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8100 | https://www.youtube.com/watch?v=ROF15eoLkrg | Consumer/Producer Surplus & Deadweight Loss
Mr. Sinn
282000 subscribers
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Posted: 29 Mar 2019
This video explains consumer surplus, producer surplus, 7 deadweight loss. By watching this video you will better understand these concepts and how to find them in a chart.
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GovernmentRegulation #Microeconomics #Economics
19 comments
Transcript:
Intro good morning afternoon evening night whenever you're watching this welcome to the mr. sin Channel today we're going to be talking about consumer surplus producer surplus and deadweight loss before watching this video make sure you understand what a price floor and price ceiling is if you haven't seen my video on it yet you can click on the card on the top right it'll help you understand these concepts that's going to be important for you to understand when we get to deadweight loss now in this video we're gonna go over consumer surplus producer surplus and deadweight loss and we're gonna be looking at some charts throughout it make sure you're following along with your notes if you want use my guided notes you can find them in the description below they'll go along with the video and help you understand everything on the screen ConsumerProducer Surplus right now you can see that I have a supply and demand chart for rubber ducks we can see that our equilibrium is at ten dollars and at ten dollars I'll be selling six rubber ducks now the first concept we're gonna go into is consumer surplus consumer surplus is the difference between what people were willing to pay and then what they actually paid now when we're just looking at our chart we can see we have multiple consumers that value rubber ducks more than what the markets selling them for so for example right now I can see someone's willing to pay $15 for this duck but they're not going to have to they'll only have to pay ten now at the same point we all saw people who are willing to pay less anyone that is below the equilibrium point on the demand but they would not be factored into this they are not valuing the product enough and there would be no sale there at the same time - we can also see our producer surplus now the producer surplus is the difference between what producers were willing to sell the product for and what they actually get to sell the product for so for here I can look at my supply line when I'm looking at the supply line I will take anything that is below the equilibrium and that's gonna be where our producer surplus is so these are companies that are willing to make the products and sell it for cheaper what about producers that were willing to make more supply if the price was raised above the equilibrium well they would not be factored into this they're going to be up here on our supply and demand chart but there's not going to be a demand so we're not going to have a sale so they would not be part of the producer surplus because no transaction will happen so consumer surplus is the difference between what people were willing to pay and what they actually had to pay so if I needed to find my total consumer surplus I would find it in this triangle right here now for producer surplus I'm gonna do the same thing however I need to go from where my producers were willing to sell it for and what they actually got to sell it for so we can see that our producer surplus is represented by a triangle here now on a test or quiz you might get a question that asks for the total surplus to find our total surplus we're just going to take our consumer surplus and add it to our producer surplus we can see it represented in the triangle here we're taking the two total consumer and producer surplus a--'s and combining them together and that's producer and consumer surplus let's pretend for a minute that people really want these rubber ducks but they don't like how high the prices $10 is just too much so people write to their congressman and the government and eventually get the government to pass a price ceiling now the price ceiling remember goes below the equilibrium this is going to be a new maximum amount that companies could sell rubber ducks for and the new price ceiling is gonna be set at $7 we can see here it's gonna change our supply and demand graph then because the price has been lowered our supply will go down and because the price is down our demand will go up now our new consumer surplus is going to be represented by this area here and our producer surplus is going to be represented by the area here now these are our total consumer and our total producer surplus now the question though is where is our deadweight loss deadweight loss is lost efficiency whenever we manipulate the market we're going to have a loss of efficiency and that's gonna be our deadweight loss while it might seem like a really good idea to try and lower prices artificially through government regulation what's gonna happen is we'll have a shortage we're not going to be able to have enough rubber ducks for everyone so in the end people actually lose out now our deadweight loss here is going to be represented by this area here this is the lost efficiency the lost area that we could have had if the market had not been changed and manipulated by the government Price Floor now let's pretend one more time that companies eventually got upset about these price changes and they got the government to change them and they got the government to put a price floor in effect a price floor remember is going to be a minimum amount that companies can charge so that will be above our equilibrium I can see on my supply and demand chart over here this new price floor is set at $12 now our consumer surplus is going to be represented by the area here our producer surplus now is going to be represented by the area right here and now we're gonna have a new deadweight loss remember this was gonna be our lost efficiency I can see my deadweight loss is represented by this area here now this is showing us what we could have had if the market again hadn't been manipulated I hope by now you have a good understanding of how to find our consumer surplus our producer surplus our total surplus and our deadweight loss if you have any questions at all make sure to post them in the comments below and I'll try to help you out thank you very much for watching this make sure to subscribe on your way out and check out some of my other videos on the channel I'm mister sin and until next time I'll see you online |
8101 | https://www.withturtled.com/blog/how-to-find-a-common-denominator | Blog - How to Find a Common Denominator: A TurtlEd Math Guide
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July 22, 2025
Math for Kids
How to Find a Common Denominator
Understanding fractions is a big milestone for many learners, and one of the trickiest parts is learning how to work with different denominators. If you've ever tried adding or subtracting fractions with different denominators, you know it’s not as simple as just lining them up and doing the math. To combine these fractions correctly, you need a common denominator. In this post, we’ll explore what a common denominator is, when you need to use one, and how to find it, all with examples to guide you.
What is a Common Denominator?
In a fraction, a denominator is the bottom number. It tells you how many equal parts the whole is divided into. A common denominator is a shared multiple of two or more denominators. We need it any time we add or subtract fractions that don’t already share the same denominator.
For example, if you want to add 1/8 and 5/12:
The denominators here are 8 and 12. They’re not the same, so we can’t add them yet!
To add 1/8 + 5/12, we need to find a number that both 8 and 12 can divide into.
How Do You Find a Common Denominator?
Step 1: List Multiples
List the multiples of both denominators.
Find the smallest number they have in common.
Example:
To find a common denominator for 1/8 and 5/12, let's list the multiples:
Multiples of 8: 8, 16, 24, 32...
Multiples of 12: 12, 24, 36, 48...
Here, the Least Common Multiple (LCM) = 24 → So the common denominator is 12!
Step 2: Rewrite Each Fraction with the Common Denominator
Now we convert each fraction:
1/8 becomes 3/24 because 8 × 3 = 24, and we multiply the numerator by the same number: 1 × 3 = 3.
5/12 becomes 10/24 because 12 × 2 = 24, and 5 × 2 = 10.
Step 3: Add the Fractions
Now that the denominators match, simply add together 3/24 + 10/24 to equal 13/24.
When Do You Need a Common Denominator?
You need a common denominator any time you’re:
Adding or subtracting fractions with different denominators.
Comparing fractions to see which is greater or smaller.
Tips for Finding Common Denominators
Look for the least common multiple (LCM) of both denominators.
Use times tables to find common multiples faster.
If you can’t find the LCM easily, you can always multiply the two denominators together, it may not be the smallest, but it works!
For example: For 1/5 and 1/6, 5 × 6 = 30 → use 30 as a common denominator.
Want to see these strategies in action?
Watch our full video on YouTube for a step-by-step walkthrough and more helpful tips to support your learning:
Remember, fractions don’t have to be frustrating! At TurtlEd, we support K–12 students in building a strong foundation in math, including tricky concepts like finding common denominators. Our approach helps students feel confident and capable every step of the way. Many of our students improve by more than one grade level in just 10 weeks. Contact us to learn how our effective tutoring strategies can help your student thrive in math.
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8102 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.02%3A_Mole_Ratios | Skip to main content
12.2: Mole Ratios
Last updated
: Mar 20, 2025
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12.1: Everyday Stoichiometry
12.3: Mass-Mole Stoichiometry
Page ID
: 53790
( \newcommand{\kernel}{\mathrm{null}\,})
What does this porch need?
You want to add some sections to the porch seen above. Before you go to the hardware store to buy lumber, you need to determine the unit composition (the material between two large uprights). You count how many posts, how many boards, how many rails – then you decide how many sections you want to add before you calculate the amount of building material needed for your porch expansion.
Mole Ratios
Stoichiometry
problems can be characterized by two things: (1) the information given in the problem, and (2) the information that is to be solved for, referred to as the unknown. The given and the unknown may both be reactants, both be
products
, or one may be a reactant while the other is a product. The amounts of the substances can be expressed in moles. However, in a laboratory situation, it is common to determine the amount of a
substance
by finding its mass in grams. The amount of a gaseous
substance
may be expressed by its volume. In this concept, we will focus on the type of problem where both the given and the unknown quantities are expressed in moles.
Chemical equations express the amounts of reactants and
products
in a reaction. The coefficients of a
balanced
equation can represent either the number of
molecules
or the number of moles of each
substance
. The production of ammonia from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber.
The
balanced
equation can be analyzed in several ways, as shown in the figure below.
We see that 1 molecule of nitrogen reacts with 3
molecules
of hydrogen to form 2
molecules
of ammonia. This is the smallest possible relative amount of the reactants and
products
. To consider larger relative amounts, each
coefficient
can be multiplied by the same number. For example, 10
molecules
of nitrogen would react with 30
molecules
of hydrogen to produce 20
molecules
of ammonia.
The most useful quantity for counting particles is the
mole
. So if each
coefficient
is multiplied by a
mole
, the
balanced
chemical equation tells us that 1
mole
of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any
balanced
chemical equation.
Finally, if each
mole
quantity is converted to grams by using the molar mass, we can see that the
law
of conservation of mass is followed. of nitrogen has a mass of , while of hydrogen has a mass of , and of ammonia has a mass of .
Mass and the number of atoms must be conserved in any chemical reaction. The number of
molecules
is not necessarily conserved.
A mole
ratio is a
conversion factor
that relates the amounts in moles of any two substances in a chemical reaction. The numbers in a
conversion factor
come from the coefficients of the
balanced
chemical equation. The following six
mole
ratios can be written for the ammonia forming reaction above.
In a
mole
ratio problem, the given
substance
, expressed in moles, is written first. The appropriate
conversion factor
is chosen in order to convert from moles of the given
substance
to moles of the unknown.
Example : Mole Ratio
How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?
Solution
Step 1: List the known quantities and plan the problem.
Known
given: H2 = 4.20 mol
Unknown
mol of NH3
The conversion is from to . The problem states that there is an excess of nitrogen, so we do not need to be concerned with any
mole
ratio involving . Choose the
conversion factor
that has the in the numerator and the in the denominator.
Step 2: Solve.
The reaction of of hydrogen with excess nitrogen produces of ammonia.
Step 3: Think about your result.
The result corresponds to the 3:2 ratio of hydrogen to ammonia from the
balanced
equation.
Summary
Mole
ratios allow comparison of the amounts of any two materials in a
balanced
equation.
Calculations can be made to predict how much product can be obtained from a given number of moles of reactant.
Review
If a reactant is in excess, why do we not worry about the
mole
ratios involving that reactant?
What is the
mole
ratio of H to N in the ammonia molecule?
The formula for ethanol is CH3CH2OH. What is the
mole
ratio of H to C in this molecule?
12.1: Everyday Stoichiometry
12.3: Mass-Mole Stoichiometry |
8103 | https://blog.evanchen.cc/2016/11/11/notes-on-publishing-my-textbook/ | Skip to content
Power Overwhelming
Notes on Publishing My Textbook
Posted on by Evan Chen (陳誼廷)
Hmm, so hopefully this will be finished within the next 10 years.
— An email of mine at the beginning of this project
My Euclidean geometry book was published last March or so. I thought I’d take the time to write about what the whole process of publishing this book was like, but I’ll start with the disclaimer that my process was probably not very typical and is unlikely to be representative of what everyone else does.
Writing the Book
The Idea
I’m trying to pin-point exactly when this project changed from “daydream” to “let’s do it”, but I’m not quite sure; here’s the best I can recount.
It was sometimes in the fall of 2013, towards the start of the school year; I think late September. I was a senior in high school, and I was only enrolled in two classes. It was fantastic, because it meant I had lots of time to study math. The superintendent of the school eventually found out, though, and forced me to enroll as an “office assistant” for three periods a day. Nonetheless, office assistant is not a very busy job, and so I had lots of time, all the time, every day.
Anyways, I had written a bit of geometry material for my math club the previous year, which was intended to be a light introduction. But in doing so I realized that there was much, much more I wanted to say, and so somewhere on my mental to-do list I added “flesh these notes out”. So one day, sitting in the office, after having spent another hour playing StarCraft, I finally got down to this item on the list. I hadn’t meant it to be a book; I was just wanted to finish what I had started the previous year. But sometimes your own projects spiral out of your control, and that’s what happened to me.
Really, I hadn’t come up with a brilliant idea that no one had thought of before. To my knowledge, no one had even tried yet. If I hadn’t gone and decided to write this book, someone else would have done it; maybe not right away, but within many years. Indeed, I was honestly surprised that I was the first one to make an attempt. The USAMO has been a serious contest since at least the 1990’s and 2000’s, and the demand for this book certainly existed well before my time. Really, I think this all just goes to illustrate that the Efficient Market Hypothesis is not so true in these kind of domains.
Setting Out
Initially, this text was titled A Voyage in Euclidean Geometry and the filename Voyage.pdf would persist throughout the entire project even though the title itself would change throughout.
The beginning of the writing was actually quite swift. Like everyone else, I started out with an empty LaTeX file. But it was different from blank screens I’ve had to deal with in my life; rather than staring in despair (think English essay mode), I exploded. I was bursting with things I wanted to write. It was the result of having years of competitive geometry bottled up in my head. In fact, I still have the version 0 of the table of contents that came to life as I started putting things together.
Angle Chasing (include “Fact 5”)
Centers of the Triangle
The Medial Triangle
The Euler Line
The Nine-Point Circle
Circles
Incircles and Excircles
The Power of a Point
The Radical Axis
Computational Geometry
All the Areas (include Extended Sine Law, Ceva/Menelaus)
Similar Triangles
Homothety
Stewart’s Theorem
Ptolemy’s Theorem
Some More Configurations (include symmedians)
Simson lines
Incircles and Excenters, Revisited
Midpoints of Altitudes
Circles Again
Inversion
Circles Inscribed in Segments
The Miquel Point (include Brokard, this could get long)
Spiral Similarity
Projective Geometry
Harmonic Division
Brokard’s Theorem
Pascal’s Theorem
Computational Techniques
Complex Numbers
Barycentric Coordinates
Of course the table of contents changed drastically over time, but that wasn’t important. The point of the initial skeleton was to provide a bucket sort for all the things that I wanted to cover. Often, I would have three different sections I wanted to write, but like all humans I can only write one thing at a time, so I would have to create section headers for the other two and try to get the first section done as quickly as I could so that I could go and write the other two as well.
I did take the time to do some things correctly, mostly LaTeX. Some examples of things I did:
Set up proper amsthm environments: earlier versions of the draft had “lemma”, “theorem”, “problem”, “exercise”, “proposition”, all distinct
Set up an organized master LaTeX file with \include’s for the chapters, rather than having just one fat file.
Set up shortcuts for setting up diagrams and so on.
Set up a “hints” system where hints to the problems would be printed in random order at the end of the book.
Set up a special command for new terms (\vocab). At the beginning all it did was made the text bold, but I suspected that later I might it do other things (like indexing).
In other words, whenever possible I would pay O(1) cost to get back O(n) returns. Indeed the point of using LaTeX for a long document is so that you can “say what you mean”: you type \begin{theorem} … \end{theorem}, and all the formatting is taken care of for you. Decide you want to change it later, and you only have to change the relevant code in the beginning.
And so, for three hours a day, five days a week, I sat in the main office of Irvington High School, pounding out chapter after chapter. I was essentially typing up what had been four years of competition experience; when you’re 17 years old, that’s a big chunk of your life.
I spent surprisingly little time revising (before first submission). Mostly I just fired away. I have always heard things about how important it is to rewrite things and how first drafts are always terrible, but I’m glad I ignored that advice at least at the beginning. It was immensely helpful to have the skeleton of the book laid out in a tangible form that I could actually see. That’s one thing I really like about writing; helps you collect your thoughts together.
It’s possible that this is part of my writing style; compared to what everyone says I should do, I don’t do very much rewriting. My first and final drafts tend to look pretty similar. I think this is just because when I write something that’s not an English essay, I already have a reasonably good idea what I want to say, and that the process of writing it out does much of the polishing for me. I’m also typically pretty hesitant when I write things: I do a lot of pausing for a few minutes deciding whether this sentence is really what I want before actually writing it down, even in drafts.
Some Encouragement
By late October, I had about 80 or so pages content written. Not that impressive if you think about it; I think it works out to something like 4 pages per day. In fact, looking through my data, I’m pretty sure I had a pretty consistent writing rate of about 30 minutes per page. It didn’t matter, since I had so much time.
At this point, I was beginning to think about possibly publishing the book, so it was coming out reasonably well. It was a bit embarrassing, since as far as I could tell, publishing books was done by people who were actually professionals in some way or another. So I reached out to a couple of teachers of mine (not high school) who I knew had published textbooks in one form or another; I politely asked them what their thoughts were, and if they had any advice. I got some gentle encouragement, but also a pointer to self-publishing: turns out in this day and age, there are services like Lulu or CreateSpace that will just let you publish… whatever you want. This gave me the guts to keep working on this, because it meant that there was a minimal floor: even if I couldn’t get a traditional publisher, the worst I could do was self-publish through Amazon, which was at any rate strictly better than the plan of uploading a PDF somewhere.
So I kept writing. The seasons turned, and by February, the draft was 200 pages strong. In April, I had staked out a whopping 333 pages.
The Review Process
Entering the MAA’s Queue
I was finally beginning to run out of things I wanted to add, after about six months of endless typing. So I decided to reach out again; this time I contacted a professor (henceforth Z) that I knew, whom I knew well from time at the Berkeley Math Circle. After some discussion, Z agreed to look briefly at an early draft of the manuscript to get a feel for what it was like. I must have exceeded his expectations, because Z responded enthusiastically suggesting that I submit it to the Problem Book Series of the MAA. As it turns out, he was on the editorial board, so in just a few days my book was in the official queue.
This was all in April. The review process was scheduled to begin in June, and likely take the entirety of the summer. I was told that if I had a more revised draft before the review that I should also send it in.
It was then I decided I needed to get some feedback. So, I reached out to a few of my close friends asking them if they’d be willing to review drafts of the manuscript. This turned out to not go quite as well as I hoped, since
Many people agreed eagerly, but then didn’t actually follow through with going through and reading chapter by chapter.
I was stupid enough to send the entire manuscript rather than excerpts, and thus ran myself a huge risk of getting the text leaked. Fortunately, I have good friends, but it nagged at me for quite a while. Learned my lesson there.
That’s not to say it was completely useless; I did get some typos fixed. But just not as many as I hoped.
The First Review
Not very much happened for the rest of the summer while I waited impatiently; it was a long four month wait for me. Finally, in the end of August 2014, I got the comments from the board; I remember I was practicing the piano at Harvard when I saw the email.
There had been six reviews. While I won’t quote the exact reviews, I’ll briefly summarize them.
There is too much idiosyncratic terminology.
This is pretty impressive, but will need careful editing.
This project is fantastic; the author should be encouraged to continue.
This is well developed; may need some editing of contents since some topics are very advanced.
Overall I like this project. That said, it could benefit from some reading and editing. For example, here are some passages in particular that aren’t clear.
This manuscript reads well, written at a fairly high level. The motivation provided are especially good. It would be nice if there were some solutions or at least longer hints for the (many) problems in the text. Overall the author should be encouraged to continue.
The most surprising thing was how short the comments were. I had expected that, given the review had consumed the entire summer, the reviewers would at least have read the manuscript in detail. But it turns out that mostly all that had been obtained were cursory impressions from the board members: the first four reviews were only a few sentences long! The fifth review was more detailed, but it was essentially a “spot check”.
I admit, I was really at a loss for how I should proceed. The comments were not terribly specific, and the only real action-able item were to use less extravagant terms in response to 1 (I originally had “configuration”, “exercise” vs “problem”, etc.) and to add solutions (in response to 5). When I showed the comments to Z, he commented that while they were positive, they seemed to suggest that the publication may not be anytime soon. So I decided to try submitting a second draft to the MAA, but if that didn’t work I would fall back on the self-publishing route.
The reviewers had commented about finding a few typos, so I again enlisted the help of some friends of mine to eliminate them. This time I was a lot smarter. First, I only sent the relevant excerpts that I wanted them to read, and watermarked the PDF’s with the names of the recipients. Secondly, this time I paid them as well: specifically, I gave dollars for each chapter read, where was the number of errors found. I also gave a much clearer “I need this done by X” deadline. This worked significantly better than my first round of edits. Note to self: people feel more obliged to do a good job if you pay them!
All in all my friends probably eliminated about 500 errors.
I worked as rapidly as I could, and within four weeks I had the new version. The changes that I made were:
In response to the first board comment, I eliminated some of the most extravagant terminology (“demonstration”, “configuration”, etc.) in favor of more conventional terms (“example”, “lemma”).
I picked about 5-10 problems from each chapter and added full solutions for them. This inflated the manuscript by another 70 pages, for a new total of 400 pages.
Many typos and revisions were corrected, thanks to my team of readers.
Some formatting changes; most notably, I got the idea to put theorems and lemmas in boxes using mdframed (most of my recent olympiad handouts have the same boxes).
Added several references.
I sent this out and sat back.
The Second Review
What followed was another long waiting process for what again were ended up being cursory comments The delay between the first and second review was definitely the most frustrating part — there seemed to be nothing I could do other than sit and wait. I seriously considered dropping the MAA and self-publishing during this time.
I had been told to expect comments back in the spring. Finally, in early April I poked the editorial board again asking whether there had been any progress, and was horrified to find out that the process hadn’t even started out due to a miscommunication. Fortunately, the editor was apologetic enough about the error that she asked the board to try to expedite the process a little. The comments then arrived in mid-May, six weeks afterwards.
There were eight reviewers this time. In addition to some stylistic changes suggested (e.g. avoid contractions), here were some of the main comments.
The main complaint was that I had been a bit too informal. They were right on all accounts here: in the draft I had sent, the chapters had opened with some quotes from years of MOP (which confused the board, for obvious reasons) and I had some snarky comments about high school geometry (since I happen to despise the way Euclidean geometry is taught in high school.) I found it amusing that no one had brought it up yet, and happily obliged to fix them.
Some reviewers had pointed out that some of the topics were very advanced. In fact, one of the reviewers actually recommend against the publication of the book on the account that no one would want to buy it. Fortunately, the book ended up getting accepted anyways.
In that vein, there were some remarks that this book, although it serves its target audience well, is written at a fairly advanced level.
Some of the reviews were cursory like before, but some of them were line-by-line readings of a random chapter, and so this time I had something more tangible to work with.
So I proceeded to make the changes. For the first time, I finally had the brains to start using git to track the changes I made to the book. This was an enormously good idea, and I wish I had done so earlier.
Here are some selected changes that were made (the full list of changes is quite long).
Eliminate a bunch of snarky comments about high school, and the MOP quotes.
Eliminate about 250 contractions.
Eliminate about 50 instances of unnecessary future tense.
Eliminate the real product from the text.
Added in about seven new problems.
Added and improved significantly on the index of the book, making it far more complete.
Fix more references.
Change the title to “Euclidean Geometry in Mathematical Olympiads” (it was originally “Geometra Galactica”).
Change the name of Part II from “Dark Arts” to “Analytic Techniques”. (Hehe.)
Added people to the acknowledgments.
Changes in formatting: most notably I change the font size from 11pt to 10pt to decrease the page count, since my book was already twice as long as many of the other books in the series. This dropped me from about 400 pages back to about 350 pages.
Fix about 200 more typos. Thanks to those of you who found them!
I sent out the third draft just as June started, about three weeks after I had received the comments. (I like to work fast.)
The Last Revisions
There were another two rounds afterwards. In late June, I got a small set of about three pages of additional typos and clarifying suggestions. I sent back the third draft one day later.
Six days later, I got back a list of four remaining edits to make. I sent an updated fourth draft 17 minutes after receiving those comments. Unfortunately, it then took another five weeks for the four changes I made to be acknowledged. Finally, in early August, the changes were approved and the editorial board forwarded an official recommendation to MAA to publish the book.
Summary of Review Timeline
In summary, the timeline of the review process was
First draft submitted: April 6, 2014
Feedback received: August 28, 2014Second draft submitted: November 5, 2014
Feedback received: May 19, 2015Third draft submitted: June 23, 2015
Feedback received: June 29, 2015Fourth draft submitted: June 29, 2015
Official recommendation to MAA made: August 2015
I think with traditional publishers there is a lot of waiting; my understanding is that the editorial board largely consists of volunteers, so this seems inevitable.
Approval and Onwards
On September 3, 2015, I got the long-awaited message:
It is a pleasure to inform you that the MAA Council on Books has approved the recommendation of the MAA Problem Books editorial board to publish your manuscript, Euclidean Geometry in Mathematical Olympiads.
I got a fairly standard royalty contract from the publisher, which I signed off without much thought.
Editing
I had a total of zero math editors and one copy editor provided. It shows through on the enormous list of errors (and this is after all the mistakes my friends helped me find).
Fortunately, my copy editor was quite good (and I have a lot of sympathy for this poor soul, who had to read every word of the entire manuscript). My Git history indicates that approximately 1000 corrections were made; on average, this is about 2 per page, which sounds about right. I got the corrections on hard copy in the mail; the entire printout of my book, except well marked with red ink.
Many of the changes fell into general shapes:
Capitalization. I was unwittingly inconsistent with “Law of Cosines” versus “Law of cosines” versus “law of cosines”, etc and my copy editor noticed every one of these. Similarly, cases of section and chapter titles were often not consistent; should I use “Angle Chasing” or “Angle chasing”? The main point is to pick one convention and stick with it.
My copy editor pointed out every time I used “Problems for this section” and had only one problem.
Several unnecessary “quotes” and italics were deleted.
Oxford commas. My god, so many Oxford commas. You just don’t notice when the IMO Shortlist says “the circle through the points E, G, and H” but the European Girls’ Olympiad says “show that KH, EM and BC are concurrent”. I swear there were at least 100 of these in the book. I tried to write a regular expression to find such mistakes, but there were lots of edge cases that came up, and I still had to do many of these manually.
Inconsistency of em dashes and en dashes. This one worked better with regular expressions.
But of course there were plenty of other mistakes like missing spaces, missing degree spaces, punctuation errors, etc.
Cover Art
This was handled for me by the publisher: they gave me a choice of five or so designs and I picked one I liked.
(If you are self-publishing, this is actually one of the hardest parts of the publishing logistics; you need to design the cover on your own.)
Proofs
It turns out that after all the hard work I spent on formatting the draft, the MAA has a standard template and had the production team re-typeset the entire book using this format. Fortunately, the publisher’s format is pretty similar to mine, and so there were no huge cosmetic changes.
At this point I got the proofs, which are essentially the penultimate drafts of the book as they will be sent to the printers.
Affiliation and Miscellani
There was a bit more back-and-forth with the publisher towards the end. For example, they asked me if I would like my affiliation to be listed as MIT or to not have an affiliation. I chose the latter. I also send them a bio and photograph, and an author questionaire, asking me for some standard details.
Marketing was handled by the publisher based on these details.
The End
Without warning, I got an email on March 25 announcing that the PDF versions of my book were now available on MAA website. The hard copies followed a few months afterwards. That marked the end of my publication process.
If I were to do this sort of thing again, I guess the main decision would be whether to self-publish or go through a formal publisher. The main disadvantage seems to be the time delay, and possibly also that the royalties are lesser than in self-publishing. On the flip side, the advantages of a formal publisher were:
Having a real copy editor read through the entire manuscript.
Having a committee of outsiders knock some common sense into me (e.g. not calling the book “Geometra Galactica”).
Having cover art and marketing completely done for me.
It’s more prestigious; having a real published book is (for whatever reason) a very nice CV item.
Overall I think publishing formally was the right thing to do for this book, but your mileage may vary.
Other advice I would give to my past self, mentioned above already: keep paying O(1) for O(n), use git to keep track of all versions, and be conscious about which grammatical conventions to use (in particular, stay consistent).
Here’s a better concluding question: what surprised me about the process, i.e, what was different than what I expected? Here’s a partial list of answers:
It took even longer than I was expecting. Large committees are inherently slow; this is no slight to the MAA, it is just how these sorts of things work.
I was surprised that at no point did anyone really check the manuscript for mathematical accuracy. In hindsight this should have been obvious; I expect reading the entire book properly takes at least 1-2 years.
I was astounded by how many errors there were in the text, be it math or grammatical or so on. During the entire process something like 2000 errors were corrected (admittedly several were minor, like Oxford commas). Yet even as I published the book, I knew that there had to be errors left. But it was still irritating to hear about them post-publication.
All in all, the entire process started in September 2013 and ended in March 2016, which is 30 months. The time was roughly 30% writing, 50% review, and 20% production.
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Published by Evan Chen (陳誼廷)
Evan is a math olympiad coach who completed his number theory PhD at MIT. View all posts by Evan Chen (陳誼廷)
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12 thoughts on “Notes on Publishing My Textbook”
Did you have to get permission to use olympiad problems or was it sufficient to just cite them? Did the publisher handle legal matters for you such as obtaining copyright or did you do it yourself?
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My understanding is that problem statements are fine to just cite, but (official) solutions are protected by copyright. But the publisher handled all the legal matters.
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Evan, it’s crazy that you spent all this time- just writing a book and publishing it.I love how good you are at math
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2. […] had the inspiration from Evan Chen. He made it sound like you just have some stuff in your head, you carve some time out each day for […]
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3. I noticed “boko”. Did you mean “book”?
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Yes, fixed, thanks.
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4. You mentioned “When I showed he comments to Z”, I think you meant “the comments”.
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Thanks, will edit.
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5. Hello, can I ask how did you draw the diagrams in your EGMO book? What software(s) did you use?
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Is EGMO coming soon in Chinese? I heard something like that online.
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I believe that Harbin Institute of Technology Press submitted a proposal for a simplified Chinese rights to my textbook. I am not involved with this process at all (my publisher takes care of all this stuff for me) so I don’t know many details, but if all goes well, it might be published in a few years.
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8104 | https://mathworld.wolfram.com/ZsigmondyTheorem.html | Zsigmondy Theorem
If and (i.e., and are relatively prime), then has at least one primitive prime factor with the following two possible exceptions:
.
and is a power of 2.
Similarly, if , then has at least one primitive prime factor with the exception .
A specific case of the theorem considers the th Mersenne number , then each of , , , ... has a prime factor that does not occur as a factor of an earlier member of the sequence, except for . For example, , , , ... have the factors 3, 7, 5, 31, (1), 127, 17, 73, 11, , ... (OEIS A064078) that do not occur in earlier . These factors are sometimes called the Zsigmondy numbers .
Zsigmondy's theorem is often useful, especially in group theory, where it is used to prove that various groups have distinct orders except when they are known to be the same (Montgomery 2001).
See also
Mersenne Number
Explore with Wolfram|Alpha
More things to try:
prime factorization
(1 - 1/3 + 1/5) / (1/2 - 1/4 + 1/6)
complete the square x^2+10x+28
References
Chabaud, F. and Vaudenay, S. "Links between Differential and Linear Cryptanalysis." EUOROCRYPT 94, pp. 356-365, 1994.Montgomery, H. "Divisibility of Mersenne Numbers." 17 Sep 2001. P. The Little Book of Big Primes. New York: Springer-Verlag, p. 27, 1991.Sloane, N. J. A. Sequence A064078 in "The On-Line Encyclopedia of Integer Sequences."Zsigmondy, K. "Zur Theorie der Potenzreste." Monatshefte für Math. u. Phys. 3, 265-284, 1882.
Referenced on Wolfram|Alpha
Zsigmondy Theorem
Cite this as:
Weisstein, Eric W. "Zsigmondy Theorem." From MathWorld--A Wolfram Resource.
Subject classifications |
8105 | https://coolconversion.com/weight/454-gram-to-pound | 454 Grams to Pounds
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How many pounds in 454 grams?
454 grams equals 1.0009 pounds
Grams To Pounds Converter
Physics Chemistry Recipes
⇆
454 grams = 1.0009 pounds Formula: multiply the value in grams by the conversion factor '0.0022046226218502'. So, 454 grams = 454 × 0.0022046226218502 = 1.00089867032 pounds. #### Conversion of 454 grams to other weight (mass) units 454 grams = 0.454 kilograms 454 grams = 454000 milligrams 454 grams = 0.000454 tonnes 454 grams = 16 ounces 454 grams = 0.0005 tons [short, US] 454 grams = 0.000447 tons [long, UK] 454 grams = 0.0715 stones 454 grams = 14.6 troy ounces 454 grams = 1.22 troy pounds 454 grams = 256 drams 454 grams = 119 drachmes 454 grams = 7010 grains 454 grams = 0.00894 hundredweights [long, UK] 454 grams = 0.01 hundredweights [short, US] 454 grams = 227000 points 454 grams = 0.04 quarters [US] 454 grams = 0.0357 quarters [UK] 454 grams = 45.4 Robies By coolconversion.com
Website Map
To calculate a gram value to the corresponding value in pound, multiply the quantity in gram by 0.0022046226218488 (the conversion factor). Here is the formula:
Value in pounds = value in grams × 0.0022046226218488
Suppose you want to convert 454 gram into pounds. Using the conversion formula above, you will get:
Value in pound = 454 × 0.0022046226218488 = 1.0009 pounds
Comprehensive guide to mass unit conversion: grams to pounds and more
Understanding how to convert mass measurements like grams to pounds, and vice versa, is important to achieve exactness in your calculations.
How to convert grams (g) to pounds (lb)
To convert from grams to pounds:
Use the conversion factor: 1 gram equals 0.00220462 pounds.
For example, to convert 454g to lb, calculate 454 x 0.00220462lb, which is 1.0009lb.
The formula is: mass in lb = mass in g x 0.00220462
How to convert lb to grams (g)
For converting pounds to grams:
Remember, 1 pound is approximately 453.592 grams.
To convert 1.0009lb to g, multiply 1.0009 x 453.592, resulting in 454g.
The conversion formula is: mass in g = mass in lb x 453.592
These formulas provide a reliable method for speed unit conversions across different systems.
This converter can help you to get answers to questions like:
How many grams are in 454 pounds?
454 grams are equal to how many pounds?
How much are 454 gram in pounds?
How to convert grams to pounds?
What is the conversion factor to convert from grams to pounds?
How to transform grams in pounds?
What is the formula to convert from grams to pounds? Among others.
Conversion Table: Grams to Pounds near 454 grams
Conversion Table: Grams to Pounds
| Grams | = | Pounds |
---
| 364 grams | = | 0.80 pounds |
| 374 grams | = | 0.82 pounds |
| 384 grams | = | 0.85 pounds |
| 394 grams | = | 0.87 pounds |
| 404 grams | = | 0.89 pounds |
| 414 grams | = | 0.91 pounds |
| 424 grams | = | 0.93 pounds |
| 434 grams | = | 0.96 pounds |
| 444 grams | = | 0.98 pounds |
| 454 grams | = | 1.00 pounds |
| 464 grams | = | 1.02 pounds |
| 474 grams | = | 1.04 pounds |
| 484 grams | = | 1.07 pounds |
| 494 grams | = | 1.09 pounds |
| 504 grams | = | 1.11 pounds |
| 514 grams | = | 1.13 pounds |
| 524 grams | = | 1.16 pounds |
| 534 grams | = | 1.18 pounds |
| 544 grams | = | 1.20 pounds |
| 554 grams | = | 1.22 pounds |
⬇ CSV
Note: values are rounded to 2 decimal places for clarity.
Sample Weight / Mass Conversions
90 mg to st
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8106 | https://pubmed.ncbi.nlm.nih.gov/15938605/ | Plating of the distal radius - PubMed
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J Am Acad Orthop Surg
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. 2005 May-Jun;13(3):159-71.
doi: 10.5435/00124635-200505000-00003.
Plating of the distal radius
Arvind D Nana1,Atul Joshi,David M Lichtman
Affiliations Expand
Affiliation
1 Department of Orthopaedic Surgery, JPS Health Network, Fort Worth, TX 76104, USA.
PMID: 15938605
DOI: 10.5435/00124635-200505000-00003
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Plating of the distal radius
Arvind D Nana et al. J Am Acad Orthop Surg.2005 May-Jun.
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. 2005 May-Jun;13(3):159-71.
doi: 10.5435/00124635-200505000-00003.
Authors
Arvind D Nana1,Atul Joshi,David M Lichtman
Affiliation
1 Department of Orthopaedic Surgery, JPS Health Network, Fort Worth, TX 76104, USA.
PMID: 15938605
DOI: 10.5435/00124635-200505000-00003
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Abstract
Distal radius fractures are common injuries that can be treated by a variety of methods. Restoration of the distal radius anatomy within established guidelines yields the best short- and long-term results. Guidelines for acceptable reduction are (1) radial shortening < 5 mm, (2) radial inclination > 15 degrees , (3) sagittal tilt on lateral projection between 15 degrees dorsal tilt and 20 degrees volar tilt, (4) intra-articular step-off < 2 mm of the radiocarpal joint, and (5) articular incongruity < 2 mm of the sigmoid notch of the distal radius. Treatment options range from closed reduction and immobilization to open reduction with plates and screws; options are differentiated based on their ability to reinforce and stabilize the three columns of the distal radius and ulna. Plating allows direct restoration of the anatomy, stable internal fixation, a decreased period of immobilization, and early return of wrist function. Buttress plates reduce and stabilize vertical shear intra-articular fractures through an antiglide effect, where-as conventional and locking plates address metaphyseal comminution and/or preserve articular congruity/reduction. With conventional and locking plates, intra-articular fractures are directly reduced; with buttress plates, the plate itself helps reduce the intra-articular fracture. Complications associated with plating include tendon irritation or rupture and the need for plate removal.
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Digital SAT Math
Course: Digital SAT Math>Unit 4
Lesson 3: Operations with polynomials: foundations
Operations with polynomials | Lesson
Operations with polynomials — Basic example
Operations with polynomials — Harder example
Operations with polynomials: foundations
Test prep>
Digital SAT Math>
Foundations: Advanced math>
Operations with polynomials: foundations
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Operations with polynomials | Lesson
Google Classroom
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A guide to operations with polynomials on the digital SAT
What are polynomial expressions?
A polynomial expression has one or more terms with a coefficient, a variable base, and an exponent.
3 x 4coefficientbaseexponent
3 x 4 is a
monomial
. We'll also frequently see
binomials
and
trinomials
.
