP
model is proposed to tackle the difficulty resulted from high-dimensional feature space, and we develop an efficient algorithm for building such a model from the perspective of multiple kernel
learning. The proposed approach outperforms traditional MLE-based GP models on both computational efficiency and prediction accuracy.
Show less
In many areas of science and engineering, discovering the governing differential equations from the noisy experimental data is an essential...
Show moreIn many areas of science and engineering, discovering the governing differential equations from the noisy experimental data is an essential challenge. It is also a critical step in
understanding the physical phenomena and prediction of the future behaviors of the systems. However, in many cases, it is expensive or time-consuming to collect experimental data. This article
provides an active learning approach to estimate the unknown differential equations accurately with reduced experimental data size. We propose an adaptive design criterion combining the
D-optimality and the maximin space-filling criterion. The D-optimality involves the unknown solution of the differential equations and derivatives of the solution. Gaussian process models are
estimated using the available experimental data and used as surrogates of these unknown solution functions. The derivatives of the estimated Gaussian process models are derived and used to
substitute the derivatives of the solution. Variable-selection-based regression methods are used to learn the differential equations from the experimental data. The proposed active learning
approach is entirely data-driven and requires no tuning parameters. Through three case studies, we demonstrate the proposed approach outperforms the standard randomized design in terms of model
accuracy and data economy.
Show less
This thesis consists of two major parts, and contributes to two areas of research in stochastic analysis: (i) Wiener-Hopf factorization (WHf)...
Show moreThis thesis consists of two major parts, and contributes to two areas of research in stochastic analysis: (i) Wiener-Hopf factorization (WHf) for Markov Chains, (ii) statistical
inference for Stochastic Partial Differential Equations (SPDEs).WHf for Markov chains is a methodology concerned with computation of expectation of some types of functionals of the underlying
Markov chain. Most results in WHf for Markov chains are done in the framework of time-homogeneous Markov chains. The major contribution of this thesis in the area of WHf for Markov chains are: •
We extend the classical theory to the framework of time-inhomogeneous Markov chains. • In particular, we establish the existence and uniqueness of solutions for a new class of operator Riccati
equations. • We connect the solution of the Riccati equation to some expectations of interest related to a time-inhomogeneous Markov chain. Statistical inference for SPDEs regards estimating
parameters of a SPDE based on available and relevant observations of the underlying phenomenon that is modeled by the given SPDE. We summarize the contribution of this thesis in the area
statistical inference for SPDEs as follows: • We conduct the statistical inference for a diagonalizable SPDE driven by a multiplicative noise of special structure, using spectral approach. We
show that the corresponding statistical model fits the classical uniform asymptotic normality (UAN) paradigm. • We prove a Bernstein-Von Mises type result that strengthens the existing results in
the literature. • We prove the asymptotic consistency, asymptotic normality and asymptotic efficiency of two Bayesian type estimators.
Show less
Automatic cubatures approximate multidimensional integrals to user-specified error tolerances. In many real-world integration problems, the...
Show moreAutomatic cubatures approximate multidimensional integrals to user-specified error tolerances. In many real-world integration problems, the analytical solution is either unavailable or
difficult to compute. To overcome this, one can use numerical algorithms that approximately estimate the value of the integral. For high dimensional integrals, quasi-Monte Carlo (QMC) methods are
very popular. QMC methods are equal-weight quadrature rules where the quadrature points are chosen deterministically, unlike Monte Carlo (MC) methods where the points are chosen randomly.The
families of integration lattice nodes and digital nets are the most popular quadrature points used. These methods consider the integrand to be a deterministic function. An alternative approach,
called Bayesian cubature, postulates the integrand to be an instance of a Gaussian stochastic process. For high dimensional problems, it is difficult to adaptively change the sampling pattern.
But one can automatically determine the sample size, $n$, given a fixed and reasonable sampling pattern. We take this approach using a Bayesian perspective. We assume a Gaussian process
parameterized by a constant mean and a covariance function defined by a scale parameter and a function specifying how the integrand values at two different points in the domain are related. These
parameters are estimated from integrand values or are given non-informative priors. This leads to a credible interval for the integral. The sample size, $n$, is chosen to make the credible
interval for the Bayesian posterior error no greater than the desired error tolerance. However, the process just outlined typically requires vector-matrix operations with a computational cost of
$O(n^3)$. Our innovation is to pair low discrepancy nodes with matching kernels, which lowers the computational cost to $O(n \log n)$. We begin the thesis by introducing the Bayesian approach to
calculate the posterior cubature error and define our automatic Bayesian cubature. Although much of this material is known, it is used to develop the necessary foundations. Some of the major
contributions of this thesis include the following: 1) The fast Bayesian transform is introduced. This generalizes the techniques that speedup Bayesian cubature when the kernel matches low
discrepancy nodes. 2) The fast Bayesian transform approach is demonstrated using two methods: a) rank-1 lattice sequences and shift-invariant kernels, and b) Sobol' sequences and Walsh kernels.
These two methods are implemented as fast automatic Bayesian cubature algorithms in the Guaranteed Automatic Integration Library (GAIL). 3) We develop additional numerical implementation
techniques: a) rewriting the covariance kernel to avoid cancellation error, b) gradient descent for hyperparameter search, and c) non-integer kernel order selection.The thesis concludes by
applying our fast automatic Bayesian cubature algorithms to three sample integration problems. We show that our algorithms are faster than the basic Bayesian cubature and that they provide
answers within the error tolerance in most cases. The Bayesian cubatures that we develop are guaranteed for integrands belonging to a cone of functions that reside in the middle of the sample
space. The concept of a cone of functions is also explained briefly.
Show less
2016, 2016-05
In statistics, we ask whether some statistical model ts observed data. We use a Markov chain proposed by Gross, Petrovi c, and Stasi to...
Show moreIn statistics, we ask whether some statistical model ts observed data. We use a Markov chain proposed by Gross, Petrovi c, and Stasi to perform exact testing for the p1 random graph
model. By comparing it to the simple switch Markov chain, we prove that it mixes rapidly on many classes of degree sequences, and we discuss why it is sometimes better suited than the simple
switch chain, and try to easily introduce the concepts from the general theory along the way.
M.S. in Applied Mathematics, May 2016
Show less
This thesis focuses on exploring and solving several problems based on partiallyobserved diffusion models. The thesis has two parts....
Show moreThis thesis focuses on exploring and solving several problems based on partiallyobserved diffusion models. The thesis has two parts. In the first part we present a tractable sufficient
condition for the consistency of maximum likelihood estimators (MLEs) in partially observed diffusion models, stated in terms of stationary distributions of the associated test processes, under
the assumption that the set of unknown parameter values is finite. We illustrate the tractability of this sufficient condition by verifying it in the context of a latent price model of market
microstructure. Finally, we describe an algorithm for computing MLEs in partially observed diffusion models and test it on historical data to estimate the parameters of the latent price model. In
the second part we provide a thorough analysis of the particle filtering algorithm for estimating the conditional distribution in partially observed diffusion models. Specifically, we focus on
estimating the distribution of unobserved processes using observed data. The algorithm involves several steps and assumptions, which are described in detail. We also examine the convergence of
the algorithm and identify the sufficient conditions under which it converges. Finally, we derive an explicit upper bound of the convergence rate of the algorithm, which depends on the set of
parameters and the choice of time frequency. This bound provides a measure of the algorithm’s performance and can be used to optimize its parameters to achieve faster convergence.
Show less
Deep learning has revolutionized many machine learning tasks in recent years.Successful applications range from computer vision, natural...
Show moreDeep learning has revolutionized many machine learning tasks in recent years.Successful applications range from computer vision, natural language processing to speech recognition, etc.
The success is partially due to the availability of large amounts of data and fast growing computing resources (i.e., GPU and TPU), and partially due to the recent advances in deep learning
technology. Neural networks, in particular, have been successfully used to process regular data such as images and videos. However, for many applications with graph-structured data, due to the
irregular structure of graphs, many powerful operations in deep learning can not be readily applied. In recent years, there is a growing interest in extending deep learning to graphs. We first
propose graph convolutional networks (GCNs) for the task of classification or regression on time-varying graph signals, where the signal at each vertex is given as a time series. An important
element of the GCN design is filter design. We consider filtering signals in either the vertex (spatial) domain, or the frequency (spectral) domain. Two basic architectures are proposed. In the
spatial GCN architecture, the GCN uses a graph shift operator as the basic building block to incorporate the underlying graph structure into the convolution layer. The spatial filter directly
utilizes the graph connectivity information. It defines the filter to be a polynomial in the graph shift operator to obtain the convolved features that aggregate neighborhood information of each
node. In the spectral GCN architecture, a frequency filter is used instead. A graph Fourier transform operator or a graph wavelet transform operator first transforms the raw graph signal to the
spectral domain, then the spectral GCN uses the coe"cients from the graph Fourier transform or graph wavelet transform to compute the convolved features. The spectral filter is defined using the
graph’s spectral parameters. There are additional challenges to process time-varying graph signals as the signal value at each vertex changes over time. The GCNs are designed to recognize
di↵erent spatiotemporal patterns from high-dimensional data defined on a graph. The proposed models have been tested on simulation data and real data for graph signal classification and
regression. For the classification problem, we consider the power line outage identification problem using simulation data. The experiment results show that the proposed models can successfully
classify abnormal signal patterns and identify the outage location. For the regression problem, we use the New York city bike-sharing demand dataset to predict the station-level hourly demand.