3 x 4+2 x is a binomial. The exponent of the term 2 x is 1 (x=x 1).
3 x 4+2 x+7 is a trinomial. 7 is a constant term. We can also think of 7 as an exponential term with an exponent of 0. Since x 0=1, 7 is equivalent to 7 x 0.
In this lesson, we'll learn to add, subtract, and multiply polynomials.
You can learn anything. Let's do this!
How do I add and subtract polynomials?
Adding polynomials
Khan Academy video wrapper
Adding polynomialsSee video transcript
What should I be careful of when adding and subtracting polynomials?
While we can add and subtract any polynomials, we can only combine like terms, which must have:
The same variable base
The same exponent
For example, we can combine the terms 2 x 3 and 4 x 3 because they have the same variable base, x, and the same exponent, 3. However, we cannot combine the terms 2 x 2 and 2 x 3 because they have different exponents, 2 and 3.
When we combine like terms, only the coefficients change. Both the base and the exponent remain the same. For example, when adding 2 x 3 and 4 x 3, the x 3 part of the terms remain the same, and we add only 2 and 4 when combining the terms:
2 x 3+4 x 3=(2+4)x 3=6 x 3
When subtracting polynomials, make sure to distribute the negative sign as needed. For example, when subtracting the polynomial −2 x 2−7, the negative sign from the subtraction is distributed to both −2 x 2 and −7, which means:
5 x 2−(−2 x 2−7)=5 x 2+(−1)(−2 x 2)+(−1)(−7)=5 x 2+2 x 2+7=(5+2)x 2+7=7 x 2+7
Subtracting −2 x 2−7 is equivalent to adding 2 x 2+7!
To add or subtract two polynomials:
Group like terms.
For each group of like terms, add or subtract the coefficients while keeping both the base and the exponent the same.
Write the combined terms in order of decreasing power.
Examples: Adding and subtracting two binomials
(x 2+2 x+3)+(4 x 2+5 x+6)=x 2+2 x+3+4 x 2+5 x+6=x 2+4 x 2+2 x+5 x+3+6=5 x 2+7 x+9
(4 x 2−7 x+1)−(2 x 2−5 x−1)=4 x 2−7 x+1+(−1)(2 x 2)+(−1)(−5 x)+(−1)(−1)=4 x 2−7 x+1−2 x 2+5 x+1=4 x 2−2 x 2−7 x+5 x+1+1=2 x 2−2 x+2
Try it!
TRY: Match the equivalent expressions
Because 9 x 2 and 3 x 2 have
description of bases
and
description of exponents
, the two terms
can/cannot be combined
into a single term.
Because 4 y 4 and 4 y have
description of bases/exponents
, the two terms
can/cannot be combined
into a single term.
Check Explain
TRY: Match the equivalent expressions
Match each polynomial expression below with an equivalent expression.
| Expression | is equivalent to... |
--- |
| (x 2+6 x)+(−x 2−x) (x 2+6 x)+(x 2+x) (x 2+6 x)−(−x 2+x) (x 2+6 x)−(x 2−x) | 5 x 7 x 2 x 2+5 x 2 x 2+7 x |
Check Explain
How do I multiply polynomials?
Multiplying binomials
Khan Academy video wrapper
Multiplying binomialsSee video transcript
What should I be careful of when multiplying polynomials?
When multiplying two polynomials, we must make sure to distribute each term of one polynomial to all the terms of the other polynomial. For example:
(a x+b)(c x+d)=(a x)(c x+d)+(b)(c x+d)=(a x)(c x)+(a x)(d)+(b)(c x)+(b)(d)
The total number of products we need to calculate is equal to the product of the number of terms in each polynomial. Multiplying two binomials requires 2⋅2=4 products, as shown above. Multiplying a monomial and a trinomial requires 1⋅3=3 products; multiplying a binomial and a trinomial requires 2⋅3=6 products.
When multiplying two binomials, we can also use the mnemonic FOIL to account for all four multiplications. For (a x+b)(c x+d):
Multiply the First terms (a x⋅c x)
Multiply the Outer terms (a x⋅d)
Multiply the Inner terms (b⋅c x)
Multiply the Last terms (b⋅d)
When multiplying terms of polynomial expressions with the same base:
Multiply the coefficients, or multiply the coefficient and the constant.
Keep the base the same.
Add the exponents.
a x m⋅b x n=a b⋅x m+n a⋅b x n=a b⋅x n
To multiply two polynomials:
Distribute the terms.
Multiply the distributed terms according to the exponent rules above.
Group like terms.
For each group of like terms, add or subtract the coefficients while keeping both the base and the exponent the same.
Write the combined terms in order of decreasing power.
Let's look at some examples!
What is the product of 2 x−1 and x−5 ?
Show me!
(2 x−1)(x−5)=(2 x)(x)+(2 x)(−5)+(−1)(x)+(−1)(−5)=2⋅x 1+1+(−10 x)+(−x)+5=2 x 2+(−10 x−x)+5=2 x 2+(−11 x)+5=2 x 2−11 x+5
What is the product of 3 x and x 2−4 x+9 ?
Show me!
(3 x)(x 2−4 x+9)=(3 x)(x 2)+(3 x)(−4 x)+(3 x)(9)=3⋅x 1+2+(3⋅−4)x 1+1+(3⋅9)x=3 x 3+(−12 x 2)+27 x=3 x 3−12 x 2+27 x
Try it!
TRY: multiply two terms
When multiplying 3 x and 2 x 2, we
add/multiply
the coefficients of the terms and
add/multiply
the exponents of x.
(3 x)(2 x 2)=
Check Explain
TRY: Multiply two binomials using FOIL
Use the table below to FOIL (8 x−3)(x 2+1).
| Term | Expression | Product |
---
| First | 8 x⋅x 2 | |
| Outer | | 8 x |
| Inner | −3⋅x 2 | |
| Last | | −3 |
Check Explain
Your turn!
Practice: add two binomials
Which of the following is the sum of x 2+5 and 2 x 2+4 x ?
Choose 1 answer:
Choose 1 answer:
(Choice A) 3 x 2+4 x+5 A 3 x 2+4 x+5
(Choice B) 3 x 2+9 x B 3 x 2+9 x
(Choice C) 7 x 2+4 x C 7 x 2+4 x
(Choice D) 12 x 2 D 12 x 2
Check Explain
Practice: subtract two polynomials to find a coefficient
(9 x 2+5 x−1)−(6 x 2−4 x)=a x 2+b x+c
The equation above is true for all x, where a, b and c are constants. What is the value of b ?
Check Explain
Practice: multiply two binomials
Which of the following is equivalent to (x+3)(2 x−5) ?
Choose 1 answer:
Choose 1 answer:
(Choice A) x 2−2 x−15 A x 2−2 x−15
(Choice B) 2 x 2−2 x−15 B 2 x 2−2 x−15
(Choice C) 2 x 2+x−15 C 2 x 2+x−15
(Choice D) 2 x 2+6 x−5 D 2 x 2+6 x−5
Check Explain
Practice: multiply two binomials with symbolic coefficients
(a x+3)(b x+2)=9 x 2+21 x+6
In the equation above, a and b are constants. What is the value of a b ?
Check Explain
Things to remember
The mnemonic FOIL for multiplying two binomials:
Multiply the First terms
Multiply the Outer terms
Multiply the Inner terms
Multiply the Last terms
a x m⋅b x n=a b⋅x m+n a⋅b x n=a b⋅x n
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8108 | https://www.webqc.org/balanced-equation-C8H18+O2=CO2+H2O | C8H18 + O2 = CO2 + H2O - Balanced chemical equation, limiting reagent and stoichiometry
Printed from
Balance Chemical Equation - Online Balancer
Enter a chemical equation to balance:
Balanced equation:
2 C 8 H 18 + 25 O 2 = 16 CO 2 + 18 H 2 O
Reaction type: combustion
Discover more
Water
H₂O
H2O
water
| Reaction stoichiometry | Limiting reagent |
--- |
| Compound | Coefficient | Molar Mass | Moles | Weight |
| Reagents |
| C 8 H 18 | 2 | 114.23 | | |
| O 2 | 25 | 32.00 | | |
| Products |
| CO 2 | 16 | 44.01 | | |
| H 2 O | 18 | 18.02 | | |
Units: molar mass - g/mol, weight - g.
| Word equation |
| 2 Octane + 25 Oxygen = 16 Carbon dioxide + 18 Water |
| Balancing step by step using the inspection method |
| Let's balance this equation using the inspection method. First, we set all coefficients to 1: 1 C 8 H 18 + 1 O 2 = 1 CO 2 + 1 H 2 O For each element, we check if the number of atoms is balanced on both sides of the equation. C is not balanced: 8 atoms in reagents and 1 atom in products. In order to balance C on both sides we: Multiply coefficient for CO 2 by 8 1 C 8 H 18 + 1 O 2 = 8 CO 2 + 1 H 2 O H is not balanced: 18 atoms in reagents and 2 atoms in products. In order to balance H on both sides we: Multiply coefficient for H 2 O by 9 1 C 8 H 18 + 1 O 2 = 8 CO 2 + 9 H 2 O O is not balanced: 2 atoms in reagents and 25 atoms in products. In order to balance O on both sides we: Multiply coefficient for O 2 by 25 Multiply coefficient(s) for CO 2, H 2 O, C 8 H 18 by 2 2 C 8 H 18 + 25 O 2 = 16 CO 2 + 18 H 2 O C is balanced: 16 atoms in reagents and 16 atoms in products. H is balanced: 36 atoms in reagents and 36 atoms in products. All atoms are now balanced and the whole equation is fully balanced: 2 C 8 H 18 + 25 O 2 = 16 CO 2 + 18 H 2 O |
| Balancing step by step using the algebraic method |
| Let's balance this equation using the algebraic method. First, we set all coefficients to variables a, b, c, d, ... a C 8 H 18 + b O 2 = c CO 2 + d H 2 O Now we write down algebraic equations to balance of each atom: C: a 8 = c 1 H: a 18 = d 2 O: b 2 = c 2 + d 1 Now we assign a=1 and solve the system of linear algebra equations: a 8 = c a8 = d 2 b 2 = c 2 + d a = 1 Solving this linear algebra system we arrive at: a = 1 b = 12.5 c = 8 d = 9 To get to integer coefficients we multiply all variable by 2 a = 2 b = 25 c = 16 d = 18 Now we substitute the variables in the original equations with the values obtained by solving the linear algebra system and arrive at the fully balanced equation: 2 C 8 H 18 + 25 O 2 = 16 CO 2 + 18 H 2 O |
Direct link to this balanced equation:
Please tell about this free chemistry software to your friends!
Instructions on balancing chemical equations:
Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
To enter an electron into a chemical equation use {-} or e
To enter an ion, specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are, enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink.
Examples of complete chemical equations to balance:
Fe + Cl 2 = FeCl 3
KMnO 4 + HCl = KCl + MnCl 2 + H 2 O + Cl 2
K 4 Fe(CN)6 + H 2 SO 4 + H 2 O = K 2 SO 4 + FeSO 4 + (NH 4)2 SO 4 + CO
C 6 H 5 COOH + O 2 = CO 2 + H 2 O
K 4 Fe(CN)6 + KMnO 4 + H 2 SO 4 = KHSO 4 + Fe 2(SO 4)3 + MnSO 4 + HNO 3 + CO 2 + H 2 O
Cr 2 O 7{-2} + H{+} + {-} = Cr{+3} + H 2 O
S{-2} + I 2 = I{-} + S
PhCH 3 + KMnO 4 + H 2 SO 4 = PhCOOH + K 2 SO 4 + MnSO 4 + H 2 O
CuSO 45H 2 O = CuSO 4 + H 2 O
calcium hydroxide + carbon dioxide = calcium carbonate + water
sulfur + ozone = sulfur dioxide
Examples of the chemical equations reagents (a complete equation will be suggested):
H 2 SO 4 + K 4 Fe(CN)6 + KMnO 4
Ca(OH)2 + H 3 PO 4
Na 2 S 2 O 3 + I 2
C 8 H 18 + O 2
hydrogen + oxygen
propane + oxygen
Understanding chemical equations
A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction). For example, in the reaction of hydrogen (H₂) with oxygen (O₂) to form water (H₂O), the chemical equation is:
H 2 + O 2 = H 2 O
However, this equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.
Balancing with inspection or trial and error method
This is the most straightforward method. It involves looking at the equation and adjusting the coefficients to get the same number of each type of atom on both sides of the equation.
Best for: Simple equations with a small number of atoms.
Process: Start with the most complex molecule or the one with the most elements, and adjust the coefficients of the reactants and products until the equation is balanced.
Example:H 2 + O 2 = H 2 O
1. Count the number of H and O atoms on both sides. There are 2 H atoms on the left and 2 H atom on the right. There are 2 O atoms on the left and 1 O atom on the right.
2. Balance the oxygen atoms by placing a coefficient of 2 in front of H 2 O: H 2 + O 2 = 2H 2 O
3. Now, there are 4 H atoms on the right side, so we adjust the left side to match: 2H 2 + O 2 = 2H 2 O
4. Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced.
Balancing with algebraic method
This method uses algebraic equations to find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, y, z), and a series of equations are set up based on the number of each type of atom.
Best for: Equations that are more complex and not easily balanced by inspection.
Process: Assign variables to each coefficient, write equations for each element, and then solve the system of equations to find the values of the variables.
Example: C 2 H 6 + O 2 = CO 2 + H 2 O
1. Assign variables to coefficients:a C 2 H 6 + b O 2 = c CO 2 + d H 2 O
2. Write down equations based on atom conservation:
2 a = c
6 a = 2 d
2 b = 2c + d
Assign one of the coefficients to 1 and solve the system.
a = 1
c = 2 a = 2
d = 6 a / 2 = 3
b = (2 c + d) / 2 = (2 2 + 3) / 2 = 3.5
Adjust coefficient to make sure all of them are integers. b = 3.5 so we need to multiply all coefficient by 2 to arrive at the balanced equation with integer coefficients: 2 C 2 H 6 + 7 O 2 = 4 CO 2 + 6 H 2 O
Balancing with oxidation number method
Useful for redox reactions, this method involves balancing the equation based on the change in oxidation numbers.
Best For: Redox reactions where electron transfer occurs.
Process: identify the oxidation numbers, determine the changes in oxidation state, balance the atoms that change their oxidation state, and then balance the remaining atoms and charges.
Example: Ca + P = Ca 3 P 2
1. Assign oxidation numbers:
Calcium (Ca) has an oxidation number of 0 in its elemental form.
Phosphorus (P) also has an oxidation number of 0 in its elemental form.
In Ca 3 P 2, calcium has an oxidation number of +2, and phosphorus has an oxidation number of -3.
Identify the changes in oxidation numbers:
Calcium goes from 0 to +2, losing 2 electrons (oxidation).
Phosphorus goes from 0 to -3, gaining 3 electrons (reduction).
Balance the changes using electrons: Multiply the number of calcium atoms by 3 and the number of phosphorus atoms by 2.
Write the balanced Equation:3 Ca + 2 P = Ca 3 P 2
Balancing with ion-electron half-reaction method
This method separates the reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined.
Best for: complex redox reactions, especially in acidic or basic solutions.
Process: split the reaction into two half-reactions, balance the atoms and charges in each half-reaction, and then combine the half-reactions, ensuring that electrons are balanced.
Example: Cu + HNO 3 = Cu(NO 3)2 + NO 2 + H 2 O
1. Write down and balance half reactions:Cu = Cu{2+} + 2{e} H{+} + HNO 3 + {e} = NO 2 + H 2 O
2. Combine half reactions to balance electrons. To accomplish that we multiple the second half reaction by 2 and add it to the first one:Cu + 2H{+} + 2HNO 3 + 2{e} = Cu{2+} + 2NO 2 + 2H 2 O + 2{e}
3. Cancel out electrons on both sides and add NO 3{-} ions. H{+} with NO 3{-} makes HNO 3 and Cu{2+} with NO 3{-} makes Cu(NO 3)3:Cu + 4HNO 3 = Cu(NO 3)2 + 2NO 2 + 2H 2 O
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8109 | https://cdn.aaai.org/ojs/7617/7617-13-11147-1-2-20201228.pdf | Fixing a Tournament Virginia Vassilevska Williams Computer Science Division UC Berkeley Berkeley, California 94720 Abstract We consider a very natural problem concerned with game manipulation. Let G be a directed graph where the nodes represent players of a game, and an edge from u to v means that u can beat v in the game. (If an edge (u, v) is not present, one cannot match u and v.) Given G and a “favorite” node A, is it possible to set up the bracket of a balanced single-elimination tourna-ment so that A is guaranteed to win, if matches occur as predicted by G? We show that the problem is NP-complete for general graphs. For the case when G is a tournament graph we give several interesting condi-tions on the desired winner A for which there exists a balanced single-elimination tournament which A wins, and it can be found in polynomial time.
Introduction Many championships use a single-elimination (also called knockout) format: the tournament proceeds in rounds; in each round players are paired up to play a game; the round winners move on to the next round, whereas the losers leave the tournament. This format is very common in sports. It also appears in the area of voting protocols where it is stud-ied as the (binary) cup voting rule (Chevaleyre et al. 2007).
How far a particular player can go in a single-elimination tournament can vary vastly depending on the initial tourna-ment bracket set-up. This work investigates the extent to which a tournament designer can influence the final tour-nament outcome by manipulating the initial brackets. We focus on the following set-up. Suppose we are to design a single-elimination tournament for some set of players. For each pair of players we have some information (obtained from history or by some other means) about which player is more likely to win in a match-up between the two. We have a favorite player A in mind, and we want to set the bracket for the tournament so that A has a very high chance of winning. We call this problem tournament fixing.
In this paper we consider two versions of the tournament fixing problem. In both versions we assume knowledge of the exact outcome of all match-ups between players.
1. In the first version only certain matches between players are allowed.
Copyright c ⃝2010, Association for the Advancement of Artificial Intelligence (www.aaai.org). All rights reserved.
2. In the second version we can match any two players against each other.
This is a well known formulation of the problem also known as agenda control (Bartholdi, Tovey, and Trick 1992).
In the first, more general version we show that tournament fixing is NP-hard; i.e., if P̸=NP even if we have perfect in-formation about the match outcomes, if some players cannot be matched with each other, it is infeasible to find a winning bracket layout for a given player.
This problem version is a special case of the following problem: we are given complete match outcome informa-tion and in addition, for every pair of players we are given an integer weight, corresponding to the revenue that would be generated if the two players are matched in the tourna-ment. The problem is to find a single-elimination tourna-ment which maximizes the total tournament revenue. If the weights capture “interestingness”, the problem is to find the most interesting single-elimination tournament. This rev-enue maximization problem has been shown by (Lang et al.
2007) to be NP-hard, and the NP-hardness of general tour-nament fixing provides an alternative proof of this.
The second version of the problem allows us to relate single-elimination tournaments to round-robin tournaments: tournaments in which every pair of players has played and we know the outcome for each pair. Round-robin tourna-ments are widely studied. A significant body of work con-cerns the problem of optimally ranking players: finding a linear order of the players, a ranking, which minimizes the number of pairs (u, v) such that u beat v but v is before u in the ranking. This objective dates back to Slater (Slater 1961) and has only recently been proven NP-hard (Alon 2006; Charbit, Thomass´ e, and Yeo 2007).
Nevertheless, it has been shown (Coppersmith, Fleischer, and Rudra 2006) that the simple heuristic of ranking the players by the number of matches they have won approximates the optimal ranking within a factor of 5.
Our study uncovers interesting relationships between round-robin and single-elimination tournaments.
For in-stance, we show that any player that beats a maximal number of players in a round-robin tournament (and is hence highly ranked), can win a single-elimination tournament, given the same match outcomes. We focus on the special case of the tournament fixing problem in which the favorite node is a king: for any player b that it cannot beat, there is some player 895 Proceedings of the Twenty-Fourth AAAI Conference on Artificial Intelligence (AAAI-10) that it can beat that beats b. We give several sufficient con-ditions for which a king is a single-elimination tournament winner. One of our results is for the case when A is a very strong king: for any player b that A cannot beat, there are at least log n players that A can beat who beat b. We show that such super-kings can win a single-elimination tournament, even though they may only be able to beat very few players head-to-head. An interesting consequence of this is that any tournament graph generated using the noisy sampling model of (Braverman and Mossel 2008) with error rate as low as Ω( q log n n ) has all its players as potential single-elimination tournament winners, with high probability. All of our ar-guments are constructive and allow us to design tournament brackets efficiently.
Prior and Related Work.
There is considerable prior re-search on manipulating the outcome of a single-elimination tournament. Typically, the input to the studied problem is a tournament graph with probabilities on the edges. One then seeks to find a single-elimination tournament which maxi-mizes the probability that a particular node will win. A ma-jor focus in such research is to maximize the winning proba-bility of the best player under some assumptions (e.g., (Ap-pleton 1995; Horen and Riezman 1985)). A common as-sumption is that the probability of a player beating another depends on the intrinsic abilities of the players which are hidden values. Moreover, the probability is monotone in the sense that every player has a higher probability of beating a weaker player than a stronger one. With these assumptions some positive results are possible, especially for small tour-naments (Appleton 1995; Horen and Riezman 1985). On the other hand, if the given tournament graph has arbitrary prob-abilities and one wishes to determine whether a given node can win a single-elimination tournament with high proba-bility, then the known results are mostly negative: this gen-eral problem is known to be NP-hard (Lang et al. 2007; Hazon et al. 2008). Even in the case when the probabili-ties are in {0, 1, 1/2}, the problem is NP-hard, as shown by Vu et al. (Vu, Altman, and Shoham 2008; 2009). The major open problem is whether tournament fixing for unweighted tournament graphs is NP-hard.
Tournament manipulation is also studied under the con-text of voting (Brams and Fishburn 2002; Hemaspaandra, Hemaspaandra, and Rothe 2007; Lang et al. 2007). In this context, the candidates are competing in an election based on majority comparisons along a binary voting tree. In each comparison, the candidate with fewer votes is eliminated, and the winner moves on to the next comparison. This set-ting is essentially a single-elimination tournament, the re-sult of each match of which is known in advance: the corre-sponding tournament graph is unweighted. It is known that if the voting tree has no prescribed structure, then there is a polynomial time algorithm to decide whether there exists a voting tree for which a given candidate wins the election.
Finding a balanced voting tree in polynomial time is a major open problem and is known in this area as the agenda con-trol problem (Bartholdi, Tovey, and Trick 1992). (Fischer, Procaccia, and Samorodnitsky 2009) consider whether (po-tentially unbalanced) voting trees can be used to always elect a candidate preferred by close to the majority of voters; i.e.
how well binary tree protocols approximate the Copeland winner. Fischer et al. give hardness results for deterministic trees, and show that by randomizing over voting trees one can obtain much better approximations.
Preliminaries For any graph G = (V, E), n = |V | and m = |E|, unless otherwise noted. For any node v ∈V , let Nin(v) = {x ∈V | (x, v) ∈E} and Nout(v) = {x ∈ V | (v, x) ∈E}. If v ∈V and S ⊆V , let Nin,S(v) = {x ∈ S | (x, v) ∈E} and Nout,S(v) = {x ∈S | (v, x) ∈E}.
The length of a path in a graph is the number of edges on the path. An arborescence is a rooted tree such that all edges are directed away from the root. The height of an arborescence is the length of the longest path from the root to a leaf. For integer n ≥1, let [n] = {1, . . ., n}. A tournament graph is a directed graph G = (V, E) such that for every pair u, v ∈ V , exactly one of (u, v) or (v, u) is in E. A node a in a directed graph G = (V, E) is a king if for any node b ∈ V \ {a}, either (a, b) ∈E, or there exists a node c ∈V with (a, c) ∈E and (c, b) ∈E.
A binomial arborescence T = (V (T ), E(T )) rooted at a ∈V (T ) is defined recursively as follows: • a single node a is a binomial arborescence rooted at a; • if |V (T )| = 2i for some i > 0, then T is a binomial ar-borescence if a has a child b, (i.e., (a, b) ∈E(T )) such that if Tb is the subarborescence of T rooted at b and Ta = T \ Tb, then Ta and Tb are |V (T )|/2 = 2i−1–node binomial arborescences rooted at a and b respectively.
T = (V (T ), E(T )) is a binomial spanning arborescence for a graph G = (V, E) if V (T ) = V, E(T ) ⊆E and T is a binomial arborescence.
Binomial Arborescences and Hardness Tournament graphs are widely used in the study of round-robin tournaments. The nodes of a tournament graph typi-cally represent players, and an edge from u to v in the graph states that u beats v. We use the same representation. How-ever, since in our more general formulation of tournament fixing some match outcomes are not needed, we do not need an edge between each pair of vertices. Our abstraction of the tournament fixing problem (TFP) can be formalized as follows: Let G = (V, E) be a directed graph on n nodes such that if (u, v) ∈E then (v, u) / ∈E. We are given a node A ∈V and we are asked whether in G there exists a spanning binomial arborescence rooted at A. If this is the case, we call A a binomial winner in G.
Binomial arborescences on n nodes represent single-elimination tournaments in the following sense: the node at which the arborescence is rooted is the winner of the tour-nament and its log n children are the players the root beats in the log n rounds; for i = 1, . . . , log n the ith child of the root has n/2log n−i+1 descendents, and the subarborescence rooted at that child inductively represents the subtournament which that child had to win to get to round i in which it lost to the root player. The edge between the root and the log n-th child represents the final of the tournament.
896 TFP is a spanning arborescence isomorphism problem for directed graphs. A spanning tree isomorphism problem in an undirected graph has the following form: given a fixed prop-erty P of trees checkable in polynomial time and a graph G, find a spanning tree of G satisfying P. (Papadimitriou and Yannakakis 1982) gave necessary and sufficient condi-tions under which a spanning tree isomorphism problem in an undirected graph is NP-complete. Their result implies that finding a binomial spanning tree in an undirected graph is NP-hard. The Papadimitriou and Yannakakis construction does not immediately imply that TFP is NP-complete, as TFP is a problem on directed graphs. It is possible that their proofs can be modified to work for rooted arborescences, al-though we do not know of such a result for directed graphs.
We include a simple NP-completeness proof for TFP below.
Theorem 1. TFP is NP-complete.
Proof. TFP is clearly in NP. To show NP-hardness, we give a reduction from Exact Cover by 4-Sets which is well known to be NP-complete (Karp 1972; Garey and Johnson 1979).
A pictorial representation of the reduction appears in Fig-ure 1. Let (U, {S1, . . . , Sk}) be an instance of Exact Cover by 4-sets, where |U| = 4n and |Si| = 4 for all i ∈[k]. The problem is to determine whether there is a way to pick n sets from S which are disjoint and cover all elements of U.
W.l.o.g. k is a power of 2: we can always add copies of the same set Si to the instance. We will construct an instance of TFP: a digraph G on 16k nodes, and a node x1 of G for which one would need to fix the tournament.
First, create a node u for each u ∈U, slightly abus-ing notation. For every set Si = {u1, u2, u3, u4} create 4 nodes, si 1, . . . , si 4. Add edges (si 1, si 2), (si 1, si 3), (si 2, si 4).
This creates a binomial arborescence ¯ Si of size 4 rooted at si 1.
Further, add edges (si 2, u1), (si 3, u2), (si 4, u3), si 1, u4). This makes (si 1, si 2, si 3, si 4, u1, u2, u3, u4) a bino-mial arborescence of size 8 rooted at si 1.
Create k −n disjoint binomial arborescences of size 4, T1, . . . , Tk−n, rooted at v1, . . . , vk−n respectively.
Add edges (vi, sj 1) for all i = 1, . . . , k −n and j = 1, . . . , k.
Then create k disjoint binomial arborescences of size 8, X1, . . . , Xk, with roots x1, . . . , xk, respectively. Add edges (xi, vi) for i = 1, . . . , k −n, and (xj, st 1) for j = (k − n + 1), . . . , k and t = 1, . . . , k. Finally, create a binomial arborescence Bk on {x1, . . . , xk} rooted at x1. This com-pletes the construction of graph G. The number of nodes in G is 4n + 4k + 4(k −n) + 8k = 16k, a power of 2.
The construction ensures that for any binomial arbores-cence B spanning G: • for i = 1, . . . , k, ¯ Si is a subarborescence of B, • for i = 1, . . . , k −n, Ti is a subarborescence of B, • for i = 1, . . . , k, Xi is a subarborescence of B, • Bk is a subarborescence of B, and B is rooted at x1, • in B, the elements of U are partitioned into groups of 4 each of which is attached to some ¯ Si forming a binomial arborescence of size 8, • in B, each subarborescence ¯ Si which does not have ele-ments of U attached to it, is linked to some Tj to form a binomial arborescence of size 8, Figure 1: This is a pictorial description of the reduction from Exact Cover by 4-Sets to TFP.
• each Tj is linked to some arborescence ¯ Si to form a bino-mial arborescence of size 8, • each Xj with j ≤k−n is linked to Tj forming a binomial arborescence of size 16 rooted at xj, • each Xj with j > k −n is linked to some arborescence ¯ Si which itself is linked to 4 elements of U, so that this forms a binomial arborescence of size 16 rooted at xj.
The above points state that if there is a binomial spanning arborescence B in G rooted at x1, then there is an exact cover of U using the sets Si for which the arborescences (si 1, si 2, si 3, si 4) are linked to 4 elements of U in B. Vice versa, if there is an exact cover C = {Si1, . . . , Sin}, we can create a binomial arborescence as follows: for each Sij = {u1, u2, u3, u4}, attach the 4 nodes u1, u2, u3, u4 to ¯ Sij. Match the k −n sets Si / ∈C to distinct arborescences Tj and attach ¯ Si to its corresponding Ti. We have a col-lection of k size 8 binomial arborescences. We can attach each of them to some Xt to form k binomial arborescences of size 16. By adding the edges of Bk we obtain a full size 16k binomial arborescence rooted at x1.
It is unclear whether it is possible to modify this proof to show that TFP is NP-complete for tournament graphs. How-ever, we can modify it to show such a result for complete binary arborescences (the proof appears in the full version of the paper). Binomial arborescences are similar to binary arborescences in that their height is logarithmic; perhaps this is an indication that tournament-TFP is NP-complete.
However, binomial spanning arborescences are also similar to Hamiltonian paths, in that both structures always exist in a tournament graph. A Hamiltonian path rooted at a given node in a tournament can be found in polynomial time; per-haps this is an indication that tournament-TFP is in P.
Theorem 2. Given a tournament graph G and a node A, deciding whether there is a complete binary arborescence spanning G rooted at A is NP-complete.
897 When All Matches Are Allowed In this section we consider the problem of fixing a tourna-ment when we are allowed to match any pair of players. In the graph representation we are given a tournament graph and a node A and we need to find a binomial spanning ar-borescence rooted at A. Here we consider some conditions under which a king in a tournament graph is a binomial win-ner. We focus on kings because they are strong players in the sense that they either have a very high winning record, or can beat some very highly ranked players. Moreover, for any graph G, if G contains a binomial spanning arborescence rooted at a node A, then any breadth first search arbores-cence starting from A must span G and must have height at most log n. The simplest nontrivial case for which this necessary condition is satisfied is when A is a king.
Tournament graphs are typically used to represent the out-comes of round-robin tournaments, and the nodes/players are often ranked by their outdegree, also called score. As mentioned in the introduction, the outdegree ranking is a good approximation to the optimal tournament ranking in the sense that it minimizes (within a constant factor) the number of matches in which a lower ranked player beat a higher ranked player. A node which has maximal outdegree, i.e., outdegree at least as high as that of any other vertex, is among the top players in the round-robin ranking. We show that we can design a single-elimination tournament for any maximal degree node so that it would win given the same match outcomes. This gives an interesting and intuitive re-lation between round-robin and single-elimination tourna-ments – a very strong player should be able to win either.
We give proofs of two more general statements which both imply the result for maximal degree nodes.
When a node has maximal outdegree, it is also a king in the tournament graph (this is the usual proof that a king al-ways exists). Yet a node A is also a king if it does not nec-essarily have maximal outdegree, but for every node b that A cannot beat, the outdegree of A is at least as large as that of b. The first statement we prove states that any such node A is a binomial winner. We note that the above outdegree condition allows for the outdegree of A to be as low as n/3.
In contrast, a node with maximal outdegree has outdegree at least n/2.
In the second part of this section we show that any king with outdegree at least n/2 is a binomial winner. Further-more, we show that this condition is tight in the sense that for any n power of 2 there exists a tournament graph and a king A with outdegree n/2 −1 such that A cannot win any single-elimination tournament. Finally, we turn our at-tention to very strong kings, called super-kings, and show that such nodes are binomial winners even if they have low outdegree (as low as log n).
Nodes Stronger than the Nodes that Beat Them.
Theorem 3. Let G = (V, E) be a tournament graph. Let A ∈V such that for all v ∈Nin(A), degout(v) ≤ degout(A). Then one can construct in polynomial time a binomial spanning arborescence of G rooted at A.
Proof. Suppose we have partitioned the vertices of B into binomial arborescences (of possibly different sizes) rooted at nodes of A. Lemma 1 below allows us to do that. Con-sider the vertices of A which were not used in creating these arborescences by the Lemma. Partition these vertices (arbi-trarily) into sets of sizes powers of 2, creating corresponding binomial arborescences (arbitrarily). Doing this, we parti-tion A∪B into disjoint binomial arborescences {S1, S2, . . .} rooted at nodes of A, each of a size power of 2. Now, if there are two arborescences of the same size 2k, link them by adding the edge between their roots to create a bino-mial arborescence of size 2k+1. Continue doing this until there is at most one arborescence of each size.
Because |A ∪B| = 2log n −1 = Plog n−1 i=0 2i, there will be a bino-mial arborescence of size 2i for every i from 0 to log n −1.
Link the root of each of these arborescences to A. Since the arborescences are rooted at vertices from A this will form a binomial arborescence of size n rooted at A.
Now it remains to prove Lemma 1. We begin by showing the claim below.
Claim 1. Let A = Nout(A) and B = Nin(A) in a tourna-ment graph. Suppose for all b ∈B, degout(b) ≤degout(A).
Then ∀b ∈B, degout,B(b) < degin,A(b).
Proof of Claim 1: For any b ∈ B, degout(b) ≤ degout(A). Also, degout(b) = degout,B(b) + degout,A(b) + 1, degout(A) = |A|, and degout,A(b) = |A| −degin,A(b).
Hence, 1 + degout,B(b) + |A| −degin,A(b) ≤|A|, and degout,B(b) < degin,A(b).
Lemma 1. Given nonempty sets A′ ⊆A and B′ ⊆B such that for all b ∈B′, degout,B′(b) < degin,A′(b), one can pick a node a′ ∈A′ and a subset S ⊆Nout,B′(a′) so that 1. |S ∪{a′}| = 2k for some integer k ≥1, and 2. ∀b′ ∈B′ \ S, degout,B′\S(b′) < degin,A′{a′}(b′).
Proof. Since B′ is nonempty and for all b ∈ B′, degout,B′(b) < degin,A′(b), there exists some a′ ∈A′ which has an out-neighbor in B′. Pick one such a′ and let N = Nout,B′(a′). For some integers k ≥1 and r we have 0 ≤r ≤2k −1 and |N| = 2k −1 + r ≥2r. We can find a matching in N of size at least r. Pick a submatch-ing of size r, consisting of a set R of r vertices uniquely pointing to some other r vertices in N. Let S = N \ R.
Clearly, |S| = 2k −1 + r −r = 2k −1, and hence |S ∪{a′}| = 2k. We will show that for all b′ ∈B′ \ S, degout,B′\S(b′) < degin,A′{a′}(b′). Let b′ ∈B′ \ S. If b′ / ∈N, then degout,B′\S(b′) < degin,A′{a′}(b′) since degout,B′\S(b′) ≤degout,B′(b′) < degin,A′(b′). If b′ ∈ N, then degin,A′{a′}(b′) = degin,A′(b′) −1. Yet, since b′ ∈N ∩{B′ \ S}, we must have b′ ∈R, and b′ must have at least one outneighborin S. Hence degout,B′\S(b′) ≤ degout,B′(b′) −1, and since degout,B′(b′) < degin,A′(b′), degout,B′\S(b′) < degin,A′{a′}(b′).
Kings who Beat Half the Players.
Suppose now that we have a king who is not necessarily the strongest player but is 898 still relatively strong – he can beat at least half of the players.
It is not immediately clear that this king is binomial winner.
In fact, if the king can only beat n 2 −1 players, there are tournament graphs for which such a king may not be able to win at all: Claim 2. For any n, power of 2, there exists a tournament graph on n nodes with a king A with outdegree n/2−1 such that there is no binomial spanning arborescence rooted at A.
Proof. Consider the following graph G and king node A: let A be the out-neighborhood of A and let B be the in-neighborhood of A, so that |A| = n 2 −1 and |B| = n 2 . Let a′ ∈A be a node so that a′ beats every node in B. This makes A a king. If x ∈A \ {a′}, then every node b ∈B beats x. The edges within A and B are arbitrary.
Then the only way for a binomial spanning arborescence T to be rooted at A is if all nodes of B are in the subarbores-cence rooted at a′: an element b ∈B can only be beaten by another element of B, or by a′. However, |B ∪{a′}| > n/2 and hence the height of the subarborescence of T rooted at a′ is at least log n. This means that this subarborescence is the entire T and T cannot be rooted at A.
We now show that if a king can beat at least half of the players then he is a binomial winner. This result is tight by Claim 2.
Theorem 4. Let A be a king in an n-node tournament graph G so that outdeg(A) ≥n/2. Then A is a binomial winner, and a binomial spanning arborescence rooted at A can be found in polynomial time.
Proof. The proof will proceed by induction on n. The base case is n = 2 and then it is trivially true. We will keep the invariant I that if N players are left in the tournament, A is one of these players and A is a king with outdegree at least N/2 in the induced tournament graph. The induction hypothesis for n is that if I holds for a node A in a tourna-ment graph G′ on n/2 nodes, then A is a binomial winner in G′. The proof will proceed by fixing a match-up for the first round of the tournament and showing that I holds for the first round winners and A.
Let A = Nout(A) and B = Nin(A). If B is empty, we are done: A will win any tournament. Otherwise, in Round 1 create a maximal matching ¯ M from A to B, and let K be all elements of A in ¯ M; k = | ¯ M| = |K| ≤|B|. The matching ¯ M ensures that all elements of K survive round 1.
For the rest of the |B| −k elements of B, pick any max-imum matching M of them; |M| = ⌊|B|−k 2 ⌋. We will show that A \ K is nonempty and we can pick an element a′ from it and match it with A, so that A survives round 1. For this it suffices to show that |A| −k ≥1.
If |B| −k is odd, there is one element b′ of B which remains unmatched. We will show that A \ (K ∪{a′}) is nonempty if |B| −k is odd, and hence there is at least one element a′′ which we can match with b′. For this it suffices to show that |A| −k −1 ≥1 if |B| −k is odd.
There is an even number of remaining unmatched ele-ments of A as n is even. Pick any matching on them. This completes round 1. Note that A and all elements of K are winners in this round, and hence A remains a king. We must show two things: 1. |A| −k ≥1, and if |B| −k is odd, |A| −k −1 ≥1, 2. the number of elements in A which survive round 1 is at least n/4.
First, |A| + |B| = n −1 is odd, and hence |B| is even iff |A| is odd. Hence, |B| −k is odd iff |A| −k −1 is odd.
Moreover, k ≤|B| ≤|A| −1 as |A| ≥n/2, and hence |A| −k −1 ≥0 and |A| −k ≥1. If |B| −k is odd, then |A| −k −1 ≥1 since |A| −k −1 is odd. We are done with part 1 above.
The number of elements in A which survive round 1 is at least k+ |A|−k−1 2 if |B|−k is even and k+ |A|−k−2 2 if |B|−k is odd. This number is always ⌊|A|+k−1 2 ⌋. Since k ≥1, at least ⌊|A| 2 ⌋≥⌊n 4 ⌋= n 4 elements of A survive round 1.
After this round we have a tournament on n/2 elements in which A is a king with outdegree at least n/4. By induction, A is a binomial winner.
Super-Kings.
We now define a super-king in a tourna-ment graph as a node A with the constraint that for any b ∈Nin(A), |Nin,Nout(A)(b)| ≥log n.
Theorem 5. Let G be a tournament graph and let A be a super-king in G. Then A is a binomial winner in G.
Proof. Let A = Nout(A) and let B = Nin(A). We will proceed by induction. If n = 1 we are done. Otherwise if n ≥2, notice that |A| ≥log n ≥1. Pick any node a′ ∈A and match it with A. Now, create a maximum matching M from A \ {a′} to B. The number of remaining unmatched nodes is n −2 −2|M|, which is even, hence we can pick some perfect matching M ′. Call the final matching N = {(A, a′)} ∪M ∪M ′.
Consider the set S of n/2 nodes which are sources in N.
If b ∈S ∩B, then |Nin,A(b) \ S| ≤1. This is since if a′′ ∈(Nin,A(b) \ S) and a′′ ̸= a′, then after the creation of M, a′′ was unmatched. Furthermore, since b ∈S, b must have been unmatched after the creation of M. This is a con-tradiction to the maximality of M. Thus, every b in the new inneighborhood of A has at least log n −1 = log(n/2) in-neighbors in A. The theorem follows by induction on n.
The Braverman–Mossel Noisy Sampling Model.
The following model was proposed in (Braverman and Mossel 2008). We are given a parameter q < 1/2 and an ordered list of n players, v1, . . . , vn which represents a sorted order of the players by their intrinsic abilities. The parameter q represents a noise rate. A tournament graph Tq is generated as follows: for every pair of nodes vi, vj with j > i indepen-dently at random one places an edge (vi, vj) with probability 1 −q, and with probability q the reverse edge (vj, vi). The smaller q is, the closer Tq is to a transitive tournament; the larger q is, the closer Tq is to a completely random tourna-ment. Here we show that q can be very small and still every player in Tq is a binomial winner, with high probability. In 899 other words, almost all tournaments generated in this model (for slightly large q) can be fixed for any player.1 Theorem 6. Let q > 4 q log n n . Then with high probability, all nodes in a random tournament Tq generated using the Braverman–Mossel model are binomial winners.2 Proof. Let q = c · q log n n−1 for some c > 4. We show that with high probability all nodes in Tq are super-kings. Let V = {v1, . . . , vn}. Call a node v “bad” if either • degout(v) ≤(c log n)/q, or • degout(v) > (c log n)/q and there exists some node b ̸= v such that |Nin,Nout(v)(b)| < log n.
Let U = V \ {v}. Let ui be the ith node of U. Let Zi be an indicator variable which is 1 if (v, ui) is an edge and 0 otherwise. Then Pr[Zi = 1] ≥q. Let Z = P i Zi; Z = degout(v). Then E[Z] = P i E[Zi] ≥q(n −1) = c p (n −1) log n. By a Chernoff bound: Pr[Z ≤(c log n)/q] ≤e−EZ2/2 ≤ e−(c(1−1/c)2/2)√ (n−1) log n = 1/2Ω(√n log n).
Now fix some b.
We are interested in the probability that b has < log n inneighbors from Nout(v), given that degout(v) > (c log n)/q.
Consider the random variable Y = degin,Nout(v)(b) = P u∈Nout(v) Yu, where Yu is 1 iff (u, b) is an edge. Pr[Yu = 1] ≥q and so E[Y ] = q degout(v) > c log n. Again by a Chernoff bound: Pr[Y < log n] ≤Pr[Y < (1/c)E[Y ]] ≤ e−(1−1/c)2/2E[Y ] ≤1/nc(1−1/c)2/(2 ln 2).
Let C = −2+c(1−1/c)2/(2 ln 2). By a union bound, the probability that v is bad is at most 1/2Ω(√n log n) +1/nC+1.
The probability that there exists a bad v is at most 1/2Ω(√n log n) + 1/nC ≤O(1/nC).
Hence with probability at least 1 −O(1/nC) all nodes are super-kings. C is at least (c −4)/(2 ln 2), and so C > 0 for any c > 4.
Acknowledgments.
The author would like to thank David Abraham, Ryan Williams, Guy Blelloch, Maverick Woo, Anupam Gupta, Noga Alon and Piotr Faliszewski for some helpful discussions, and the anonymous reviewers for their valuable feedback.
This material is based upon work supported by the Na-tional Science Foundation under Grant CCF-0832797 at Princeton University/IAS and Award #0937060 to the Com-puting Research Association for the CIFellows Project.
1This result crucially relies on knowing the match outcomes beforehand: for any tournament bracket that we pick before the coin flips, vn has only a qlog n chance of winning the tournament which is at most 1/poly(n) even when q = Θ(1).
2The statement of this theorem was suggested to the author by an anonymous reviewer.
References Alon, N. 2006. Ranking tournaments. SIAM J. Discret.
Math. 20(1):137–142.
Appleton, D. R. 1995. May the best man win? The Statisti-cian 44(4):529–538.
Bartholdi, J.; Tovey, C.; and Trick, M. 1992. How hard is it to control an election?
Mathematical and Computer Modeling 16(8/9):27–40.
Brams, S. J., and Fishburn, P. C. 2002. Voting procedures.
K. J. Arrow, A. K. Sen, and K. Suzumura, editors, Handbook of Social Choice and Welfare.
Braverman, M., and Mossel, E. 2008. Noisy sorting without resampling. In Proc. SODA, 268–276.
Charbit, P.; Thomass´ e, S.; and Yeo, A. 2007. The mini-mum feedback arc set problem is NP-hard for tournaments.
Combinatorics, Probability & Computing 16(1):1–4.
Chevaleyre, Y.; Endriss, U.; Lang, J.; and Maudet, N.
2007. A short introduction to computational social choice.
SOFSEM 2007: Theory and Practice of Computer Science 4362:51–69.
Coppersmith, D.; Fleischer, L.; and Rudra, A. 2006. Order-ing by weighted number of wins gives a good ranking for weighted tournaments. In Proc. SODA, 776–782.
Fischer, F.; Procaccia, A.; and Samorodnitsky, A. 2009. A new perspective on implementation by voting trees. In Proc.
EC, 31–40.
Garey, M., and Johnson, D.
1979.
Computers and In-tractability: A Guide to the Theory of NP-Completeness.
W.H. Freeman and Company.
Hazon, N.; Dunne, P.; Kraus, S.; and Wooldridge, M. 2008.
How to rig elections and competitions. In Proc. COMSOC.
Hemaspaandra, E.; Hemaspaandra, L. A.; and Rothe, J.
2007. Anyone but him: The complexity of precluding an alternative. Artif. Intell. 171(5/6):255–285.
Horen, J., and Riezman, R.
1985.
Comparing draws for single elimination tournaments.
Operations Research 33(2):249–262.
Karp, R. M. 1972. Reducibility among combinatorial prob-lems. In Complexity of Computer Computations, R.E. Miller and J.W. Thatcher (eds.), Plenum Press, New York, 85–104.
Lang, J.; Pini, M. S.; Rossi, F.; Venable, K. B.; and Walsh, T.
2007. Winner determination in sequential majority voting.
In Proc. IJCAI, 1372–1377.
Papadimitriou, C. H., and Yannakakis, M.
1982.
The complexity of restricted spanning tree problems. J. ACM 29(2):285–309.
Slater, P. 1961. Inconsistencies in a schedule of paired com-parisons. Biometrika 48(3/4):303–312.
Vu, T.; Altman, A.; and Shoham, Y. 2008. On the agenda control problem in knockout tournaments. In Proc. COM-SOC.
Vu, T.; Altman, A.; and Shoham, Y. 2009. On the complex-ity of schedule control problems for knockout tournaments.
In Proc. AAMAS, 225–232.
900 |
8110 | https://proofwiki.org/wiki/Quotient_Ring_is_Ring | Quotient Ring is Ring - ProofWiki
Quotient Ring is Ring
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Contents
1 Theorem
2 Proof
2.1 Well-definition of ++
2.2 Well-definition of ∘∘
2.3 Ring Axiom A A: Addition forms an Abelian Group
2.4 Ring Axiom M 0 M 0: Closure under Product
2.5 Ring Axiom M 1 M 1: Associativity of Product
2.6 Ring Axiom D D: Distributivity of Product over Addition
3 Sources
Theorem
Let (R,+,∘)(R,+,∘) be a ring.
Let J J be an ideal of R R.
Let (R/J,+,∘)(R/J,+,∘) be the quotient ring of R R by J J.
Then R/J R/J is also a ring.
Proof
First, it is to be shown that ++ and ∘∘ are in fact well-defined operations on R/J R/J.
Well-definition of ++
From Ideal is Additive Normal Subgroup that J J is a normal subgroup of R R under ++.
Thus, the quotient group(R/J,+)(R/J,+) is defined, and as a Quotient Group is Group, ++ is well-defined.
□◻
Well-definition of ∘∘
From Left Cosets are Equal iff Product with Inverse in Subgroup, we have:
x 1+J x 1+J==x 2+J x 2+J
⇝⇝x 1+(−x 2)x 1+(−x 2)∈∈J J
and:
y 1+J y 1+J==y 2+J y 2+J
⇝⇝y 1+(−y 2)y 1+(−y 2)∈∈J J
Hence from the definition of ideal:
(x 1+(−x 2))∘y 1(x 1+(−x 2))∘y 1∈∈J J
x 2∘(y 1+(−y 2))x 2∘(y 1+(−y 2))∈∈J J
Thus:
(x 1+(−x 2))∘y 1+x 2∘(y 1+(−y 2))(x 1+(−x 2))∘y 1+x 2∘(y 1+(−y 2))∈∈J Jas (J,+)(J,+) is a group
⇝⇝x 1∘y 1+(−(x 2∘y 2))x 1∘y 1+(−(x 2∘y 2))∈∈J JVarious ring properties
⇝⇝x 1∘y 1+J x 1∘y 1+J==x 2∘y 2+J x 2∘y 2+JLeft Cosets are Equal iff Product with Inverse in Subgroup
□◻
Now to prove that (R/J,+,∘)(R/J,+,∘) is a ring, proceed by verifying the ring axioms in turn:
Ring Axiom A A: Addition forms an Abelian Group
From:
Ideal is Additive Normal SubgroupThe definition of a quotient groupQuotient Group is Group
it follows that (R/J,+)(R/J,+) is a group.
□◻
Ring Axiom M 0 M 0: Closure under Product
By definition of ∘∘ in R/J R/J, it follows that (R/J,∘)(R/J,∘) is closed.
□◻
Ring Axiom M 1 M 1: Associativity of Product
Associativity can be deduced from the fact that ∘∘ is associative on R R:
∀x,y,z∈R:∀x,y,z∈R:(x+J)∘((y+J)∘(z+J))(x+J)∘((y+J)∘(z+J))
==(x+J)∘(y∘z+J)(x+J)∘(y∘z+J)
==x∘y∘z+J x∘y∘z+J
==(x∘y+J)∘(z+J)(x∘y+J)∘(z+J)
==((x+J)∘(y+J))∘(z+J)((x+J)∘(y+J))∘(z+J)
□◻
Ring Axiom D D: Distributivity of Product over Addition
Distributivity can be deduced from the fact that ∘∘ is distributive on R R:
∀x,y,z∈R:∀x,y,z∈R:((x+J)+(y+J))∘(z+J)((x+J)+(y+J))∘(z+J)
==(x+y+J)∘(z+J)(x+y+J)∘(z+J)
==((x+y)∘z)+J((x+y)∘z)+J
==((x∘z)+(y∘z))+J((x∘z)+(y∘z))+J
==((x∘z)+J)+((y∘z)+J)((x∘z)+J)+((y∘z)+J)
==((x+J)∘(z+J))+((y+J)∘(z+J))((x+J)∘(z+J))+((y+J)∘(z+J))
□◻
Having verified all of the ring axioms, it follows that (R/J,+,∘)(R/J,+,∘) is a ring.
■◼
Sources
1955:John L. Kelley: General Topology... (previous)... (next): Chapter 0 0: Algebraic Concepts
1965:Seth Warner: Modern Algebra... (previous)... (next): Chapter IV IV: Rings and Fields: 22 22. New Rings from Old: Theorem 22.2 22.2
1978:Thomas A. Whitelaw: An Introduction to Abstract Algebra... (previous)... (next): §60.2§60.2 Factor rings: (i)(i)
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8111 | https://www.makingmolecules.com/blog/chirality | An Introduction to Chirality
Introduction
Let's clear one thing up from the start, chirality is not a chemical concept, it is based on the geometry or shape of an object. The world can be divided into two sets of objects, those that are chiral and those that are achiral. Most students seem to think that chirality is something special. I’d argue that it isn’t, most objects, especially if they have writing on them, are chiral, and it is achiral objects that are in the minority. So what is chirality?
An object is chiral if it is non-identical toits mirror image while an achiral object is identical to its mirror image. Hands are chiral (if you ignore fingerprints and scars), with you left hand being a non-identical mirror image of the right hand (in fact, the word chiral is derived from the Greek for ‘hand’). Most people intuitively know that left and right-handed objects are different, and what is true of ‘real world’ objects is equally true of chemistry and molecules. This is why you often hear chemists talk about left and right-handed molecules when talking about chiral compounds. The inherent difference between the left and right hands is why chirality is important.
Below there is a cartoon of hands as an example of chiral objects and a sphere, such as a snooker ball, as an achiral object due to it being identical to its mirror image.
Chiral versus Achiral
What exactly do we mean when we say chiral objects are non-identical mirror images? Isn’t the whole point of a mirror that it gives you a perfect reflection of the original object? Not quite. When discussing chirality, being non-identical mirror images means the original object and its reflection cannot be superposed or they cannot occupy exactly the same space.
Your hands are chiral because the left and right hands are mirror images that cannot occupy the same space. You could presses your hands so that your fingers and thumbs match up (like the prayer gesture or 🙏) but you can see the knuckles of both hands, they are not occupying the same space. Alternatively, overlap your hands so that both sets of knuckles face up. Your hands are still not superposed, the thumbs are pointing in different directions. Your hands are chiral.
As an aside, most books have it wrong! They say that chiral objects are non-superimposable. This is incorrect. Superimposable means to overlap or place on top (normally so that both objects are still visible). Superpose means to overlap so that all parts coincide. This imprecision is really frustrating (almost as frustrating as the fact I still say superimposable even though I know it is wrong).
An achiral object can be superposed or is identical to its mirror image. The ball in the example above is achiral as it can be superposed so that it occupies the same volume.
The more technical definition of chirality is based on symmetry. Chiral objects cannot have any element of symmetry except for an axis of rotation (another way of putting this is that chiral objects cannot have any reflective symmetry elements). That last clause is important. It explains why chiral does not necessarily mean asymmetric (a very successful class of chiral ligand are the bis(oxazolines) or BOX ligands. These are chiral but they are also symmetric, they have a C2 rotational axis (you can spin them through a 180° and they look the same)).
It is hard to draw three dimensional shapes in two dimensions and even harder to convince the viewer that they are the same or different. I’ve tried and I hope some of you find these pictures useful.
Achiral objects will have some form of symmetry (in addition to an axis of rotation), they are symmetric. Often you will see it stated that chiral objects will not have a plane of symmetry (also known as a mirror plane), while achiral objects will have a plane of symmetry. This isn’t true, but is rather a generalization as a plane of symmetry is the most important or common form of symmetry but it is not the only form. Achiral objects might have a center of inversion or a rotation/reflection axis (axis of improper rotation).
Examples of chiral objects includes hands, gloves, feet, shoes, propellers, corkscrews, golf clubs, snail shells and the horns of certain sheep. Such objects have no symmetry (see above for the exception) as exemplified by the cartoon snail below. Achiral objects (if they have no writing one them) include balls (these have an infinite number of planes of symmetry), tennis rackets (without writing), nails (as in, hammer and nails), and most cutlery (but not fish knifes!). Most cutlery has a plane of symmetry as shown in the diagram below. In my drawing forks and spoons having a vertical plane of symmetry running through the center while the knife has a horizontal plane (this is the same as if you had set a table normally).
Building an Understanding of Chirality
Over the years, I have found that some students just get chirality. They understand why our hands are mirror images but are different and, more importantly, they can apply that understanding to the more abstract idea of a three-dimensional shape or a molecule. Others can't. They cannot picture how one tetrahedral atom can be achiral while another can have 'handedness' simply by changing a single group on it. Throwing students in at the deep end with three-dimensional chirality can be daunting. Some students find it useful to start in one-dimension and work up to three dimensional objects.
I’m not the only person who has spotted this, but, more importantly, I’m not the first to write about it using the examples below … honestly, I thought I’d come up with this myself but it is so eerily similar to the earlier paper referenced here J. Chem. Ed. 2005, 82, 1009, that I suspect that I have unconsciously rephrased it (I own Robert Gawley’s book on asymmetric synthesis so I clearly like his explanations). Luckily, I spotted this before publishing this blog and can reference it as a source!
Chirality in one-dimension.
A line is a one-dimensional object. In the diagram below, that dimension is left-right (the x axis). The object can only move in one-dimension, it can only move left and right. If you start with a line with a black dot at either end, it will be achiral. It will be identical to its mirror image (technically, the mirror should be a mirror point as we’re in one dimension but I’ve drawn mirror to emphasize the action).
Now change one of the dots to a color. The object is now chiral (in one dimension), it is no longer the same as its reflection. The two lines cannot overlap (remember you can only move the line left or right and cannot rotate it, that would be a second dimension) so that the color dots are in the same place.
The first line has a point of symmetry (an internal mirror point) and is achiral. The line with two different ends has no internal mirror point and is chiral.
Chirality in two dimensions
To make a two dimensional object, add a third dot and form a triangle. This now occupies both the x and y axes. This object is allowed to move in two dimensions. It can move anywhere in the plane of the page, which means it can move from side to side, up and down or rotate (anti-)clockwise. It cannot flip over as this would involve the third dimension or the z axis.
The triangle with three black dots is clearly achiral. Its mirror image is identical to the original object (the reflection should be along a mirror line and not the three dimensional mirror I have drawn but again, I’m doing this to stress what I have done and not to be mathematically correct). What happens of you replace one dot with a colored dot? The object is still achiral. Rotating the reflection allows the mirror image to coincide with the original triangle as shown below.
If you add a third dot of a different color the triangle is finally chiral (in two dimensions). Now the mirror image is different. It cannot coincide with the original object no matter how many times you move or rotate it (remember, in two dimensions you cannot flip the object).
A (equilateral) triangle with three identical corners has three lines of symmetry (three internal mirror lines) and is achiral. A triangle with two different corners has one line of symmetry and is still achiral. With three different corners, there is no symmetry (in two dimensions) and the triangle is chiral.
For someone of my age, a classic example of chirality in two dimensions (it is only in the x and y axes) is Tetris. This game allows movement only in the plane of the screen, either from side-to-side, downwards (up-down but, annoyingly never up) or rotation but no flipping. The latter is important as it means the pieces can be divided into those that are achiral (I, O & T) and have at least one line of symmetry, and those that are chiral (J, L, S & Z), which have no symmetry.
I’ll use Tetris to show you why chirality is important a little later (but if you have ever played Tetris, you already know the reason).
Chirality in three-dimensions
By adding a point, a vertex, attached to the three corners of the triangle you create the three dimensional shape called a tetrahedron (like chemists don’t know what a tetrahedron is). Finally, we can use all three axes, x, y and z, and movement is now possible in all three dimensions. The shape can be translated (moved), rotated or flipped. At this point, I hope that you can see that if all the corners are the same color, the shape will be achiral and will have multiple internal mirror planes or a planes of symmetry. The same is true if there are two different colored corners and even three different colors. To remove all planes of symmetry requires four different corners as shown below.
I hope that by building from one dimension to three, it is clear why undergraduate courses are obsessed with four different substituents on an sp3 or tetrahedral atom. But I must repeat my initial comment from the introduction, having four different groups on a tetrahedron is not the reason why molecules are chiral. Shape (and the symmetry of that shape) is the only important thing. Can the original object be superposed on its mirror image or not?
Why is Chirality Important?
Intuitively you already know the answer to this question, just think about putting your shoes on in the morning. You left foot only fits in the left shoe (comfortably). This is the result of chirality.
An achiral object and its mirror image are identical. They will interact with other objects in an identical manner. This should come as no surprise, as I said, they are identical. There is no difference between a person catching a ball or catching its mirror image. All the properties are the same. You can think of the reverse as well. There is no difference in holding a ball in your left or right hand. All the properties are the same.
Replacing the ball with a glove changes everything. Instead of an achiral object interacting with your hands, there is a second chiral object. Now it matters which mirror image of the glove you have. Only the left glove will fit on your left hand. The mirror image glove, the right-hand glove will not. Two chiral objects have different interactions depending on the mirror image involved.
An alternative example is the interaction between a nut and a bolt. If you have a bolt with a clockwise thread you can only attach or thread a nut with an identical clockwise thread. If you take the mirror image of the bolt, it will have an anti-clockwise thread. It will not fit in the original nut.
The two paragraphs below and the diagram below are attempts to sum this key point up (I probably should have edited this out of the summary but the picture looks so nice (and took me so long) that I want an excuse to keep it in!):
Achiral objects will have the same interaction with either mirror image of a second object (regardless as to whether the second object was chiral or not). But chiral objects interact differently with each of the mirror images of a second chiral object.
So the key to this summary is that the mirror images of achiral objects are identical and will behave the same whatever they interact with. Mirror images of chiral objects are almost identical but behave differently when they interact with other chiral objects (or things).
If the analogy of your feet interacting with shoes, or nuts and bolts, doesn’t convince you that chiral objects behave differently with other chiral objects then I can try again with something a little more fun and return to our discussion of Tetris. Earlier you saw that the Tetris pieces can be divided in two; those that have a mirror line and are achiral, I, O and T and those that have no symmetry and are chiral, J, L, S & Z. If you are playing a game of Tetris and you are given a ’T’ piece it will fit on the board below.
The mirror image of the ’T’ piece is another ’T’ piece and it fits equally well (of course, it does, it is the same. It doesn’t matter that the board, in this case, is chiral (it is not the same as its mirror image).
The frustration of Tetris arises when the game keeps giving you ’S’ pieces which will not fit into this chiral board. You need the mirror image, the ‘Z’ piece if you want to complete the two lines. The ’S’ and ‘Z’ pieces are mirror images but they don't behave the same. They are chiral. They have different interactions with other chiral pieces, such as the board.
Chirality and Chemistry
Molecules behave in exactly the same way as other objects, and are either achiral or chiral. Achiral molecules have an identical mirror image while chiral molecules are non-identical to their mirror image.
Chiral molecules and their mirror images are stereoisomers. Stereoisomers are a class of isomer meaning they have all the same atoms but these atoms are arranged differently. Stereoisomers have all the same atoms and all the same bonds, they have same constitution, but they are different due to their shape which means they are non-superposable or cannot occupy the same space. Two non-identical mirror images are known as enantiomers. Two stereoisomers that are not mirror images are known as diastereomers. I have written a summary of stereoisomers HERE.
As chirality is the result of a physical property (shape) and not a chemical characteristic, all the statements above about chiral and achiral objects also apply to molecules. Chiral molecules cannot have an element of symmetry other than a possible axis of rotation. In other words, achiral molecules will have one, or more of the following, a plane of symmetry (mirror plane), a center of inversion or an axis of improper rotation (rotation/reflection). Fortunately, the plane of symmetry is most common as this is easier to see than the other forms of symmetry.
The diagram below shows an achiral molecule, isopropyl alcohol, with the internal plane of symmetry cutting through the hydroxyl group and bisecting two methyl groups. It then shows that adding a methyl group to one side to give butan-2-ol breaks the internal plane of symmetry and results in butan-2-ol being chiral. Finally, there is an example of a bis(oxazoline). This is chiral. It has no plane of symmetry, no center of inversion, and no improper rotation, but it is not asymmetric. It does have an axis of rotation. And for those of you that like symmetry, it is C2 symmetric.
One property of molecules makes them more complicated (until you ignore it … which you will at undergraduate). Molecules are not static and there is constant rotation around single bonds, or the molecules are said to be changing conformation. This leads to even the simplest molecules, such as ethane, existing as non-superposable mirror images in the majority of their potential conformations. Should we consider these molecules as chiral? No. A good way to think about chirality is through the concept of residual stereoisomers. Residual stereoisomers are the shapes of molecules that can be distinguished by a given analytical technique. For example, proton NMR cannot differentiate the different conformations of ethane so they are not stereoisomers. 1H NMR can differentiate the geometry of certain amides, and so they can be said to exist as E or Z isomers even though C–N bond rotation is possible. In practice, this means you can consider any molecule that has at least one accessible symmetric conformation as achiral, or if the static representation of a molecule (its drawing) is achiral then the molecule is considered achiral.
Yeah, I know, this is a simplification and I’m ignoring compounds like atropisomers (restricted rotation) for the time being. Let’s try and keep the introduction simple!
A single molecule can be chiral but a substance, or a collection of molecules, cannot (a collection of molecule is unlikely to be a mirror image). A collection of chiral molecules will have one of three compositions: It can be enantiomerically pure meaning it consists of a single mirror image of the compound. It can be enantiomerically enriched if it contains more of one mirror image than the other or it can be a racemate or racemic mixture, which means it contains an equal quantity of the two enantiomers.
The properties of the two enantiomers of a molecule are almost identical. This is to be expected, they are mirror images of each other. They will have the same physical properties in terms of boiling point, melting point, infrared spectra, NMR etc. The two enantiomers will only differ when they interact with another chiral species.
There are two major upshots from this. The first is that the two enantiomers rotate plane polarized light in opposite directions. This gives rise to what is known as optical rotation. For most undergraduates, it is sufficient to know that a pure enantiomer rotates plane polarized light by a certain number of degrees in one direction. The mirror image will rotate plane polarized light the same amount but in the opposite direction. The size of the rotation is proportional to the purity of the same so that an enantiomerically enriched sample that contains 75% of one enantiomer and 25% of the other will rotate light by 50% the amount of a pure sample (25% is cancelled out by the other enantiomer or is effectively a racemic mixture). A racemic mixture or a 50:50 mixture of enantiomers does not rotate plane polarized light as each enantiomer cancels out the action of the other.
If you are lucky, I might one day write a summary on this subject alone (somewhere I already have a diagram showing the helical path of the electric field component of light).
The second important consequence is that chiral molecules can react differently with other chiral molecules. The most obvious example of this is the interaction of chiral molecules with biological systems. The machinery that controls living organisms, enzymes, are composed of amino acids. All bar one of the amino acids are chiral (the exception is glycine, which is achiral). This means that enzymes are chiral and that the two mirror images of a chiral molecule may have different interactions with the enzyme. This leads to enantiomers having different properties. The examples below show some of the surprising differences that can be found between enantiomers.
An overlooked consequence of enantiomers becoming different in the presence of other chiral molecules is that a sample of an enantiomerically pure compound does not necessarily have the same physical properties as a sample containing a mixture of enantiomers. For instance, an enantiomerically pure sample of tartaric acid has a melting point of 172 °C but the racemic mixture (both mirror images) melts at 206 °C. Why is the ‘same’ compound behaving differently? In a pure sample, there is only one mirror image. In a racemic mixture, there is a second chiral molecule, one enantiomer can interact with itself or a different chiral molecule, their mirror image and this allows different intermolecular interactions.
Chirality and Stereocenters
At undergraduate level, chiral compounds are normally identified by the presence of a stereogenic center, also know as a stereocenter or, far too frequently, chiral center (I'm not a fan of this term due to the confusion it causes when we have meso-compounds. How can an achiral compound have two, or more, chiral centers? It just seems wrong). In this context, a stereocenter is a tetrahedral atom with four different groups or substituents coming off it. But, please be aware that molecules without a stereocenter can be chiral and molecules with multiple stereocenters can be achiral. The only real definition of a chiral molecule is that it cannot be superposed upon its mirror image (or a compound lacking any form of symmetry other than a rotational axis).
It should be clear from the discussion of chiral objects above why a tetrahedral atom requires four different groups to be considered a stereocenter but below is a diagram that again shows that four different groups is a necessity to remove any plane of symmetry. If all the vertices are the same there are multiple planes of symmetry. If there are two different vertices there are at least two planes of symmetry (this would be two white circles and two blue circles) or three planes of symmetry for the molecule shown. With three different vertices, there is a single plane of symmetry. It is only with four different corners that we break the symmetry.
it should be noted that this paragraph is only correct if none of the groups coming off the central tetrahedral carbon are connected to one another. As soon as the molecule contains a ring all bets are off!
While it is a useful skill to be able to identify stereogenic centers within a molecule (identify tetrahedral atoms with four different groups), I must stress that this is not the same as identifying a chiral molecule. A molecule with a single stereogenic center will be chiral but a molecule with multiple stereogenic centers might not be chiral. With multiple stereoentres, it becomes possible for the molecule to have an element of symmetry and hence be achiral. Molecules that contain stereogenic centers but are achiral are called meso compounds. The classic example is tartaric acid.
“Other” Chiral Molecules
A molecule is either chiral or it is not. It is chiral if the shape or geometry of the molecule results in a lack of certain symmetry elements. But, chemists like to think in terms of bonds, and this means we often classify chiral molecules by the stereochemical element that breaks the symmetry. Or, in English, we we use stereochemistry/stereoisomers to categorize different “kinds” of chirality (this annoys many people). All this means is that the absence or presence of a stereogenic centre, a tetrahedral atom with four different groups coming off it is not a necessity for chirality.
In this introduction to chirality I don’t want to go into detail about the “other” kinds of chirality (the other stereochemical motifs that break the symmetry of a molecule), but I do want to show examples of these molecules so that you are aware of their existence (this is also the reason I spent most of this summary not discussing molecules). Below are chiral molecules that do not have a stereogenic center but are chiral. If you are interested I have published various posts on how chemists name or give a stereochemical descriptor to so-called planar chirality, axial chirality or helical chirality.
Conclusion
There are two kinds of object, those that are chiral and those that are achiral. Achiral objects are identical to their mirror image. The mirror images are the same thing. Chiral objects are not the same as their mirror image. The object and its reflection are said to be non-superposable as they cannot occupy the same volume of space. Achiral objects include balls, spoons and chairs (as long as there is no writing on any of these objects). Chiral objects include shoes, gloves, and screws. Basically, anything that can be described as either left- or right-handed, is chiral. The properties of the pair of chiral objects will be almost identical - they are mirror images so they are identical in almost every way. Two chiral objects will only differ in one way, how they interact with another chiral object. Think about your left shoe. This is identical to your right shoe (or at least it was when you bought it) except it will only fit on your left foot. It is different when it interacts with your chiral foot.
Molecules are exactly the same, they are either chiral or achiral. If they are achiral then they are identical to their mirror image. If they are chiral then the molecule and its mirror image will be non-identical. They are said to be non-superposable. The chiral molecule and its mirror image are known as enantiomers. A molecule is chiral if it has no element of symmetry (it can have a rotational axis but it cannot have a plane of symmetry, a center of inversion, or an improper rotation axis (rotation and reflection). Achiral molecules will have symmetry above and beyond an axis of rotation. The most common form of symmetry is a internal plane of symmetry.
A molecule can be chiral but a substance or collection of molecules cannot. A substance comprising of chiral molecules will be enantiomerically pure if only contains one enantiomer. It will be enantiomerically enriched if it is predominantly one enantiomer but is contaminated with the mirror image. If there is an equal quantity of both enantiomers it is called a racemic mixture.
As enantiomers are mirror images they are almost identical. All their physical properties, such as melting point or solubility will be the same as will their reactions with achiral reagents. The only difference between the two enantiomers occurs when they interact with another chiral species. This means each enantiomer will rotate plane polarized light in opposite directions. It also means that enantiomers can have different interactions with biological systems such as our bodies.
Chirality is a fascinating topic and this is just a simple introduction.
gjr
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Hi,
Great work, in particular drawings. Will recommend to my students as a must for reading!!!
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Thanks for the kind words. Determination of enantiomeric excess/ratio by NMR is most commonly achieved by one of two methods, the use of chiral derivatizing agents or the use of chiral solvating/shift agents/reagents. The former requires covalent attachment of a chiral subunit (Mosher's esters are probably the most well known example), while the latter uses temporary interactions (coordination to a metal centre, commonly Eu, or hydrogen bonding).
try:
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Intresting and lucid explanation about chiral and achiral. I have a question. How to determination of chiral molecules by NMR?
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Numbers having in decimal representation no common digits with all their proper divisors
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Let us call a positive integer having in decimal representation no common digits with all its proper divisors "a good number".
54 is a good number : 1,2,3,6,9,18,27.
48 is not a good number : 1,2,3,4,6,8,12,16,24
Good numbers : 2,3,4,5,6,7,8,9,23,27,29,34,37,38,43,46,47,49,53,54,⋯
I've been interested in these numbers, and I noticed that OEIS has this sequence. One can easily see the followings :
1 is not included in the digits of good numbers.
The right-most digit of a good number n≥6 is neither 0,1,2 nor 5.
Let d(n) be the number of divisors of n. I noticed that d(n) is relatively small for (smaller) good numbers n. The following shows d(n) for good numbers n except prime numbers.
nd(n)43648493274344384464493548568574584686694
nd(n)76678886487424742494259426742893323432943344
It seems that there exists the maximum of d(n) for good numbers n. Also, it seems that the number of sets of four consecutive good numbers, such as {56,57,58,59}, is finite. However, though I've been trying to prove/disprove the two conjectures, I can neither prove nor disprove them. So, here are my questions.
Question 1 : Does there exist the maximum of d(n) for good numbers n? If yes, what is it? If no, how can we show that?
Question 2 : Is the number of sets of four consecutive good numbers finite? If yes, what is the biggest such set? If no, how can we show that?
Added : For Question 1, a user san showed that the answer is yes and that an upper bound for d(n) is 16. However, the maximum of d(n) has not been obtained yet.
For Question 2, san showed that the answer is yes, and that the biggest such set is {56,57,58,59}.
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edited Feb 23, 2023 at 22:20
Bill Dubuque
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just an observation, I do not have a computer now (tablet) so I can not check it, but looking to the examples you show, there is the possibility that if d(n)∣n then nd(n)=p∈P. There is not enough data in your samples so probably there is a counterexample, or maybe there are not more n than {8,9,56}, but it could be an interesting property if it happens for all n.
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iadvd
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@iadvd: That's interesting if n>3 because n=2 is a counterexample (It seems true that there exist finitely many good numbers n such that d(n) divides n). (btw, it'd be nice if you could show some relation between your conjecture and my questions).
mathlove
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mathlove
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that observation came from another observation I did about the Totient function. Every time I see a set of numbers like this I tend to look for those kind of patterns automatically. This one: math.stackexchange.com/questions/1235390/…
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For the first question the answer is yes, and an upper bound is 16.
We define a nearly good number to be such that the last digit is not the last digit
of any of its proper divisors. We have the following result:
Every nearly good number is the product of at most 4 prime numbers. (∗)
Hence every nearly good number (and also every good number) has at most 16 divisors.
The proof of () is very long, but straightforward. Here a sketch of the proof:
Assume that n>10 is a nearly good number. Write n¯ for the last digit (or, by abuse of notation, also for the class in Z/10Z).
We first prove that if n¯=9, then it has only two prime factors at most.
In fact, the only valid factorizations of 9¯ in two factors are
9¯=3¯⋅3¯and9¯=7¯⋅7¯
Since the only valid factorization of 3¯ is 3¯=9¯⋅7¯ and the only
valid factorization of 7¯ is 7¯=9¯⋅3¯, both factors must be prime factors.
Arguing similarly, one obtains that if n¯=3, then it has at most three prime factors, and in that case
n¯=7¯⋅7¯⋅7¯. And it is also straightforward to show that if n¯=7, then it has
at most three prime factors, and in that case
n¯=3¯⋅3¯⋅3¯.
Now comes the clumsy part:
If n¯=8, then the only valid factorizations in two factors are
n¯=2¯⋅4¯,n¯=3¯⋅6¯,n¯=2¯⋅9¯andn¯=4¯⋅7¯.
In the four cases one shows that there are only three prime factors possible. For example, if we assume 8¯=2¯⋅4¯, then the factor 2¯ correspond to the prime factor 2 (Else the factor is even and of the form m¯⋅2 with m>1, hence 8¯=m¯⋅2¯⋅4¯ and the proper factor
2¯⋅4¯ has 8 as last digit).
Since the factor with 4¯ is even, it has a factorization 4¯=m¯⋅2¯, and once again 2¯ is the prime factor 2, and m¯=2 or m¯=7. In the first case all three factors are 2 and in the second case
we cannot factorize 7¯ (else 7¯=3¯⋅9¯, and the original number has a proper factor corresponding to 2¯⋅9¯, with last digit 8). The other cases are similar and prove that if n¯=8, then there are at most three factors.