The prediction accuracy is superior to other models. We next study graph neural network (GNN) models, which have been widely used for learning graph-structured data. Due to the
permutation-invariant requirement of graph learning tasks, a basic element in graph neural networks is the invariant and equivariant linear layers. Previous work by Maron et al. (2019) provided a
maximal collection of invariant and equivariant linear layers and a simple deep neural network model, called k-IGN, for graph data defined on k-tuples of nodes. It is shown that the expressive
power of k-IGN is equivalent to k-Weisfeiler-Lehman (WL) algorithm in graph isomorphism tests. However, the dimension of the invariant layer and equivariant layer is the k-th and 2k-th bell
numbers, respectively. Such high complexity makes it computationally infeasible for k-IGNs with k > 3. We show that a much smaller dimension for the linear layers is su"cient to achieve the same
expressive power. We provide two sets of orthogonal bases for the linear layers, each with only 3(2k & 1) & k basis elements. Based on these linear layers, we develop neural network models GNN-a
and GNN-b, and show that for the graph data defined on k-tuples of data, GNN-a and GNN-b achieve the expressive power of the k-WL algorithm and the (k + 1)-WL algorithm in graph isomorphism
tests, respectively. In molecular prediction tasks on benchmark datasets, we demonstrate that low-order neural network models consisting of the proposed linear layers achieve better performance
than other neural network models. In particular, order-2 GNN-b and order-3 GNN-a both have 3-WL expressive power, but use a much smaller basis and hence much less computation time than known
neural network models. Finally, we study generative neural network models for graphs. Generative models are often used in semi-supervised learning or unsupervised learning. We address two types
of generative tasks. In the first task, we try to generate a component of a large graph, such as predicting if a link exists between a pair of selected nodes, or predicting the label of a
selected node/edge. The encoder embeds the input graph to a latent vector space via vertex embedding, and the decoder uses the vertex embedding to compute the probability of a link or node label.
In the second task, we try to generate an entire graph. The encoder embeds each input graph to a point in the latent space. This is called graph embedding. The generative model then generates a
graph from a sampled point in the latent space. Di↵erent from the previous work, we use the proposed equivariant and invariant layers in the inference model for all tasks. The inference model is
used to learn vertex/graph embeddings and the generative model is used to learn the generative distributions. Experiments on benchmark datasets have been performed for a range of tasks, including
link prediction, node classification, and molecule generation. Experiment results show that the high expressive power of the inference model directly improves latent space embedding, and hence
the generated samples.
Show less
The increasing evidence has shown that having a sensitive detection method for Listeria monocytogenes in food products is critical for public...
Show moreThe increasing evidence has shown that having a sensitive detection method for Listeria monocytogenes in food products is critical for public health as well as industrial economics. L.
monocytogenes was associated with foodborne illness outbreaks linked to ice cream in the United States from 2010 to 2015, with another recent outbreak under investigation. The FDA Bacteriological
Analytical Manual (BAM) method was commonly used for L. monocytogenes detection. However, the performance characteristics of the chromogenic methods (MOX, RLM, and R&F agars) remain to be
elucidated. The factorial effect on Level of Detection (LOD) as an essential element of the International Organization for Standardization (ISO) approach for qualitative method validation was
investigated in this study.For examining the LOD of L. monocytogenes in ice cream, fractional contaminated samples were prepared with the ice cream obtained from the 2015 outbreak and enumerated
using the FDA BAM Most Probable Number (MPN) method for Listeria. The effect of test portion size was determined by comparing 10g and 25g using the BAM method with chromogenic agars (MOX, RLM,
and R&F). The ISO single-lab validation requirement was followed for the factorial effect study, including four different factors: sample size (10g and 25g), ice cream types (commercially
available regular vanilla ice cream and vanilla ice cream with low fat and no added sugar), re-freezing process (with re-freezing and without re-freezing process), and thawing process (slow thaw
and fast thaw). LOD and relative LOD (RLOD) were computed using MiBiVal software to compare the sensitivity of the three chromogenic agars and the different factors. For all of the detection
experiments, presumptive colonies were identified using the API listeria kit. The 2015 naturally contaminated ice cream was enumerated and resulted in an average contamination level of 2.15 MPN/
g. At fractional levels of 0.25 MPN/10g and 0.75 MPN/10g, the positive rates of L. monocytogenes detected from 10g and 25g of sample portions were consistent with the statistically theoretical
positive rates. The RLOD values for the reference method (MOX) and the alternative methods (RLM, R&F) were above 1 in both portion sizes, which suggested that MOX was slightly more sensitive than
RLM and R&F. The factorial effect study indicated that the four factors have no significant influence on the LOD of L. monocytogenes detection at the fractional contamination levels. However, the
test portion size of 25g provided more consistent results among the chromogenic media than the 10g portion size. Fat content was shown to have an effect on L. monocytogenes detection in a large
test portion. The information from this study will be useful for the improvement of the reproducibility of a qualitative detection method and can also be used for data analysis standards such as
ISO 16140 in method validation studies.
Show less
This thesis develops mathematical background for the design of algorithms for discrete-data problems, two in statistics and one in operations...
Show moreThis thesis develops mathematical background for the design of algorithms for discrete-data problems, two in statistics and one in operations research. Chapter 1 gives some background on
what chapters 2 to 4 have in common. It also defines some basic terminology that the other chapters use.Chapter 2 offers a general approach to modeling longitudinal network data, including
exponential random graph models (ERGMs), that vary according to certain discrete-time Markov chains (The abstract of chapter 2 borrows heavily from the abstract of Schwartz et al., 2021). It
connects conditional and Markovian exponential families, permutation- uniform Markov chains, various (temporal) ERGMs, and statistical considerations such as dyadic independence and
exchangeability. Markovian exponential families are explored in depth to prove that they and only they have exponential family finite sample distributions with the same parameter as that of the
transition probabilities. Many new statistical and algebraic properties of permutation-uniform Markov chains are derived. We introduce exponential random ?-multigraph models, motivated by our
result on replacing ? observations of a permutation-uniform Markov chain of graphs with a single observation of a corresponding multigraph. Our approach simplifies analysis of some network and
autoregressive models from the literature. Removing models’ temporal dependence but not interpretability permitted us to offer closed-form expressions for maximum likelihood estimators that
previously did not have closed-form expression available. Chapter 3 designs novel, exact, conditional tests of statistical goodness-of-fit for mixed membership stochastic block models (MMSBMs) of
networks, both directed and undirected. The tests employ a ?²-like statistic from which we define p-values for the general null hypothesis that the observed network’s distribution is in the MMSBM
as well as for the simple null hypothesis that the distribution is in the MMSBM with specified parameters. For both tests the alternative hypothesis is that the distribution is unconstrained, and
they both assume we have observed the block assignments. As exact tests that avoid asymptotic arguments, they are suitable for both small and large networks. Further we provide and analyze a
Monte Carlo algorithm to compute the p-value for the simple null hypothesis. In addition to our rigorous results, simulations demonstrate the validity of the test and the convergence of the
algorithm. As a conditional test, it requires the algorithm sample the fiber of a sufficient statistic. In contrast to the Markov chain Monte Carlo samplers common in the literature, our
algorithm is an exact simulation, so it is faster, more accurate, and easier to implement. Computing the p-value for the general null hypothesis remains an open problem because it depends on an
intractable optimization problem. We discuss the two schools of thought evident in the literature on how to deal with such problems, and we recommend a future research program to bridge the gap
those two schools. Chapter 4 investigates an auctioneer’s revenue maximization problem in combinatorial auctions. In combinatorial auctions bidders express demand for discrete packages of
multiple units of multiple, indivisible goods. The auctioneer’s NP-complete winner determination problem (WDP) is to fit these packages together within the available supply to maximize the bids’
sum. To shorten the path practitioners traverse from from legalese auction rules to computer code, we offer a new wdp formalism to reflect how government auctioneers sell billions of dollars of
radio-spectrum licenses in combinatorial auctions today. It models common tie-breaking rules by maximizing a sum of bid vectors lexicographically. After a novel pre-solving technique based on
package bids’ marginal values, we develop an algorithm for the WDP. In developing the algorithm’s branch-and-bound part adapted to lexicographic maximization, we discover a partial explanation of
why classical WDP has been successful in using the linear programming relaxation: it equals the Lagrangian dual. We adapt the relaxation to lexicographic maximization. The algorithm’s
dynamic-programming part retrieves already computed partial solutions from a novel data structure suited specifically to our WDP formalism. Finally we show that the data structure can “warm
start” a popular algorithm for solving for opportunity-cost prices.
Show less | {"url":"https://repository.iit.edu/islandora/search?type=dismax&f%5B0%5D=mods_subject_topic_ms%3AStatistics","timestamp":"2024-11-06T05:34:48Z","content_type":"text/html","content_length":"140350","record_id":"<urn:uuid:f8fe5ff7-60a7-4426-abde-7a7f5fea996f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00513.warc.gz"} |
density_matrix_infidelity | Quantum information | Nodes | Graphs | Boulder Opal | References | Q-CTRL Documentation
Graph.density_matrix_infidelity(x, y, *, name=None)
Calculate the infidelity between two states represented by density matrices.
• x (np.ndarray or Tensor) – The density matrix $x$ with shape (..., D, D). The last two dimensions must have the same size for both matrices, and its batch dimensions must be broadcastable.
• y (np.ndarray or Tensor) – The density matrix $y$ with shape (..., D, D). The last two dimensions must have the same size for both matrices, and its batch dimensions must be broadcastable.
• name (str or None , optional) – The name of the node.
The state infidelity of the density matrices with respect to each other. Its shape is the broadcasted value of the batch shapes of the two input parameters.
This function assumes that the parameters are density matrices and therefore are positive definite. Passing matrices that have negative or complex eigenvalues will result in wrong values for the
Graph.infidelity_pwc : Total infidelity of a system with a piecewise-constant Hamiltonian.
Graph.infidelity_stf : Total infidelity of a system with a sampleable Hamiltonian.
Graph.state_infidelity : Infidelity between two quantum states.
Graph.unitary_infidelity : Infidelity between a unitary and target operators.
The general formula for the infidelity of two density matrices is
$I = 1 - \left[ \mathrm{Tr}\left( \sqrt{\sqrt{x} y \sqrt{x}} \right) \right]^2$
>>> infidelity = graph.density_matrix_infidelity(
... np.array([[0.5, 0], [0, 0.5]]),
... np.array([[1, 0], [0, 0]]),
... name="infidelity",
... )
>>> result = bo.execute_graph(graph=graph, output_node_names="infidelity")
>>> result["output"]["infidelity"]["value"] | {"url":"https://docs.q-ctrl.com/references/boulder-opal/boulderopal/graph/Graph/density_matrix_infidelity","timestamp":"2024-11-13T09:41:53Z","content_type":"text/html","content_length":"73492","record_id":"<urn:uuid:f2f1c466-9b1a-4610-87c5-7118710533a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00632.warc.gz"} |
HW1 PageRank Solved - Programming Help
In this assignment, you will use the python programming to implement variations of the classic PageRank algorithm. You will use an adjacency matrix to represent edges and compute the PageRank scores
of the nodes.