Finally one deals with the cases n¯=4 and n¯=6. Here
the valid factorizations for n¯=4 in two factors are
n¯=2¯⋅2¯=7¯⋅2¯=3¯⋅8¯=9¯⋅6¯.
One verifies that there are at most 4 prime factors, with the only possible configuration
n¯=3¯⋅3¯⋅3¯⋅2¯.
Similarly
the valid factorizations for n¯=6 in two factors are
n¯=2¯⋅3¯=8¯⋅2¯=4¯⋅4¯=9¯⋅4¯=7¯⋅8¯.
One verifies that there are at most 4 prime factors, and the only possible configurations are then
n¯=2¯⋅2¯⋅2¯⋅7¯,n¯=2¯⋅2¯⋅7¯⋅7¯andn¯=2¯⋅7¯⋅7¯⋅7¯.
So we have proved: If n is a (nearly) good number, then d(n)≤k(n¯) where
k(9)=4,k(3)=8,k(7)=8,k(8)=8,k(4)=16,andk(6)=16.
Clearly n¯=0,1,2,5 is not possible if n>10.
Although there are good numbers with four prime factors, I have never seen any with more than 8 divisors. It is highly probable that the maximum d(n) for good numbers is 8. It is nearly impossible that it is 16, and there is a small chance that it is 9 or 12: If d(n)>8, then n must have 4 prime factors and the valid configurations are
n¯=2¯⋅3¯⋅3¯⋅3¯,n¯=2¯⋅7¯⋅7¯⋅7¯andn¯=2¯⋅2¯⋅7¯⋅7¯.
In the first case 1,2,3,6,7,8,9≠Dig(n), and the combinations 50 and 40 in the string of digits of n are impossible, hence the decimal representation of n is a string with only 5's and 4's. How many of these numbers are product of 4 primes ending in 3,3,3,2? At least 54 is one of them, but as I said, I find it nearly impossible that there is a bigger good number like that.
In the second case 1,2,3,4,7,8,9≠Dig(n), and the combinations 50 and 60 in the string of digits of n are impossible, hence the decimal representation of n is a string with only 5's and 6's. How many of these numbers are product of 4 primes ending in 7,7,7,2? I find it nearly impossible that there is number like that.
In the third case 1,2,4,7,8,9≠Dig(n), hence the decimal representation of n is a string with only 0's,3's,5's and 6's. If the two primes ending in 7 are equal, then n has 9 divisors, else it has 12. Although the chance seems bigger than in the two previous cases, the bigger the number are that one tests, the more digits can yield a failure in being a good number, so the chance is very small.
Second question: For the second question the answer is yes, and 56,57,58,59 is the biggest such set. In particular, the only sets
of four consecutive good numbers are
{2,3,4,5},{3,4,5,6},{4,5,7,8},{5,6,7,8},{6,7,8,9},{56,57,58,59}.
Note that if m>0, then
10m+5 and 10m+1 are not good numbers, so the only possible configuration for four consecutive numbers greater than 10 is
10m+6,10m+7,10m+8,10m+9. Moreover, 1,2,3 and 4 are divisors of at least one of them, so the set Dig(m) of the digits of m,
cannot contain 1,2,3 or 4, in particular m¯≠1,2,3,4. Now we assume m>0 and that 10m+6,10m+7,10m+8,10m+9
are good numbers. We have to prove that m=5 is the only possible case.
Remark:
a) 3 divides at least one of the four numbers, and it cannot divide 10m+6, since then 6 would be a proper divisor of 10m+6. Hence it divides
10m+7 or 10m+8.
b) If 4|10m+8, then 10m+8¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=2⋅2⋅7¯, since there are at most three prime factors, and
10m+8=2¯⋅2¯⋅2¯ leads to m=0. In particular, in this case 3/|10m+8.
We will analyse all possible cases of m¯, the last digit of m.
CASE m¯=9:
Clearly m¯≠9, since then 10m+8 is not a good number:
10m+8=100k+98andd:=(10m+8)/2=50k+49,
so 9 is the last digit of the proper divisor d and 9∈Dig(10m+8), a contradiction.
CASE m¯=0:
In this case 4|10m+8=100k+8, and, since m>0, we have that k must be odd (otherwise 8|10m+8).
Write k=2j+1, then (10m+8)/2=100j+54, hence k¯≠5 and (10m+8)/4=50j+27, hence k¯≠7. Since
k¯≠1,3, we have only to discard the case k¯=9. But 3|10m+7 (by a) and b) above), and so
10m+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=3⋅9¯, which implies in this case
9∈Dig(10m+7)∩Dig((10m+7)/3).
CASE m¯=8:
In this case 4|10m+8=100k+88, and, since m>0, we have that k must be odd (otherwise 8|10m+8).
Write k=2j+1, then (10m+8)/2=100j+94, hence k¯≠9 and (10m+8)/4=50j+47, hence k¯≠7.
Since k¯≠1,3, we have only to discard the case k¯=5. But 3|10m+7 (by a) and b) above), hence,
if k¯=5, we have 10m+7=1000(3r+1)+587 and so (10m+7)/3=1000r+529 which implies
5∈Dig(10m+7)∩Dig((10m+7)/3), a contradiction.
CASE m¯=6:
In this case 4|10m+8=100k+68, and, since m>0, we have that k must be even (otherwise 8|10m+8).
Since k¯≠2,4, we have to discard the cases k¯=0,6,8. As above 3|10m+7, hence,
if k¯=8, we have 10m+7=3000r+867 and so (10m+7)/3=1000r+289 which implies
8∈Dig(10m+7)∩Dig((10m+7)/3), a contradiction. Similarly,
if k¯=0, we have 10m+7=1000(3r+2)+67 and so (10m+7)/3=1000r+689 which implies
6∈Dig(10m+7)∩Dig((10m+7)/3), a contradiction. Finally, if k¯=6, we have
10m+8=1000r+668. Then (10m+8)/2=500r+334. If r is odd (r=2j+1), then
(10m+8)/2=1000j+834, impossible (8∈Dig(10m+8)∩Dig((10m+8)/2)), and if
r is even, then (10m+8)/2=1000j+334 and so (10m+8)/4=500j+167, which leads to the contradiction
6∈Dig(10m+8)∩Dig((10m+8)/4).
CASE m¯=7:
In this case 4|10m+6.
Write m=10k+7. We will discard all possibilities for k¯. Clearly k¯≠1,2,3,4.
If k¯=9, then 9∈Dig(10m+8)∩Dig((10m+8)/2), if k¯=8, then
8∈Dig(10m+6)∩Dig((10m+6)/2) and if k¯=7, then
8∈Dig(10m+8)∩Dig((10m+8)/2) (note that (1000j+778)/2=500j+389).
If k¯=6, we have
10m+6=1000r+676. Then (10m+6)/2=500r+338. If r is odd (r=2j+1), then
(10m+8)/2=1000j+839, impossible (8∈Dig(10m+8)∩Dig((10m+8)/2)), and if
r is even, then (10m+6)/2=1000j+338 and so (10m+6)/4=500j+169, which leads to the contradiction
6∈Dig(10m+6)∩Dig((10m+6)/4).
If k¯=5 we have two options:
If 3|10m+7, then we have 10m+7=1000(3r+2)+577 and so (10m+7)/3=1000r+859 which implies
5∈Dig(10m+7)∩Dig((10m+7)/3), a contradiction.
If 3|10m+8, then we have 10m+8=1000(3r+1)+578 and so (10m+7)/3=1000r+526 which implies
5∈Dig(10m+8)∩Dig((10m+8)/3), a contradiction.
So we are left with the case k¯=0. Here the only digits that can appear in m are 0,5,6,7:
if 9∈Dig(m), then 9∈Dig(10m+8)∩Dig((10m+8)/2, since 10m+8=1000r+78. Similarly,
if 8∈Dig(m), then 8∈Dig(10m+6)∩Dig((10m+6)/2, since 10m+6=1000r+76.
Now we discard combinations of two consecutive digits of m. Clearly the combination 50 can be discarded, since
then m/2 has the combination 25 or the combination 75, leading to 5∈Dig(10m+6)∩Dig((10m+6)/2.
We can also discard 70: In that case m/2 admits the combination 35 (the other possibility 85 leads to
8∈Dig(10m+8)∩Dig((10m+8)/2)), and so m/4 admits the combination 17 or 67, leading to
7∈Dig(10m+6)∩Dig((10m+6)/4), a contradiction.
The combinations 56 and 76 can be also discarded, since they lead to 8∈Dig(10m+8)∩Dig((10m+8)/2.
Resuming the restrictions, we have that on the left of a zero there can be only zero or 6, and to the left of a
6 there can be only a 0 or a 6. Since the two last digits of m are 07, the only digits in the expansion to the left
are 0 and 6. But then m=600r+7 and so 3|10m+7 is impossible and 3|10m+8 leads to (10m+8)/3=(6000r+78)/3=2000r+26,
which implies 6∈Dig(10m+8)∩Dig((10m+8)/3).
CASE m¯=5:
In this case 4|10m+6.
Write m=10k+5. We will discard all possibilities for k¯. Clearly k¯≠1,2,3,4.
If k¯=9, then 9∈Dig(10m+8)∩Dig((10m+8)/2), if k¯=8, then
8∈Dig(10m+6)∩Dig((10m+6)/2) and if k¯=7, then
7∈Dig(10m+8)∩Dig((10m+8)/2) (note that (1000j+758)/2=500j+379).
If k¯=6, we have
10m+6=1000r+656. Then (10m+6)/2=500r+328. If r is odd (r=2j+1), then
(10m+8)/2=1000j+829, impossible (8∈Dig(10m+8)∩Dig((10m+8)/2)), and if
r is even, then (10m+6)/2=1000j+328 and so (10m+6)/4=500j+164, which leads to the contradiction
6∈Dig(10m+6)∩Dig((10m+6)/4).
If k¯=5 we have two options:
If 3|10m+7, then we have 10m+7=1000(3r+1)+557 and so (10m+7)/3=1000r+519 which implies
5∈Dig(10m+7)∩Dig((10m+7)/3), a contradiction.
If 3|10m+8, then we have 10m+8=3000r+558 and so (10m+7)/3=1000r+186 which implies
8∈Dig(10m+8)∩Dig((10m+8)/3), a contradiction.
So we are left with the case k¯=0. Here the only digits that can appear in m are 0,5,6,7:
if 9∈Dig(m), then 9∈Dig(10m+8)∩Dig((10m+8)/2), since 10m+8=1000r+58. Similarly,
if 8∈Dig(m), then 8∈Dig(10m+6)∩Dig((10m+6)/2), since 10m+6=1000r+56.
Now we discard combinations of two consecutive digits of m. Clearly the combination 50 can be discarded, since
then m/2 has the combination 25 or the combination 75, leading to 5∈Dig(10m+6)∩Dig((10m+6)/2).
Similarly the combination 70 can be discarded, since
then m/2 has the combination 35 or the combination 85, leading to 5∈Dig(10m+6)∩Dig((10m+6)/2).
The combination 60 can be discarded, since
then m/2 has the combination 30 (the other possibility 80 leads to
8∈Dig(10m+8)∩Dig((10m+8)/2)), and then (10m+6)/4 has the combination 15 or the combination 65,
which both imply 5∈Dig(10m+6)∩Dig((10m+6)/2).
Since the two last digits of m are 05, we are left we the only case m=5, as desired.
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edited May 27, 2015 at 1:32
answered May 21, 2015 at 3:41
sansan
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That's an interesting idea, but maybe I don't understand well what you mean. A few questions : (1) Why isn't 8¯=2¯⋅4¯ valid? (2) Can you elaborate on how to show that for n¯=8 there are only three prime factors possible? (why isn't 8¯=2¯⋅2¯⋅3¯⋅7¯⋅7¯, for example, valid?) (3) Similarly, why isn't 4¯=(3¯)4m−1⋅2¯, for example, valid?
mathlove
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mathlove
05/21/2015 12:35:33
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May 21, 2015 at 12:35
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8¯=2¯⋅4¯ is valid (mistake corrected). If you write 8¯=2¯⋅2¯⋅3¯⋅7¯⋅7¯, then there is a proper factor 2¯⋅2¯⋅7¯, which has 8 as last digit. Similarly, if you write 4¯=(3¯)4m−1⋅2¯ with m>1, then there is a proper factor 4¯=3¯3⋅2¯ with 4 as last digit.
san
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san
05/21/2015 15:29:44
Commented
May 21, 2015 at 15:29
Ah, I think I can get what you mean. Thanks. (btw, for n¯=4, we also have 8¯⋅8¯, right? Also, the valid factorization you wrote for n¯=6 in two factors are the same as those for n¯=4. Are they typos, right?)
mathlove
–
mathlove
05/21/2015 21:18:54
Commented
May 21, 2015 at 21:18
n¯=8¯⋅8¯ is not a valid configuration, since then 4 would be a factor of n, since 2 divides 8¯. The other was a typo (fixed).
san
–
san
05/21/2015 21:42:05
Commented
May 21, 2015 at 21:42
I would +10 this, if I could.
Conor O'Brien
–
Conor O'Brien
05/21/2015 21:45:36
Commented
May 21, 2015 at 21:45
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8113 | https://en.wikipedia.org/wiki/Lever_rule | Jump to content
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Contents
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1 Derivation
2 Calculations
2.1 Binary phase diagrams
2.2 Eutectic phase diagrams
3 References
Lever rule
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From Wikipedia, the free encyclopedia
Formula for determining the mole or mass fraction of phases in a binary phase diagram
In chemistry, the lever rule is a formula used to determine the mole fraction (xi) or the mass fraction (wi) of each phase of a binary equilibrium phase diagram. It can be used to determine the fraction of liquid and solid phases for a given binary composition and temperature that is between the liquidus and solidus line.
In an alloy or a mixture with two phases, α and β, which themselves contain two elements, A and B, the lever rule states that the mass fraction of the α phase is
where
is the mass fraction of element B in the α phase
is the mass fraction of element B in the β phase
is the mass fraction of element B in the entire alloy or mixture
all at some fixed temperature or pressure.
Derivation
[edit]
Suppose an alloy at an equilibrium temperature T consists of mass fraction of element B. Suppose also that at temperature T the alloy consists of two phases, α and β, for which the α consists of , and β consists of . Let the mass of the α phase in the alloy be so that the mass of the β phase is , where is the total mass of the alloy.
By definition, then, the mass of element B in the α phase is , while the mass of element B in the β phase is . Together these two quantities sum to the total mass of element B in the alloy, which is given by . Therefore,
By rearranging, one finds that
This final fraction is the mass fraction of the α phase in the alloy.
Calculations
[edit]
Binary phase diagrams
[edit]
Before any calculations can be made, a tie line is drawn on the phase diagram to determine the mass fraction of each element; on the phase diagram to the right it is line segment LS. This tie line is drawn horizontally at the composition's temperature from one phase to another (here the liquid to the solid). The mass fraction of element B at the liquidus is given by wBl (represented as wl in this diagram) and the mass fraction of element B at the solidus is given by wBs (represented as ws in this diagram). The mass fraction of solid and liquid can then be calculated using the following lever rule equations:
where wB is the mass fraction of element B for the given composition (represented as wo in this diagram).
The numerator of each equation is the original composition that we are interested in is +/- the opposite lever arm. That is if you want the mass fraction of solid then take the difference between the liquid composition and the original composition. And then the denominator is the overall length of the arm so the difference between the solid and liquid compositions. If you're having difficulty realising why this is so, try visualising the composition when wo approaches wl. Then the liquid concentration will start increasing.
Eutectic phase diagrams
[edit]
There is now more than one two-phase region. The tie line drawn is from the solid alpha to the liquid and by dropping a vertical line down at these points the mass fraction of each phase is directly read off the graph, that is the mass fraction in the x axis element. The same equations can be used to find the mass fraction of alloy in each of the phases, i.e. wl is the mass fraction of the whole sample in the liquid phase.
References
[edit]
^ a b Smith.h, William F. halkias; Hashemi, Javad (2006), Foundations of Materials Science and Engineering (4th ed.), McGraw-Hill, pp. 318–320, ISBN 0-07-295358-6.
^ Callister, William D.; Rethwisch, David (2009), Materials Science and Engineering An Introduction (8th ed.), Wiley, pp. 298–303, ISBN 978-0-470-41997-7.
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8114 | https://pressbooks.bccampus.ca/collegephysics/chapter/nuclear-radioactivity/ | 31.1 Nuclear Radioactivity – College Physics: OpenStax
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Contents
Preface to College Physics
I. Chapter 1 The Nature of Science and Physics
1.1.0 Introduction
2.1.1 Physics: An Introduction
3.1.2 Physical Quantities and Units
4.1.3 Accuracy, Precision, and Significant Figures
5.1.4 Approximation
II. Chapter 2 One-Dimensional Kinematics
6.2.0 Introduction
7.2.1 Displacement
8.2.2 Vectors, Scalars, and Coordinate Systems
9.2.3 Time, Velocity, and Speed
10.2.4 Acceleration
11.2.5 Motion Equations for Constant Acceleration in One Dimension
12.2.6 Problem-Solving Basics for One-Dimensional Kinematics
13.2.7 Falling Objects
14.2.8 Graphical Analysis of One-Dimensional Motion
III. Chapter 3 Two-Dimensional Kinematics
15.3.0 Introduction
16.3.1 Kinematics in Two Dimensions: An Introduction
17.3.2 Vector Addition and Subtraction: Graphical Methods
18.3.3 Vector Addition and Subtraction: Analytical Methods
19.3.4 Projectile Motion
20.3.5 Addition of Velocities
IV. Chapter 4 Dynamics: Force and Newton's Laws of Motion
21.4.0 Introduction
22.4.1 Development of Force Concept
23.4.2 Newton’s First Law of Motion: Inertia
24.4.3 Newton’s Second Law of Motion: Concept of a System
25.4.4 Newton’s Third Law of Motion: Symmetry in Forces
26.4.5 Normal, Tension, and Other Examples of Forces
27.4.6 Problem-Solving Strategies
28.4.7 Further Applications of Newton’s Laws of Motion
29.4.8 Extended Topic: The Four Basic Forces—An Introduction
V. Chapter 5 Further Applications of Newton's Laws: Friction, Drag and Elasticity
30.5.0 Introduction
31.5.1 Friction
32.5.2 Drag Forces
33.5.3 Elasticity: Stress and Strain
VI. Chapter 6 Uniform Circular Motion and Gravitation
34.6.0 Introduction
35.6.1 Rotation Angle and Angular Velocity
36.6.2 Centripetal Acceleration
37.6.3 Centripetal Force
38.6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
39.6.5 Newton’s Universal Law of Gravitation
40.6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
VII. Chapter 7 Work, Energy, and Energy Resources
41.7.0 Introduction
42.7.1 Work: The Scientific Definition
43.7.2 Kinetic Energy and the Work-Energy Theorem
44.7.3 Gravitational Potential Energy
45.7.4 Conservative Forces and Potential Energy
46.7.5 Nonconservative Forces
47.7.6 Conservation of Energy
48.7.7 Power
49.7.8 Work, Energy, and Power in Humans
50.7.9 World Energy Use
VIII. Chapter 8 Linear Momentum and Collisions
51.8.0 Introduction
52.8.1 Linear Momentum and Force
53.8.2 Impulse
54.8.3 Conservation of Momentum
55.8.4 Elastic Collisions in One Dimension
56.8.5 Inelastic Collisions in One Dimension
57.8.6 Collisions of Point Masses in Two Dimensions
58.8.7 Introduction to Rocket Propulsion
IX. Chapter 9 Statics and Torque
59.9.0 Introduction
60.9.1 The First Condition for Equilibrium
61.9.2 The Second Condition for Equilibrium
62.9.3 Stability
63.9.4 Applications of Statics, Including Problem-Solving Strategies
64.9.5 Simple Machines
65.9.6 Forces and Torques in Muscles and Joints
X. Chapter 10 Rotational Motion and Angular Momentum
66.10.0 Introduction
67.10.1 Angular Acceleration
68.10.2 Kinematics of Rotational Motion
69.10.3 Dynamics of Rotational Motion: Rotational Inertia
70.10.4 Rotational Kinetic Energy: Work and Energy Revisited
71.10.5 Angular Momentum and Its Conservation
72.10.6 Collisions of Extended Bodies in Two Dimensions
73.10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
XI. Chapter 11 Fluid Statics
74.11.0 Introduction
75.11.1 What Is a Fluid?
76.11.2 Density
77.11.3 Pressure
78.11.4 Variation of Pressure with Depth in a Fluid
79.11.5 Pascal’s Principle
80.11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
81.11.7 Archimedes’ Principle
82.11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
83.11.9 Pressures in the Body
XII. Chapter 12 Fluid Dynamics and Its Biological and Medical Applications
84.12.0 Introduction
85.12.1 Flow Rate and Its Relation to Velocity
86.12.2 Bernoulli’s Equation
87.12.3 The Most General Applications of Bernoulli’s Equation
88.12.4 Viscosity and Laminar Flow; Poiseuille’s Law
89.12.5 The Onset of Turbulence
90.12.6 Motion of an Object in a Viscous Fluid
91.12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
XIII. Chapter 13 Temperature, Kinetic Theory, and the Gas Laws
92.13.0 Introduction
93.13.1 Temperature
94.13.2 Thermal Expansion of Solids and Liquids
95.13.3 The Ideal Gas Law
96.13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
97.13.5 Phase Changes
98.13.6 Humidity, Evaporation, and Boiling
XIV. Chapter 14 Heat and Heat Transfer Methods
99.14.0 Introduction
100.14.1 Heat
101.14.2 Temperature Change and Heat Capacity
102.14.3 Phase Change and Latent Heat
103.14.4 Heat Transfer Methods
104.14.5 Conduction
105.14.6 Convection
106.14.7 Radiation
XV. Chapter 15 Thermodynamics
107.15.0 Introduction
108.15.1 The First Law of Thermodynamics
109.15.2 The First Law of Thermodynamics and Some Simple Processes
110.15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
111.15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
112.15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
113.15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
114.15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
XVI. Chapter 16 Oscillatory Motion and Waves
115.16.0 Introduction
116.16.1 Hooke’s Law: Stress and Strain Revisited
117.16.2 Period and Frequency in Oscillations
118.16.3 Simple Harmonic Motion: A Special Periodic Motion
119.16.4 The Simple Pendulum
120.16.5 Energy and the Simple Harmonic Oscillator
121.16.6 Uniform Circular Motion and Simple Harmonic Motion
122.16.7 Damped Harmonic Motion
123.16.8 Forced Oscillations and Resonance
124.16.9 Waves
125.16.10 Superposition and Interference
126.16.11 Energy in Waves: Intensity
XVII. Chapter 17 Physics of Hearing
127.17.0 Introduction
128.17.1 Sound
129.17.2 Speed of Sound, Frequency, and Wavelength
130.17.3 Sound Intensity and Sound Level
131.17.4 Doppler Effect and Sonic Booms
132.17.5 Sound Interference and Resonance: Standing Waves in Air Columns
133.17.6 Hearing
134.17.7 Ultrasound
XVIII. Chapter 18 Electric Charge and Electric Field
135.18.0 Introduction
136.18.1 Static Electricity and Charge: Conservation of Charge
137.18.2 Conductors and Insulators
138.18.3 Coulomb’s Law
139.18.4 Electric Field: Concept of a Field Revisited
140.18.5 Electric Field Lines: Multiple Charges
141.18.6 Electric Forces in Biology
142.18.7 Conductors and Electric Fields in Static Equilibrium
143.18.8 Applications of Electrostatics
XIX. Chapter 19 Electric Potential and Electric Field
144.19.0 Introduction
145.19.1 Electric Potential Energy: Potential Difference
146.19.2 Electric Potential in a Uniform Electric Field
147.19.3 Electrical Potential Due to a Point Charge
148.19.4 Equipotential Lines
149.19.5 Capacitors and Dielectrics
150.19.6 Capacitors in Series and Parallel
151.19.7 Energy Stored in Capacitors
XX. Chapter 20 Electric Current, Resistance, and Ohm's Law
152.20.0 Introduction
153.20.1 Current
154.20.2 Ohm’s Law: Resistance and Simple Circuits
155.20.3 Resistance and Resistivity
156.20.4 Electric Power and Energy
157.20.5 Alternating Current versus Direct Current
158.20.6 Electric Hazards and the Human Body
159.20.7 Nerve Conduction–Electrocardiograms
XXI. Chapter 21 Circuits and DC Instruments
160.21.0 Introduction
161.21.1 Resistors in Series and Parallel
162.21.2 Electromotive Force: Terminal Voltage
163.21.3 Kirchhoff’s Rules
164.21.4 DC Voltmeters and Ammeters
165.21.5 Null Measurements
166.21.6 DC Circuits Containing Resistors and Capacitors
XXII. Chapter 22 Magnetism
167.22.0 Introduction
168.22.1 Magnets
169.22.2 Ferromagnets and Electromagnets
170.22.3 Magnetic Fields and Magnetic Field Lines
171.22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
172.22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
173.22.6 The Hall Effect
174.22.7 Magnetic Force on a Current-Carrying Conductor
175.22.8 Torque on a Current Loop: Motors and Meters
176.22.9 Magnetic Fields Produced by Currents: Ampere’s Law
177.22.10 Magnetic Force between Two Parallel Conductors
178.22.11 More Applications of Magnetism
XXIII. Chapter 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
179.23.0 Introduction
180.23.1 Induced Emf and Magnetic Flux
181.23.2 Faraday’s Law of Induction: Lenz’s Law
182.23.3 Motional Emf
183.23.4 Eddy Currents and Magnetic Damping
184.23.5 Electric Generators
185.23.6 Back Emf
186.23.7 Transformers
187.23.8 Electrical Safety: Systems and Devices
188.23.9 Inductance
189.23.10 RL Circuits
190.23.11 Reactance, Inductive and Capacitive
191.23.12 RLC Series AC Circuits
XXIV. Chapter 24 Electromagnetic Waves
192.24.0 Introduction
193.24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
194.24.2 Production of Electromagnetic Waves
195.24.3 The Electromagnetic Spectrum
196.24.4 Energy in Electromagnetic Waves
XXV. Chapter 25 Geometric Optics
197.25.0 Introduction
198.25.1 The Ray Aspect of Light
199.25.2 The Law of Reflection
200.25.3 The Law of Refraction
201.25.4 Total Internal Reflection
202.25.5 Dispersion: The Rainbow and Prisms
203.25.6 Image Formation by Lenses
204.25.7 Image Formation by Mirrors
XXVI. Chapter 26 Vision and Optical Instruments
205.26.0 Introduction
206.26.1 Physics of the Eye
207.26.2 Vision Correction
208.26.3 Color and Color Vision
209.26.4 Microscopes
210.26.5 Telescopes
211.26.6 Aberrations
XXVII. Chapter 27 Wave Optics
212.27.0 Introduction
213.27.1 The Wave Aspect of Light: Interference
214.27.2 Huygens's Principle: Diffraction
215.27.3 Young’s Double Slit Experiment
216.27.4 Multiple Slit Diffraction
217.27.5 Single Slit Diffraction
218.27.6 Limits of Resolution: The Rayleigh Criterion
219.27.7 Thin Film Interference
220.27.8 Polarization
221.27.9 Extended Topic Microscopy Enhanced by the Wave Characteristics of Light
XXVIII. Chapter 28 Special Relativity
222.28.0 Introduction
223.28.1 Einstein’s Postulates
224.28.2 Simultaneity And Time Dilation
225.28.3 Length Contraction
226.28.4 Relativistic Addition of Velocities
227.28.5 Relativistic Momentum
228.28.6 Relativistic Energy
XXIX. Chapter 29 Introduction to Quantum Physics
229.29.0 Introduction
230.29.1 Quantization of Energy
231.29.2 The Photoelectric Effect
232.29.3 Photon Energies and the Electromagnetic Spectrum
233.29.4 Photon Momentum
234.29.5 The Particle-Wave Duality
235.29.6 The Wave Nature of Matter
236.29.7 Probability: The Heisenberg Uncertainty Principle
237.29.8 The Particle-Wave Duality Reviewed
XXX. Chapter 30 Atomic Physics
238.30.0 Introduction
239.30.1 Discovery of the Atom
240.30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
241.30.3 Bohr’s Theory of the Hydrogen Atom
242.30.4 X Rays: Atomic Origins and Applications
243.30.5 Applications of Atomic Excitations and De-Excitations
244.30.6 The Wave Nature of Matter Causes Quantization
245.30.7 Patterns in Spectra Reveal More Quantization
246.30.8 Quantum Numbers and Rules
247.30.9 The Pauli Exclusion Principle
XXXI. Chapter 31 Radioactivity and Nuclear Physics
248.31.0 Introduction
249.31.1 Nuclear Radioactivity
250.31.2 Radiation Detection and Detectors
251.31.3 Substructure of the Nucleus
252.31.4 Nuclear Decay and Conservation Laws
253.31.5 Half-Life and Activity
254.31.6 Binding Energy
255.31.7 Tunneling
XXXII. Chapter 32 Medical Applications of Nuclear Physics
256.32.0 Introduction
257.32.1 Medical Imaging and Diagnostics
258.32.2 Biological Effects of Ionizing Radiation
259.32.3 Therapeutic Uses of Ionizing Radiation
260.32.4 Food Irradiation
261.32.5 Fusion
262.32.6 Fission
263.32.7 Nuclear Weapons
XXXIII. Chapter 33 Particle Physics
264.33.0 Introduction
265.33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
266.33.2 The Four Basic Forces
267.33.3 Accelerators Create Matter from Energy
268.33.4 Particles, Patterns, and Conservation Laws
269.33.5 Quarks: Is That All There Is?
270.33.6 GUTs: The Unification of Forces
XXXIV. Chapter 34 Frontiers of Physics
271.34.0 Introduction
272.34.1 Cosmology and Particle Physics
273.34.2 General Relativity and Quantum Gravity
274.34.3 Superstrings
275.34.4 Dark Matter and Closure
276.34.5 Complexity and Chaos
277.34.6 High-temperature Superconductors
278.34.7 Some Questions We Know to Ask
Appendix A Atomic Masses
Appendix B Selected Radioactive Isotopes
Appendix C Useful Information
Appendix D Glossary of Key Symbols and Notation
College Physics: OpenStax
Chapter 31 Radioactivity and Nuclear Physics
249 31.1 Nuclear Radioactivity
Summary
Explain nuclear radiation.
Explain the types of radiation—alpha emission, beta emission, and gamma emission.
Explain the ionization of radiation in an atom.
Define the range of radiation.
The discovery and study of nuclear radioactivity quickly revealed evidence of revolutionary new physics. In addition, uses for nuclear radiation also emerged quickly—for example, people such as Ernest Rutherford used it to determine the size of the nucleus and devices were painted with radon-doped paint to make them glow in the dark (see Figure 1). We therefore begin our study of nuclear physics with the discovery and basic features of nuclear radioactivity.
Figure 1. The dials of this World War II aircraft glow in the dark, because they are painted with radium-doped phosphorescent paint. It is a poignant reminder of the dual nature of radiation. Although radium paint dials are conveniently visible day and night, they emit radon, a radioactive gas that is hazardous and is not directly sensed. (credit: U.S. Air Force Photo)
Discovery of Nuclear Radioactivity
In 1896, the French physicist Antoine Henri Becquerel (1852–1908) accidentally found that a uranium-rich mineral called pitchblende emits invisible, penetrating rays that can darken a photographic plate enclosed in an opaque envelope. The rays therefore carry energy; but amazingly, the pitchblende emits them continuously without any energy input. This is an apparent violation of the law of conservation of energy, one that we now understand is due to the conversion of a small amount of mass into energy, as related in Einstein’s famous equation E=m c 2. It was soon evident that Becquerel’s rays originate in the nuclei of the atoms and have other unique characteristics. The emission of these rays is called nuclear radioactivity or simply radioactivity. The rays themselves are called nuclear radiation. A nucleus that spontaneously destroys part of its mass to emit radiation is said to decay (a term also used to describe the emission of radiation by atoms in excited states). A substance or object that emits nuclear radiation is said to be radioactive.
Two types of experimental evidence imply that Becquerel’s rays originate deep in the heart (or nucleus) of an atom. First, the radiation is found to be associated with certain elements, such as uranium. Radiation does not vary with chemical state—that is, uranium is radioactive whether it is in the form of an element or compound. In addition, radiation does not vary with temperature, pressure, or ionization state of the uranium atom. Since all of these factors affect electrons in an atom, the radiation cannot come from electron transitions, as atomic spectra do. The huge energy emitted during each event is the second piece of evidence that the radiation cannot be atomic. Nuclear radiation has energies of the order of 10 6 eV per event, which is much greater than the typical atomic energies (a few eV), such as that observed in spectra and chemical reactions, and more than ten times as high as the most energetic characteristic x rays. Becquerel did not vigorously pursue his discovery for very long. In 1898, Marie Curie (1867–1934), then a graduate student married the already well-known French physicist Pierre Curie (1859–1906), began her doctoral study of Becquerel’s rays. She and her husband soon discovered two new radioactive elements, which she named _polonium_ (after her native land) and _radium_ (because it radiates). These two new elements filled holes in the periodic table and, further, displayed much higher levels of radioactivity per gram of material than uranium. Over a period of four years, working under poor conditions and spending their own funds, the Curies processed more than a ton of uranium ore to isolate a gram of radium salt. Radium became highly sought after, because it was about two million times as radioactive as uranium. Curie’s radium salt glowed visibly from the radiation that took its toll on them and other unaware researchers. Shortly after completing her Ph.D., both Curies and Becquerel shared the 1903 Nobel Prize in physics for their work on radioactivity. Pierre was killed in a horse cart accident in 1906, but Marie continued her study of radioactivity for nearly 30 more years. Awarded the 1911 Nobel Prize in chemistry for her discovery of two new elements, she remains the only person to win Nobel Prizes in physics and chemistry. Marie’s radioactive fingerprints on some pages of her notebooks can still expose film, and she suffered from radiation-induced lesions. She died of leukemia likely caused by radiation, but she was active in research almost until her death in 1934. The following year, her daughter and son-in-law, Irene and Frederic Joliot-Curie, were awarded the Nobel Prize in chemistry for their discovery of artificially induced radiation, adding to a remarkable family legacy.
Alpha, Beta, and Gamma
Research begun by people such as New Zealander Ernest Rutherford soon after the discovery of nuclear radiation indicated that different types of rays are emitted. Eventually, three types were distinguished and named alpha (α), beta (β), and gamma (γ), because, like x-rays, their identities were initially unknown. Figure 2 shows what happens if the rays are passed through a magnetic field. The γ s are unaffected, while the $latex\boldsymbol{\alpha} $ s and β s are deflected in opposite directions, indicating the α s are positive, the β s negative, and the γ s uncharged. Rutherford used both magnetic and electric fields to show that α s have a positive charge twice the magnitude of an electron, or +2|q e|. In the process, he found the α s charge to mass ratio to be several thousand times smaller than the electron’s. Later on, Rutherford collected α s from a radioactive source and passed an electric discharge through them, obtaining the spectrum of recently discovered helium gas. Among many important discoveries made by Rutherford and his collaborators was the proof that α radiation is the emission of a helium nucleus. Rutherford won the Nobel Prize in chemistry in 1908 for his early work. He continued to make important contributions until his death in 1934.
Figure 2. Alpha, beta, and gamma rays are passed through a magnetic field on the way to a phosphorescent screen. The α s and β s bend in opposite directions, while the γ s are unaffected, indicating a positive charge for α s, negative for β s, and neutral for γ s. Consistent results are obtained with electric fields. Collection of the radiation offers further confirmation from the direct measurement of excess charge.
Other researchers had already proved that β s are negative and have the same mass and same charge-to-mass ratio as the recently discovered electron. By 1902, it was recognized that β radiation is the emission of an electron. Although β s are electrons, they do not exist in the nucleus before it decays and are not ejected atomic electrons—the electron is created in the nucleus at the instant of decay.
Since γ s remain unaffected by electric and magnetic fields, it is natural to think they might be photons. Evidence for this grew, but it was not until 1914 that this was proved by Rutherford and collaborators. By scattering γ radiation from a crystal and observing interference, they demonstrated that γ radiation is the emission of a high-energy photon by a nucleus. In fact, γ radiation comes from the de-excitation of a nucleus, just as an x ray comes from the de-excitation of an atom. The names “γ ray” and “x ray” identify the source of the radiation. At the same energy, γ rays and x rays are otherwise identical.
| Type of Radiation | Range |
--- |
| α -Particles | A sheet of paper, a few cm of air, fractions of a mm of tissue |
| β -Particles | A thin aluminum plate, or tens of cm of tissue |
| γ Rays | Several cm of lead or meters of concrete |
| Table 1: Properties of Nuclear Radiation |
Ionization and Range
Two of the most important characteristics of α, β, and γ rays were recognized very early. All three types of nuclear radiation produce ionization in materials, but they penetrate different distances in materials—that is, they have different ranges. Let us examine why they have these characteristics and what are some of the consequences.
Like x rays, nuclear radiation in the form of α s, β s, and γ s has enough energy per event to ionize atoms and molecules in any material. The energy emitted in various nuclear decays ranges from a few keV to more than 10 MeV, while only a few eV are needed to produce ionization. The effects of x rays and nuclear radiation on biological tissues and other materials, such as solid state electronics, are directly related to the ionization they produce. All of them, for example, can damage electronics or kill cancer cells. In addition, methods for detecting x rays and nuclear radiation are based on ionization, directly or indirectly. All of them can ionize the air between the plates of a capacitor, for example, causing it to discharge. This is the basis of inexpensive personal radiation monitors, such as pictured in Figure 3. Apart from α, β, and γ, there are other forms of nuclear radiation as well, and these also produce ionization with similar effects. We define ionizing radiation as any form of radiation that produces ionization whether nuclear in origin or not, since the effects and detection of the radiation are related to ionization.
Figure 3. These dosimeters (literally, dose meters) are personal radiation monitors that detect the amount of radiation by the discharge of a rechargeable internal capacitor. The amount of discharge is related to the amount of ionizing radiation encountered, a measurement of dose. One dosimeter is shown in the charger. Its scale is read through an eyepiece on the top. (credit: L. Chang, Wikimedia Commons)
The range of radiation is defined to be the distance it can travel through a material. Range is related to several factors, including the energy of the radiation, the material encountered, and the type of radiation (see Figure 4). The higher the _energy_, the greater the range, all other factors being the same. This makes good sense, since radiation loses its energy in materials primarily by producing ionization in them, and each ionization of an atom or a molecule requires energy that is removed from the radiation. The amount of ionization is, thus, directly proportional to the energy of the particle of radiation, as is its range.
Figure 4. The penetration or range of radiation depends on its energy, the material it encounters, and the type of radiation. (a) Greater energy means greater range. (b) Radiation has a smaller range in materials with high electron density. (c) Alphas have the smallest range, betas have a greater range, and gammas penetrate the farthest.
Radiation can be absorbed or shielded by materials, such as the lead aprons dentists drape on us when taking x rays. Lead is a particularly effective shield compared with other materials, such as plastic or air. How does the range of radiation depend on _material_? Ionizing radiation interacts best with charged particles in a material. Since electrons have small masses, they most readily absorb the energy of the radiation in collisions. The greater the density of a material and, in particular, the greater the density of electrons within a material, the smaller the range of radiation.
Collisions
Conservation of energy and momentum often results in energy transfer to a less massive object in a collision. This was discussed in detail in Chapter 7 Work, Energy, and Energy Resources, for example.
Different _types_ of radiation have different ranges when compared at the same energy and in the same material. Alphas have the shortest range, betas penetrate farther, and gammas have the greatest range. This is directly related to charge and speed of the particle or type of radiation. At a given energy, each α, β, or γ will produce the same number of ionizations in a material (each ionization requires a certain amount of energy on average). The more readily the particle produces ionization, the more quickly it will lose its energy. The effect of _charge_ is as follows: The α has a charge of +2 q e , the β has a charge of −q e , and the γ is uncharged. The electromagnetic force exerted by the α is thus twice as strong as that exerted by the β and it is more likely to produce ionization. Although chargeless, the γ does interact weakly because it is an electromagnetic wave, but it is less likely to produce ionization in any encounter. More quantitatively, the change in momentum Δ p given to a particle in the material is Δ p=F Δ t, where F is the force the α, β, or γ exerts over a time Δ t. The smaller the charge, the smaller is F and the smaller is the momentum (and energy) lost. Since the speed of alphas is about 5% to 10% of the speed of light, classical (non-relativistic) formulas apply.
The speed at which they travel is the other major factor affecting the range of α s, β s, and γ s. The faster they move, the less time they spend in the vicinity of an atom or a molecule, and the less likely they are to interact. Since α s and β s are particles with mass (helium nuclei and electrons, respectively), their energy is kinetic, given classically by 1 2 m v 2. The mass of the β particle is thousands of times less than that of the α s, so that β s must travel much faster than α s to have the same energy. Since β s move faster (most at relativistic speeds), they have less time to interact than α s. Gamma rays are photons, which must travel at the speed of light. They are even less likely to interact than a β, since they spend even less time near a given atom (and they have no charge). The range of γ s is thus greater than the range of β s.
Alpha radiation from radioactive sources has a range much less than a millimeter of biological tissues, usually not enough to even penetrate the dead layers of our skin. On the other hand, the same α radiation can penetrate a few centimeters of air, so mere distance from a source prevents α radiation from reaching us. This makes α radiation relatively safe for our body compared to β and γ radiation. Typical β radiation can penetrate a few millimeters of tissue or about a meter of air. Beta radiation is thus hazardous even when not ingested. The range of β s in lead is about a millimeter, and so it is easy to store β sources in lead radiation-proof containers. Gamma rays have a much greater range than either α s or β s. In fact, if a given thickness of material, like a lead brick, absorbs 90% of the γ s, then a second lead brick will only absorb 90% of what got through the first. Thus, γ s do not have a well-defined range; we can only cut down the amount that gets through. Typically, γ s can penetrate many meters of air, go right through our bodies, and are effectively shielded (that is, reduced in intensity to acceptable levels) by many centimeters of lead. One benefit of γ s is that they can be used as radioactive tracers (see Figure 5).
Figure 5. This image of the concentration of a radioactive tracer in a patient’s body reveals where the most active bone cells are, an indication of bone cancer. A short-lived radioactive substance that locates itself selectively is given to the patient, and the radiation is measured with an external detector. The emitted γ radiation has a sufficient range to leave the body—the range of α s and β s is too small for them to be observed outside the patient. (credit: Kieran Maher, Wikimedia Commons)
PhET Explorations: Beta Decay
Watch beta decay occur for a collection of nuclei or for an individual nucleus.
Figure 6.Beta Decay
Section Summary
Some nuclei are radioactive—they spontaneously decay destroying some part of their mass and emitting energetic rays, a process called nuclear radioactivity.
Nuclear radiation, like x rays, is ionizing radiation, because energy sufficient to ionize matter is emitted in each decay.
The range (or distance traveled in a material) of ionizing radiation is directly related to the charge of the emitted particle and its energy, with greater-charge and lower-energy particles having the shortest ranges.
Radiation detectors are based directly or indirectly upon the ionization created by radiation, as are the effects of radiation on living and inert materials.
Conceptual Questions
1: Suppose the range for 5.0 MeV α ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeV α a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.
2: What is the difference between γ rays and characteristic x rays? Is either necessarily more energetic than the other? Which can be the most energetic?
3: Ionizing radiation interacts with matter by scattering from electrons and nuclei in the substance. Based on the law of conservation of momentum and energy, explain why electrons tend to absorb more energy than nuclei in these interactions.
4: What characteristics of radioactivity show it to be nuclear in origin and not atomic?
5: What is the source of the energy emitted in radioactive decay? Identify an earlier conservation law, and describe how it was modified to take such processes into account.
6: Consider Figure 2. If an electric field is substituted for the magnetic field with positive charge instead of the north pole and negative charge instead of the south pole, in which directions will the α, β , and γ rays bend?
7: Explain how an α particle can have a larger range in air than a β particle with the same energy in lead.
8: Arrange the following according to their ability to act as radiation shields, with the best first and worst last. Explain your ordering in terms of how radiation loses its energy in matter.
(a) A solid material with low density composed of low-mass atoms.
(b) A gas composed of high-mass atoms.
(c) A gas composed of low-mass atoms.
(d) A solid with high density composed of high-mass atoms.
9: Often, when people have to work around radioactive materials spills, we see them wearing white coveralls (usually a plastic material). What types of radiation (if any) do you think these suits protect the worker from, and how?
Glossary
alpha rays one of the types of rays emitted from the nucleus of an atom beta rays one of the types of rays emitted from the nucleus of an atom gamma rays one of the types of rays emitted from the nucleus of an atom ionizing radiation radiation (whether nuclear in origin or not) that produces ionization whether nuclear in origin or not nuclear radiation rays that originate in the nuclei of atoms, the first examples of which were discovered by Becquerel radioactivity the emission of rays from the nuclei of atoms radioactive a substance or object that emits nuclear radiation range of radiation the distance that the radiation can travel through a material
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8115 | https://www.nagwa.com/en/videos/936138434385/ | Question Video: Finding the Unknown Lengths in a Right Triangle Using the Pythagorean Theorem | Nagwa
Question Video: Finding the Unknown Lengths in a Right Triangle Using the Pythagorean Theorem | Nagwa
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Question Video: Finding the Unknown Lengths in a Right Triangle Using the Pythagorean Theorem Mathematics • First Year of Preparatory School
Given that 𝐴𝐵𝐶𝐷 is a trapezoid, determine the length of line segment 𝐷𝐶.
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Given that 𝐴𝐵𝐶𝐷 is a trapezoid, determine the length of line segment 𝐷𝐶.
We’re told that the quadrilateral 𝐴𝐵𝐶𝐷 is a trapezoid, which means it has one pair of parallel sides. We can see these parallel sides marked on the figure. They’re the sides 𝐴𝐷 and 𝐵𝐶. We’re asked to determine the length of the line segment 𝐷𝐶, which is one of the nonparallel sides of the trapezoid, also known as its legs.
Now, it may appear at first that we don’t have enough information to be able to do this. However, if we sketch in the perpendicular from point 𝐷 to the side 𝐵𝐶, meeting this side at a point we’ll call 𝐸, then we create a right triangle in which 𝐷𝐶 is the hypotenuse. The length of this new line segment is the same as the length of line segment 𝐴𝐵, as both are perpendicular to the two parallel sides of the trapezoid. Hence, 𝐷𝐸 equals 12 centimeters.
We can also determine the length of line segment 𝐸𝐶. As 𝐵 is vertically below 𝐴 and 𝐸 is vertically below 𝐷, line segments 𝐴𝐷 and 𝐵𝐸 are of equal length. Hence, 𝐵𝐸 is also 12.1 centimeters. And the length of 𝐸𝐶 is the difference between the lengths of 𝐵𝐶 and 𝐵𝐸. That’s 21.1 minus 12.1, which is nine centimeters.
We now have a right triangle in which we know the lengths of two of the sides and wish to calculate the length of the third side. We can do this by applying the Pythagorean theorem. This states that in any right triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides. If we denote the lengths of the two shorter sides as 𝑎 and 𝑏 and the length of the hypotenuse as 𝑐, then this can be expressed as 𝑎 squared plus 𝑏 squared equals 𝑐 squared.
In our triangle, 𝐷𝐶 is the hypotenuse as it is directly opposite the right angle. So we can form the equation 𝐷𝐶 squared equals nine squared plus 12 squared. To solve for 𝐷𝐶, we evaluate the squares and find their sum, giving 𝐷𝐶 squared equals 225. Finally, we take the square root of both sides of the equation, giving 𝐷𝐶 equals 15. Note that we’re only interested in the positive solution here as 𝐷𝐶 represents a length.
We may notice that as all three side lengths of this triangle are integer values, it is a Pythagorean triple. In fact, it is an enlargement of perhaps the most commonly known Pythagorean triple, the right triangle with side lengths of three, four, and five units. Including the length units, we’ve found that the length of line segment 𝐷𝐶 is 15 centimeters.
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8116 | https://www.sciencedirect.com/science/article/abs/pii/S2173578619300861 | Validation of the CUETO scoring model for predicting recurrence and progression in T1G3 urothelial carcinoma of the bladder - ScienceDirect
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Actas Urológicas Españolas (English Edition)
Volume 43, Issue 8, October 2019, Pages 445-451
Original article
Validation of the CUETO scoring model for predicting recurrence and progression in T1G3 urothelial carcinoma of the bladder Validación de las tablas del grupo CUETO para predecir la recurrencia y progresión del carcinoma urotelial de vejiga T1G3☆
Author links open overlay panel W.Krajewski a, O.Rodríguez-Faba b, A.Breda b, F.Pisano b, S.Poletajew c, A.Tukiendorf d, R.Zdrojowy a, A.Kołodziej a, J.Palou b
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Abstract
Introduction and objectives
Various studies tried to validate Club Urológico Español de Tratamiento Oncológico (CUETO) tables, yet, none of this papers focused on the high and very high risk bladder cancers. The aim of the study was to externally validate the CUETO model for predicting disease recurrence and progression in group of T1G3 tumours treated with BCG immunotherapy.
Patients or materials and methods
Data from 414 patients with primary T1G3 bladder cancer were analyzed. To evaluate the model discrimination, Cox proportional hazard regression models were created and concordance indexes were calculated.
Results
The median follow-up was 68 months. The recurrence was observed in 212 (51.2%) and 64 patients (15.5%) experienced the recurrence more than once during the study follow-up. Progression of the cancer was observed in 106 patients (25.6%). Radical cystectomy was performed in 115 patients (27.8%) and there were 64 (15.5%) cancer specific deaths. For recurrence and progression probability, the concordance index of the CUETO models was 0.633 and 0.697 respectively. CUETO tables underestimated significantly the risk of recurrence and marginally the risk of progression in the first year of observation. For 5 years of observation, the trend for the recurrence was much less clear. On the contrary, there was slight overestimation in the risk of progression. The study is limited by retrospective nature.
Conclusions
It was shown that the CUETO risk tables exhibit a fair discrimination for both disease recurrence and progression in T1G3 patients treated with BCG. CUETO scoring model underestimates the risk of tumour recurrence, but predicts well risk of progression.
Resumen
Introducción y objetivos
Existen varios estudios con el objetivo de validar las tablas del Club Urológico Español de Tratamiento Oncológico (CUETO). Sin embargo, ninguno de estos estudios se ha centrado en el cáncer de vejiga de alto y muy alto riesgo. El objetivo del presente estudio fue validar externamente el modelo CUETO para predecir la recidiva y la progresión de la enfermedad en el grupo de tumores T1G3 tratados con bacilo Calmette-Guérin (BCG).
Pacientes o materiales y métodos
Se analizaron los datos de 414 pacientes con cáncer de vejiga T1G3 primario. Para evaluar la discriminación del modelo se usaron modelos de riesgos proporcionales de Cox y se calcularon los índices de concordancia.
Resultados
La mediana de seguimiento fue de 68 meses. Se observó recidiva en 212 (51,2%) y 64 pacientes (15,5%) experimentaron más de un episodio de recurrencia durante el periodo de seguimiento. La progresión del cáncer se observó en 106 pacientes (25,6%), 115 pacientes (27,8%) fueron tratados con cistectomía radical, y hubo 64 (15,5%) muertes por tumor. Para la probabilidad de recidiva y progresión, el índice de concordancia de los modelos CUETO fue de 0,633 y 0,697, respectivamente. Las tablas de CUETO subestimaron significativamente el riesgo de recidiva y marginalmente el riesgo de progresión en el primer año de observación. Durante los 5 años de observación, la tendencia de la recidiva fue mucho menos clara. Por el contrario, hubo una ligera sobreestimación en el riesgo de progresión. El estudio está limitado por su naturaleza retrospectiva.
Conclusiones
Se demostró que las tablas de riesgo del grupo CUETO logran una discriminación correcta, tanto para la recidiva de la enfermedad como para la progresión, en pacientes con T1G3 tratados con BCG. El modelo de puntuación (CUETO) subestima el riesgo de recidiva del tumor, pero acierta al predecir el riesgo de progresión.
Introduction
Bladder cancer (BC) is one of the most common tumours worldwide.1 Basing on infiltration of histological structures, BCs are divided into muscle invasive (MIBC) and non-muscle invasive cancers (NMIBC). At initial diagnosis, NMIBC represents majority of bladder tumours.2 Due to non- advanced stage, correctly managed NMIBCs present relatively good cancer-specific survival rates, however, with high perpetual risk of recurrence or progression occurrence.3, 4 For that reason, an accurate prognostic estimation for each individual patient is crucial. To enable this, the European Organization for Research and Treatment of Cancer (EORTC) developed a scoring system and risk tables to predict the short- and long-term probability of disease recurrence and progression. The tool was created based on data of 2596 patients diagnosed with Ta/T1 tumours from seven previous EORTC trials.3 However, despite a great total number of patients included in the analysis, only 171 patients were treated with bacillus Calmette-Guerin (BCG) immunotherapy. To overcome this drawback, the Club Urológico Español de Tratamiento Oncológico (CUETO) created another scoring tool based only on patients treated with BCG.4 Nevertheless, the number of instillations was low in both studies, including patients receiving only one induction regimen. Additionally, a relatively small number of T1G3 tumours were included in the EORTC trial.3 Various studies tried to validate EORTC and CUETO tables, yet, none of this papers focused on the high and very high risk cancers.5, 6, 7
The aim of the study was to externally validate the CUETO scoring model for predicting disease recurrence and progression in a large, homogenous group of T1G3 tumours treated with BCG immunotherapy.
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Section snippets
Material and methods
Data from 414 patients with T1G3 NMIBC treated with intravesical BCG therapy were retrospectively reviewed. Patients were treated between 1986 and 2016 in two academic centres. All the patients underwent complete transurethral resection of bladder tumour (TURBT) and random biopsies of the normal looking mucosa. One hundred and twelve (27%) patients received a single immediate postoperative instillation of chemotherapy. A re- staging resection was performed in 93 patients (22.5%) according to
Results
The study included 414 patients with mean age of 65.9 years. Baseline patients’ characteristics are presented in Table 1. Recurrence was observed in 212 (51.2%) and 64 pts (15.5%) experienced the recurrence more than once during the study follow-up. Progression was observed in 106 pts (25.6%). Cystectomy was performed in 115 pts (27.8%) and there were 64 (15.5%) cancer specific deaths. The estimated HRs for the analyzed Cox's regressions are reported in Table 2. On the basis of the results
Discussion
NMIBC is a heterogeneous disease with highly variable behaviour and outcomes. The T1G3 group is considered to be a high-risk subgroup of NMIBC. Despite the fact, that BCG immunotherapy has significantly improved the outcome of T1G3 tumours, the risk of progression may reach up to 30% of patients.11, 12 What is more, these patients present worse prognosis after radical cystectomy than patients treated with RC because of primary MIBC.13 For those reasons, a precise clinical risk assessment and
Conclusions
In this study it was shown, that the CUETO risk tables exhibit a fair discrimination for both disease recurrence and progression in high and very high risk T1G3 patients treated with BCG immunotherapy. CUETO scoring model underestimates the risk of tumour recurrence but predicts well risk of progression.
Ethical approval
All procedures performed in studies involving human participants were in accordance with the ethical standards of the institutional and/or national research committee and with the 1964 Helsinki declaration and its later amendments or comparable ethical standards.
Informed consent
This is retrospective observation. Informed consent was obtained from all individual participants at the time of their BCG therapy onset stating that in future their data may be included in the scientific studies.
Study data availability
Data are available for bona fide researchers who request it from the authors. Wojciech Krajewski had full access to all the data in the study and takes responsibility for the integrity of the data and the accuracy of the data analysis.
Compliance with ethical standards
None.
Funding
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
Conflict of interest
The authors declare that they have no conflict of interest.
Recommended articles
References (17)
M. Babjuk et al.
EAU guidelines on non-muscle-invasive urothelial carcinoma of the bladder: update 2016
Eur Urol (2017)
J. Fernandez-Gomez et al.
Predicting nonmuscle invasive bladder cancer recurrence and progression in patients treated with bacillus Calmette-Guerin: The CUETO scoring model
J Urol (2009)
T. Xu et al.
Predicting recurrence and progression in Chinese patients with nonmuscle-invasive bladder cancer using EORTC and CUETO scoring models
Urology (2013)
P. Gontero et al.
Prognostic factors and risk groups in T1G3 non-muscle-invasive bladder cancer patients initially treated with Bacillus Calmette-Guérin: results of a retrospective multicenter study of 2451 patients
Eur Urol (2015)
S. Cambier et al.
EORTC nomograms and risk groups for predicting recurrence progression, and disease-specific and overall survival in non-muscle-invasive stage Ta-T1 urothelial bladder cancer patients treated with 1-3 years of maintenance bacillus Calmette-Guérin
Eur Urol (2016)
S. van den Bosch et al.
Long-term cancer-specific survival in patients with high-risk, non-muscle-invasive bladder cancer and tumour progression: a systematic review
Eur Urol (2011)
S. Chavan et al.
International variations in bladder cancer incidence and mortality
Eur Urol (2014)
R.L. Siegel et al.
Cancer statistics, 2017
CA Cancer J Clin (2017)
There are more references available in the full text version of this article.
Cited by (0)
☆
Please cite this article as: Krajewski W, Rodríguez-Faba O, Breda A, Pisano F, Poletajew S, Tukiendorf A, et al. Validación de las tablas del grupo CUETO para predecir la recurrencia y progresión del carcinoma urotelial de vejiga T1G3. Actas Urol Esp. 2019;43:445–451.
View full text
© 2019 AEU. Published by Elsevier España, S.L.U. All rights reserved.
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8117 | https://medium.com/data-science/simpsons-paradox-d2f4d8f08d42 | An Introduction to Simpson’s Paradox | by David Ramsay | TDS Archive | Medium
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An Introduction to Simpson’s Paradox
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Simpson’s paradox is a fascinating phenomena that illustrates the importance of causality in reasoning. If you haven’t already read it, you should check out The Book of Why — it’s one of the most important books of this decade.
I think Simpson’s paradox is a great introduction to the value of causal reasoning. In this post I’m going to walk through a few examples to elucidate how it arises, and how we can start to formalize our thinking around it.
Batting Averages
A common example of Simpson’s Paradox comes from Ken Ross. He showed that when looking at Derek Jeter’s and David Justice’s batting averages, a curious contradiction arises:
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David Justice has a better batting average every year, but Derek Jeter has a better batting average overall!
Derek Jeter has a worse batting average every season, but somehow has a better batting average overall!
If we stare at this for a moment, the reason becomes clear as a few things stand out. In the first year, Derek and David both had low scores; in the second year they both had high scores. Most importantly, Derek’s low score was only from a few at bats, and thus doesn’t factor into his overall average very strongly. David, on the other hand, is exactly the opposite — his lower score makes up the vast majority of his attempts at bat. Even though David did slightly better in both years without weighting, when we combine this imbalance of at bats with their similarly large difference in score, we get this result. Simpson’s paradox shows us that a properly weighted average for an overall statistic can contradict the trends we see in every subgroup.
Another simple example — this time graphical — is below. Let’s say very young people (who tend to be shorter than adults) practice a lot of basketball and are pretty good. Let’s say older folks (who tend to be taller) quit practicing and generally are actually pretty bad at shooting hoops.
Now let’s say we measure their performance shooting hoops, with the hopes of finally proving that taller people are better at basketball. And what we get looks like this:
Press enter or click to view image in full size
A graphical example of Simpson’s paradox. Taller kids and taller adults are both better at basketball, but in general being tall makes you worse!
Another conundrum! It appears that — for each age group — being taller means you are a better basketball player. Overall, though, being taller makes you worse!
We all know that being tall makes you better, and the right way to interpret this result is separated by age instead of using the overall population. But why and how do we know that? And how do we generalize that rule of thumb to trickier situations?
Causal Reasoning
Here’s a more difficult example, which mirrors a real life medical study. Imagine you’re sick, and you’re offered a drug that had the following results for previous patients:
Press enter or click to view image in full size
Uh oh. The drug is good for men, good for women, and bad for people.
You read it correctly. This drug improves your odds of recovering if you’re a man; it also improves your odds of recovering if you’re a woman. But if you’re a person, it’s more likely to harm you. Do you take this drug?
Thankfully, the above statement is logically incoherent and false. But it’s easy to read the data and see it that way. How can we address this problem?
The Answer
We know that there are three things at work here to give us Simpson’s paradox. There’s (1) the gender of the patient, (2) how effective the drug is, and (3) how likely a patient is to be in a treatment or non-treatment group. Remember, it’s the imbalance of the groups that makes it possible for the overall average to be weighted differently then the trend in each subgroup.
The question then becomes, where are the causal relationships? It’s safe to say that your gender isn’t going to change based on your treatment or recovery. On the other hand, gender does seem to influence how likely you are to choose the treatment — women chose it 75% of the time, while only 12% of the men did.
Given we’re examining the data in this way, we clearly believe that gender might influence recovery. Finally (and obviously), treatment will affect your chance of recovery and not the other way around. We can draw a nice little causal diagram to summarize all of our causal knowledge.
A simple causal diagram of what’s going on. Gender alters the likelihood of accepting treatment. We also expect the likelihood of recovery to be a function of both receiving treatment AND gender.
Where do things go wrong? Hopefully it’s clear that the ‘gender-influenced treatment selection’ part of the diagram — that arrow linking gender to treatment — is what is causing our paradox.
We don’t want that information to alter our decision. How likely someone is to choose the treatment doesn’t tell us anything about how effective the treatment ultimately is for them. To get an accurate account for how effective this drug is for people as a whole, we must eliminate that link. In other words, we should just average the percentages of each group, without weighting them by size. This removes the effect of uneven treatment selection by the two genders.
Press enter or click to view image in full size
Simpson’s paradox is resolved!
Does It Generalize?
In the case above, it made sense to trust the sub-groupings and not the overall average. In other cases, it can be the opposite.
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Take an example (this one from Judea Pearl’s book) where we look at a drug designed for heart attack prevention, and we assess its effect on low-blood pressure and high-blood pressure individuals. We see a similar issue — the drug decreases heart attacks in people with low- and high- blood pressure, but increases heart attacks overall! What do we do? In this case, when we draw a causal diagram, it might look something like this:
Causal Diagram for our second example. In this case, the blood pressure is a mediating variable in the path from treatment to heart attack, and thus we should trust the overall statistic.
This looks very similar to our first example, but it’s not. Look closely. The locations of our subgroups and the treatment have switched.
In the first case, our subgroup (gender) affected who receives treatment. In the second, our subgroup (blood pressure) is affected by who receives treatment.
In this case, both the risk of heart attack and the patient’s blood pressure are causally downstream from treatment, so we should notsubdivide by blood pressure. Blood pressure is a mechanism by which the treatment is working — it’s a mediating variable along the causal path from treatment to heart attack risk.
Contrast that to the prior case, where the relationship between treatment and recovery was confounded by gender. Gender was upstream of both treatment selection and recovery; to get at the real relationship between treatment and recovery we were forced to condition on it. When the treatment has no graphical parents, we don’t need the secondary variable. When it does (and that parent has a ‘backdoor path’ to the effect of interest), we do.
Another Approach
Another way to think about Simpson’s paradox is through the language of pure statistics. First let’s take a small detour back through the basics.
A joint probability can be easily factored into a product of conditionals. Here’s a fun example — you’re getting a new dog, and you’re interested in their personality! Some of the dogs are (Y)oung, some are (I)ntelligent (and some aren’t), and some are quite (E)nergetic.
Below is the box that represents the distribution of dog personality probabilities. We set the area of the square equal to 1. Drop a pin, and the dog it represents will have the personality traits underneath.
Amongst all the possible dogs, you can see some are young, some are smart, some are energetic, and a few are all three. The likelihood of a random dog with some specific combination of traits can be factored easily — below is a simple example to illustrate how when we’re looking for young, intelligent, and energetic dogs (Y=1, I=1, and E=1).
Press enter or click to view image in full size
Factoring a joint probability into the product of conditionals.
Why would you want to factor a joint into conditionals like that? In the context of Simpson’s paradox, we’ll answer that by revisiting our first example above.
In that example, we were looking at the joint probability of (T) Treatment, (G) Gender, and (R) Recovery. We want to query the likelihood of R=1 (a successful recovery) given other combinations of gender and treatment. Let’s factor our joint so we can work with it:
Press enter or click to view image in full size
All possible ways to rewrite that joint as a product of conditioned variables.
There are six ways to factor a joint of three variables. However, in this case we are trying to take an observation of relationships(in which it’s okay to say things like ‘the probability someone is a man given they recovered successfully’) and instead query causal relationships between them(‘if we force the probability of treatment to equal 1, what happens to a man’s recovery?’).Here we introduce Pearl’s do-operator.
Instead of p(R|T), an observation about the correlation between treatment and recovery, we actually care about p(R|do(T)) — the probability of a recovery given we causally intervene and force treatment (setting T=1 by force of will). The do operator limits how we can factor our joint, because we must keep do(T) on the right side of the conditional line with respect to recovery to answer a causal question about how treatment affects recovery.
So we’re interested in p(R|G, do(T)). That leaves us with two options:
Press enter or click to view image in full size
In the second choice we find p(G|do(T)), a nonsensical relationship.We’re now thinking causally, and this term begs the question ‘What is the probability that your gender changes given you had treatment?’ If gender doesn’t change when we do(T), we can’t use a term like this coherently in a causal setting.
The first equation, on the other hand, doesn’t have logical flaws. We notice that p(do(T)|G) is just 1 — the probability we do(T) conditioned on anything rational will be 1, as that’s the definition of the do operator. We can remove this term and simplify our expression.
Another way to say it is that the likelihood of accepting treatment based on gender is not something we want affecting our assessment of the probability of treatment success, so the term has to go. We know this term is bad news because of our causal intuition. Our conditionals must follow our causal instincts, matching our causal graph such that the probability of child nodes are always (and only) conditioned on their parents.
We’re left with p(R | do(T), G) p(G). It shows us that the likelihood of recovery is related to the fraction of people in that gender, not the overall population. This gives us the same answer as our causal diagram through a more formal process.
Real Life, Man
Simpson’s Paradox can come up in real life — like in the famously contentious gender bias issue at UC Berkeley. Its real beauty, though, lies in illustrating and introducing the rationale behind causal reasoning in statistics.
None of what we did makes sense in the context of standard statistics, but it does make sense. I hope this introduction to Simpson’s paradox gave you a taste for how to start formalizing your causal logic.
For a long time, statistics has lacked the language to discuss causal reasoning and relationships. Thankfully it seems the world of statistics is actively shifting. With it will come a huge boon for science.
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Bob L
Apr 1, 2020
This analysis begs the question of what is the right partitioning of the data. In the sports analysis it is obvious we are talking apples and apples so the number adjustments are obviously correct. The medical example correction is entirely bogus…more
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Thomas Benes
Jan 26, 2020
I have a different perception of this paradox considering your examples.
For me the paradox emerges with a special partitioning of the initial data set: Lets say you have two basketball players throwing the ball consecutively with the following…more
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jaskij
Dec 3, 2019
In this case, both the risk of heart attack and the patient’s blood pressure are causally downstream from treatment, so we should not subdivide by blood pressure. Blood pressure is a me...
This contrived blood pressure example makes me scream internally: wrong data collection. Or at least oversimplification in analysis.
Shouldn’t there be, like, one more variable? Blood pressure prior to treatment as independent and by how much it was lowered during treatment as mediating?
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8118 | https://www.youtube.com/watch?v=Kas0tIxDvrg | Exponential growth and epidemics
3Blue1Brown
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Posted: 8 Mar 2020
A primer on exponential and logistic growth
Help fund future projects:
An equally valuable form of support is to simply share some of the videos.
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Excellent visualization of this kind of growth from Minutephysics and Aatish Bhatia:
Data source:
Some have (quite rightfully) commented on how you shouldn't look at the R^2 of linear regressions on cumulative data since even if the changes from one day to the next are completely random, the totals they add up to wouldn't be independent of each other. Since the derivative of an exponential should also be an exponential, we could instead run the same test on the logarithms of the differences from day to day, which in this case gives R^2 = 0.91.
While this video uses COVID-19 as a motivating example, the main goal is simply a math lesson on exponentials and logistic curves. If you're looking for a video more focused on COVID-19 itself, I'd recommend taking a look at this one from Osmosis:
Extrapolation xkcd:
Thanks to these viewers for their contributions to translations
Hebrew: Omer Tuchfeld
These animations are largely made using manim, a scrappy open source python library:
If you want to check it out, I feel compelled to warn you that it's not the most well-documented tool, and it has many other quirks you might expect in a library someone wrote with only their own use in mind.
Music by Vincent Rubinetti.
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If you want to contribute translated subtitles or to help review those that have already been made by others and need approval, you can click the gear icon in the video and go to subtitles/cc, then "add subtitles/cc". I really appreciate those who do this, as it helps make the lessons accessible to more people.
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Intro The phrase exponential growth is familiar to most people, and yet human intuition has a hard time really recognizing what it means sometimes. We can anchor on a sequence of small seeming numbers and then become surprised when suddenly those numbers look big, even if the overall trend follows an exponential perfectly consistently. This right here is the data for the recorded cases of COVID-19, aka the coronavirus, at least at the time I'm writing this. Never one to waste an opportunity for a math lesson, I thought this might be a good time for all of us to go back to the basics on what exponential growth really is, where it comes from, what it implies, and maybe most pressingly how to know when it's coming to an end. Exponential growth means that as you go from one day to the next, it involves multiplying by some constant. In our data, the number of cases in each day tends to be a multiple of about 1.15 to 1.25 of the number of cases the previous day. Viruses Viruses are a textbook example of this kind of growth, because what causes new cases are the existing cases. If the number of cases on a given day is n, and we say that each individual with the virus is exposed to, on average, e people on a given day, and each one of those exposures has a probability p of becoming a new infection, then the number of new cases on a given day is e times p times n. The fact that n itself is a factor in its own change is what really makes things go fast, because if n gets big, it means the rate of growth itself is getting big. One way to think about this is that as you add the new cases to get the next day's growth, you can factor out the n, so it's just the same as multiplying by some constant that's bigger than 1. logarithmic scale This is sometimes easier to see if we put the y-axis of our graph on a logarithmic scale, which means that each step of a fixed distance corresponds to multiplying by a certain factor, in this case each step is another power of 10. On this scale, exponential growth should look like a straight line. Looking at our data, it seems like it took 20 days to go from 100 to 1000, and 13 days to go from that to 10,000, and if you do a simple linear regression to find the best fit line, you can look at the slope of that line to draw a conclusion like we tend to multiply by 10 every 16 days on average. This regression also lets us be a little more quantitative about exactly how close the exponential fit really is, and to use the technical statistical jargon here, the answer is that it's really freaking close. But it can be hard to digest exactly what that means if true. When you see one country with, say, 6000 cases and another with 60, it's easy to think that the second is doing 100 times better, and hence fine. But if you're actually in a situation where numbers multiply by 10 every 16 days, another way to view the same fact is that the second country is about a month behind the first. This is of course rather worrying if you draw out the line. I'm recording this on March 6th, and if the present trend continues it would mean hitting a million cases in 30 days, hitting 10 million in 47 days, 100 million in 64 days, and 1 billion in 81 days. Needless to say, though, you can't just draw out a line like this forever, it clearly has to start slowing down at some point. But the crucial question is when. Is it like the SARS outbreak of 2002 which capped out around 8000 cases, or the Spanish flu of 1918 which ultimately infected about 27% of the world's population? In general, with no context, just drawing a line through your data is not a great way to make predictions, but remember, there's an actual reason to expect an exponential here. If the number of new cases each day is proportional to the number of existing cases, it necessarily means each day you multiply by some constant, so moving forward d days is the same as multiplying by that constant d times. The only way that stops is if either the number E or P goes down. It's inevitable that this will eventually happen. Even in the most perfectly pernicious model for a virus, which would be where every day each person with the infection is exposed to a random subset of the world's population, at some point most of the people they're exposed to would already be sick, and so they couldn't become new cases. In our equation, that would mean that the probability of an exposure becoming a new infection would have to include some kind of factor to account for the probability that someone you're exposed to is already infected. For a random shuffling model like this, that could mean including a factor like 1 minus the proportion of people in the world who are already infected. Logistic Curve Including that factor, and then solving for how N grows, you get what's known in the model. ss as a logistic curve, which is essentially indistinguishable from an exponential at the beginning, but ultimately levels out once you're approaching the total population size, which is what you would expect. True exponentials essentially never exist in the real world, every one of them is the start of a logistic curve. This point right here, where that logistic curve goes from curving upward to instead curving downward, is known as the inflection point. There, the number of new cases each day, represented by the slope of this curve, stops increasing and stays roughly constant before it starts decreasing. One number that people often follow with epidemics is the growth factor, which is defined as the ratio between the number of new cases one day and the number of new cases the previous day. Growth Factor Just to be clear, if you were looking at all of the totals from one day to the next, then tracking the changes between those totals, the growth factor is a ratio between two successive changes. While you're on the exponential part, this factor stays consistently above one, whereas as soon as your growth factor looks closer to one, it's a sign that you've hit the inflection. This can make for another counterintuitive fact while following the data. Think about what it would feel like for the number of new cases one day to be about 15% more than the number of new cases the previous day, and contrast that with what it would feel like for it to be about the same. Just looking at the totals they result in, they don't really feel that different. But if the growth factor is one, it could mean you're at the inflection point of a logistic, which would mean the total number of cases is going to max out at about two times wherever you are now. But a growth factor bigger than one, subtle though that might seem, means you're on the exponential part, which could imply there are orders of magnitude of growth still waiting ahead of you. Fractal Pattern Now, while it's true that in the worst-case situation the saturation point is around the total population, it's of course not at all true that people with the virus are randomly shuffled around the world's population like this. People are clustered in local communities. However, if you run simulations where there's even a little bit of travel between clusters like this, the growth is actually not that much different. What you end up with is a kind of fractal pattern, where communities themselves function like individuals. Each one has some exposure to others, with some probability of spreading the infection, so the same underlying and exponential-inducing laws apply. Fortunately, saturating the whole population is not the only thing that can cause the two factors we care about to go down. The amount of exposure can also go down when people stop gathering and traveling, and the infection rate can go down when people just wash their hands more. The other thing that's counterintuitive about exponential growth, this time in a more optimistic sense, is just how sensitive it is to this constant. For example, if it's 15%, like it is as I'm recording this, and we're at 21,000 cases now, that would mean that 61 days from now you hit over 100 million. But if through a bit less exposure and infection, that rate drops down to 5%, it doesn't mean the projection also drops down by a factor of 3, it actually drops down to around 400,000. So if people are sufficiently worried, there's a lot less to worry about. But if no one is worried, that's when you should worry. |
8119 | https://robincornecki.com/the-1-method-for-finding-slope-without-using-a-formula/ | The #1 method for finding slope without using a formula!
I’m so excited to talk about the #1 method for finding a slope without using a formula! Through my years of teaching, I have found that my students struggled with the slope formula. So instead of using just the formula, I disguised the formula in a slightly different way that students would understand! We can use this #1 method to find slope without a formula and apply it to finding slope from a graph, two points, from an equation, and so on! If you need to find a slope, this method is for you!
Finding Slope
Let’s talk about finding a slope and what a slope is! First, we need a line! The slope is the steepness of a line as it moves from LEFT to RIGHT. The slope is the ratio of the rise, the vertical change, to the run, the horizontal change of a line. The slope of a line is always constant (it never changes) no matter what 2 points on the line you choose.
When would slope be important in the real world? Have you ever seen a spiral staircase in a house? The reason may be for design features, but in reality, a staircase most likely wouldn’t fit because the slope would be too steep to climb up the stairs!
Want to skip reading and watch instead? Don’t worry I got you covered!
Formula for Finding Slope
The formula or the equation for finding slope is m=y2-y1/x2-x1
Where m = slope and (x1, y1) and (x2, y2) are points found on the line.
Finding Slope with Two Points
So how are we going to disguise this formula to make it easier to remember how to find the slope? We are going to put the formula into a table like this ⬇️
Next, we will find the change in the y values by using a vertical number line. I like to just use the words “what is the change in y values and does the number go up or does the number go down”?
Then repeat this for the change in x values.
Then you simply write your change in y over your change in x. Simplify if necessary and then voila you now have your slope!