PageRank is an algorithm developed by Sergey Brin and Larry Page that built the early foundation of the Google search engine. We will use a simplified version.
The intuition behind PageRank is that the web can be represented as a graph, with links between pages as directed edges. The importance or “popularity” of a page on the web can be determined by how
many other pages link to it. But should every link be treated equally? A link from a popular site like Google’s homepage should be more important than a link from an unpopular blog. We want to weigh
links from more important pages more than edges links from less important edges when computing the PageRank score of a page. We need PageRank values to determine other PageRank values, and since all
nodes depend on each other and the graph isn’t necessarily acyclic, this becomes a chicken and egg problem!
• Problem 3: 30 points (10 points for each subquestion)
Debugging your program using python unit tests (nosetests)
You can use the provided unit tests to debug your code. For example, to debug the code of your solution to problem 3, type:
nosetests test3.py
If the code is correct, you will pass the unit tests.
How to submit?
Please submit problem[1-6].py (six python files) in Canvas system under “HW1 assignment”. | {"url":"https://www.edulissy.org/product/in-this-assignment-you-will-use-the-python-programming-to-implement-variations-of-the-classic-pagerank-algorithm/","timestamp":"2024-11-09T10:33:37Z","content_type":"text/html","content_length":"172024","record_id":"<urn:uuid:beae325e-c19e-4634-a484-7d61d60ec8cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00593.warc.gz"} |
StudyQA — Master: Mathematics — University of Copenhagen
Mathematics is the science of structures in a broad sense. They may be numerical structures, spatial structures, colour structures, musical structures, logical structures or a thousand other things.
Mathematicians study these structures: they form them, stretch them, bend them, play with them and form connections between them.
When you start looking, you find structures everywhere. And mathematics is the language that we use to express many of our deepest thoughts about the world. Mathematical applications can be used as a
way to get inside the structures. Or you can become engrossed in the abstract game. You're allowed - in fact, you're encouraged! - to use both approaches as a graduate student in mathematics.
If you are interested in the way Mathematics is shared and learned in various institutions, you can also study Didactics of Mathematics at the University of Copenhagen.
Profile and Career
Mathematics is an exact science, and a mathematical theorem is not accepted until a stringent proof has been produced for it. Acquiring the requisite precision is demanding, and continual practice is
Before a mathematician can produce a stringent proof, he or she must undergo a creative process in order to achieve an understanding of what the theorem is about and how it can be proved. At this
stage, the mathematician draws especially on imagination and experience.
In some cases, computers can also be used to develop ideas about what is right or wrong, but only in rare cases can they be used to prove a theorem.
Mathematics is also communication. Mathematicians talk about things outside the experience of daily life, and a strong talent for storytelling is needed to make it comprehensible. Presentation is a
high priority among mathematicians, and a key aspect of the programme.
Competence Description
A graduate in mathematics will be able to:
• Conduct independent, stringent argumentation
• Structure a study of open mathematical issues
• Define mathematical disciplines in relation to each other, but also to take advantage of interdisciplinary skills.
• Independently take responsibility for his or her own professional development and specialisation.
A graduate in mathematics will also have acquired the skills to:
• Read and understand original mathematical literature
• Convey and communicate mathematical issues and problems on a scientific basis
• Explain, orally and in writing, mathematical studies of open issues.
Career Opportunities
MSc graduates in mathematics have many different job opportunities, and there is basically no unemployment. Many graduates find employment in the private sector, where they either work specifically
with applying mathematics to specialised problems, for instance within economics or telecommunications, or act as "trouble shooters" in a broader sense.
When we ask employers why they hire our graduates, they often emphasise the mathematician's abilities to see patterns in problems that often arise and to solve them once and for all.
There is also a high demand for mathematics teachers in upper secondary education and other post compulsory education programmes. Moreover, there is the opportunity to continue conducting research in
mathematics after earning a PhD.
You can choose from a wide variety of courses to design a course of study that adresses issues that arouse your curiosity. Mathematics has many disciplines: algebra, analysis, geometry, topology...
It also includes application-oriented disciplines such as mathematical physics, probability theory and optimisation, as well as cultural disciplines like the history of mathematics and the didactics
of mathematics.
Although these disciplines are studied independently, it is remarkable how closely related they are when you take a closer look. Mathematics as a whole can be approached in many ways.
In addition to the elective courses of study, a number of recommended courses of study have been designed for those with teaching or industrial ambitions.
If you are interested in the way Mathematics is shared and learned in various institutions, you can also study Didactics of Mathematics at the University of Copenhagen.
Master's thesis
Your studies will conclude with a thesis project in which you work in more detail within one of the themes that you focused on in your course of study. Often the thesis will be written in association
with one of the Department's research teams, where you have access to both a supervisor and the entire team's knowledge and involvement. In other cases, you might instead be investigating something
entirely new. In recent years, solutions have been found to profound problems that have existed for hundreds of years or more (Poincaré's conjecture and Fermat's last theorem), and the enthusiasm
that accompanies these new breakthroughs has led to many thesis topics.
Examples of thesis topics:
• Algebra and the theory of numbers
• Geometric analysis and mathematical physics
• Noncommutative geometry
• Topology
• Didactics of mathematics
Study Abroad
It is also possible to study abroad during your degree. You can choose to study abroad for one or two semesters or for a shorter period of time; e.g. take a summer school course.
(1) Applicants with a Bachelor's degree in Mathematics from the University of Copenhagen may be admitted to the MSc in Mathematics.(2) Applicants with a Bachelor's degree in Mathematics from other
universities in Denmark or the Nordic Region may be admitted to the Master's programme in Mathematics.(3) Applicants with a Bachelor's degree in actuarial mathematics, statistics or
mathematics-economics may be admitted if their programme included courses in:Linear algebra, corresponding to min. 7.5 ECTS creditsGeometry, corresponding to min. 7.5 ECTS credits.(4) Applicants with
a Bachelor's degree in Natural Science may be admitted if their BSc incorporates min. 67.5 ECTS credits from courses in the following mathematical subject areas:Mathematical analysis, corresponding
to min. 30 ECTS creditsLinear algebra and algebra, corresponding to min. 22.5 ECTS creditsGeometry and topology, corresponding to min. 15 ECTS credits.(5) In addition, the Faculty may admit
applicants who, after a thorough assessment, are deemed to possess educational qualifications equivalent to those required in subsections (1) (4).Prioritisation criteriaApplicants with Bachelor
degrees in Mathematics from the University of Copenhagen are guaranteed admission on the first MSc intake after graduation.Second priority will be accorded to applicants with Bachelor degrees in
mathematics from other universities in Denmark or the Nordic Region. After that, priority will be accorded to other applicants with a BSc in actuarial mathematics, statistics or
mathematics-economics, as per (3) and then applicants with a Bachelor's degree in science as per (4).Language RequirementsTo gain admission to an MSc in the English language, non-Danish applicants
must document qualifications on par with the Danish secondary school English level B'. The Faculty of Science accepts the following 3 ways of documenting this:English is your native language. The
Faculty of Science at the University of Copenhagen, accepts The University of Purdue view that citizens of the following countries are exempted from taking an English Language Proficiency exam:
Anguilla, Antigua, Australia, Bahamas, Barbados, Barbuda, Belize, British Virgin Islands, Canada (except Quebec), Dominica, Grand Cayman Islands, Grenada, Guyana, Irish Republic, Jamaica, Montserrat,
New Zealand, St. Kitts & Nevis, St. Lucia, St. Vincent & the Grenadines, Trinidad & Tobago, Turks & Caicos Islands, United Kingdom: England, Northern Ireland, Scotland, & Wales and United States of
AmericaIf you are citizen of one of the above countries, you are not required to submit any proof of English proficiency. Your copy of your passport will suffice as sufficient proof.Prior studies
completed in the English language/in an English speaking country. For example, if you have studied your Bachelor degree in England you are not required to complete an English language proficiency
exam. We ask such students to provide a signed statement from the Educational institution (with the institution's stamp on it) stating that English is the main language of instruction. Furthermore,
applicants from Nordic countries (Denmark, Sweden, Norway, Finland and Iceland) do not need to provide proof of English language proficiency.Applicants with English as their second language (except
Scandinavians) must pass an IELTS, TOEFL or Cambridge Advanced English test before being admitted. The Faculty of Science, University of Copenhagen, accepts the following tests and scores:IELTS-test
(British Council) with a minimum score of 6.5Computer-based TOEFL-test with a minimum score of 213 pointsPaper-based TOEFL-test with a minimum score of 560 pointsInternet-based TOEFL-test with a min
score of 83 pointsLanguage tests older than 2 years are not accepted (from the application deadline) English Language Requirements IELTS band: 6.5 TOEFL paper-based test score : 560 TOEFL iBT® test:
The Faculty of Science will not be awarding any scholarships for the academic year 2014/2015.
Please note that many scholarships are offered by companies or organisations, it can be worthwhile to research your particular options from your home country.
Similar programs: | {"url":"https://studyqa.com/program/mathematics-master-in-university-of-copenhagen","timestamp":"2024-11-09T22:37:32Z","content_type":"text/html","content_length":"241671","record_id":"<urn:uuid:80c35720-4a6e-44b6-bdb3-48aaacab09f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00095.warc.gz"} |
what comes after 999
How will understanding of attitudes and predisposition enhance teaching? Grades K-8 Worksheets. How do you put grass into a personification? I thought it was 1000 but then if it is what comes before
1100.So confused. any number of thousands, from 1 to 999, e.g. 999 trillion and one, to be exact. How long was Margaret Thatcher Prime Minister? Copyright © 2020 Multiply Media, LLC. What are the
disadvantages of primary group? Why is melted paraffin was allowed to drop a certain height and not just rub over the skin? There are a bunch more, if you go to the site listed below you can see all
of them. A bit trivial I … Many numbers come after 999 but the number that comes directly after it is 1,000. Place Value: Numbers to 999. Reply Reply Author. All Rights Reserved. Thread Starter. What
is the contribution of candido bartolome to gymnastics? All Rights Reserved. What is the conflict of the story of sinigang? Quadrillion - and of course any number followed by twelve zeros. Looking
for high-quality Math worksheets aligned to Common Core standards for Grades K-8? Topic closed. Who is the longest reigning WWE Champion of all time? 999 + 1 = 1000 Therefore, 999 comes before 1000.