How to Find Slope from a Graph
You can either pick 2 points from the line and use the table method or you can count how many units you rise over how many units you run. You can “rise” up or down, but I always like to go up because rise means up! Then you can “run” right if you are moving towards the positive numbers or you can “run” left if you are moving towards the negative numbers.
You can always check with Slope Dude if you need to remember whether your slope is positive or negative.
Finding Slope of an Equation
If you want to find the slope of an equation, the equation needs to be in Slope-Intercept Form. Slope-Intercept Form: y = mx + b
M = slope
B = y-intercept.
If your equation is in Standard Form Ax + By = C, your slope = -A/B or you can solve for y. Then all you do is identify your slope in the equation y = mx + b and that is it!
Do you need any more slope practice? Check out these slope resources!
Love, Robin
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8120 | https://www.linkedin.com/posts/organisation-eco-cooperation-development-organisation-cooperation-developpement-eco_taxing-wages-2025-activity-7351597401628893185-F-io | Taxing Wages 2025 | OECD - OCDE
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Taxing Wages 2025
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Labour taxes edge up in the OECD as real wages recover in 2024. The latest edition of Taxing Wages reveals that the post-tax income of a single worker earning the average wage increased in real terms in 28 of the 38 OECD countries in 2024, having declined in 21 countries in 2023 and 33 countries in 2022. Taxing Wages compares the labour tax wedge across OECD countries, the share of labour costs paid in taxes by both employers and employees, minus cash benefits for working families. It examines eight household types, varying by income and family structure. The 2025 edition shows that effective tax rates rose slightly for seven of the eight household types in 2024, with the average tax wedge for each returning to pre COVID-19 pandemic levels. Additionally, this year’s edition of Taxing Wages contains a special feature that analyses the impact of tax credits and allowances on personal income tax rates in OECD countries. Access the full report & key findings in the highlights brochure ➡️ Read the press release ➡️ | OECD Tax
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List of Thermal Expansion Coefficients (CTE) for Natural and Engineered Materials | MSE Supplies LLC
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Home›List of Thermal Expansion Coefficients (CTE) for Natural and Engineered Materials
List of Thermal Expansion Coefficients (CTE) for Natural and Engineered Materials
The thermal expansion coefficients of materials are essential metrics that reveal how materials respond to temperature changes. The materials span a range of engineered and natural substances, and their coefficients shed light on their thermal behaviors. The thermal expansion coefficient is represented by α and quantifies the fractional size change of a substrate per unit temperature variation. Positive coefficients indicate expansion and negative ones showcase contraction.
The thermal expansion coefficients of materials vary significantly and influence structural durability and performance in industries like electronics and construction. The material coefficient of thermal expansion for steel, for example, is determined by how much the structure's dimensions change in response to temperature variations.
Natural materials like copper or aluminum have noteworthy coefficients, while natural stones like granite and marble have more distinct terminal expansion characteristics. That makes these materials essential for use in construction. Polymers like polyethylene's coefficient properties make them a good choice for use in daily products. Composite materials, with their nuanced expansion behaviors, are suited for use in automotive and aerospace applications.
What Is Linear Thermal Expansion?
Linear thermal expansion refers to a material's capacity to expand in length when subjected to an increase in temperature. This happens because the kinetic energy of the material's molecules surges, leading to a vibration that creates more space between them. Conversely, a reduction in temperature, which reduces vibrational energy, causes the material to contract.
The expansion is quantified by using the coefficient of linear expansion, a formula representing the fractional change in length per unit temperature change. The coefficient of linear expansion looks as follows:
ΔL = L 0 αΔT
ΔL represents the change in length, α is the coefficient of linear expansion, L 0 refers to the start length and ΔT is the temperature adjustment. The linear thermal expansion property is essential to various engineering applications. These applications often include designing components or structures exposed to temperature variations, like bridges.
What Is the Coefficient of Thermal Expansion?
The coefficient of thermal expansion (CTE) is defined as the fractional change in length or volume per unit temperature change. It quantifies a material's propensity to contract or expand as temperatures change. CTE for an area is mathematically expressed as:
α = ΔL/(L 0 ΔT)
ΔL refers to the change in length, L 0 is the original length and ΔT represents the change in temperature.
It is vital for engineers to predict how materials may behave under various temperature fluctuations correctly. This prediction offers critical insights into material choice and structural integrity design considerations. It allows engineers to accommodate these dimensional changes in CTE materials. Materials with lower CTE offer decent dimensional stability. Higher CTE indicates more contraction or expansion in a substrate, which aligns with use in structures that need more flexibility to adjust to temperature changes.
How Are MSE Supplies Products Used?
MSE Supplies is a leading supplier of high-quality materials, equipment and materials characterization services for advanced materials research and manufacturing. Our Analytical Services provide professional Dilatometry and Thermal Expansion Coefficient (CTE) Testing. Alternatively, make use ofour dilatometry testing services, which use NETZSCH DIL 402 PC or similar instruments.
The following is a list of the most popular applications of our products, including LED, ferroelectrics, piezoelectric, electro-optical, photonics, high-power electronics and high-frequency power devices.
| Materials | Linear Thermal Expansion Coefficient -α- |
--- |
| (10-6 m/(m K))) |
| ABS (Acrylonitrile butadiene styrene) thermoplastic | 73.8 |
| ABS -glass fiber-reinforced | 30.4 |
| Acetals | 106.5 |
| Acetal - glass fiber-reinforced | 39.4 |
| Acrylic | 75 |
| Alumina (aluminium oxide, Al 2 O 3)(Sputtering Target)(powders)(Milling media)(Sapphire/Single Crystal)(Planetary Milling Jars)(Roller Milling Jars)(High Purity Powders) | 8.1 |
| Aluminum (Sputtering Target) | 21 - 24 |
| Aluminum Alloy AlSi7Mg (Additive Manufacturing/3D Printing Powder) | 21-22 |
| Aluminum Alloy AlSi10Mg (Additive Manufacturing/3D Printing Powder) | 20 |
| Aluminum Alloy AlSi12(Additive Manufacturing/3D Printing Powder) | 20 |
| Aluminum Metaphosphate, Al(PO 3)3 (High Purity Powder) | 9.6 |
| Aluminum nitride (Sputtering Target)(AlN template on Sapphire) | 5.3 |
| Amber | 50 - 60 |
| Antimony (Sputtering Target) | 9 - 11 |
| Arsenic | 4.7 |
| Bakelite, bleached | 22 |
| Barium | 20.6 |
| Barium Fluoride, BaF 2 (High Purity Powder)(Single crystal and Substrates)(Sputtering Targets) | 18.1 |
| Barium ferrite, | 10 |
| Barium Titanate, BaTiO 3(Sputtering Target) (Single Crystal Substrate) | 9.4 |
| Benzocyclobutene | 42 |
| Beryllium | 11.5 |
| Bismuth(Sputtering Target) | 13 |
| Bismuth Germanate, Bi 4 Ge 3 O 12 (Scintillation Crystals) | 7 |
| Bismuth Oxide, Bi 2 O 3 (High Purity Powder)(Sputtering Target) | Parallel to a-26.7, b-6.6, c-9.0 |
| Boron Carbide,B 4 C (Sputtering Target) | 4.5 |
| Boron Nitride, BN (Sputtering Target) | 1.0 |
| Brass | 18.7 |
| Brick masonry | 5.5 |
| Bronze | 18.0 |
| Cadmium (Sputtering Target) | 30 |
| Cadmium Difluoride,CdF 2(Sputtering Target) | 21.8 |
| Cadmium Selenide, CdSe (Sputtering Target) | Parallel to a-6.26, c-4.28 |
| Cadmium Sulfide, CdS (Sputtering Target) | Parallel to a-6.26, c-3.5 |
| Cadmium Telluride, CdTe (Sputtering Target) | 4.8 |
| Calcium | 22.3 |
| Calcium Fluoride, CaF 2 (Single Crystal)(Sputtering Target)(Eu doped Scintillation Crystal) | 18.85 |
| Calcium Fluoride Europium doped, Eu:CaF 2(Eu doped Scintillation Crystal) | 19.5 |
| Cast Iron Gray | 10.8 |
| Caoutchouc | 66 - 69 |
| Celluloid | 100 |
| Cellulose acetate (CA) | 130 |
| Cellulose acetate butynate (CAB) | 96 - 171 |
| Cellulose nitrate (CN) | 80 - 120 |
| Cement | 10.0 |
| Cesium Iodide, CsI (Scintillation Crystals) | 54 |
| Cerium (Sputtering Target) | 5.2 |
| Cerium Dioxide, CeO 2(Sputtering Target) | 11.2 |
| Cerium Fluoride, CeF 3 (Scintillation Crystal) (Sputtering Target) | Parallel to a-12.9, c-16.5 |
| Cerium Oxide, Ce 2 O 3 (High Purity Powder) | 12 |
| Chlorinated polyether | 80 |
| Chlorinated polyvinylchloride (CPVC) | 66.6 |
| Chromium (Sputtering Target) | 6.2 |
| Chromium Boride,Cr 5 B 3(Sputtering Target) | 13.7 |
| Chromium Carbide,Cr 3 C 2 (Sputtering Target) | 10.3 |
| Chromium Diboride CrB 2(Sputtering Target) | 6.2 - 7.5 |
| Chromium Disilicide,CrSi 2(Sputtering Target) | Parallel to a-8.2, c-9.0 |
| Chromium Monoboride, CrB | 12.3 |
| Chromium Nitride,CrN | 2.3 |
| Chromium Nitride,Cr 2 N | 9.4 |
| Chromium Oxide, Cr 2 O 3 (High Purity Powder)(Sputtering Target) | 9.0 |
| Clay tile structure | 5.9 |
| Cobalt (Sputtering Target) | 12 |
| Cobalt Chrome Molybdenum alloy, CoCrMo (Additive Manufacturing/3D Printing Powder) | 11.5 |
| Cobalt Chrome Tungsten alloy, CoCrW (Additive Manufacturing/3D Printing Powder) | 21 |
| Cobalt Disilicide,CoSi 2 | 10 |
| Concrete | 14.5 |
| Concrete structure | 9.8 |
| Constantan | 18.8 |
| Copper (Sputtering Target) | 16.6 |
| Copper, Beryllium 25 | 17.8 |
| Copper Tin alloy, CuSn10(Additive Manufacturing/3D Printing Powder) | 18 |
| Corundum, sintered | 6.5 |
| Cupronickel 30% | 16.2 |
| Diamond (Carbon) | 1.18 |
| Duralumin | 23 |
| Dysprosium (Sputtering Target) | 9.9 |
| Dysprosium Oxide, Dy 2 O 3 (High Purity Powder)(Sputtering Targets) | 6.6 |
| Ebonite | 76.6 |
| Epoxy, cast resins & compounds, unfilled | 45 - 65 |
| Erbium (Sputtering Target) | 12.2 |
| Erbium Oxide, Er 2 O 3 (High Purity Powder)(Sputtering Targets) | 6.6 |
| Ethylene ethyl acrylate (EEA) | 205 |
| Ethylene vinyl acetate (EVA) | 180 |
| Europium (Sputtering Target) | 35 |
| Europium Oxide, Eu 2 O 3 (High Purity Powder) | 6.4 |
| Fluoroethylene propylene (FEP) | 135 |
| Fluorspar, CaF 2 (Single Crystal) (Sputtering Target) | 19.5 |
| Gadolinium (Sputtering Target) | 9 |
| Gadolinium Gallium Garnet (GGG), Gd 3 Ga 5 O 12 (Single Crystal)(Nd doped Single Crystal) | 8 |
| Gadolinium Oxide, Gd 2 O 3 (High Purity Powder) | 7.6 |
| Gadolinium Tetraboride,GdB 4 | 7.0 |
| Gallium | 37 |
| Gallium Arsenide, GaAs (Sputtering Target) (Single Crystal Substrates) | 5.73 |
| Gallium Oxide, Ga 2 O 3 (Single Crystal Wafer)(High Purity Powder)(Sputtering Target) | Parallel to a-3.77, b-7.8, c-6.34 |
| Gallium Nitride, GaN (Single Crystal) | 3.17 |
| Germanium (Sputtering Target)(Single Crystal) | 5.9 |
| Germanium Oxide,GeO 2(High Purity Powder) | 7.5 |
| Germanium Telluride, GeTe (Sputtering Target) | 11.2 |
| German silver | 18.4 |
| Glass, hard | 5.9 |
| Glass, Pyrex | 4.0 |
| Glass, plate | 9.0 |
| Gold (Sputtering Target) | 14.2 |
| Gold - copper | 15.5 |
| Gold - platinum | 15.2 |
| Granite | 7.9 |
| Graphite, pure (Carbon) (Sputtering Target) (Anode Materials) | 2 - 6 |
| Gunmetal | 18 |
| Gutta percha | 198 |
| Hafnium (Sputtering Target) | 5.9 |
| Hafnium Carbide, HfC (Sputtering Target) | 6.9 |
| Hafnium Dioxide,HfO 2(Sputtering Target) | 6.0 |
| Hafnium Disilicide,HfSi 2(Sputtering Target) | 16.4 |
| Hafnium Nitride, HfN (Sputtering Target) | 6.9 |
| Hard alloy K20 | 6 |
| Hastelloy C22 (Sputtering Target) | 12.4 |
| Hastelloy C276 (Sputtering Target) | 11.2 |
| Hastelloy N(Sputtering Target) | 12.3 |
| Holmium (Sputtering Target) | 11.2 |
| Holmium Oxide, Ho 2 O 3 (High Purity Powder)(Sputtering Targets) | 7.4 |
| Ice, 0 o C water | 51 |
| Inconel 625 (Additive Manufacturing/3D Printing Powder) | 13.1 |
| Inconel 718 (Additive Manufacturing/3D Printing Powder) | 12.8 |
| Indium (Sputtering Target) | 33 |
| Indium Antimonide, InSb (Sputtering Target) | 5.37 |
| Indium Arsinide, InAs (Single Crystal)(Sputtering Target) | 4.52 |
| Indium Gallium Zinc Oxide (IGZO), In 2 Ga 2 ZnO 7 (Sputtering Target) | 4.31 |
| Indium Oxide,In 2 O 3(High Purity Powder)(Sputtering Target) | 7.2 |
| Indium Phosphide, InP (Single Crystal) | 4.5 |
| Indium Sulfide,In 2 S 3(Sputtering Target) | Tetragonal Form: Parallel to a-11.7, c-6.7 Trigonal Form: Parallel to a-14.1, c-26.7 |
| Indium Tin Oxide (ITO),In 2-x Sn x O 3(Sputtering Target)(ITO coated glass and film) | 8 |
| Invar | 1.5 |
| Iridium (Sputtering Target) | 6.4 |
| Iron, pure (Sputtering Target) | 12.0 |
| Iron, cast | 10.4 |
| Iron, forged | 11.3 |
| Iron Boride, FeB (Sputtering Target) | 23 |
| (Di)Iron Boride Fe 2 B(Sputtering Target) | 7.65 |
| (Tri)Iron Carbide,Fe 3 C(Sputtering Target) | 8.6 |
| Kapton | 20 |
| Langasite (LGSO), La 3 Ga 5 SiO 14 (Single Crystal) | 16 |
| Langatate (LGT), La 3 Gao 5.5 Ta 0.5 O 14 (Single Crystal) | 7.65 |
| Lanthanum (Sputtering Target) | 12.1 |
| Lanthanum Aluminate,LaAlO 3(Single Crystal)(Sputtering Target) | 9.4 |
| Lanthanum Fluoride,LaF 3(Sputtering Target) | Parallel to a-15.8, c-11 |
| Lanthanum Hexaboride,LaB 6(Sputtering Target) | 6.2 |
| Lanthanum Manganate,LaMnO 3 (Sputtering Target) | 11.62 |
| Lanthanum Oxide, La 2 O 3 (High Purity Powder) | 8.6 |
| Lanthanum-Strontium Aluminium Tantalate (LSAT),(La 0.18 Sr 0.82)(Al 0.59 Ta 0.41)O 3 (Single Crystal) | 10 |
| Lead (Sputtering Target) | 28.9 |
| Lead Fluoride, PbF 2(Single Crystal)(Sputtering Target) | 29 |
| Lead Selenide, PbSe (Sputtering Target) | 19 |
| Lead Sulfide, PbS (Sputtering Target) | 20.1 |
| Lead Telluride, PbTe (Sputtering Target) | 20.4 |
| Lead Tungstate,PbWO 4(Single Crystal) | Parallel to a-29.5, c-12.8 |
| Limestone | 8 |
| Lithium (Chips for Batteries) | 46 |
| Lithium Aluminate,LiALO 2(Single Crystal) | Parallel to a-10.1, c-16.5 |
| Lithium Fluoride, LiF (Sputtering Target)(High Purity Powder)(Single Crystal) | 37 |
| Lithium Niobate,LiNbO 3(Single Crystal) (Sputtering Target) | Parallel to a-15.4, c-7.5 |
| Lithium Tantalate,LiTaO 3(Single Crystal) | Parallel to a-16, c-4 |
| Lutetium (Sputtering Target) | 9.9 |
| Lutetium Aluminum Garnet (LuAG),Y 3 Al 5 O 12 | 6.0 |
| Lutetium Oxide, Lu 2 O 3(High Purity Powder) | 7.7 |
| Macor | 9.3 |
| Magnalium | 23.8 |
| Magnesium (Sputtering Target) | 25 |
| Magnesium Aluminate, MgAl 2 O 4(Single Crystal) | 6.72 |
| Magnesium Fluoride,MgF 2 (Sputtering Target)(Single Crystal)(High Purity Powder) (Single Crystal Substrates) | Parallel to C axis 13.7 Perpendicular to C axis 8.48 |
| Magnesium Oxide, MgO (Sputtering Target)(Crucible)(Single Crystal Substrate)(High Purity Powder) | 10.8 |
| Manganese (Sputtering Target) | 22 |
| Manganin | 18.1 |
| Marble | 5.5 - 14.1 |
| Masonry | 4.7 - 9.0 |
| Mercury | 61 |
| Mica | 3 |
| Molybdenum (Sputtering Target) | 5 |
| Molybdenum Boride, Mo 2 B (Sputtering Target) | 4.78 |
| Molybdenum Carbide,Mo 2 C (Sputtering Target) | 7.8 |
| Molybdenum Disilicide, MoSi 2 (Sputtering Target) | Parallel to a = 5.6, c =4.1 |
| Molybdenum Oxide, MoO 3(Sputtering Target) | 5 |
| Monel metal | 13.5 |
| Mortar | 7.3 - 13.5 |
| Neodymium (Sputtering Target) | 9.6 |
| Neodymium Copper Oxide,Nd 2 CuO 4 (Sputtering Target) | 10.1 |
| Neodymium Fluoride,NdF 3(Sputtering Target) | Parallel to a-17.4, c-14.7 |
| Neodymium Hexaboride,NdB 6(Sputtering Target) | 7.3 |
| Neodymium Oxide, Nd 2 O 3 (High Purity Powder) | 14.7 |
| Nickel (Sputtering Target) | 13.0 |
| Nickel Chromium (Nichrome), NiCr(Sputtering Target) | 14 |
| Nickel Oxide, NiO (Sputtering Target) | 10 |
| Niobium (Sputtering Target) | 7.3 |
| Niobium Carbide, NbC (Sputtering Target) | 7.81 |
| Niobium Disilicide,NbSi 2 | Parallel to a = 9.1, c =8.7 |
| Niobium Nitride, NbN (Sputtering Target) | 10.1 |
| Niobium Oxide, Nb 2 O 5 (Sputtering Target) (High Purity Powder) | 5.6 |
| Niobium Selenide, NbSe 2(Sputtering Target) | Parallel to a = 6.6, c = 19.9 |
| Niobium Silicide, Nb 5 Si 3(Sputtering Target) | 7.6 |
| Nylon, general purpose | 72 |
| Nylon, Type 11, molding and extruding compound | 100 |
| Nylon, Type 12, molding and extruding compound | 80.5 |
| Nylon, Type 6, cast (Planetary Milling Jar)(Roller Milling Jar) | 85 |
| Nylon, Type 6/6, molding compound | 80 |
| Oak, perpendicular to the grain | 54 |
| Osmium | 5 |
| Palladium (Sputtering Target) | 11.8 |
| Paraffin | 106 - 480 |
| Phenolic resin without fillers | 60 - 80 |
| Phosphor bronze | 16.7 |
| Plaster | 16.4 |
| Plastics | 40 - 120 |
| Platinum (Sputtering Target) | 9.0 |
| Plutonium | 54 |
| Polyacrylonitrile | 70 |
| Polyallomer | 91.5 |
| Polyamide (PA) | 110 |
| Polybutylene (PB) | 130 |
| Polycarbonate (PC) | 70.2 |
| Polycarbonate - glass fiber-reinforced | 21.5 |
| Polyester | 123.5 |
| Polyester - glass fiber-reinforced | 25 |
| Polyethylene (PE) | 200 |
| Polyethylene (PE) - High Molecular Weight | 108 |
| Polyethylene terephthalate (PET) (Indium Tin Oxide on PET)(Monolayer Graphene on PET) | 59.4 |
| Polyphenylene - glass fiber-reinforced | 35.8 |
| Polypropylene (PP), unfilled (Single Crystal, Wafer Substrate Cases) | 100 - 200 |
| Polypropylene - glass fiber-reinforced | 32 |
| Polystyrene (PS) | 70 |
| Polysulfone (PSO) | 55.8 |
| Polyurethane (PUR), rigid (Planetary and Roller Milling Jars) | 57.6 |
| Polyvinyl chloride (PVC) | 50.4 |
| Polyvinylidene fluoride (PVDF) | 127.8 |
| Porcelain, Industrial | 6.5 |
| Potassium | 83 |
| Potassium Bromide, KBr (Single Crystal) | 43 |
| Potassium Chloride, KCl (Single Crystal) | 36 |
| Potassium Tantalate, KTaO 3(Single Crystal) | 4.03 |
| Praseodymium (Sputtering Target) | 6.7 |
| Praseodymium Fluoride,PrF 3(Sputtering Target) | Parallel to a-16.4, c-14 |
| Praseodymium Oxide, Pr 6 O 11 (High Purity Powder)(Sputtering Targets) | 12 |
| Promethium | 11 |
| Quartz (Single Crystal Wafer)(Monolayer Graphene on Quartz) | 0.77 - 1.4 |
| Rhenium (Sputtering Target) | 6.7 |
| Rhenium Disilicide,ReSi 2 | Parallel to a = 4.2, c =7.5 |
| Rhodium | 8 |
| Rock salt | 40.4 |
| Rubber, hard | 77 |
| Ruthenium (Sputtering Target) | 9.1 |
| Samarium (Sputtering Target) | 12.7 |
| Samarium Oxide, Sm 2 O 3 (High Purity Powder)(Sputtering Targets) | 8.5 |
| Sandstone | 11.6 |
| Sapphire (Single Crystal and Wafers) | Please see this page |
| Scandium (Sputtering Target) | 10.2 |
| Scandium Oxide(High Purity Powder)(Sputtering Targets) | 6.7 |
| Selenium (Sputtering Target) | 3.8 |
| Silicon (Single Crystal Wafer)(Sputtering Target)(N-type Sputtering Target)(P-type Sputtering Target) | Please see this page |
| Silicon Carbide, SiC (Single Crystal Wafer)(Sputtering Target) | Please see this page |
| Silicon Dioxide (Fused Silica, amorphous), SiO 2 (Sputtering Target)(Wafers and Substrates)(High Purity Powder) | 5.6 |
| Silicon Nitride,Si 3 N 4 (Sputtering Target) | 4.3 |
| Silver (Sputtering Target) | 19.5 |
| Sitall | 0.15 |
| Slate | 10.4 |
| Sodium | 70 |
| Sodium Chloride, NaCl (Single Crystal) | 44 |
| Sodium Fluoride, NaF (Sputtering Target) | 34 |
| Solder lead - tin, 50% - 50% | 24.0 |
| Speculum metal | 19.3 |
| Steatite | 8.5 |
| Steel | 12.0 |
| Steel 18 Ni(300) (Maraging)(Additive Manufacturing/3D Printing Powder) | 10.3 |
| Steel Invar 36(Additive Manufacturing/3D Printing Powder) | 1.3 |
| Steel M2 (Tool)(Additive Manufacturing/3D Printing Powder) | 11 |
| Steel Stainless Austenitic (304) (Milling media)(Planetary Milling Jars)(Roller Milling Jars) | 17.3 |
| Steel Stainless Austenitic (310) | 14.4 |
| Steel Stainless Austenitic (316)(Additive Manufacturing/3D Printing Powder)(Milling media)(Roller Milling Jars)(Planetary Milling Jars) | 16.0 |
| Steel Stainless Ferritic (410)(Additive Manufacturing/3D Printing Powder) | 9.9 |
| Steel Stainless 17-4PH(Additive Manufacturing/3D Printing Powder) | 10.8 |
| Steel Stainless 440C(Additive Manufacturing/3D Printing Powder) | 10.2 |
| Strontium | 22.5 |
| Strontium Fluoride,SrF 2(Sputtering Target) | 18.4 |
| Strontium Titanate, SrTiO 3 (Sputtering Target)(Single Crystal Wafers& Substrates) (Fe doped single Crystal)(Nb doped single Crystal) | 9 |
| Tantalum (Sputtering Target) | 6.5 |
| Tantalum Carbide, TaC (Sputtering Target) | 6.3 |
| Tantalum Diboride, TaB 2 | 8.2 |
| Tantalum Disilicide,TaSi2(Sputtering Target) | Parallel to a = 6.8, c =6.1 |
| Tantalum Nitride, TaN (Sputtering Target) | 3.6 |
| Tantalum Pentoxide,Ta 2 O 5(Sputtering Target)(High Purity Powder) | 6.7 |
| Tantalum Sulfide,TaS 2(Sputtering Target) | 12 - 15 |
| Teflon, (Sputtering Target)(Planetary Milling Jars)(Roller Milling Jars) | 120 - 170 |
| Tellurium (Sputtering Target) | 36.9 |
| Tellurium Dioxide,TeO 2(Single Crystal) | Parallel to a = 19.5, c =6.1 |
| Terbium (Sputtering Target) | 10.3 |
| Terne | 11.6 |
| Thallium | 29.9 |
| Thorium | 12 |
| Thulium (Sputtering Target) | 13.3 |
| Thulium Oxide,Tm 2 O 3(High Purity Powder)(Sputtering Targets) | 6.6 |
| Tin (Sputtering Target) | 23.4 |
| Tin Oxide SnO 2(Sputtering Targets) | 4.0 |
| Titanium (Sputtering Target) | 8.6 |
| Titanium Alloy Ti-6Al-4v (TC4)(Additive Manufacturing/3D Printing Powder) | 8.7 - 9.1 |
| Titanium Carbide, TiC (Sputtering Target) | 7.4 |
| Titanium Diboride, TiB 2(Sputtering Targets) | 7 |
| Titanium Dioxide (Rutile), TiO 2(Single Crystal)(High Purity Powder)(Sputtering Target) | Parallel to a = 7.14, c =9.19 |
| Titanium Disilicide,TiSi 2(Sputtering Target) | Parallel to a = 6.9, c =5.4 |
| Titanium Monoxide, TiO (Sputtering Target) | 6.6 |
| Titanium Nitride, TiN (Sputtering Target) | 9.4 |
| Titanium (III) Oxide,Ti 2 O 3(High Purity Powder) | 7.6 |
| Titanium Silicide,Ti 5 Si 3 | 7.5 |
| Topas | 5 - 8 |
| Tungsten (Sputtering Target) | 4.4 |
| Tungsten Carbide (Cobalt Stabilized), WC-6%Co (Sputtering Target)(Milling Media)(Planetary Milling Jars)(Pellet Pressing Die) | 5.5(pure) 4.9 (6% Co) |
| Tungsten Diboride,WB 2 | Parallel to a-6.5, c-8.8 |
| Tungsten Diselenide,WSe 2(Sputtering Target) | 7 |
| Tungsten Disilicide,WSi 2(Sputtering Target) | Parallel to a-6.04, c-9.39 |
| Tungsten Disulfide,WS 2(Sputtering Target) | 7 - 10 |
| Tungsten Ditelluride,WTe 2(Sputtering Target) | Parallel to a-10.1, b-7.5, c-4.5 |
| Tungsten (VI) Oxide,WO 3(High purity Powder)(Sputtering Target) | 8 - 15 |
| Uranium | 13.9 |
| Vanadium (Sputtering Target) | 8 |
| Vanadium Carbide, VC (Sputtering Target) | 7.3 |
| Vanadium Disilicide,VSi 2(Sputtering Target) | Parallel to a-8.0, c-7.5 |
| Vanadium Nitride, VN | 8.1 |
| Vanadium Oxide,V 2 O5(Sputtering Target) | Parallel a-3.3, b--1.7, c-42.2 |
| Vinyl Ester | 16 - 22 |
| Vulcanite | 63.6 |
| Wax | 2 - 15 |
| Wedgwood ware | 8.9 |
| Wood, fir | 3.7 |
| Wood, parallel to grain | 3 |
| Wood, across (perpendicular) to grain | 30 |
| Wood, pine | 5 |
| YCOB, Ca 4 YO(BO 3)3 (NLO Crystal) | Parallel a-9.9, b-8.2, c-12.8 |
| Ytterbium (Sputtering Target) | 26.3 |
| Ytterbium Fluoride,YbF 3(Sputtering Target) | 8.5 |
| Ytterbium Oxide, Yb 2 O 3(High Purity Powder)(Sputtering Targets) | 8.5 |
| Yttrium (Sputtering Target) | 10.6 |
| Yttrium Aluminum Garnet (YAG),Y 3 Al 5 O 3 (YAG Single Crystal)(Ce:YAG single Crystal)(Er:YAG Single Crystal)(Nd:YAG Single Crystal)(Yb:YAG Single Crystal)(Ce:YAG LED Phosphor) | 6.1(undoped) |
| Yttrium Fluoride,YF 3(Sputtering Target) | 28.5 |
| Yttrium Oxide, Y 2 O 3(High Purity Powder)(Sputtering Targets) | 8.1 |
| Yttrium Orthovanadate,YVO 4(Single Crystal)(Nd doped Single Crystal) | Parallel to a = 4.43, c =11.37 |
| Zinc (Sputtering Target) | 29.7 |
| Zinc Oxide, ZnO (Single Crystal)(Sputtering Target)(High Purity Powder) | Parallel to a = 6.5, c =3.7 |
| Zinc Selenide, ZnSe (Sputtering Target) | 7.1 |
| Zinc Sulfide, ZnS (Sputtering Target) | 6.5 |
| Zinc Telluride, ZnTe (Sputtering Target) | 8.19 |
| Zirconium (Sputtering Target) | 5.7 |
| Zirconium Carbide, ZrC (Sputtering Target) | 6.7 |
| Zirconium Diboride,ZrB 2(Sputtering Target) | Parallel to a = 6.7, c =6.9 |
| Zirconium Disilicide,ZrSi 2(Sputtering Target) | Parallel to a = 1.1, c =8.6 |
| Zirconium Nitride, ZrN (Sputtering Target) | 7.2 |
| Zirconia, ZrO 2(High Purity Powder) | 7 |
| Yttria Stabilized Zirconia (YSZ), 3-8% Y 2 O 3/ZrO 2(Sputtering Target)(Single Crystal)(Milling Media)(Planetary Milling Jars)(High Purity Powder) | 3% = 10.8, 8% = 10.5 |
) m/m = meter per meter
Most values for temperature are set at 25 o C (77 o F):
t K = t C + 273.16
1 in (inch) = 25.4 mm
1 ft (foot) = 0.3048 m
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Algebra 2
Course: Algebra 2>Unit 3
Lesson 6: Polynomial identities
Polynomial identities introduction
Analyzing polynomial identities
Polynomial identities
Describing numerical relationships with polynomial identities
Math>
Algebra 2>
Polynomial factorization>
Polynomial identities
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Polynomial identities introduction
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About About this video Transcript
Polynomial identities are equal expressions involving polynomials. We can prove these identities through algebraic manipulation, like expanding and simplifying. This helps us see if two expressions are the same for all values of a variable, making them true polynomial identities.
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claire 2 years ago Posted 2 years ago. Direct link to claire's post “these people on the stree...” more these people on the street need to mind their business
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Success The Best💞 2 years ago Posted 2 years ago. Direct link to Success The Best💞's post “I know right! They're jus...” more I know right! They're just asking questions they themselves are unable to answer, making us tryna crack our brains.
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Logar a year ago Posted a year ago. Direct link to Logar's post “Gotta hate when people wa...” more Gotta hate when people walk up to you randomly one day and ask you a polynomial identity question
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Amy Kay a year ago Posted a year ago. Direct link to Amy Kay's post “I'd ask, "Do i get money ...” more I'd ask, "Do i get money if I answer right?" Then proceed to look for a camera. lol. Thanks!! : D
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Anastasia 3 years ago Posted 3 years ago. Direct link to Anastasia's post “i'd walk away fastly” more i'd walk away fastly
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Matthew 6 years ago Posted 6 years ago. Direct link to Matthew's post “Tell me if my definition ...” more Tell me if my definition of a mathematical identity is correct: an equality between two expressions in which they are not the same but, upon manipulating one or the other or both, they can be the same.
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Christopher Blake 6 years ago Posted 6 years ago. Direct link to Christopher Blake's post “You have to be careful ab...” more You have to be careful about using phrases like "the same" in math. You need to be very precise in math for good reasons. If two expressions are equal, that means they are always "the same." They might look different, but an identity can be expressed as an equation which is always true. For example 5 + 1 = 4 + 2.
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devadharson 5 years ago Posted 5 years ago. Direct link to devadharson's post “Is there a real-world pur...” more Is there a real-world purpose to this concept? I tend to retain this type of information better if I know how it is used in the world.
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Austin 5 years ago Posted 5 years ago. Direct link to Austin's post “One example I can think o...” more One example I can think of is that polynomial identities are used to find variable(s) to find where to support a curve in a bridge or building for an engineer. Statistics in graphs could also use polynomials in general to find trends in stocks or other economic related topics.
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SOHA BHUIYAN 4 years ago Posted 4 years ago. Direct link to SOHA BHUIYAN's post “3:40 Where'd the 6n come ...” more 3:40 Where'd the 6n come from?? Wouldn't it just be n^2 + 9?
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Jerry Nilsson 4 years ago Posted 4 years ago. Direct link to Jerry Nilsson's post “(𝑛 + 3)² = (𝑛 + 3)(𝑛 +...” more (𝑛 + 3)² = (𝑛 + 3)(𝑛 + 3)
Using the distributive property twice, aka "FOIL",
we get 𝑛² + 3𝑛 + 3𝑛 + 9, which simplifies to 𝑛² + 6𝑛 + 9
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Marc Qin a year ago Posted a year ago. Direct link to Marc Qin's post “Why doesn't Sal subtract ...” more Why doesn't Sal subtract 9 from both sides of the equation at 4:45 ? Wouldn't we be able to solve for n after that as n^2 = 4, n = 2?
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “Yes that's the right step...” more Yes that's the right steps to solving for x, except for the last step which you must add ± to the other side of the equation: n = ±2.
Sal's looking for polynomial identities though
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sarra 2 years ago Posted 2 years ago. Direct link to sarra's post “what is a Polynomial iden...” more what is a Polynomial identity
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Tanner P 2 years ago Posted 2 years ago. Direct link to Tanner P's post “A polynomial identity is ...” more A polynomial identity is an equation involving polynomials that is always true.
For example, this is a polynomial identity:
(x+1)^2 = x^2+2x+1. No matter what x is, both sides are always equal.
You can verify a polynomial identity if you can rewrite one side to look exactly the same as the other side.
Hope this helps!
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Ulrik Lipscomb 3 months ago Posted 3 months ago. Direct link to Ulrik Lipscomb's post “Smack the person asking t...” more Smack the person asking the question
Yup sounds good.
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Video transcript
[Instructor] What we're going to do in this video is talk a little bit about polynomial identities, and this is really just a fancy way of seeing whether an expression that involves a polynomial is equal to another expression. So for example, you're familiar with x squared plus two x plus one, we've seen polynomials like this multiple times, this is a quadratic, and you might recognize that this would be equal to x plus one squared, that for any value of x, x squared plus two x plus one is the same thing as adding one to that x, and then squaring the whole thing. And we saw this when we first got, when we first learned how to multiply binomials and we took the square of binomials, but now we're going to do this with slightly complicated expressions, things that aren't just simple quadratics or that might not be as obvious as this. And the way that we're gonna prove whether they are true or not is just with a little bit of algebraic manipulation. So for example, if someone walked up to you on the street and said, all right, m to the third minus one, is it equal to m minus one times one plus m plus m squared? Pause this video and see what you would tell that person, whether you could prove whether it is or is not a true polynomial identity. Okay, let's do it together, and the way I would tackle this is I would expand out, I would multiply out what we have on the right-hand side, so this is going to be equal to, so first, I could take this m and then multiply it times every term in this second expression, so m times one is m, m times m is m squared, and then m times m squared is m to the third power, and then I would take this negative one, and then multiply and then distribute that times every term in that other expression, so negative one times one is negative one, negative one times m is negative m, and negative one times m squared is negative m squared, and now let's see if we can simplify this. We have an m and a negative m, so those are going to cancel out, we have an m squared and a negative m squared, so those cancel out, and so we are going to be left with m to the third power, minus one. Now, clearly, m to the third power minus one is going to be equal to m to the third power minus one for any value of m, these are identical expressions. So this is, this is indeed a polynomial identity. Let's do another example. Let's say someone were to walk up to you on the street and said, quick, n plus three squared plus two n, is that equal to eight n plus 13, is this a polynomial identity? Pause this video and see if you can figure that out. All right, now we're gonna work on that together. And I would do it the exact same way. I would try to simplify with a little bit of algebra, maybe the easiest thing to do first, and you could do this in multiple ways is, I have this, I have these n terms, two n's here, eight n's over here, well, what if I were to get these two n's out of the left-hand side, so if I were to just subtract two n from both sides of this equation, I'm going to get on the left-hand side, n plus three squared, and on the right-hand side, I'm going to get six n, eight n minus two n, plus 13. Now, what's n plus three squared? Well, that's going to be n squared plus two times three times n. And if what I just did does not seem familiar to you, I encourage you to look at the videos about squaring binomials, but this is going to be plus six n, plus three squared, which is nine, and is this going to be equal to six n plus 13? Well, already this is starting to look a little bit, a little bit sketchy, but let's just keep going with the algebra. So let's see if we subtract six n from both sides, what do you get? Well, on the left-hand side, you're just going to have n squared plus nine, and on the right-hand side, you're going to get 13. Now, are there values of n for which this is not always true? Well, sure. I can find a lot of values of n for which this is not always true. If n is a zero, this is not going to be true. If n is one, this is not going to be true. If n is two, this actually would be true, but if n is three, this is not going to be true. If n is four, five, et cetera, so for actually most values of n, this is not going to be true. So in order for it to be a polynomial identity, it has to be true for all of the values that are legitimate values that you can evaluate for those, for the variable in question. So this one right over here is not a polynomial, polynomial identity, and we are done.