What is the reflection of the story the mats by francisco arcellana? How long does a fresh turkey last in the refrigerator? When did organ music become associated with baseball? what follows 999. For
larger n (between 10 and 999), prefixes can be constructed based on a system described by Conway and Guy. 10,000,000,000 = 9,999,999,999+1 or simply explained and probably the most common answer, in
base 10 math in the American English, 10 billion comes after nine billion nine hundred ninety nine million nine hundred ninety nine thousand nine hundred ninety nine. How will understanding of
attitudes and predisposition enhance teaching? Page 1 of 1: Print E-mail Link: LILBIGMOMMA. How do you put grass into a personification? What is the reflection of the story the mats by francisco
arcellana? What comes just after 9547 999 Get the answers you need, now! I'm bad at numbers. What comes after the Porsche 999? How long will the footprints on the moon last? unfortunately this number
cant be decide because the zero is repeating! Last post 17 years ago by OldHank. How long does a fresh turkey last in the refrigerator? t1grm. What is the conflict of the story of sinigang? How long
will the footprints on the moon last? Tuesday 22nd July 2003. its just a real number, and its the absolute next number after 999 billion. Who is the longest reigning WWE Champion of all time? What
number comes before one trillion? What comes before 1000? Quadrillion comes after trillion. quintillion. If you add one to a trillion, the number becomes a trillion and one, a trillion and two, etc.
Why is melted paraffin was allowed to drop a certain height and not just rub over the skin? Copyright © 2020 Multiply Media, LLC. 999,999,999,999, 999 billion. United States Member #1023 January 11,
… 999 = 10 5994 (long scale) may be named. Our premium worksheet bundles contain 10 activities and answer key to challenge your students and help them understand each … It is not accurate, but I
suppose you are asking what comes after trillions, and the answer is a quadrillions. Is there a way to search all eBay sites for different countries at once? Why don't libraries smell like
bookstores? Infinity itself isn't a number in the conventional sence therefore not only is there everything before it there is nothing Ie. When did organ music become associated with baseball? The
material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. What is 999999999999999999999999999999
9999999999999999999999999999999999999999999999999? Is it ok to eat a frozen turkey with black spots on it? Is there a way to search all eBay sites for different countries at once? What are the
disadvantages of primary group? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Really bad (so
don't mention it). Thread Starter. That’s in short scale. 3 replies. there is a bar on top of the zero, so its 999 billion with zero repeating and when done repeating there is a one at the end. Is it
ok to eat a frozen turkey with black spots on it? But I think you want to know the order of the larger numbers, so next is Quadrillion. 4,621 posts. It's over 999 docilion, 999 … What is the number
after 999 quadrillion? Original Poster. 999 - followers: 987 or 901 usually falls in less than 15 days. What is the contribution of candido bartolome to gymnastics? Adding another quadrillion to 999
quadrillion gives 1 Discussion. Why don't libraries smell like bookstores? How long was Margaret Thatcher Prime Minister? what number comes after 999 trillion. ATLANTA, GA United States Member #78956
August 24, 2009 5292 Posts Offline. WILLWRK4NUMBERS. Thanks 243 months. The choice of roots and the concatenation procedure is that of the standard dictionary numbers if n is 9 or smaller. 1000 comes
after 999 What comes after a trillion?
How To Spawn Animals In Minecraft Command Block
Image Formed By Convex Mirror Table
Types Of Clothing Styles With Pictures
Augmented Reality Vs Virtual Reality
Poole Pottery Collectors Guide | {"url":"http://primeapps.com/review/viewtopic.php?id=85665e-what-comes-after-999","timestamp":"2024-11-03T17:24:39Z","content_type":"text/html","content_length":"16898","record_id":"<urn:uuid:be8c1d5b-61f6-4240-b1c3-56a20682afbc>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00397.warc.gz"} |
925 kilometers per square second to millimeters per square second
4,925 Kilometers per square second = 4,925,000,000 Millimeters per square second
Acceleration Converter - Kilometers per square second to millimeters per square second - 4,925 millimeters per square second to kilometers per square second
This conversion of 4,925 kilometers per square second to millimeters per square second has been calculated by multiplying 4,925 kilometers per square second by 1,000,000 and the result is
4,925,000,000 millimeters per square second. | {"url":"https://unitconverter.io/kilometers-per-square-second/millimiters-per-square-second/4925","timestamp":"2024-11-14T22:16:43Z","content_type":"text/html","content_length":"27309","record_id":"<urn:uuid:888e73c8-e449-49cb-8257-5bb08394cee0>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00011.warc.gz"} |
Math Exam Resources/Courses/MATH152/April 2022/Question B5 (b)/Solution 1
We are given ${\textstyle C_{2}=3F}$ in the question statement, so what remains is to write ${\textstyle J_{2}}$ in terms of ${\textstyle V_{1}}$, ${\textstyle V_{2}}$, and ${\textstyle I}$.
First use Kirchoff’s current law on the upper middle junction: ${\textstyle J_{1}=J_{2}+I}$. Rearranging gives ${\textstyle J_{2}=J_{1}-I}$.
Now, to determine ${\textstyle J_{1}}$, we use Kirchoff’s voltage law on the left loop: ${\textstyle -V_{1}+J_{1}\cdot 1-V_{2}=0}$. Rearranging this and substituting it back into the equation for ${\
textstyle J_{2}}$ gives ${\textstyle J_{2}=V_{1}+V_{2}-I}$.
Thus, the second differential equation is | {"url":"https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH152/April_2022/Question_B5_(b)/Solution_1","timestamp":"2024-11-15T00:39:09Z","content_type":"text/html","content_length":"34499","record_id":"<urn:uuid:9f266f18-6d97-43c3-a418-f65a01c7f82b>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00262.warc.gz"} |
Download Graphing Calculator + Math PRO APK 2023.07.165 for Android
Graphing Calculator + Math PRO APK 2023.07.165
Name Graphing Calculator + Math PRO APK
Publisher Mathlab Apps, LLC
Version 2023.07.165
Size 9M
Category Education
MOD Features N/A
Support Android 6.0+
Get it on Google Play
Introduce MOD APK Graphing Calculator + Math PRO
Graphing Calculator + Math PRO MOD APK will be an essential and valuable tool for those who are passionate about math and require a professional graphing calculator. A program that can help users
easily graph complex problems quickly and accurately thanks to its graphing capabilities. The advanced math tools provided by Graphing Calculator + Math PRO go beyond the tools of standard graphing
calculators, including the ability to calculate derivatives, integrals, solve equations and systems of equations, and calculate integrals: Numbers and more. Users can now solve complex math problems
that require multi-step operations faster and with less effort.
This will be a valuable tool for students, teachers and people working in mathematics who often have to work with complex calculations. Everything a user needs to perform mathematical calculations
accurately and efficiently is provided by Graphing Calculator + Math PRO.
Download Graphing Calculator + Math PRO MOD APK – Multifunctional Math Calculator
A smartphone app, Graphing Calculator + Math PRO, helps users solve complex problems and create graphs. This application provides many powerful tools and high-speed calculation capabilities, ensuring
that it can meet all users’ needs for solving mathematical calculations from basic to advanced. Quickly and easily answering algebra, geometry and system of equations, problems are just some of the
most outstanding things this app offers its users; its potential includes functions that solve equations, calculate derivatives and integrals, and work with complex numbers. In addition, the ability
to draw the map of this program is not even discussed when it is possible to draw accurate and detailed graphs and customize the appearance.
Solve complex math problems
Users can use the Graphing Calculator + Math PRO program to perform complex algebra, geometry and trigonometry calculations. Thanks to its fast and accurate calculation ability, it will be a valuable
assistant in solving complex problems, allowing it to handle multivariable equations, derivatives, integrals, and many other complex mathematical operations. At the same time, it supports users to
perform multiplication, division, addition, subtraction and many basic algebra operations in just one note. In addition, Graphing Calculator + Math PRO provides approximations and solutions to help
users solve complex multivariable equations. So calculating and solving mathematical calculations will no longer be as tricky or laborious as before when users equip this program on their devices.
Graph complex functions and customize them
In addition to having intelligent and reliable calculations, Graphing Calculator + Math PRO is a valuable resource for problems requiring graphing. Users of this application can graph various types
of complex functions, such as biomechanical, tangent, exponential, and trigonometric functions. But first, the user’s task is to enter complex functions using ordinary mathematical syntax and complex
digital symbols. Then, the function will be automatically graphed and displayed visually on the device’s screen by the program. To draw attention to the critical components of the graph that Graphing
Calculator + Math PRO MOD APK provides, users can also change several properties, including colour, thickness, and line style.
Save calculation results and function graphs
The Graphing Calculator + Math PRO program makes it easy for users to store and share information by allowing users to save calculation results and function graphs in memory or share them with
others. With the calculations and function graphs performed, users can completely store them right on the device’s SD memory and then review them whenever they need them in the future. This
application will classify the saved data into a separate folder and organize them clearly and neatly to ensure that the user’s search is always as fast as possible. In addition, Graphing Calculator +
Math PRO also helps users to share their calculations and graphs with other people or specific applications very quickly; with just a few taps, they can immediately transition.
Graphing Calculator + Math PRO MOD APK is a user-friendly, robust application for solving complex calculations and graphing functions while providing powerful, reliable computing capabilities in a
short time.
How to Download & Install Graphing Calculator + Math PRO APK for Android
Please login to comment
0 Comments
Inline Feedbacks
View all comments
| Reply | {"url":"https://gamedva.com/graphing-calculator-math-pro","timestamp":"2024-11-14T04:19:45Z","content_type":"text/html","content_length":"85256","record_id":"<urn:uuid:476fe637-e823-4418-b497-650e07cd522b>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00693.warc.gz"} |
What is
What is base 21 called?