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8123 | https://math.stackexchange.com/questions/3512517/how-to-deal-with-negative-area-when-evaluating-a-definite-integral | integration - How to deal with negative area when evaluating a definite integral - Mathematics Stack Exchange
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How to deal with negative area when evaluating a definite integral
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Asked 5 years, 8 months ago
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Find the area bounded by the curves y=2−x 2 y=2−x 2 and x+y=0 x+y=0
−x=2−x 2⟹x 2−x−2=(x−2)(x+1)=0−x=2−x 2⟹x 2−x−2=(x−2)(x+1)=0
My Attempt
A 1:A 1: Area above the x-axis and A 2:A 2: Area below the x-axis
A 1=∫2√−1(2−x 2)d x−∫0−1(−x)d x=[2 x−x 3 3]2√−1−[−x 2 2]0−1=2 2–√−2 2–√3+2−1 3−(1 2)=4 2–√3+7 6=8 2–√+7 6 A 2=∣∣∫2 0(−x)d x∣∣−∣∣∫2 2√(2−x 2)∣∣=∣∣[−x 2 2]2 0∣∣−∣∣[2 x−x 3 3]2 2√∣∣=|−2|−|4−8 3−2 2–√+2 2–√3|=2−∣∣(4−4 2–√3)∣∣=2−(4 2–√−4 3)=∣∣∣10−4 2–√3∣∣∣=20−8 2–√6 A=A 1+A 2=8 2–√+7+20−8 2–√6=27 6=9 2 A 1=∫−1 2(2−x 2)d x−∫−1 0(−x)d x=[2 x−x 3 3]−1 2−[−x 2 2]−1 0=2 2−2 2 3+2−1 3−(1 2)=4 2 3+7 6=8 2+7 6 A 2=|∫0 2(−x)d x|−|∫2 2(2−x 2)|=|[−x 2 2]0 2|−|[2 x−x 3 3]2 2|=|−2|−|4−8 3−2 2+2 2 3|=2−|(4−4 2 3)|=2−(4 2−4 3)=|10−4 2 3|=20−8 2 6 A=A 1+A 2=8 2+7+20−8 2 6=27 6=9 2
Reference
A=∫2−1(2−x 2+x)d x=[2 x−x 3 3+x 2 2]2−1=4−8 3+2+2−1 3−1 2=5−1 2=9/2 A=∫−1 2(2−x 2+x)d x=[2 x−x 3 3+x 2 2]−1 2=4−8 3+2+2−1 3−1 2=5−1 2=9/2
Isn't the attempt in my reference factually incorrect ?
yet why am I getting a same solutions in my attempt, ie. after splitting the areas and subtracting absolute values ?
Another Example Another Example
Area bounded by the curve y=x 3 y=x 3, x-axis at x=−2 x=−2 and x=1 x=1
Method 1
A=|∫0−2(x 3)d x|+|∫1 0 x 3 d x|=|[x 4 4]0−2|+|[x 4 4]1 0|=|−4|+|1 4|=4+1 4=17/4 A=|∫−2 0(x 3)d x|+|∫0 1 x 3 d x|=|[x 4 4]−2 0|+|[x 4 4]0 1|=|−4|+|1 4|=4+1 4=17/4
Method 2
A=|∫1−2(x 3)d x|=|[x 4 4]1−2|=|1 4−4|=|−15 4|=15 4 A=|∫−2 1(x 3)d x|=|[x 4 4]−2 1|=|1 4−4|=|−15 4|=15 4
Here I think we are not getting the correct answer in method 2 because the area is counted negative, right ?
integration
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edited Jan 17, 2020 at 17:10
SOORAJ SOMANSOORAJ SOMAN
asked Jan 17, 2020 at 15:16
SOORAJ SOMANSOORAJ SOMAN
8,068 4 4 gold badges 56 56 silver badges 98 98 bronze badges
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it might be a stupid doubt or may be not upto the standard, but it'd be helpful to comment before down voting the post !!SOORAJ SOMAN –SOORAJ SOMAN 2020-01-17 15:41:41 +00:00 Commented Jan 17, 2020 at 15:41
The simplest approach for the first one is that, if I have curves y=f(x)y=f(x) and y=g(x)y=g(x) such that f(x)<g(x)f(x)<g(x) for all a<x<b a<x<b, then the area between the curves on this interval is ∫b a[g(x)−f(x)]d x∫a b[g(x)−f(x)]d x.Semiclassical –Semiclassical 2020-01-17 16:13:02 +00:00 Commented Jan 17, 2020 at 16:13
@Semiclassical could you please confirm which one is the correct solution ?SOORAJ SOMAN –SOORAJ SOMAN 2020-01-17 16:18:29 +00:00 Commented Jan 17, 2020 at 16:18
@Semiclassical In the second quadrant, as you said it is very clear as both areas are above x axis. But in the second quadrant, some portions are below x axis, so I worry about the negative area if I directly integrate it, can you clarify ?SOORAJ SOMAN –SOORAJ SOMAN 2020-01-17 16:27:54 +00:00 Commented Jan 17, 2020 at 16:27
"Isn't the attempt in my reference factually incorrect ?" No. It's completely valid. "why am I getting a same solutions in my attempt, ie. after splitting the areas and subtracting absolute values ?" Because (M+B)−(N+B)=M−N(M+B)−(N+B)=M−N. You were both asked "What is U−V U−V" and had an image were one could see if one wanted to that U=M+B U=M+B and V=N+B V=N+B. You studied the picture and worried about the B B and reworded the question and solved for M−N M−N. They simply solve U−V U−V directly knowing any overlap would be "cancelled out".fleablood –fleablood 2020-01-17 17:46:30 +00:00 Commented Jan 17, 2020 at 17:46
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The simplest way to do it I see is
∫2−1(2−x 2+x)d x=2 x−x 3 3+x 2 2∣∣∣2−1 4−8 3+4 2−(−2)−1 3−1 2==9 2∫−1 2(2−x 2+x)d x=2 x−x 3 3+x 2 2|−1 2 4−8 3+4 2−(−2)−1 3−1 2==9 2
This is the approach in your reference, but there is a typo in the upper limit of the second integral. Your A 1 A 1 is trying to get the area above the x x axis, but the −1 3−1 3 should be positive. Your A 2 A 2 has no term involving the −x−x integral from 2–√2 to 2 2.
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edited Jan 17, 2020 at 16:37
answered Jan 17, 2020 at 16:30
Ross MillikanRoss Millikan
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It is correct because it evaluates the vertical distance between the curves, which does not care about crossing the axis. I have added a couple errors I found in your work.Ross Millikan –Ross Millikan 2020-01-17 16:38:17 +00:00 Commented Jan 17, 2020 at 16:38
okay I think I fixed many mistakes in the post. It seems to get same 9/2 9/2 in both attempts. In my second example, it does not work because the given function is not always greater than or less than the other, which is the x axis, right ?SOORAJ SOMAN –SOORAJ SOMAN 2020-01-17 17:24:56 +00:00 Commented Jan 17, 2020 at 17:24
That is correct. Your second counts the area from x=0 x=0 to x=1 x=1 as negative. There are times you want the integral to count area below the axis as negative, but areas are always taken to be positive.Ross Millikan –Ross Millikan 2020-01-17 18:52:05 +00:00 Commented Jan 17, 2020 at 18:52
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Example 1:
When trying to find the area between curves f(x)f(x) and g(x)g(x) you can achieve this by integrating the function H(x)=|f(x)−g(x)|H(x)=|f(x)−g(x)|. However integration of an absolute value function is piecewise. In this case though f(x)≥g(x)f(x)≥g(x) for −1≤x≤2−1≤x≤2 so integrating ∫2−1(f(x)−g(x))d x∫−1 2(f(x)−g(x))d x is valid.
Now we can go into greater (but unnecessary) analysis and do what you did and note that for −1≤x≤0−1≤x≤0 we have f(x)≥g(x)>0 f(x)≥g(x)>0. While for 0g(x)f(x)≥0>g(x) and for 2–√f(x)≥g(x)0>f(x)≥g(x). and we can break it into sections of ∫2√−1 f(x)d(x)∫−1 2 f(x)d(x) (where f(x)≥0 f(x)≥0) and subtract −∫0−1 g(x)d x−∫−1 0 g(x)d x (where g(x)≥0 g(x)≥0) . The we can add then absolute value of g(x)g(x) where g(x)<0 g(x)<0 so +∫2 0[−g(x)]d x+∫0 2[−g(x)]d x and the subtract the absolute value of f(x)f(x) where f(x)<0 f(x)<0 so −∫2 2√[−f(x)]d x−∫2 2[−f(x)]d x.
This is exactly the same result.
Example 2:
Here f(x)=x 3 f(x)=x 3 and g(x)=0 g(x)=0 and we are trying to integrate H(x)=|f(x)−g(x)|=|x 3|H(x)=|f(x)−g(x)|=|x 3|.
Here it is NOT the case that f(x)≥g(x)f(x)≥g(x) and it is not the case that |f(x)−g(x)|=f(x)−g(x)|f(x)−g(x)|=f(x)−g(x).
So we can solve this by
Method 1: ∫1−2|x 3|d x=∫1−2{−x 3 x 3 x<0 x≥0 d x=[{−x 4 4 x 4 4 x<0 x≥0]1−2=(1 4−(−16 4))∫−2 1|x 3|d x=∫−2 1{−x 3 x<0 x 3 x≥0 d x=[{−x 4 4 x<0 x 4 4 x≥0]−2 1=(1 4−(−16 4))
Method 2: for −2≤x<0−2≤x<0 we have x 3<0 x 3<0 and for 0≤x≤1 0≤x≤1 we have x 3≥0 x 3≥0 so |∫0−2 x 3 d x|=|0−16 4||∫−2 0 x 3 d x|=|0−16 4| and |∫1 0 x 3 d x|=|1 4−0||∫0 1 x 3 d x|=|1 4−0|. And the result follows.
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answered Jan 17, 2020 at 18:45
fleabloodfleablood
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8124 | https://testbook.com/question-answer/find-the-number-of-triangles-that-can-be-made-from--60a511ad9d2aec46347ca6a3 | [Solved] Find the number of triangles that can be made from 15 points
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Permutation and Combination
Question
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Find the number of triangles that can be made from 15 points out of which 6 are collinear.
455
445
435
420
380
Answer (Detailed Solution Below)
Option 3 : 435
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Detailed Solution
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Concept:-
Basic combination concept.
Triangle cannot be made from 3 collinear points.
Formula:-
Number of triangles = nC3 - mC3
where n is the total number of points and m the number of collinear points.
Calculation:-
Number of triangles = 15C3 - 6C3 = 455 - 20 = 435
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8125 | https://www.medicalnewstoday.com/articles/322661 | Valsalva maneuver: How to do it, uses, and warnings
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What is the Valsalva maneuver?
Medically reviewed by Nancy Hammond, M.D. — Written by Rachel Nall, MSN, CRNA on August 2, 2018
How to do the Valsalva maneuver
Uses
Risks and considerations
Takeaway
The Valsalva maneuver is a particular way of breathing that increases pressure in the chest. It causes various effects in the body, including changes in the heart rate and blood pressure.
People may perform the maneuver regularly without knowing it. For example, they may use it when they push to initiate a bowel movement.
However, this technique can also be beneficial when people use it intentionally as it can regulate heart rhythms and help the ears to pop.
The physician Antonio Maria Valsalva first described the technique in the 1700s as a way to clear pus out of the ears.
How to do the Valsalva maneuver
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The Valsalva maneuver involves holding the breath.
To do the Valsalva maneuver, follow these steps:
Inhale deeply and then hold your breath.
Imagine that the chest and stomach muscles are very tight and bear down as though straining to initiate a bowel movement.
Hold this position for a short time, usually about 10 seconds.
Breathe out forcibly to release the breath rapidly.
Resume normal breathing.
An alternative method involves lying down and blowing into an empty syringe for 15 seconds.
The Valsalva maneuver creates numerous effects in the body because it builds up the pressure in the pleural cavity, known as the intrapleural pressure.
This increased pressure can lead to the compression of the chambers of the heart and key blood vessels in the body, including:
the aorta, which is the major artery that pumps oxygen-rich blood through the body
the vena cava, which is the major vein that returns blood to the heart
The compression of the aorta initially causes the blood pressure to rise. A sensor in the carotid artery, called the baroreceptor, detects the increased blood pressure.
This activates parasympathetic fibers, which quickly reduce the heart rate and blood pressure. Doctors sometimes refer to this effect as vagaling.
The Valsalva maneuver reduces cardiac output, which is the amount of blood that the heart puts out with every beat. The individual may feel lightheaded or dizzy as a result.
Once the baroreceptor senses the decrease in heart rate and blood pressure, it will stimulate the sympathetic nervous system. This can cause a person’s heart rate and blood pressure to increase, offsetting the parasympathetic effects.
However, when a person breathes out, this releases the compression on the heart, allowing it to fill back up with blood. Breathing out increases the pressure inside the aorta, stimulating the parasympathetic nervous system and decreasing the heart rate again.
Uses
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The Valsalva maneuver helps treat a rapid heart rate.
Doctors can use the Valsalva maneuver to treat people with supraventricular tachycardia (SVT).
SVT is a rapid heart rate that is typically over 100 beats per minute at rest. It can cause symptoms including heart palpitations, chest pain, and shortness of breath.
A heart rate of this speed can be dangerous because the heart cannot pump enough blood when it beats so quickly.
SVT often requires emergency treatment. Once emergency responders have identified a person’s heart rhythm and determined that their blood pressure is stable, they may demonstrate how to perform the Valsalva maneuver.
According to a report in the Journal of the American College of Cardiology, the Valsalva maneuver was more effective in slowing down the heart rate than other similar procedures, such as carotid sinus massage or applying an ice-cold towel to the face.
A meta-analysis found that the Valsalva maneuver can restore regular heart rate between 19.4 and 54.3 percent of the time.
If the Valsalva maneuver does not normalize the heart rhythm, a person will need to receive electric shocks, called cardioversion, or medications.
These approaches can cause unpleasant side effects, such as chest pain, pressure, and flushing.
The Valsalva maneuver has other uses too. These include:
clearing the ears when scuba diving or a change in altitude increases pressure
increasing colonic pressure to induce a bowel movement
Weightlifters also tend to perform the Valsalva maneuver when they lift heavy weights. Some may try this technique deliberately while others will not realize that they are using it.
It is a common belief that using the Valsalva maneuver when lifting heavy weights can provide momentum and trunk stabilization.
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Risks and considerations
The primary side effect of the Valsalva maneuver is hypotension, which is a sudden, persistent decrease in blood pressure.
People performing the maneuver may also feel lightheaded or experience syncope, which is a brief loss of consciousness.
Takeaway
The Valsalva maneuver can help reduce rapid heart rhythms for some people. It can also help people to pop their ears, lift weights, or have a bowel movement.
While the Valsalva maneuver does have some potential side effects, it is a non-invasive alternative to other treatments, such as medication or electric shocks.
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Ekinci, S., Akgül, G., Arş, E., Aydin, A., Musalar, E., & Aktaş, C. (2017, November 10). Valsalva maneuver techniques for supraventricular tachycardias: Which and how? Hong Kong Journal of Emergency Medicine, 24(6), 298–302
Hackett, D. A., & Chow, C.-M. (2013, August). The Valsalva maneuver: Its effect on intra-abdominal pressure and safety issues during resistance exercise. Journal of Strength and Conditioning Research, 27(8), 2338–2345
Klabunde, R. E. (2014, April 28). Hemodynamics of a Valsalva maneuver
Page, R. L., Joglar, J. A., Caldwell, M. A., Calkins, H., Conti, J. B., Deal, B. J., … Al-Khatib, S. M. (2016, April). 2015 ACC/AHA/HRS guideline for the management of adult patients with supraventricular tachycardia. Journal of the American College of Cardiology, 67(13)
Smith, G. D., Fry, M. M., Taylor, D., Morgans, A., & Cantwell, K. (2015, February 18). The effectiveness of the Valsalva manoeuvre for stopping an abnormal heart rhythm. Cochrane Database of Systematic Reviews, 2015(2)
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Definition
An asymmetric carbon atom, also known as a chiral carbon atom, is a carbon atom that is bonded to four different substituents, resulting in a non-superimposable mirror image. This structural feature gives rise to the concept of optical activity, which is the ability of a molecule to rotate the plane of polarized light.
5 Must Know Facts For Your Next Test
The presence of an asymmetric carbon atom is a necessary and sufficient condition for a molecule to be chiral.
Chiral molecules have non-superimposable mirror images, meaning they cannot be rotated or translated to match their mirror image.
Enantiomers are a pair of stereoisomers that are non-superimposable mirror images of each other, and they exhibit opposite optical activity.
The degree of optical rotation depends on the concentration of the chiral substance and the path length of the light through the sample.
Racemic mixtures are equal mixtures of two enantiomers and do not exhibit optical activity, as the opposing rotations cancel each other out.
Review Questions
Explain the relationship between asymmetric carbon atoms and chirality.
The presence of an asymmetric carbon atom is a necessary and sufficient condition for a molecule to be chiral. An asymmetric carbon atom is one that is bonded to four different substituents, resulting in a non-superimposable mirror image. This structural feature gives rise to the concept of chirality, where the molecule and its mirror image are non-identical and cannot be superimposed on one another. The asymmetric carbon atom is the key to understanding the origins of chirality in organic molecules.
Describe how the optical activity of enantiomers is related to the presence of asymmetric carbon atoms.
Enantiomers, which are a pair of stereoisomers that are non-superimposable mirror images of each other, exhibit opposite optical activity due to the presence of asymmetric carbon atoms. Chiral molecules with asymmetric carbon atoms are able to rotate the plane of polarized light, with one enantiomer rotating it clockwise and the other rotating it counterclockwise. This differential rotation of polarized light is the basis of optical activity and is a key characteristic of chiral molecules with asymmetric carbon atoms.
Evaluate the importance of asymmetric carbon atoms in the context of optical activity and their implications for the pharmaceutical industry.
Asymmetric carbon atoms are of critical importance in the context of optical activity, as they are the structural feature that gives rise to chirality in organic molecules. The ability of chiral molecules with asymmetric carbon atoms to rotate the plane of polarized light in opposite directions has significant implications for the pharmaceutical industry. Many drugs are chiral molecules, and the two enantiomers can have drastically different biological activities, with one enantiomer being the desired therapeutic agent and the other potentially being inactive or even toxic. Understanding the role of asymmetric carbon atoms in determining the optical activity and stereochemistry of drug molecules is essential for the development of safe and effective pharmaceuticals.
Related terms
Chirality:
Chirality is a geometric property of certain molecules and ions. A chiral molecule or ion is non-superimposable on its mirror image, meaning it cannot be rotated or translated to match its mirror counterpart.
Enantiomers:
Enantiomers are a pair of stereoisomers that are non-superimposable mirror images of each other. Enantiomers have the same chemical formula and connectivity, but differ in their spatial arrangement of atoms.
Optical Activity:
Optical activity is the ability of a chiral molecule to rotate the plane of polarized light. Enantiomers of a chiral molecule will rotate the plane of polarized light in opposite directions, one clockwise and the other counterclockwise.
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Capacitor Energy Storage Math Problems
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Capacitor Energy Storage Math Problems
Thread starterAustin Clark
Start dateAug 30, 2012
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Austin Clark
Joined Dec 28, 2011 412
Aug 30, 2012
#1
If I try and calculate the energy (J) stored in a capacitor algebraically, I get J = CV^2
However, I know that in reality J = (CV^2)/2
I know this because that's the answer I get when I solve it graphically.
What's causing this contradiction? Correct mathematics will always come out consistently, so what am I doing wrong?
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Austin Clark
Joined Dec 28, 2011 412
Aug 30, 2012
#2
I think I answered my own question.
I was confusing instantaneous voltage with average voltage, or something along those lines. I realized that, even though J = QV (where Q is charge), the Voltage across the capacitor isn't constant, so each charge receives a different amount of energy as the capacitor charges.
Even still, I'd like to see the math worked out, It'd make more sense then. Anyone care to give it a shot?
Last edited: Aug 30, 2012
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t_n_k
Joined Mar 6, 2009 5,455
Aug 30, 2012
#3
If a source is charging a capacitor and at any instant the current is i(t) and the voltage is e(t) then the instantaneous power delivered by the source would be
p(t)=e(t)×i(t)p(t)=e(t)×i(t)
We also know the relationship between capacitance C, capacitor current i(t) & instantaneous capacitor terminal voltage e(t) is ...
i(t)=C d e d t i(t)=C d e d t
The energy delivered by the charging source to the capacitor Wc is the integral of the instantaneous power over time or ...
W c=∫t−∞p(t)d t=∫t−∞e(t)×i(t)d t=∫t−∞e(t)×C d e d t d t W c=∫−∞t p(t)d t=∫−∞t e(t)×i(t)d t=∫−∞t e(t)×C d e d t d t
It's normally reasonable to permit a change in the limits of integration with the change of variable [from t to e] within the integration. If the capacitor voltage at time t=-∞ is zero and at time t it is E volts, then one may write ...
W c=∫E 0 C e(t)d e=1 2 C E 2 J o u l e s W c=∫0 E C e(t)d e=1 2 C E 2 J o u l e s
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The scientist discovers a new type of material or energy and the engineer discovers a new use for it.”
Gordon Lindsay Glegg |
8128 | https://www.statology.org/correlation-coefficient-by-hand/ | How to Calculate a Pearson Correlation Coefficient by Hand
A Pearson Correlation Coefficient measures the linear association between two variables.
It always takes on a value between -1 and 1 where:
The formula to calculate a Pearson Correlation Coefficient, denoted r, is:
This tutorial provides a step-by-step example of how to calculate a Pearson Correlation Coefficient by hand for the following dataset:
Step 1: Calculate the Mean of X and Y
First, we’ll calculate the mean of both the X and Y values:
Step 2: Calculate the Difference Between Means
Next, we’ll calculate the difference between each of the individual X and Y values and their respective means:
Step 3: Calculate the Remaining Values
Next, we’ll calculate the remaining values needed to complete the Pearson Correlation Coefficient formula:
Step 4: Calculate the Sums
Next, we’ll calculate the sums of the the last three columns:
Step 5: Calculate the Pearson Correlation Coefficient
Now we’ll simply plug in the sums from the previous step into the formula for the Pearson Correlation Coefficient:
The Pearson Correlation Coefficient turns out to be 0.947.
Since this value is close to 1, this is an indication that X and Y are strongly positively correlated.
In other words, as the value for X increases the value for Y also increases in a highly predictable fashion.
Additional Resources
An Introduction to the Pearson Correlation Coefficient
How to Find a Confidence Interval for a Correlation Coefficient
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
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8129 | https://www.youtube.com/watch?v=HA6VO2S6TeY | Calculus 4.2 The Mean Value Theorem
Asher Roberts
13400 subscribers
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Posted: 22 Aug 2020
My notes are available at (so you can write along with me).
Calculus: Early Transcendentals 8th Edition by James Stewart
3 comments
Transcript:
Rolls Theorem section 4.2 the mean value theorem we first start with rolls theorem where if f is a function that satisfies these three hypothesis uh one f is continuous on the closed interval from a to b f is f is differentiable on the open interval a to b and f of a equals f b then there is a number c in this interval such that the derivative at c is zero so we're saying that let's say here's some function it goes oops no can't backtrack has some function goes crazy okay so over here is say a and over here is let's say b so then f of a is equal to f b they have the same y value it's got no you know points where it's going to be not differentiable so no no sharp points no discontinuities in the closed interval and roll's theorem guarantees that um there's a number in here somewhere where the derivative is zero okay so to prove this we break it down into a number of cases first um if the function happens to be a constant so if it happens to equal k for some constant then the derivative is zero at every single point in there because derivative of a constant is zero okay so next case suppose that f is uh bigger than f of a for some f x in the interval so going back to our picture instead of going uh down assume that it might go down but then it goes up a little bit at some point so at some point in the interval that f of x is bigger than the uh and y value okay so in that case it must have a local maximum somewhere could be over here or it could be somewhere else but it has a local maximum somewhere by the extreme value theorem we know that local maximum is inside of the open interval not the closed one because f of x is bigger than f of a so it can't be on the end point we also know it can't be in the other endpoint because f b is equal to f of a okay since f is differentiable on the open interval a b f prime of c is zero by ferma's theorem so um you can do the exact same thing um supposing the next case where instead of f of x being bigger than f of a you suppose that f of x is smaller than f a and you'll see that by the same reasoning you'll get some number c such that f prime of c equals zero so let's do an example Example how could roll's theorem be applied to a position function that models a ball thrown upward well let's say that our position function is s equals f of t okay and let's suppose that if t equals a and uh t equals b then f of a is equal to f of b in other words the n y values are equal along some interval from a to b okay so then roll's theorem comes in so rolls theorem implies that there exists i use this backwards e for there exists some values c inside of our interval from a to b such that its derivative is zero or the derivative is zero at c so that means that well that's derivative position so that means that the velocity is zero okay so this tells you that if you take a ball and you throw it upward then at some point the velocity will go down to zero which should make sense because at some point the ball when you throw it up is going to end up coming all the way back down okay let's prove that the equation x cubed plus x minus 1 equals 0 has exactly one real root remember we've used the intermediate value theorem previously to prove that equations have roots we have not done it with this exactly business but that's where rolle's theorem will come in we'll use the intermediate value theorem to prove that the root exists and then we'll use roles to prove that there can't be any other roots so let's start with uh letting f be equal to this polynomial x cubed plus x minus one um we're looking let's see we want to show that it has one real root so we want to show one y value of zero so let's choose some x values that give y values a little bit below zero and a little bit above zero notice by inspection if i plug in zero for the x's then i get minus one so um that's a little below zero so f of zero is minus one below zero and if i plug in one it's another easy number to plug in then i get two minus one equals one so f of one is one which is above zero okay this is a polynomial so it's definitely continuous on the closed interval we also have some y value a little bit below the y value we're looking at 0 and the y value a little bit above the y value we're looking at so the intermediate value theorem ivt tells us that there exists a c inside in the interval from zero to one such that f of c is equal to zero okay so we've proven that a root exists the next step is now to show that there can't be any others so let's suppose towards a contradiction that a and b are roots we're going to show that this is impossible that only one of these guys can actually be a root okay well if there are two roots then we know that each of those by definition a root is an x value that gives a y value of zero so f of a is zero and f of b is also zero because they're both roots okay so that means that we can use rolls theorem because we have a continuous function the polynomial is continuous in the closed interval it's differentiable in the open interval and we have f of a equals f b so rules implies that there exists a c inside the interval from a to b such that the derivative at c is zero okay but there's one problem with that if you take the derivative of this guy we get that it equals three x squared plus one by the power rule well x squared is strictly greater than or equal to zero so the least this thing could possibly be is one so this thing is greater than or equal to one and that's for all i'm using the upside down a for all x values so for any x value you plug in this thing is going to be greater than or equal to one so it's impossible that it equals zero so that's a contradiction so it must be that we weren't allowed to invoke roll's theorem in the first place in other words there could not have been two roots so f cannot have two real roots notice uh by the fundamental theorem of algebra it has uh three roots but uh so it has two complex roots however we're not going to get into that it's not really um part of this calculus course but you know just want to stress that this real condition matters it actually does have the other roots Theorem okay the moment we've all been waiting for the mean value theorem suppose f is a function that satisfies this hypothesis it's continuous in the closed interval and differentiable in the open so we only need the first two for rules theorem okay well then there's a number c in the interval such that the derivative is equal to this guy however that looks very familiar that is the slope of the secant line or the average rate of change whereas the derivative is the instantaneous rate of change or the slope of the tangent line so this tells you that somewhere inside of our interval if the function is continuous and the closed interval and differentiable in the open then the derivative will be equal to the average rate of change the slope of the tangent line will be equal to the slope of the secant line at some point so that's pretty cool you can multiply both sides by b minus a and if you want to get everything on one side like this and then you can get f of b minus f of a is that that's the same thing okay let's see if we can prove this what we're going to do is we're going to concoct some function h where h is the difference between f and the secant lines f on the interval the closed interval a b so here's f and here's the secant line to f notice that to build the equation of a line we can use point slope form so y let's say minus y one equals m times x minus x1 however m is this guy that's the slope of the secant line so that's where this comes in and then if i wanted to i could just move over the y value so this is the same thing as y one plus m times x minus x one so that's where this f of a is that's my y value at a and this is the x value day which is just a so that's where this comes in let's go back uh okay so then h is continuous on a b why is it continuous on a b because we're assuming that f is and this function is obviously continuous on a b this is oh i even wrote it's a sum of f in a first degree polynomial okay so that means it's also differentiable on the open interval so what we can do now is a little bit of a trick we take the value of h a so if we plug in a for all of the x's then notice what happens you end up with f of a minus f of a cancels so let's pull out uh another color so this is gone and then a minus a cancels and leaves you with zero so that actually kills off this whole thing because zero times this thing is zero so you end up with zero if you look at what h of a is if we look at what h of b is then we have b minus a b minus a cancels and gives you one so then you have f of b minus f of b zero and we have minus f of a minus a minus is plus f of a is zero so h of b is also zero okay so that's pretty cool because remember if we have these two conditions and we have this third condition where h of a equals h of b then we can use the rolls theorem so by rolle's theorem there must be a number c in this interval such that the derivative at c for h is zero okay so we know that zero is h prime c but we know that h prime of c is actual is actually equal to this thing why well go back to how we defined h notice that if we take the derivative of h we get f prime of x f of a well that's a constant so derivative of zero same thing over here with this a zero so that means i have a constant times x so when you have a constant times um of a variable x remember the derivative is just that constant you lose the x so that's why you end up with this you get f prime of x minus f of b minus f of a over b minus a and then you plug in c for the x value and it becomes this okay so that's pretty cool because i have zero equaling this so i can just add this thing to both sides i get f prime of c equals this well that's exactly what we wanted to prove we wanted to prove the derivative at c is equal to the average rate of change let's find a number c in the interval 0 2 such that the slope of the secant line is equal to the slope of the tangent line for the function f of x equals x cubed minus x notice x cubed minus x is a polynomial it's definitely continuous on the closed interval zero two and is definitely differentiable on the open edge of all zero two so by the mean value theorem this value of c definitely exists right there is there exists a c inside the interval from zero to two such that f of 2 minus f of 0 is equal to f prime of c times 2 minus 0. that's our mean value theorem when you multiply both sides by mu minus a okay so what's f of 2 well that's 6 if you plug in 2 and then f of 0 is 0. so 6 minus 0 is equal to three times c squared minus one because the derivative of x cubed minus x is three x squared minus one so we just plug in c and then times two minus zero which is times two okay we can simplify this so we have six equals six c squared minus two by distributing and then we solve for c and we get well first we solve for c squared we get c squared is four thirds so that means that c is equal to plus or minus uh the square root of that so two over rad three however if we look at minus 2 over at 3 we know that that's not in 0 2. so we can reject that we can see that c has to be positive so c must be equal to two over rad three in order to be in the interval zero two okay pretty cool if you look at a graph we can illustrate the mean value theorem so this function looks something like maybe something like that and let's say here's zero and we want to go all left 2 so i'm going to cheat a little bit and draw a secant line first and then go right underneath and say that that's 2. okay so there's x there's y and this is the graph of x cubed minus x the line i just drew is the secant line that cuts through x equals 0 and x equals 2 at the y values 0 and 6. this mean value theorem that we just did says that there is a guaranteed value for c which is a 2 over rad 3 roughly over there so that's x value c such that if i were to draw the tangent line to the curve over there then the slopes match you can't tell it from my terrible drawing but this line that secant line and this line that's the tangent line actually do have the same slope so their slopes are equal that means the derivative is equal to the average rate of change over the entire interval from zero to two pretty cool what does the mean value theorem say about the velocity of an object moving in a straight line well the velocity or the average velocity is the slope of the secant line mean value theorem says that that must be equal to the derivative which would be the instantaneous velocity so it says at some point or how about at some time the average velocity is equal to the instantaneous velocity uh you could see this as an example if um let's say a car was traveling let's say it went 180 kilometers in two hours so travel 100 kilometers traveled in two hours well then the average rate of change is 90 kilometers per hour so by the mean value theorem that speedometer on the car must have actually read exactly 90 90 kilometers per hour at least once during that journey and must have actually read that average rate of change okay suppose that f of 0 is negative 3 and f prime of x is less than or equal to 5 for all values of x how large can f of 2 possibly be let's use the mean value theorem so we've got uh f of two minus f of zero equals f prime of c times two minus zero for some c we're going to try to use the mean value theorem to get some kind of information on f of two without even knowing what the function is all right let's solve for f of two so that's f of 0 plus 2 minus 0 is 2 so 2 times f prime of c and that equals negative 3 plus 2 times f prime of c because they told us f of 0 is negative 3. so we only have one other piece of information here f prime of x is less than or equal to 5. so let's use that now well if f prime of x is less than equal to 5 that means that f prime of c is less than equal to 5 because that was true for all values of x c is the value of x okay so we needed a estimate on 2 times f prime of c so that means that 2 times f prime of c is less than or equal to 2 times 5 which is 10. so going back to what we have for f of 2 we know f of 2 is equal to negative 3 plus 2 times f prime of c and we know that that's less than or equal to negative 3 plus 10 because 2 times f prime of c is less than or equal to 10. so that's 7. so f of 2 must be less than or equal to 7. so the largest possible value for f of 2 is 7. okay time for another theorem if the derivative at x is 0 for all x in an interval a b then f is constant on a b this is the converse to what we had before where if you take the derivative of a a constant and you get zero so now we know that the reverse is true it took us a while to get to this point but it's kind of cool notice this is only true though if it's zero for all x in an interval if we look at some other functions like let's say f of x is equal to x over absolute value of x that notice is equal to 1 if x is positive because then you get x over x is 1 it's equal to minus 1 if x is negative so if you take the derivative for x being positive you get derivative 1 is 0. derivative of minus 1 if x is negative is also 0. so f prime of x is 0. but notice the domain is not an interval in this case the domain d is all of the x values such that x is not equal to 0. that's not an interval that's two intervals that's the interval from uh minus infinity to zero not including zero and it's the interval from zero to infinity notice that f is constant over here it's one over here because then you can actually use this theorem and f is constant over here it's minus 1 over there oh so i reverse that it's minus 1 over here and 1 over here but it is definitely not constant over the entire thing because the entire thing is not an interval so to be slightly careful here all right let's prove this really quick let x1 and x2 be two x values inside of the interval a b okay we want to make them so that one is smaller than the other one that way we can make a new interval out of them so we make this new interval from x1 to x2 well by the mean value theorem then we know that uh the derivative is zero for all x and interval a b so we can apply the mean value theorem it's differentiable for all x in there then it's definitely diff uh continuous on this closed interval and differentiable there so we can use the mean value theorem we get f of x2 minus f of x1 equals f prime of c times x 2 minus x 1 for sum c in the open interval but the derivative is 0 for all x in the interval so that means that this is 0 it's gone so this entire thing becomes zero so that means that this thing equals zero so if they equal zero you can add this guy to both sides and get that they equal each other so that means that any two y values any two y values will be equal to each other because we chose x one and x two completely arbitrarily so if any y values are equal to each other then that means that all the y values are equal to each other so f is constant on the entire integral corollary let's do a quick corollary if f prime of x equals g prime of x for all x in an interval a b then f minus g is constant on a b in other words uh f of x is equal to g of x plus some constant so the same function one is just moved up okay let's prove that so that's actually kind of cool but uh if two functions have the same derivative then they must be basically almost equal to each other okay so let's let big f be equal to this difference function well then the derivative is the difference of the derivatives right but we know that the two derivatives are equal to each other so this is the same as f minus f which is zero or f prime minus f prime which is zero well the derivative is zero and it's zero for all the x in an interval so that means that it's constant by the theorem we just proved so if this guy is constant that means f minus g is constant because that's just the way we defined it okay let's prove the identity tan inverse x plus cotangent inverse x equals pi over two you could prove this directly using some manipulation and you know just a standard trig proof however there is a cool calculus proof let's do that let's let f of x be equal to this function tan inverse plus cotan inverse okay so then the derivative we know is one over one plus x squared for a tangent and minus one over one plus x squared for cotangent so that's 0. so the derivative is 0 that implies that f is constant so f is some constant let's say big c so that means that that constant is equal to f for any x value we plug in so let's choose an easy x value to plug in over here to evaluate it because we know tan inverse x is constant so it doesn't matter what x value we plug in well an easy value for tan inverse is one because we know that at one sine is equal to cosine so that's at pi over four so let's plug in one for x because that's easy for us to evaluate so f of one is tan inverse of one plus cotangent inverse of one so they both have the same because that's when sine and cosine are equal so it doesn't matter the order you divide so that's pi over 4 plus pi over 4 which is pi over 2. okay so then that means that it must be that tan inverse of x in general plus cotangent of inverse and x in general must be equal to pi over 2 because we just proved that they're constant so if they equal pi over 2 for a single x value they equal pi over 2 for every x value |
8130 | https://pressbooks.bccampus.ca/collegephysics/chapter/half-life-and-activity/ | Skip to content
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Chapter 31 Radioactivity and Nuclear Physics
253 31.5 Half-Life and Activity
Summary
Define half-life.
Define dating.
Calculate age of old objects by radioactive dating.
Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore half-life and activity, the quantitative terms for lifetime and rate of decay.