Base 21 or unovigesimal numeral system is based on twenty-one. The twenty one symbols used are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I, J and K. Plural name is base-21.
What are the 3 bases in a relationship?
First Base – Kissing. Second Base – Heavy Petting/Manual Stimulation. Third Base – Oral Stimulation. Home Run – Penetrative Sex.
What is the formula in finding the rate?
However, it’s easier to use a handy formula: rate equals distance divided by time: r = d/t.
How do you convert base numbers?
The general steps for converting a base 10 or “normal” number into another base are:
1. First, divide the number by the base to get the remainder.
2. Then repeat the process by dividing the quotient of step 1, by the new base.
3. Repeat this process until your quotient becomes less than the base.
How many types of number systems are there?
The four most common number system types are:
• Decimal number system (Base- 10)
• Binary number system (Base- 2)
• Octal number system (Base-8)
• Hexadecimal number system (Base- 16)
What is the formula for base?
Formula for geometrical figures
Perimeter formula
Square 4 × side
Rectangle length × width
Parallelogram base × height
Triangle base × height / 2
How do you convert numbers?
Step 1 − Divide the decimal number to be converted by the value of the new base. Step 2 − Get the remainder from Step 1 as the rightmost digit (least significant digit) of new base number. Step 3 −
Divide the quotient of the previous divide by the new base.
What is a base shape?
more The surface a solid object stands on, or the bottom line of a shape such as a triangle or rectangle. But the top is also called a base when it is parallel to the bottom!
Is DNA a base 4?
Summary: For decades, scientists have known that DNA consists of four basic units — adenine, guanine, thymine and cytosine. Those four bases have been taught in science textbooks and have formed the
basis of the growing knowledge regarding how genes code for life.
What is 1st 2nd and 3rd base in a relationship?
Some people only consider French kissing as getting to first base. Second base is direct physical contact, usually meaning his hands to her breast. It also includes other forms of petting, touching
and groping. Third base may include manual or oral sex for either partner.
Which symbols are used in octal number system?
The octal number system says that it is a number system of base 8 which means that we require 8 different symbols in order to represent any number in octal system. The symbols are 0, 1, 2, 3, 4, 5,
6, and 7. The smallest two digit number in this system is (10)8 which is equivalent to decimal 8.
What is number base?
A number base is the number of digits or combination of digits that a system of counting uses to represent numbers. A base can be any whole number greater than 0. For example, 178 is read as 17 base
8, which is 15 in base 10. Example 7.2.
What is the application of binary number system?
Binary system is used for representing binary quantities which can be represented by any device that has only two operating states or possible conditions. For example, a switch has only two states:
open or close. In the Binary System, there are only two symbols or possible digit values, i.e., 0 and 1. | {"url":"https://www.presenternet.com/what-is-base-21-called/","timestamp":"2024-11-02T05:08:58Z","content_type":"text/html","content_length":"40632","record_id":"<urn:uuid:0d53c9fb-7ace-4f36-bfcd-e2c1b9defee0>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00814.warc.gz"} |
Representing transverse waves | Oak National Academy
Hello, my name is Mr. Fairhurst, and this lesson is about representing transverse waves.
In this lesson, what we're aiming for is at the end of the lesson for you to be able to interpret and sketch what we call displacement-time graphs of transverse waves.
These are different from the displacement-distance graphs and we need to think about a few other things first before we get to them.
During this unit, we'll be using a number of keywords.
We'll be looking at what we mean by wave medium and we'll be using this idea that we've seen before about displacement and applying that to new situations, and we'll be thinking about the period of a
wave and the frequency of a wave, which are both key terms that apply to the speed or the rate of vibration in the wave.
We'll start by looking at what we mean by frequency and period.
These are the terms related to the rate of vibration of the wave, how quickly the wave oscillates up and down, if you like.
And once we've done that, we can look at how we calculate the frequency and period of a wave, and then we're going to use those ideas together to represent transverse waves as the displacement-time
graphs, which is a new way of displaying these graphs.
So first of all, let's have a look at frequency and period.
Let's just remind ourselves what we mean by a transverse wave.
In this picture, we've got a transverse wave moving forwards from left to right along the rope, and it's a transverse wave because the rope's been shaken at right angles to the direction the wave is
In this case, it's been shaken up and down.
Now, we call the rope the wave medium because the wave is moving through the rope, and without the rope there wouldn't be a wave.
In this example, the wave medium is the water because water is what the wave is moving forwards through, and the water as the wave moves forward is oscillating or vibrating up and down at right
angles to the direction that the wave is moving.
So let's think about this wave moving through the water.
The frequency of the wave is the number of waves passing a point each second.
And frequency is measured in units called hertz, or Hz for short.
And a frequency of one hertz, as this wave is shown here, means that the source is producing one wave every second, or one complete oscillation in each second.
In this ripple tank, the waves being made by the bar vibrating up and down is moving at right angles to the direction that the wave is moving.
It's making the water move up and down, but the wave is moving forwards away from it.
So that's a transverse wave.
And in this instant it's making eight waves each second.
That means that the frequency of the wave is eight hertz.
So let's have a look at an example of how we can work out the frequency.
A ripple tank makes 24 waves in four seconds.
So what's the frequency of the wave? Well, the frequency is the number of waves in each second.
So what we need to do is to divide the total number of waves, 24, by the number of seconds to find out how many in each second.
So the sum we need to do is 24 divided by four, which will give us six hertz.
Have a look at this example and have a go at this yourself.
Just pause the video and have a go.
So how did you get on? In this case, we had 36 waves in total over a period of nine seconds.
What's the frequency? How many waves in one second? All we need to do is to divide the 36 by the number of seconds that we've got and we have this sum, 36 divided by nine, which gives us four hertz.
And we can think about this backwards, if one wave takes four, so if the frequency is four hertz, that's four ways in every second, we've got nine seconds.
So nine lots of four, that will give us 36 waves in total.
So it works both ways.
What about this example? This one just gives us a more awkward answer and maybe it takes a little bit longer to think about.
A ripple tank makes two waves in four seconds.
What's the frequency of the wave? It's exactly the same as before.
We want to know how many waves in one second.
So you've got two waves divided by the four seconds, and that gives us half a wave in each second.
So if we count four seconds, half a wave, one wave, one and a half waves, two waves, that's taken us four seconds to get a total of two waves.
So even though the frequency is less than one, it's still the right answer.
Have a go at this example yourself and just pause the video whilst you do so.
So how did you get on? Following the same process, we want to know how many waves in one second? Which is the frequency of the wave.
We've got six waves in total over a period of 24 seconds.
So in one second, we want six divided by 24.
You can perhaps do this on your calculator.
60 divided by 24 gives us a quarter, or 0.
25 hertz.
So as well as the frequency, we need to think about the period of the wave.
This is a time taken for one complete oscillation.
So if you like, it's almost the opposite of frequency.
On this rope, two waves have been made every second.
So on there you can see two complete waves in a period of one second.
Those waves moving forwards on the rope.
And that means that each wave takes half a second to be made.
The frequency is two because it's two waves per second, but the period, the time for one wave is half a second.
So in other words, the period is 0.
5 seconds.
The symbol for period is a capital T.
So it's a bit like time but it's a particular time, so we give it a capital letter, and that's 0.
5 seconds.
On this rope, there's a fixed marker and the waves moving forward past the marker, and we've counted four ways passing that marker in every second.
That means that each separate wave passes in a quarter of a second.
So a quarter, a half, three quarters, one, four waves over a second, but each wave taking a quarter of a second to pass that point.
That means that the period of the wave is a quarter of a second, or 0.
25 seconds.
So let's have a look at an example.
Two waves are made on the rope in eight seconds.
What's the period of the wave? Well, this time we need to do the sum back the other way around than when we did for frequency.
We want to know how long one wave takes.
We've had eight seconds in total and it took eight seconds to make two waves, or for two waves to pass.
So that's four seconds per wave, and the sum we did was eight seconds divided by the number of waves, which is four seconds per wave, or the period is four seconds.
And we can put the T instead of writing the word period.
Have a look at this example yourself and see what you can work out as the period.
Just pause the video again whilst you do so.
Okay, what did you get? The time taken was two seconds and we had eight waves.
So how long does it take for each wave? This time we divide the time, two seconds, by eight to get the time for each wave.
And two divided by eight gives us 0.
25 seconds, or a quarter of a second.
So the time period is 0.
25 seconds.
What about this example? Eight waves pass a fixed marker in four seconds.
What's the period? So the waves moving forward pass the marker.
We've counted eight in four seconds.
So the time for each wave to pass is four seconds divided by eight, which gives us 0.
5 seconds.
So the period again is 0.
5 seconds.
And finally, for now, just try this example and again, pause the video whilst you do so.
Okay, so this time we've got two waves passing a marker in 10 seconds.
So the time for each wave to pass is the total time, 10 seconds, divided by the number of waves passing, which is 10 divided by two.
And that gives us a time of five seconds, which is the period for one wave.
Okay, here's a set of questions for you to practise to see if you've understood what we've talked about in this section of the lesson.
Just pause the video whilst you have a go at those.
Okay, so let's have a look at the answers.
First of all, a wave's got a frequency of five hertz.
How many waves does that make in one second? Well, it's five because the frequency means how many waves in one second.
How many waves in five seconds? Well, five in one second times five gives us 25 waves.
And the period of the wave is the time it takes for one wave.
We've got five waves in one second, so we can do one second divided by five, which gives us 0.
2 seconds for the period.
You might notice it also looks exactly the same way, if you've got 25 waves in five seconds, we could do five divided by 25 and we get exactly the same time period because each wave is taking the
same amount of time each time.
Okay, what about part b? A wave has now got a period of 0.
1 second.
How many waves in one second? Well, the number of waves is one second divided by 0.
1, which is the time for one wave, which gives us 10 waves.
You might also think about this as if the time period of the wave is a 10th of a second, that's 10 per second.
And the frequency, as we said, it's 10 waves in one second, the frequency will also be 10, this time 10 hertz.
You must use the correct units for frequency.
And part c now, one water wave passes the marker every two seconds, what's the frequency? Now, the most common mistake here is to say two because you're not thinking it through properly, but
hopefully you have done one wave in two seconds.
The frequency is the number of waves one divided by the number of seconds two, which gives us half a hertz, half a wave every second.
The period of the wave, if it's one wave every two seconds, that's the period, it's one wave in two seconds.
So the period is two seconds.
So hopefully you got most of those right, if not all of them, and well done if you did.
In this part of the lesson, we'll be looking at how we can calculate the frequency and period now that we understand what the two terms mean.
So we've just seen in the last part of the lesson how we can work out the frequency and period of a wave.
It turns out that the frequency and period of the wave are connected in a different way as well.
And that way is that if we, you may have spotted this, the shorter the period of the wave, the higher its frequency.
And it turns out that if we half the period of the wave, the frequency is twice as high.
So say the period of the wave was half a second, the number of waves in one second gives us two.
So the frequency is two.
If we make the period of the wave shorter, maybe make it a quarter of a second, that means we've got four waves per second, and that means that the frequency is now four hertz.
So what we've just seen there is if we half the period of the wave, we double the frequency and we can put that together in an equation.
And the equation, if you like, it's just a shorthand way of saying that.
And what we've said is that the frequency is one divided by the period.
Let's just put those numbers into the same equation and see what we get.
So if I remember right, we said that the period that the wave starts is half a second.
So one divided by half a second on the right of this equation, one divided by a half gives us two, which was the frequency.
If we half the period to make it a quarter of a second, how many quarters in one? It's one divided by a quarter.
How many quarters in the whole one is four, the frequency is equal to four.
So if we half the period, the frequency is doubled, and this equation is a shorthand way of saying that.
And we can make it even shorter by using symbols for frequency and for period.
What this relationship is showing is that the frequency is inversely proportional to the period of the wave.
Inversely proportional means that if one thing doubles, the other value halves.
If one triples, the other will be a third as big, and so on.
We can see that if we try a few examples.
So here's the first example.
What happens if the period of the wave is doubled to the frequency? Well, I've illustrated that the time period of the wave is doubling by making the T symbol for period twice as big.
And the one divided by something that's bigger gives you a smaller fraction.
So that means that the frequency would be smaller, and in this case, if the period is doubled, the frequency is halved.
Have a look at this example and see if you can work out what happens to the frequency.
Pause the video if you need a little bit of time to do this.
Okay, so in this case, I've redrawn, rewritten the equation, but this time with the time period four times smaller, which seems to suggest that the right hand side of this equation is much bigger and
the frequency must therefore be bigger.
So if the period is four times smaller, the frequency is four times higher, it's four times greater than it was before.
Let's look at this other example, but this time we're changing the frequency.
What happens if we double the frequency? What happens to the period? Well, if the left hand side, the frequency is twice as big, the right hand side of this equation needs to be twice as big, and the
way to do that is to make the period smaller.
So one divided by something smaller will give you a bigger value, and in this case the period needs to be twice as small.
It needs to be halved.
Have a go at this equation, or this example and see what you think.
Just pause the video again if you need to.
So how did you get on? Once again I've written out the equation, and this time I've made the frequency half the size, 'cause the frequency has been halved.
And if we think about what we need to do to the right hand side to make that half as big as well, we need to double the size of the period, the big T there.
So we need to double the period in order to half the frequency.
Here's some questions for you to have a practise.
Have a look through those, write down your answers, and just pause the video whilst you're doing so.
So how did you get on? Part a, a water wave's got a period of five seconds.
What's its frequency? Well, frequency is one divided by the period, it's one divided by five, or 0.
2 hertz.
So we can use that equation to get the answer very quickly.
Part b, a wave on the rope's got a period of.
5 seconds.
What's its frequency? Frequency is one divided by the period, one divided by 0.
5 gives you two hertz.
Part c, a wave with a period of.
25 seconds.
How many waves pass a marker in one second? We're asking for a frequency, number of waves a second, so frequency is one divided by the period, and one divided by 0.
25 gives us four, four waves per second, or four hertz.
Part d, a transverse wave with a frequency of eight hertz.
It's period T is going to be eight times smaller so its period is going to be one eighth, or 0.
125 seconds.
To use the equation, you may have had to rearrange that, which is fine.
And part e, a wave on the string has got a frequency of 25 hertz.
How long is each full vibration? Well, frequency is one over the period, or period is one over the frequency.
It's one 25th of a second for each one.
And if you think about that, if it's got a frequency of 25 hertz, that's 25 in a second, or each one is a 25th of a second.
In decimals, that's 0.
04 seconds.
And now that you understand about frequency and period and are able to calculate them, we can now move on to looking at displacement-time graphs.
We can actually represent a wave by two different sorts of graph, a displacement-distance graph, or a displacement-time graph.
As you can see here, we've got both graphs and they both look very, very similar, but there are some very important differences between them.
And it's important first of all to check what sort of graph you have and what you are looking at.
And the only way to be really certain is to look at the scales of a graph.
That's always the first thing you should look at.
When you look at a graph, look at the scales.
So the right hand graph has got time on the horizontal axis rather than distance, and that is a displacement-time graph.
Again, a displacement-time graph is not a picture of a wave drawn to scale.
It's nothing at all to do with that.
What it shows is how the displacement at one point changes as the wave passes.
So the graph on the right hand side shows how point x on the ripple tank goes up and down as the wave moves past it.
And the gap between two crests on that graph is a time, because we're measuring time on the horizontal axis.
And if you think about how the wave is moving forwards as the wave goes up and down, we can think of that gap between two crests as equal to a period.
If you are watching the wave go past the point, when the crest of the wave goes past, you start timing.
When the next crest goes past you've measured the time for one wave to go past, or you've measured its period.
And that time period can be shown on a displacement-time graph like this, and it can be measured between two troughs, or between any two similar points on adjacent waves.
It's always the same amount of time.
Let's have a look at an example and something for you to have a go at.
In this example, a pupil makes a wave with a rope and her friend makes a graph to represent the same wave.
The graph she makes has time on the horizontal axis.
And what I'd like you to do is to have a think about these three statements and decide which ones you think are right and which one you think are wrong.
And in each case you decide whether you're absolutely certain you're right or wrong, or whether it is just the best guess by ticking the right box next to each statement.
Just pause the video and then have a look at these.
Okay, so let's have a look at the statements.
The first one, A, a snapshot of the wave on the rope.
Is that graph a snapshot a photograph, if you like, of the wave? And it's not, because it's not.
On the photograph, if you like, you would have distance along the horizontal axis.
Here we're measuring time.
What about statement B? The movement at one point on the rope.
Is that graph showing how one point on the graph moves? And yes, it is, it's showing how part of a rope at any one point is moving up and down as the wave moves past it.
And does the, statement C, does that graph show three complete wavelengths? Well, it does have three complete waves, but they're not wavelengths because we have time on the horizontal axis and not a
So it's showing how the displacement of a point has changed over time, it's not shown the wavelengths of that wave.
It's got the distance between the crest, if you like, is a time and not a distance, so we can't measure wavelengths from that graph.
We can also use displacement-time graphs to measure the period accurately if we can plot the graphs out.
So on this graph, we're asked what the period of the wave is.
we've got time on horizontal axis.
So all we need to do is to pick two points on the graph that are the same point on opposite waves.
Here I've chosen the crests, and the difference between the crest is five seconds minus one second, which gives a period for the wave of four seconds.
Just pause the video and have a look at this graph and work out the period of the wave.
Okay, so what did we get? We could measure from crest to crest, as you've done here, and we could measure 45 seconds, take away 25 seconds to give a period of 20 seconds, or more easily, you could
look to see where the graph crosses the zero point on the horizontal axis and you've got a complete wave between 20 and 40 seconds, and that will give you the same answer.
So you don't have to use the crest, you can use at any point and sometimes it's easier to use the zero line on the horizontal axis.
Now it's time for you to have a little practise.
Pause the video and see if you can answer these questions about this displacement-time graph.
Okay, how did you get on? First of all, you were asked to measure the time period, and we could measure that from crest to crest, but this scale is a little bit awkward there, so it's a lot easy to
measure it from where the line on the graph crosses the horizontal axis.
So between two and zero seconds, it's the same point on the graph each place and we get a time period of two seconds.
We can measure the amplitude directed from the graph as well, which is the maximum displacement, and that comes in at seven centimetres.
However, for the frequency, we can't measure that directly.
And what we have to do is use our time period and the equation we saw earlier in the lesson, frequency is one divided by the period, which is one divided by two seconds, or 0.
5 hertz.
So just to finish the lesson, let's have a look at this summary.
What we've been doing is seeing how we can represent waves with a displacement-time graph.
There's a picture of a displacement-time graph and the time period mark with a capital T is the time for a complete wave to pass any point as the wave moves forward.
It's also the same as the time needed to make one complete wave, or one complete oscillation it's caused the wave.
And as we've just said, the gap between any two crests, or any two points that are the same on adjacent waves is equal to the period of a wave, and it's equal to that period because we're measuring
time on the horizontal axis.
When we're thinking about the graph, we can also calculate the frequency from measurements of the graph from the measurements of the period.
Frequency is the number of waves produced in one second, it's measured in hertz, or Hz for short, it's got to be a capital H.
And we can use the equation, frequency is one divided by the period, to work it out, or in symbols, f equals one divided by a capital letter T for period.
I hope you've enjoyed the lesson and you've learned all you needed to know about interpreting displacement-time graphs for waves. | {"url":"https://www.thenational.academy/pupils/programmes/physics-secondary-year-10-foundation-ocr/units/measuring-waves/lessons/representing-transverse-waves/video","timestamp":"2024-11-07T08:50:41Z","content_type":"text/html","content_length":"139975","record_id":"<urn:uuid:0fa93ba4-ed82-43cb-93f8-4cb0fbc2820a>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00511.warc.gz"} |
Bipartite graphs with no K<sub>6</sub> minor
A theorem of Mader shows that every graph with average degree at least eight has a K[6] minor, and this is false if we replace eight by any smaller constant. Replacing average degree by minimum
degree seems to make little difference: we do not know whether all graphs with minimum degree at least seven have K[6] minors, but minimum degree six is certainly not enough. For every ε>0 there are
arbitrarily large graphs with average degree at least 8−ε and minimum degree at least six, with no K[6] minor. But what if we restrict ourselves to bipartite graphs? The first statement remains true:
for every ε>0 there are arbitrarily large bipartite graphs with average degree at least 8−ε and no K[6] minor. But surprisingly, going to minimum degree now makes a significant difference. We will
show that every bipartite graph with minimum degree at least six has a K[6] minor. Indeed, it is enough that every vertex in the larger part of the bipartition has degree at least six.
All Science Journal Classification (ASJC) codes
• Theoretical Computer Science
• Discrete Mathematics and Combinatorics
• Computational Theory and Mathematics
• Bipartite
• Edge-density
• Minors
Dive into the research topics of 'Bipartite graphs with no K[6] minor'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/bipartite-graphs-with-no-ksub6sub-minor","timestamp":"2024-11-04T08:19:49Z","content_type":"text/html","content_length":"47453","record_id":"<urn:uuid:3a988eaf-9fcf-4fdd-b5b3-2c7a4e3a8098>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00012.warc.gz"} |
ERROR: ReadOnlyMemoryError() when doing a basic plot?
I’m installing fresh computers, and one of the things I’m installing is Julia. I have tried so far three computers, all using the same distribution (ArchLinux), and all have the same versions of
everything as far as I can see, but in two of them I get a ReadOnlyMemoryError() when trying a basic plot. These are the versions and the error I get:
[angelv@kepa ~]$ pacman -Q python python-matplotlib julia
python 3.6.2-1
python-matplotlib 2.0.2-1
julia 2:0.6.0-3
[angelv@kepa ~]$ julia
_ _ _(_)_ | A fresh approach to technical computing
(_) | (_) (_) | Documentation: https://docs.julialang.org
_ _ _| |_ __ _ | Type "?help" for help.
| | | | | | |/ _` | |
| | |_| | | | (_| | | Version 0.6.0 (2017-06-19 13:05 UTC)
_/ |\__'_|_|_|\__'_| |
|__/ | x86_64-pc-linux-gnu
julia> Pkg.status("Plots")
- Plots 0.12.4
julia> Pkg.status("PyPlot")
- PyPlot 2.3.2
julia> using Plots
julia> backend()
julia> plot(sin)
Error showing value of type Plots.Plot{Plots.PyPlotBackend}:
ERROR: ReadOnlyMemoryError()
The exact same versions of these packages in two computers give me the error, while in the third one the plot is created no problem. ¿Any suggestions on how to find out what is going on?
Ángel de Vicente
What happens if you do the following?
julia> using PyPlot
julia> xx = linspace(-2, 2, 100)
julia> plot(xx, sin.(xx))
That way we could atleast know whether the problem is with the backend. Have you tried the same thing with other backends? For example GR or Plotly?
this is quite weird, but I get the same result consistently in both computers which give trouble (let’s call them p1 and p2): the first plot command raises the ReadOnlyMemoryError, the second one
makes the plot OK. But remember that the same libraries and versions give no trouble in the other computer, let’s call it o1 (the only thing that it is obviously different between these computers is
that the one not giving any trouble has only one core, while the other two are multi-core computers).
using GR instead of PyPlot creates the plot OK in two of the computers (o1 and p1), but couldn’t try in p2, it ends up timing out when Pkg.add(“GR”). I tried the command several times and this one
happens to be on the fastest network, and if I do a git clone of GR from outside julia it takes only seconds. In p1 Pkg.add(“GR”) also took quite much longer than the same in o1, despite being in the
same network. I don’t see how these things could be related, but it looks like Julia has some serious issues in p1 and p2, despite having the same versions of everything.
Really puzzled… Any hints?
[angelv@kepa ~]$ julia
_ _ _(_)_ | A fresh approach to technical computing
(_) | (_) (_) | Documentation: https://docs.julialang.org
_ _ _| |_ __ _ | Type "?help" for help.
| | | | | | |/ _` | |
| | |_| | | | (_| | | Version 0.6.0 (2017-06-19 13:05 UTC)
_/ |\__'_|_|_|\__'_| |
|__/ | x86_64-pc-linux-gnu
julia> using PyPlot
julia> x=collect(0:0.01:2*pi);
julia> plot(x,sin.(x))
ERROR: PyError (ccall(@pysym(:PyObject_Call), PyPtr, (PyPtr, PyPtr, PyPtr), o, arg, C_NULL)) <class 'RuntimeError'>
RuntimeError('Julia exception: ReadOnlyMemoryError()',)
File "/usr/lib/python3.6/site-packages/matplotlib/pyplot.py", line 3306, in plot
ax = gca()
File "/usr/lib/python3.6/site-packages/matplotlib/pyplot.py", line 950, in gca
return gcf().gca(**kwargs)
File "/usr/lib/python3.6/site-packages/matplotlib/pyplot.py", line 586, in gcf
return figure()
[1] pyerr_check at /home/angelv/.julia/v0.6/PyCall/src/exception.jl:56 [inlined]
[2] pyerr_check at /home/angelv/.julia/v0.6/PyCall/src/exception.jl:61 [inlined]
[3] macro expansion at /home/angelv/.julia/v0.6/PyCall/src/exception.jl:81 [inlined]
[4] #_pycall#67(::Array{Any,1}, ::Function, ::PyCall.PyObject, ::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:653
[5] _pycall(::PyCall.PyObject, ::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:641
[6] #pycall#71(::Array{Any,1}, ::Function, ::PyCall.PyObject, ::Type{PyCall.PyAny}, ::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:675
[7] pycall(::PyCall.PyObject, ::Type{PyCall.PyAny}, ::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:675
[8] #plot#85(::Array{Any,1}, ::Function, ::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyPlot/src/PyPlot.jl:172
[9] plot(::Array{Float64,1}, ::Vararg{Array{Float64,1},N} where N) at /home/angelv/.julia/v0.6/PyPlot/src/PyPlot.jl:169
[10] macro expansion at ./REPL.jl:97 [inlined]
[11] (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at ./event.jl:73
julia> plot(x,sin.(x))
1-element Array{PyCall.PyObject,1}:
PyObject <matplotlib.lines.Line2D object at 0x7f87a83c7a58>
GR comes bundled with binaries for lots of different operating systems, when you call Pkg.add("GR") it downloads the binaries. You could check their repo to see if p1 is compatible with the binaries.
What happens if you use p1 or p2 and try using matplotlib from python? That way we could see whether or not the problem is in PyPlot.jl or matplotlib.
I will try GR again in the computer that was giving trouble, but it was stuck at the git cloning step, so I don’t know if it will work yet, but I guess that is a separate problem.
As per matplotlib in python in p1 and p2, no problem with either, so it certainly looks like PyPlot is the culprit…
Today I upgraded all packages in ArchLinux and all packages in Julia, but I still have the same problem.
The first time I issue the plot command I get the ReadOnlyMemoryError(), the second time I get a matplotlib window but without any content, all black, the third time I call plot then I get the plot,
and after that all calls to plot work OK.
Whoops, I only just saw this now (sorry). The only other troubleshooting advice that I can give is to try see if you can get PyCall to work. That way you would have a really strong case for blaming
PyPlot. Try the following:
using PyCall
@pyimport matplotlib.pyplot as plt
x = linspace(0,2*pi,1000); y = sin(3*x + 4*cos(2*x));
plt.plot(x, y, color="red", linewidth=2.0, linestyle="--")
(Or you could try any other eamples on their README)
The only other general advice that I have would be to hope that somebody more experienced comes to this thread or that you open an issue on PyPlot. In the meantime, it sounds like you have GR working
on two computers, so that should hopefully help you until we can fix PyPlot.
Also, if you just click the reply button on my comment, then I will get a notification which will hopefully mean that I wont take a month to respond
thanks for the reply. There is an issue on PyPlot about this (https://github.com/JuliaPy/PyPlot.jl/issues/291) but it seems that is not assigned to anyone yet, perhaps this is not happening to many
people. Don’t know.
I was actually trying what you suggest just today and I put the results in that thread. To summarize:
• The library versions I have:
python 3.6.3-1
python-matplotlib 2.1.0-1
julia 2:0.6.1-1
qt5-3d 5.9.2-1
qt5-datavis3d 5.9.2-1
qt5-x11extras 5.9.2-1
Plots 0.13.1
PyPlot 2.3.2
• If I create a basic plot from within Python there is no issue at all.
• If I try to create a plot from Julia using Plots, I only manage to get a working plot at the third attempt (OK after that)
• If I try to use PyCall as per your suggestion I first get the ReadOnlyMemoryError(), then a blank matplotlib window, and then nothing… (see below)
julia> using PyCall
julia> PyCall.python
julia> @pyimport matplotlib.pyplot as plt
julia> x = linspace(0,2*pi,100); y = sin.(x);
julia> plt.plot(x, y)
1-element Array{PyCall.PyObject,1}:
PyObject <matplotlib.lines.Line2D object at 0x7f9570f24e48>
julia> plt.show()
ERROR: ReadOnlyMemoryError()
[1] macro expansion at /home/angelv/.julia/v0.6/PyCall/src/exception.jl:78 [inlined]
[2] #_pycall#67(::Array{Any,1}, ::Function, ::PyCall.PyObject) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:653
[3] #pycall#71(::Array{Any,1}, ::Function, ::PyCall.PyObject, ::Type{PyCall.PyAny}) at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:675
[4] #call#72 at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:678 [inlined]
[5] (::PyCall.PyObject)() at /home/angelv/.julia/v0.6/PyCall/src/PyCall.jl:678
[6] macro expansion at ./REPL.jl:97 [inlined]
[7] (::Base.REPL.##1#2{Base.REPL.REPLBackend})() at ./event.jl:73
julia> plt.show()
julia> plt.show()
type or paste code here
The second call to plt.show() creates a matplotlib window but no plot is seen. I close the window and issue a third plt.show() and then the third call returns immediately but does not create any plot
That issue looks like your best bet for getting this to work. I wouldn’t expect anyone to be assigned to the issue because it seems like nobody knows precisely what is causing the problem. Sorry that
I couldn’t be more useful, but I am fairly sure that you will be able to resolve things through that issue.
Thanks. I fill follow the issue in GitHub and see if some solution comes up. If it does I will update this thread in case it can be of interest to somebody in the future.
1 Like | {"url":"https://discourse.julialang.org/t/error-readonlymemoryerror-when-doing-a-basic-plot/6145","timestamp":"2024-11-12T10:03:05Z","content_type":"text/html","content_length":"41386","record_id":"<urn:uuid:a2e6a6e6-8de3-4341-b7e4-1d0c2ac58edf>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00581.warc.gz"} |
Tutte Polynomial -- from Wolfram MathWorld
Let undirected graph, and let cardinal number of the set of externally active edges of a spanning tree cardinal number of the set of internally active edges of
(Biggs 1993, p. 100).
An equivalent definition is given by
where the sum is taken over all subsets edge set of a graph vertex count of
Several analogs of the Tutte polynomial have been considered for directed graphs, including the cover polynomial (Chung and Graham 1995), Gordon-Traldi polynomials (Gordon and Traldi 1993), and
The Tutte polynomial can be computed in the Wolfram Language using TuttePolynomial[g, x, y
The Tutte polynomial is multiplicative over disjoint unions.
For an undirected graph on
where rank polynomial (generalizing Biggs 1993, p. 101). The Tutte polynomial is therefore a rather general two-variable graph polynomial from which a number of other important one- and two-variable
polynomials can be computed.
For not-necessarily connected graphs, the Tutte polynomial chromatic polynomial flow polynomial rank polynomial reliability polynomial
The Tutte polynomial of the dual graph
i.e., by swapping the variables of the Tutte polynomial of the original graph. A special case of this identity relates the flow polynomial planar graph chromatic polynomial of its dual graph
The Tutte polynomial of a connected graph
1. If
2. If
Closed forms for some special classes of graphs are summarized in the following table, where web graph was considered by Biggs et al. (1972) and Brennan et al. (2013).
book graph
centipede graph
cycle graph
empty graph 1
gear graph
helm graph
ladder graph
ladder rung graph
pan graph
path graph
star graph
sunlet graph
wheel graph
The following table summarizes the recurrence relations for Tutte polynomials for some simple classes of graphs.
graph order recurrence
antiprism graph 6
book graph 2
centipede graph 1
cycle graph 2
gear graph 3
helm graph 3
ladder graph 2
ladder rung graph 1
Möbius ladder 6
pan graph 2
path graph 1
prism graph 6
star graph 1
sunlet graph 2
web graph 6
wheel graph 3
An equation for the Tutte polynomial complete graph exponential generating function
(Gessel 1995, Gessel and Sagan 1996). This can be written more simply in terms of the coboundary polynomial
where vertex count of a graph
which can be converted to the corresponding Tutte polynomial using the above relationship and the substitution
A formula for the Tutte polynomial of a complete bipartite graph exponential generating function for the coboundary polynomial as
by Martin and Reiner (2005).
Nonisomorphic graphs do not necessarily have distinct Tutte polynomials. de Mier and Noy (2004) call a graph that is determined by its Tutte polynomial a wheel graphs, ladder graphs, Möbius ladders,
complete multipartite graphs (with the exception of hypercube graphs are generalized Petersen graphs line graphs
The numbers of simple graphs on A243048), while the corresponding numbers of Tutte-unique graphs are 1, 2, 4, 7, 19, 72, 496, 6717, ... (OEIS A243049). The following table summarizes some small
co-Tutte graphs.
Tutte polynomial graphs
4 claw graph path graph
5 fork graph, path graph star graph
5 paw graph
5 bull graph, cricket graph, tadpole graph
5 dart graph, kite graph | {"url":"https://mathworld.wolfram.com/TuttePolynomial.html","timestamp":"2024-11-03T22:20:48Z","content_type":"text/html","content_length":"101481","record_id":"<urn:uuid:c363db02-d304-4473-bce5-f0d300a65b7e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00809.warc.gz"} |
How Do You Make Coffee In A Brentwood Coffee Maker? | Coffee Nerd
How Do You Make Coffee In A Brentwood Coffee Maker?
Simply place the grounds, add a cup of water and hit the power button to start brewing Warming plate keeps your coffee hot. Save money and help protect the earth using the included reusable mesh
filter basket. Use the included reusable scooper to choose the number of cups and strength of your coffee.
What is the ratio of coffee to water in a coffee maker?
The standard ratio for brewing coffee is 1-2 tablespoons of ground coffee per 6 ounces of water – 1 tablespoon for lighter coffee and 2 for stronger coffee. That 6-ounce measure is equivalent to one
“cup” in a standard coffeemaker, but keep in mind that the standard mug size is closer to 12 ounces or larger.
How much coffee do I use for 2 cups of water?
When using tablespoons and an 8-ounce cup, this is the general rule. For 2 cups, 16 oz water plus four tablespoons of coffee will be enough, and for 3 cups =24 oz water plus 6 tablespoons coffee. For
4 oz water plus 8 tablespoons of coffee and for cups, 40 oz water plus 10 tablespoons coffee.
How much coffee do I use for 4 cups of water?
How much coffee for 4 cups? To make four cups of coffee at average strength, use 36 grams of coffee and 20 ounces (2 1/2 measuring cups) of water That’s about 4 level scoops of coffee, or 8 level
tablespoons. To make the coffee strong, use 41 grams of coffee (4 1/2 scoops or 9 tablespoons).
How much coffee do I use for 8 cups of water?
For making 6 cups, we recommend 10 Tablespoons or ~ 60 grams of coffee. For making 8 cups, we think 14 Tablespoons or ~80 grams of coffee is a good starting point. You may need to use more or less
coffee, depending on your preferred coffee strength.
How much coffee do I use for 6 cups of water?
For each cup of coffee you want to brew, use an equivalent number of scoops. So if you’d like to brew a 6-cup pot of coffee, use 6 scoops of coffee.
Why is my coffee maker not brewing?
The primary cause for this type of problem is any type of blockage or water clog The first thing to do is check the tube within the coffee pot. If there are obstructions here, or if the tube is
clogged, water or any other liquid will not be able to pass through.
Where do you put the mixture in a coffee maker?
To use a coffee maker, start by placing a coffee filter in the filter basket Then, fill the filter with 2 tablespoons of coffee per every 6 ounces of water you’ll be brewing. Next, fill the coffee
maker’s water compartment with however much water you want to use.
How long does it take to make coffee in a coffee maker?
Coffee brewing should take between three and five minutes on most machines, from the time the water starts dripping onto the coffee to when it drips all the way through the coffee ground.
How many tablespoons of coffee do I use for 10 oz of water?
A general guideline is called the Golden ratio – 2 tablespoons of ground coffee for every 8 ounces of water This is my preferred coffee ratio for drip, pour over and French press (I do use different
ratios for cold brew).
How much coffee do you use for one cup coffee maker?
The SCAA defines 10 grams or 0.36 oz per 6-oz (180 ml) cup as the proper measure for brewed coffee if using the American standards. If using Euro standards the measure is 7 grams per 125 ml (4.2 fl.
oz). To further confuse things I will add a few more measures of how many oz in a cup (coffee weight to water volume):.
How many tablespoons ground coffee for Pour over?
For one cup (8 fluid oz.), you will need to use about 2.5 level tablespoons or about 18 grams (more or less depending on taste) of whole bean coffee. Grind to a medium-coarse level that looks
somewhere between table salt and kosher salt. Place your pourover brewer on top of your mug.
What is the size of a coffee scoop?
1. A Coffee Scoop Is Typically About 2 Tablespoons, or 1/8 Cup A coffee scoop is a kitchen utensil used for measuring quantities of ground coffee beans, though some are also used to measure loose
tea. Traditionally, the coffee scoop holds about 15 ml (0.5 US fl oz) of liquid (~1 tablespoon).
How much coffee do I put in a 4 cup Mr Coffee?
The general rule of thumb is one to two tablespoons of coffee per six ounces of water , but exact measurements will be up to your personal preferences.
How much is a scoop of coffee?
A level coffee scoop holds approximately 2 tablespoons of coffee. So, for a strong cup of coffee, you want one scoop per cup. For a weaker cup, you might go with 1 scoop per 2 cups of coffee or 1.5
scoops for 2 cups.
How many tablespoons of coffee do you use for 12 cups?
For a strong cup use the amount of coffee you want. To reduce the bitter taste, use 95 grams (10/3 scoops or 20/3 tablespoons each). If you’re brewing coffee for a crowd, you’ll need more coffee. For
12 cups, you’ll need about 1/2 cup of ground coffee.
How much water do I put in a cup of coffee?
Measure the grounds – The standard measurement for coffee is 6 ounces of fresh water to 2 tablespoons ground coffee Most coffee lovers will quote a standard “3 tablespoons for 12 fl oz”. It’s easy to
measure out – and will save you the frustration of using up your grounds (and cash) too quickly. 5.
How much coffee do I put in a 10 cup coffee maker?
The suggested amount of coffee to use to brew one cup is 1-2 tablespoons for 6oz water. This means that for 10 6oz cups, you should expect to use 10-20 tablespoons of ground coffee. This is known as
the “Golden Ratio”.
How many tablespoons of coffee do you use for 4 cups?
If you want to prepare four cups of coffee you will need exactly 4 scoops of ground beans, or, if you prefer, 8 tablespoons If you want stronger coffee, you can go for 10 tablespoons and you will get
four delicious cups of coffee.
How long does it take to make coffee in an electric percolator?
How long do you let coffee percolate in a percolator? Depending on the desired strength level, you’ll want to percolate coffee for 7 to 10 minutes It’s important to keep even heat in the percolator
during this process (an area where electric coffee percolators definitely shine). | {"url":"https://www.coffeenerd.blog/how-do-you-make-coffee-in-a-brentwood-coffee-maker/","timestamp":"2024-11-01T22:50:25Z","content_type":"text/html","content_length":"142735","record_id":"<urn:uuid:2c258e12-3a0b-4201-b548-29c51ca2222a>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00725.warc.gz"} |