Half-Life
Why use a term like half-life rather than lifetime? The answer can be found by examining Figure 1, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life, [latex]\boldsymbol{t _{1/2}}[/latex]. Half of the remaining nuclei decay in the next half-life. Further, half of that amount decays in the following half-life. Therefore, the number of radioactive nuclei decreases from [latex]\boldsymbol{N}[/latex] to [latex]\boldsymbol{N/2}[/latex] in one half-life, then to [latex]\boldsymbol{N/4}[/latex] in the next, and to [latex]\boldsymbol{N/8}[/latex] in the next, and so on. If [latex]\boldsymbol{N}[/latex] is a large number, then many half-lives (not just two) pass before all of the nuclei decay. Nuclear decay is an example of a purely statistical process. A more precise definition of half-life is that each nucleus has a 50% chance of living for a time equal to one half-life [latex]\boldsymbol{t _{1/2}}[/latex]. Thus, if [latex]\boldsymbol{N}[/latex] is reasonably large, half of the original nuclei decay in a time of one half-life. If an individual nucleus makes it through that time, it still has a 50% chance of surviving through another half-life. Even if it happens to make it through hundreds of half-lives, it still has a 50% chance of surviving through one more. The probability of decay is the same no matter when you start counting. This is like random coin flipping. The chance of heads is 50%, no matter what has happened before.
There is a tremendous range in the half-lives of various nuclides, from as short as [latex]\boldsymbol{10^{-23}}[/latex] s for the most unstable, to more than [latex]\boldsymbol{10^{16}}[/latex] y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest half-lives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of half-life is applicable to other subatomic particles, as will be discussed in Chapter 33 Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero ([latex]\boldsymbol{N_0}[/latex]) and the number ([latex]\boldsymbol{N}[/latex]) at a later time [latex]\boldsymbol{t}[/latex]:
[latex]\boldsymbol{N = N_0 e ^{- \lambda t}}[/latex]
where [latex]\boldsymbol{e = 2.71828 \cdots }[/latex] is the base of the natural logarithm, and [latex]\boldsymbol{\lambda}[/latex] is the decay constant for the nuclide. The shorter the half-life, the larger is the value of [latex]\boldsymbol{\lambda}[/latex], and the faster the exponential [latex]\boldsymbol{e^{- \lambda t}}[/latex] decreases with time. The relationship between the decay constant [latex]\boldsymbol{\lambda}[/latex] and the half-life [latex]\boldsymbol{t_{1/2}}[/latex] is
[latex]\boldsymbol{\lambda =}[/latex][latex]\boldsymbol{ \frac{ \textbf{ln} (2)}{t_{1/2}}}[/latex][latex]\boldsymbol{\approx}[/latex][latex]\boldsymbol{\frac{0.693}{t_{1/2}}}[/latex]
To see how the number of nuclei declines to half its original value in one half-life, let [latex]\boldsymbol{t = t_{1/2}}[/latex] in the exponential in the equation [latex]\boldsymbol{N = N_0 e^{- \lambda t}}[/latex]. This gives [latex]\boldsymbol{N = N_0 e^{- \lambda t} = N_0 e^{-0.693} = 0.500N_0}[/latex]. For integral numbers of half-lives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten half-lives have passed, we divide [latex]\boldsymbol{N}[/latex] by 2 ten times. This reduces it to [latex]\boldsymbol{N/1024}[/latex]. For an arbitrary time, not just a multiple of the half-life, the exponential relationship must be used.
Radioactive dating is a clever use of naturally occurring radioactivity. Its most famous application is carbon-14 dating. Carbon-14 has a half-life of 5730 years and is produced in a nuclear reaction induced when solar neutrinos strike [latex]\boldsymbol{^{14} \textbf{N}}[/latex] in the atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the ecosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Thus, if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you multiply that number by [latex]\boldsymbol{1.3 \times 10^{-12}}[/latex] to find the number of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] nuclei in the object. When an organism dies, carbon exchange with the environment ceases, and [latex]\boldsymbol{^{14} \textbf{C}}[/latex] is not replenished as it decays. By comparing the abundance of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] nuclei in them is greater. Very old biological materials contain no [latex]\boldsymbol{^{14} \textbf{C}}[/latex] at all. There are instances in which the date of an artifact can be determined by other means, such as historical knowledge or tree-ring counting. These cross-references have confirmed the validity of carbon-14 dating and permitted us to calibrate the technique as well. Carbon-14 dating revolutionized parts of archaeology and is of such importance that it earned the 1960 Nobel Prize in chemistry for its developer, the American chemist Willard Libby (1908–1980).
One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 2). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the [latex]\boldsymbol{^{14} \textbf{C}}[/latex] found in living tissues, allowing the shroud to be dated (see Example 1).
Figure 2. Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons)
Example 1: How Old Is the Shroud of Turin?
Calculate the age of the Shroud of Turin given that the amount of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] found in it is 92% of that in living tissue.
Strategy
Knowing that 92% of the [latex]\boldsymbol{^{14} \textbf{C}}[/latex] remains means that [latex]\boldsymbol{N/N_0 = 0.92}[/latex]. Therefore, the equation [latex]\boldsymbol{N = N_0 e^{- \lambda t}}[/latex] can be used to find [latex]\boldsymbol{\lambda t}[/latex]. We also know that the half-life of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] is 5730 y, and so once [latex]\boldsymbol{\lambda t}[/latex] is known, we can use the equation [latex]\boldsymbol{\lambda = \frac{0.693}{t_{1/2}}}[/latex] to find [latex]\boldsymbol{\lambda}[/latex] and then find [latex]\boldsymbol{t}[/latex] as requested. Here, we postulate that the decrease in [latex]\boldsymbol{^{14} \textbf{C}}[/latex] is solely due to nuclear decay.
Solution
Solving the equation [latex]\boldsymbol{N = N_0 e^{- \lambda t}}[/latex] for [latex]\boldsymbol{N/N_0}[/latex] gives
[latex]\boldsymbol{\frac{N}{N_0}}[/latex][latex]\boldsymbol{= e^{- \lambda t}}[/latex]
Thus,
[latex]\boldsymbol{0.92 = e^{- \lambda t}}[/latex]
Taking the natural logarithm of both sides of the equation yields
[latex]\boldsymbol{ln \; 0.92 = - \lambda t}[/latex]
so that
[latex]\boldsymbol{-0.0834 = - \lambda t}[/latex]
Rearranging to isolate [latex]\boldsymbol{t}[/latex] gives
[latex]\boldsymbol{t=}[/latex][latex]\boldsymbol{\frac{0.0834}{\lambda}}[/latex]
Now, the equation [latex]\boldsymbol{\lambda = \frac{0.693}{t^{1/2}}}[/latex] can be used to find [latex]\boldsymbol{\lambda}[/latex] for [latex]\boldsymbol{^{14} \textbf{C}}[/latex]. Solving for [latex]\boldsymbol{\lambda}[/latex] and substituting the known half-life gives
[latex]\boldsymbol{\lambda =}[/latex][latex]\boldsymbol{\frac{0.693}{t_{1/2}}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{0.693}{5730 \;\textbf{y}}}[/latex]
We enter this value into the previous equation to find [latex]\boldsymbol{t}[/latex]:
[latex]\boldsymbol{t=}[/latex][latex]\boldsymbol{\frac{0.0834}{\frac{0.693}{5730 \;\textbf{y}}}}[/latex][latex]\boldsymbol{=690 \;\textbf{y}}[/latex]
Discussion
This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. [latex]\boldsymbol{1320 \pm 60}[/latex]. The uncertainty is typical of carbon-14 dating and is due to the small amount of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived.
There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of [latex]\boldsymbol{^{238} \textbf{U}}[/latex]. The decay series for [latex]\boldsymbol{^{238} \textbf{U}}[/latex] ends with [latex]\boldsymbol{^{206} \textbf{Pb}}[/latex], so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since [latex]\boldsymbol{^{238} \textbf{U}}[/latex] has a half-life of 4.5×1094.5×109 y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about [latex]\boldsymbol{3.5 \times 10^9}[/latex] years ago.
Activity, the Rate of Decay
What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define activity [latex]\boldsymbol{R}[/latex] to be the rate of decay expressed in decays per unit time. In equation form, this is
[latex]\boldsymbol{R =}[/latex][latex]\boldsymbol{\frac{\Delta N}{\Delta t}}[/latex]
where [latex]\boldsymbol{\Delta N}[/latex] is the number of decays that occur in time [latex]\boldsymbol{\Delta t}[/latex]. The SI unit for activity is one decay per second and is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is,
[latex]\boldsymbol{1 \;\textbf{Bq} = 1 \;\textbf{decay/s}}[/latex]
Activity [latex]\boldsymbol{R}[/latex] is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of [latex]\boldsymbol{^{226} \textbf{Ra}}[/latex], in honor of Marie Curie’s work with radium. The definition of curie is
[latex]\boldsymbol{1 \;\textbf{Ci} = 3.70 \times 10^{10} \;\textbf{Bq}}[/latex]
or [latex]\boldsymbol{3.70 \times 10^{10}}[/latex] decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. [latex]\boldsymbol{1 \;\textbf{MBq} = 100 \;\textbf{microcuries} ( \mu \textbf{Ci})}[/latex]. In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).
Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity [latex]\boldsymbol{R}[/latex] should be proportional to the number of radioactive nuclei, [latex]\boldsymbol{N}[/latex], and inversely proportional to their half-life, [latex]\boldsymbol{t_{1/2}}[/latex]. In fact, your intuition is correct. It can be shown that the activity of a source is
[latex]\boldsymbol{R =}[/latex][latex]\boldsymbol{\frac{0.693N}{t_{1/2}}}[/latex]
where [latex]\boldsymbol{N}[/latex] is the number of radioactive nuclei present, having half-life [latex]\boldsymbol{t_{1/2}}[/latex]. This relationship is useful in a variety of calculations, as the next two examples illustrate.
Example 2: How Great Is the 14C Activity in Living Tissue?
Calculate the activity due to [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.
Strategy
To find the activity [latex]\boldsymbol{R}[/latex] using the equation [latex]\boldsymbol{R = \frac{0.693N}{t_{1/2}}}[/latex], we must know [latex]\boldsymbol{N}[/latex] and [latex]\boldsymbol{t_{1/2}}[/latex]. The half-life of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] can be found in Appendix B, and was stated above as 5730 y. To find [latex]\boldsymbol{N}[/latex], we first find the number of [latex]\boldsymbol{^{12} \textbf{C}}[/latex] nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by [latex]\boldsymbol{1.3 \times 10^{-12}}[/latex] (the abundance of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in a carbon sample from a living organism) to get the number of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] nuclei in a living organism.
Solution
One mole of carbon has a mass of 12.0 g, since it is nearly pure [latex]\boldsymbol{^{12} \textbf{C}}[/latex]. (A mole has a mass in grams equal in magnitude to [latex]\boldsymbol{A}[/latex] found in the periodic table.) Thus the number of carbon nuclei in a kilogram is
[latex]\boldsymbol{N(^{12} \textbf{C}) =}[/latex][latex]\boldsymbol{\frac{6.02 \times 10^{23} \;\textbf{mol}^{-1}}{12.0 \;\textbf{g/mol}}}[/latex][latex]\boldsymbol{\times (1000 \;\textbf{g}) = 5.02 \times 10^{25}}[/latex]
So the number of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] nuclei in 1 kg of carbon is
[latex]\boldsymbol{N(^{14} \textbf{C}) = (5.02 \times 10^{25})(1.3 \times 10^{-12}) = 6.52 \times 10^{13}}[/latex]
Now the activity [latex]\boldsymbol{R}[/latex] is found using the equation [latex]\boldsymbol{R = \frac{0.693N}{t_{1/2}}}[/latex].
Entering known values gives
[latex]\boldsymbol{R =}[/latex][latex]\boldsymbol{\frac{0.693(6.52 \times 10^{13})}{5730 \;\textbf{y}}}[/latex][latex]\boldsymbol{= 7.89 \times 10^9 \;\textbf{y}^{-1}}[/latex]
or [latex]\boldsymbol{7.89 \times 10^9}[/latex] decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,
[latex]\boldsymbol{R = (7.89 \times 10^9 \;\textbf{y}^{-1})}[/latex][latex]\boldsymbol{\frac{1.00 \;\textbf{y}}{3.16 \times 10^7 \;\textbf{s}}}[/latex][latex]\boldsymbol{= 250 \;\textbf{Bq}}[/latex]
or 250 decays per second. To express [latex]\boldsymbol{R}[/latex] in curies, we use the definition of a curie,
[latex]\boldsymbol{R =}[/latex][latex]\boldsymbol{\frac{250 \;\textbf{Bq}}{3.7 \times 10^{10} \;\textbf{Bq/Ci}}}[/latex][latex]\boldsymbol{= 6.76 \times 10^{-9} \;\textbf{Ci}}[/latex]
Thus,
[latex]\boldsymbol{R = 6.76 \;\textbf{nCi}}[/latex]
Discussion
Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of [latex]\boldsymbol{^{14} \textbf{C}}[/latex] decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less [latex]\boldsymbol{^{14} \textbf{C}}[/latex], and for samples more than 50 thousand years old, it is impossible.
Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Chapter 32 Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see Figure 3). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of [latex]\boldsymbol{^{131} \textbf{I}}[/latex], [latex]\boldsymbol{^{90} \textbf{S}}[/latex], [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex], [latex]\boldsymbol{^{239} \textbf{Pu}}[/latex], [latex]\boldsymbol{^{238} \textbf{U}}[/latex], and [latex]\boldsymbol{^{235} \textbf{U}}[/latex]. Estimates are that the total amount of radiation released was about 100 million curies.
Human and Medical Applications
Figure 3. The Chernobyl reactor. More than 100 people died soon after its meltdown, and there will be thousands of deaths from radiation-induced cancer in the future. While the accident was due to a series of human errors, the cleanup efforts were heroic. Most of the immediate fatalities were firefighters and reactor personnel. (credit: Elena Filatova)
Example 3: What Mass of 137Cs Escaped Chernobyl?
It is estimated that the Chernobyl disaster released 6.0 MCi of [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex] into the environment. Calculate the mass of [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex] released.
Strategy
We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei [latex]\boldsymbol{N}[/latex] released. Since the activity [latex]\boldsymbol{R}[/latex] is given, and the half-life of [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex] is found in Appendix B to be 30.2 y, we can use the equation [latex]\boldsymbol{R= \frac{0.693N}{t_{1/2}}}[/latex] to find [latex]\boldsymbol{N}[/latex].
Solution
Solving the equation [latex]\boldsymbol{R = \frac{0.693N}{t_{1/2}}}[/latex] for [latex]\boldsymbol{N}[/latex] gives
[latex]\boldsymbol{N=}[/latex][latex]\boldsymbol{\frac{Rt_{1/2}}{0.693}}[/latex]
Entering the given values yields
[latex]\boldsymbol{N=}[/latex][latex]\boldsymbol{\frac{(6.0 \;\textbf{MCi})(30.2 \;\textbf{y})}{0.693}}[/latex]
Converting curies to becquerels and years to seconds, we get
[latex]\begin{array}{r @{{}={}}l} \boldsymbol{N} & \boldsymbol{\frac{(6.0 \times 10^6 \;\textbf{Ci})(3.7 \times 10^{10} \;\textbf{Bq/Ci})(30.2 \;\textbf{y})(3.16 \times 10^7 \;\textbf{s/y})}{0.693}} \[1em] & \boldsymbol{3.1 \times 10^{26}} \end{array}[/latex]
One mole of a nuclide [latex]\boldsymbol{^A X}[/latex] has a mass of [latex]\boldsymbol{A}[/latex] grams, so that one mole of [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex] has a mass of 137 g. A mole has [latex]\boldsymbol{6.02 \times 10^{23}}[/latex] nuclei. Thus the mass of [latex]\boldsymbol{^{137} \textbf{Cs}}[/latex] released was
[latex]\begin{array}{r @{{}={}}l} \boldsymbol{m} & \boldsymbol{(\frac{137 \;\textbf{g}}{6.02 \times 10^{23}})(3.1 \times 10^{26}) = 70 \times 10^3 \;\textbf{g}} \[1em] & \boldsymbol{70 \;\textbf{kg}} \end{array}[/latex]
Discussion
While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design.
Activity [latex]\boldsymbol{R}[/latex] decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since [latex]\boldsymbol{R = \frac{0.693N}{t_{1/2}}}[/latex], the activity decreases as the number of radioactive nuclei decreases. The equation for [latex]\boldsymbol{R}[/latex] as a function of time is found by combining the equations [latex]\boldsymbol{N = N_0 e^{- \lambda t}}[/latex] and [latex]\boldsymbol{R = \frac{0.693N}{t_{1/2}}}[/latex], yielding
[latex]\boldsymbol{R = R_0 e^{- \lambda t}}[/latex]
where [latex]\boldsymbol{R_0}[/latex] is the activity at [latex]\boldsymbol{t=0}[/latex]. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation [latex]\boldsymbol{R = R_0 e^{- \lambda t}}[/latex] must be used to find [latex]\boldsymbol{R}[/latex].
PhET Explorations: Alpha Decay
Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.
Figure 4. Alpha Decay
Section Summary
Half-life [latex]\boldsymbol{t_{1/2}}[/latex] is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei [latex]\boldsymbol{N}[/latex] as a function of time is [latex]\boldsymbol{N = N_0 e^{- \lambda t}}[/latex],where [latex]\boldsymbol{N_0}[/latex] is the number present at [latex]\boldsymbol{t=0}[/latex], and [latex]\boldsymbol{\lambda}[/latex] is the decay constant, related to the half-life by
[latex]\boldsymbol{\lambda =}[/latex][latex]\boldsymbol{\frac{0.693}{t_{1/2}}}[/latex]
One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity [latex]\boldsymbol{R}[/latex]:
[latex]\boldsymbol{R=}[/latex][latex]\boldsymbol{\frac{\Delta N}{\Delta t}}[/latex]
The SI unit for [latex]\boldsymbol{R}[/latex] is the becquerel (Bq), defined by
[latex]\boldsymbol{1 \;\textbf{Bq} = 1 \;\textbf{decay/s}}[/latex]
[latex]\boldsymbol{R}[/latex] is also expressed in terms of curies (Ci), where
[latex]\boldsymbol{1 \;\textbf{Ci} = 3.70 \times 10^{10} \;\textbf{Bq}}[/latex]
The activity [latex]\boldsymbol{R}[/latex] of a source is related to [latex]\boldsymbol{N}[/latex] and [latex]\boldsymbol{t_{1/2}}[/latex] by
[latex]\boldsymbol{R =}[/latex][latex]\boldsymbol{\frac{0.693N}{t_{1/2}}}[/latex]
Since [latex]\boldsymbol{N}[/latex] has an exponential behavior as in the equation [latex]\boldsymbol{N=N_0 e^{- \lambda t}}[/latex], the activity also has an exponential behavior, given by
[latex]\boldsymbol{R = R_0 e^{- \lambda t}}[/latex],
where [latex]\boldsymbol{R_0}[/latex] is the activity at [latex]\boldsymbol{t=0}[/latex].
Conceptual Questions
1: In a [latex]\boldsymbol{3 \times 10^9}[/latex] -year-old rock that originally contained some [latex]\boldsymbol{^{238} \textbf{U}}[/latex], which has a half-life of [latex]\boldsymbol{4.5 \times 10^{9}}[/latex] years, we expect to find some [latex]\boldsymbol{^{238} \textbf{U}}[/latex] remaining in it. Why are [latex]\boldsymbol{^{226} \textbf{Ra}}[/latex], [latex]\boldsymbol{^{222} \textbf{Rn}}[/latex], and [latex]\boldsymbol{^{210} \textbf{Po}}[/latex] also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?
2: Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.
3: Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?
4: Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?
5: Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)
6: To obtain the most precise value of BE from the equation [latex]\boldsymbol{\textbf{BE} = [ZM (^1 \textbf{H}) + Nm_n]c^2 - m(^A X)c^2}[/latex], we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?
7: How does the finite range of the nuclear force relate to the fact that [latex]\boldsymbol{\textbf{BE} /A}[/latex] is greatest for [latex]\boldsymbol{A}[/latex] near 60?
Problems & Exercises
Data from the appendices and the periodic table may be needed for these problems.
1: An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of [latex]\boldsymbol{^{14} \textbf{C}}[/latex]. Estimate the minimum age of the charcoal, noting that [latex]\boldsymbol{2^{10} = 1024}[/latex].
2: A [latex]\boldsymbol{^{60} \textbf{Co}}[/latex] source is labeled 4.00 mCi, but its present activity is found to be [latex]\boldsymbol{1.85 \times 10^7}[/latex] Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00-mCi activity?
3: (a) Calculate the activity [latex]\boldsymbol{R}[/latex] in curies of 1.00 g of [latex]\boldsymbol{^{226} \textbf{Ra}}[/latex]. (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.
4: Show that the activity of the [latex]\boldsymbol{^{14} \textbf{C}}[/latex] in 1.00 g of [latex]\boldsymbol{^{12} \textbf{C}}[/latex] found in living tissue is 0.250 Bq.
5: Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% [latex]\boldsymbol{^{232} \textbf{Th}}[/latex], with a half-life of [latex]\boldsymbol{1.405 \times 10^{10} \;\textbf{y}}[/latex]. If an average lantern mantle contains 300 mg of thorium, what is its activity?
6: Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of [latex]\boldsymbol{^{131} \textbf{I}}[/latex] per liter, to check for possible reactor leakage. What mass of [latex]\boldsymbol{^{131} \textbf{I}}[/latex] has this activity?
7: (a) Natural potassium contains [latex]\boldsymbol{^{40} \textbf{K}}[/latex], which has a half-life of [latex]\boldsymbol{1.277 \times 10^9}[/latex] y. What mass of [latex]\boldsymbol{^{40} \textbf{K}}[/latex] in a person would have a decay rate of 4140 Bq? (b) What is the fraction of [latex]\boldsymbol{^{40} \textbf{K}}[/latex] in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)
8: There is more than one isotope of natural uranium. If a researcher isolates 1.00 mg of the relatively scarce [latex]\boldsymbol{^{235} \textbf{U}}[/latex] and finds this mass to have an activity of 80.0 Bq, what is its half-life in years?
9: [latex]\boldsymbol{^{50} \textbf{V}}[/latex] has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of [latex]\boldsymbol{^{50} \textbf{V}}[/latex] is 1.75 Bq. What is the half-life in years?
10: You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium. Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it. Neglect the activity of any daughter nuclides.
11: A tree falls in a forest. How many years must pass before the [latex]\boldsymbol{^{14} \textbf{C}}[/latex] activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour?
12: What fraction of the [latex]\boldsymbol{^{40} \textbf{K}}[/latex] that was on Earth when it formed [latex]\boldsymbol{4.5 \times 10^9}[/latex] years ago is left today?
13: A 5000-Ci [latex]\boldsymbol{^{60} \textbf{Co}}[/latex] source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?
14: Natural uranium is 0.7200% [latex]\boldsymbol{^{235} \textbf{U}}[/latex] and 99.27% [latex]\boldsymbol{^{238} \textbf{U}}[/latex]. What were the percentages of [latex]\boldsymbol{^{235} \textbf{U}}[/latex] and [latex]\boldsymbol{^{238} \textbf{U}}[/latex] in natural uranium when Earth formed [latex]\boldsymbol{4.5 \times 10^9}[/latex] years ago?
15: The [latex]\boldsymbol{beta ^-}[/latex] particles emitted in the decay of [latex]\boldsymbol{^3 \textbf{H}}[/latex] (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of [latex]\boldsymbol{^3 \textbf{H}}[/latex]. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?
16: World War II aircraft had instruments with glowing radium-painted dials (see Chapter 31.1 Figure 1). The activity of one such instrument was [latex]\boldsymbol{1.0 \times 10^5}[/latex] Bq when new. (a) What mass of 226Ra226Ra was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made?
17: (a) The [latex]\boldsymbol{^{210} \textbf{Po}}[/latex] source used in a physics laboratory is labeled as having an activity of [latex]\boldsymbol{1.0 \;\mu \textbf{Ci}}[/latex] on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the [latex]\boldsymbol{\alpha}[/latex] s emitted are observed by the detector.
18: Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its [latex]\boldsymbol{^{235} \textbf{U}}[/latex] removed for reactor use and is nearly pure [latex]\boldsymbol{^{238} \textbf{U}}[/latex]. Depleted uranium has been erroneously called non-radioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure [latex]\boldsymbol{^{238} \textbf{U}}[/latex]. (b) Calculate the activity of 60.0 g of natural uranium, neglecting the [latex]\boldsymbol{^{234} \textbf{U}}[/latex] and all daughter nuclides.
19: The ceramic glaze on a red-orange Fiestaware plate is [latex]\boldsymbol{\textbf{U}_2 \textbf{O}_3}[/latex] and contains 50.0 grams of [latex]\boldsymbol{^{238} \textbf{U}}[/latex] , but very little [latex]\boldsymbol{^{235} \textbf{U}}[/latex]. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the [latex]\boldsymbol{^{238} \textbf{U}}[/latex] decay. (c) If energy is worth 12.0 cents per kW⋅h, what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.)
20: Large amounts of depleted uranium ([latex]\boldsymbol{^{238} \textbf{U}}[/latex]) are available as a by-product of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a 4000-kg block of [latex]\boldsymbol{^{238} \textbf{U}}[/latex]. (a) Find its activity. (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain.
21: The Galileo space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of [latex]\boldsymbol{^{238} \textbf{Pu}}[/latex], a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV [latex]\boldsymbol{\alpha}[/latex] particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of [latex]\boldsymbol{^{238} \textbf{Pu}}[/latex] is 87.7 years. (a) What was the original activity of the [latex]\boldsymbol{^{238} \textbf{Pu}}[/latex] in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping [latex]\boldsymbol{\gamma}[/latex] rays.
22: Construct Your Own Problem
Consider the generation of electricity by a radioactive isotope in a space probe, such as described in Chapter 31.5 Problems & Exercises 21. Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its half-life and decay energy, the power needs of the probe and the length of the flight.
23: Unreasonable Results
A nuclear physicist finds [latex]\boldsymbol{1.0 \;\mu \textbf{g}}[/latex] of [latex]\boldsymbol{^{236} \textbf{U}}[/latex] in a piece of uranium ore and assumes it is primordial since its half-life is [latex]\boldsymbol{2.3 \times 10^7 \;\textbf{y}}[/latex]. (a) Calculate the amount of [latex]\boldsymbol{^{236} \textbf{U}}[/latex] that would had to have been on Earth when it formed [latex]\boldsymbol{4.5 \times 10^9 \;\textbf{y}}[/latex] ago for [latex]\boldsymbol{1.0 \;\mu \textbf{g}}[/latex] to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible?
24: Unreasonable Results
(a) Repeat Chapter 31.5 Problems & Exercises 14 but include the 0.0055% natural abundance of [latex]\boldsymbol{^{234} \textbf{U}}[/latex] with its [latex]\boldsymbol{2.45 \times 10^5 \;\textbf{y}}[/latex] half-life. (b) What is unreasonable about this result? (c) What assumption is responsible? (d) Where does the [latex]\boldsymbol{^{234} \textbf{U}}[/latex] come from if it is not primordial?
25: Unreasonable Results
The manufacturer of a smoke alarm decides that the smallest current of [latex]\boldsymbol{\alpha}[/latex] radiation he can detect is [latex]\boldsymbol{1.00 \;\mu \textbf{A}}[/latex]. (a) Find the activity in curies of an α α emitter that produces a [latex]\boldsymbol{1.00 \;\mu \textbf{A}}[/latex] current of [latex]\boldsymbol{\alpha}[/latex] particles. (b) What is unreasonable about this result? (c) What assumption is responsible?
Glossary
becquerel
: SI unit for rate of decay of a radioactive material
half-life
: the time in which there is a 50% chance that a nucleus will decay
radioactive dating
: an application of radioactive decay in which the age of a material is determined by the amount of radioactivity of a particular type that occurs
decay constant
: quantity that is inversely proportional to the half-life and that is used in equation for number of nuclei as a function of time
carbon-14 dating
: a radioactive dating technique based on the radioactivity of carbon-14
activity
: the rate of decay for radioactive nuclides
rate of decay
: the number of radioactive events per unit time
curie
: the activity of 1g of [latex]\boldsymbol{^{226} \textbf{Ra}}[/latex], equal to [latex]\boldsymbol{3.70 \times 10^{10} \;\textbf{Bq}}[/latex]
Solutions
Problems & Exercises
1: 57,300 y
3: (a) 0.988 Ci
(b) The half-life of [latex]\boldsymbol{^{226} \textbf{Ra}}[/latex] is now better known.
5: [latex]\boldsymbol{1.22 \times 10^3 \;\textbf{Bq}}[/latex]
7: (a) 16.0 mg
(b) 0.0114%
9: [latex]\boldsymbol{1.48 \times 10^{17} \;\textbf{y}}[/latex]
11: [latex]\boldsymbol{5.6 \times 10^4 \;\textbf{y}}[/latex]
13: 2.71 y
15: (a) 1.56 mg
(b) 11.3 Ci
17: (a) [latex]\boldsymbol{1.23 \times 10^{-3}}[/latex]
(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly [latex]\boldsymbol{\alpha}[/latex] particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.
19: (a) [latex]\boldsymbol{1.68 \times 10^{-5} \;\textbf{Ci}}[/latex]
(b) [latex]\boldsymbol{8.65 \times 10^{10} \;\textbf{J}}[/latex]
(c) [latex]\boldsymbol{ \$ 2.9 \times 10^3}[/latex]
21: (a) [latex]\boldsymbol{6.97 \times 10^{15} \;\textbf{Bq}}[/latex]
(b) 6.24 kW
(c) 5.67 kW
25: (a) 84.5 Ci
(b) An extremely large activity, many orders of magnitude greater than permitted for home use.
(c) The assumption of [latex]\boldsymbol{1.00 \;\mu \textbf{A}}[/latex] is unreasonably large. Other methods can detect much smaller decay rates.
License
College Physics: OpenStax Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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Place Value: Decimal Numbers
ConceptsWhole Numbers
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How does place value work with small numbers?
Small numbers can be expressed as fractions, such as ½, or a decimal-place numbers, such as 0.5. That dot between the zero and the five in 0.5 is called a "decimal point", and the 5 stands for "five-tenths". By using that dot, we can now write small numbers using place values, rather than writing them only with fractions.
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Decimal Place Value
Think back to our fingers-and-marbles model. At each step going to the left, our units grew larger by a factor of 10. We started with 1s (being our fingers), and went to 10s (being the marbles in the first pile, standing for one instance of "all-my-fingers") and then 100s (being the marbles in the second pile, standing for ten times of "all-my-fingers").
So, at each stage leftward, we multiplied by 10. On the other hand, when we go off to the right (that is, when we go into smaller-than-one units), we divide by 10.
To picture this process with the smaller numbers, let's imagine that we're counting packs of candy bars. Each pack of candy bars contains ten individual bars, and each individual bar is formed so as to break neatly into ten pieces. How would we count single bars, or parts of bars, in our fingers-and-marbles model?
Let's say we counted up thirteen packs and two single bars. Remember that "one pack" is our unit; each finger counts one pack of bars.
Each pack is ten bars, so two single bars would be two parts of our "unit"; that is, they're two parts out of the ten parts of what we're calling "1". Using marbles, this time to the right of our hands, to indicate "parts of a pack", we'd count like this:
The picture above shows "ten packs (being the marble at the left), plus another three packs (being the three fingers), plus two parts of a pack (being the two marbles to the right)", or 10 + 3 + 2/10.
When writing this out in numbers, we could use a line to divide between the whole numbers (in this case, the one 10 and the three 1s) and the parts (in this case, the two 1/10s):
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10 + 3 + 2/10 = 13|2
...but the vertical dividing line looks too much like the digit "1", so instead, let's use a dot:
10 + 3 + 2/10 = 13.2
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This dot is called the "decimal point". The number above can be read in either of two ways: "thirteen point two" or "thirteen and two tenths". In this case, since we're counting packs of candy bars, the number stands for "thirteen and two tenths packs of candy bars".
Note: The use of "and" in this context is important. It indicates that you have a whole-number part, "and" also a smaller-than-one part. The "and" goes between the two parts.
Now let's say we got hungry and ate three pieces of one of the loose bars. One bar is one-tenth of a pack, and the seven remaining pieces are seven-tenths of a bar, so those pieces are seven-tenths of something that is one-tenth of a unit, or (7/10) × (1/10) = 7/100 of a pack. So, after snacking, we have:
10 + 3 + 1/10 + 7/100 = 13.17
That is, we have thirteen point one seven (or thirteen and seventeen hundredths) packs of candy bars left.
Just as we can count bigger and bigger numbers by using digits further and further to the left, so also we can count smaller and smaller numbers by using digits further and further to the right. Each place further to the right counts pieces that are 1/10 as big as the piece at the previous step, and the names are similar to the whole-number names, as you can see in the rows of the table below:
← swipe to view full table →
| | | | | |
--- ---
| big units | name | fraction | name | small units |
| 10 | tens | 1/10 | tenths | 0.1 |
| 100 | hundreds | 1/100 | hundredths | 0.01 |
| 1,000 | thousands | 1/1,000 | thousandths | 0.001 |
| 10,000 | ten thousands | 1/10,000 | ten thousandths | 0.000 1 |
| 100,000 | hundred thousands | 1/100,000 | hundred thousandths | 0.000 01 |
When the numbers get really small (that is, really far to the right of the dot), the names for those numbers get really big, so people usually switch to the "point" way of reading the numbers; it's easier.
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When we're working with numbers smaller than the whole numbers (that is, when we're working with numbers with the "dot" or "decimal point" in them), we have to use zeroes as placeholders after the decimal point. For instance, the fraction "7/100" is written in decimal form as "0.07". If we didn't include that zero between the dot and the 7, we'd be saying "7/10", which wouldn't be what we'd meant. Also, for however many digits we have to the right of the decimal point, that's how many "decimal places" we have.
How are big- and small-number place names related?
Take a close look at the correspondence between the numbers of zeroes in the "big units" expressions in the table above and the numbers of decimal places in the "small units" expressions; if you count them up, you'll see that they match. For instance, one thousand has three zeroes, and one thousandth has three decimal places.
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To figure out what "place" we're at when looking at a decimal-places number, we count up the digits after the decimal point. Whatever is the number of decimal places that we have, that's the number of zeroes for the number matching the decimal "place" that we're at. For instance, the number 83.295 has three digits after the decimal point (those digits begin the 2, 9, and 5), so the number has three decimal places. The number with 1 followed by three zeroes is 1,000: one thousand. That means that the 0.295 portion of the number stands for "two hundred ninety-five thousandths", and the entire number 83.295 stands for "eighty-three and two hundred ninety-five thousandths".
Remember: You should use "and" only between the whole-number part of a number and the decimal-places part. The number "108", for instance, is "one hundred eight", not "one hundred 'and' eight".
How is "and" used in decimal numbers in "real life"?
If you've ever written a check, or seen one filled out, you will have seen that the amount of the check is written in two places: in numerical characters inside a box to the right, and as words on the second line of the check. On the line where the amount is written out in words, there is an "and" between the whole-dollar (or the whole-rand, or the whole-naira, or the whole pretty much any other modern currency) portion and the cents portion. This is the only place where "and" is appropriate when saying or spelling out numbers!
(Yes, pretty much every currency has by now been "decimalized". The main currency unit will go by different names in different countries — guilders, pesos, rials, euros, kwacha, whatever — but they all are subdivided into one hundred of a smaller unit — fils, kopeks, naya paisas, metonnyas, satangs, whatever.)
Express "47.3692" (a) in expanded notation and (b) in words.
(a) In expanded notation, I need to write each "place" out separately:
40 + 7 + 0.3 + 0.06 + 0.009 + 0.0002
(b) To write this out in words, I first need to figure out what kind of number I'm working with. There are four decimal places (the 3, the 6, the 9, and the 2 that are to the right of the decimal point, or dot). The number 1 followed by four zeroes is 10,000 (ten thousand), so the decimal-places part refers to 1/10,000 (ten thousandths). In words, this is:
forty-seven and three thousand, six hundred ninety-two ten thousandths
Again, note the "and" that went between the whole-number part (being the 47) and the decimal-places portion (being the 3692). This is the only proper place for an "and" within a number name.
Given the number 16.875, state (a) the value of the 5 and (b) the digit in the tenths place.
a) The 5 is three decimal places to the right of the decimal point. The number 1 followed by three zeroes is 1,000, so:
the 5 is in the thousandths place.
b) Ten is a 1 followed by one zero, so tenths is one decimal place to the right of the decimal point. Then:
the digit in the tenths place is the 8.
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8132 | https://xaktly.com/TaylorSeries.html | MATH CHEMISTRY PHYSICS BIOLOGY EDUCATION
xaktly | Infinite series
Taylor series
This section makes heavy use of differential calculus. It is also related to the material in these sections:
Linear approximation
Infinite series
Geometric series
Alternating series
p-series
Power series
Test for convergence
Comparison test
Limit comparison test
Integral test
Ratio & Root tests
Building a power-series representation of a function
Taylor series are infinite series of a particular type. They are extremely important in practical and theoretical mathematics. Very often we are faced with using functions that aren't that easy to work with in practice, like exponential and logarithmic functions, or trigonometric functions, or tricky combinations of those. Life gets much simpler if we can replace them with something that's easier to work with.
The accuracy and precision of a Taylor series approximation of a function depend on how many terms we have in our series. In most cases we can get to arbitrary precision (as high as we want), but we balance that with the difficulty of deriving and calculating more terms. Taylor series aren't difficult to come up with, either – you'll see.
We'll start by reviewing linear approximations of functions. We'll learn how to form Taylor-series representations of functions beginning at some central point, and we'll discuss the errors involved in approximating functions in this way.
Linear approximations — a review
We showed that the linear approximation of a function $f$ near a point $x = a$ in its domain is
$$f(x) \approx f'(a)(x - a) + f(a)$$
That's really just $y = mx + b,$ the slope-intercept form of a line, but let's just review how it is derived from a function and its derivative. Look at the figure below. It shows a function, $f(x)$, and its linear approximation, $L(x)$ at the point $(a, f(a))$. Remember that any function, when we view it closely enough, looks approximately linear. We can approximate a curve in the region around a point, $a$, in its domain by a line that has the same slope as the curve at $x = a$, and is tangent to $f(x)$ there, too. Using the approximation, we can estimate values of the function in the domain around $x = a$.
Here's how we derive $L(x)$. Start with the linear equation $y = mx + b$, and plug in the slope and the y-intercept.
$$y = mx + b \phantom{000} \text{ at } \: (a, f(a))$$
Notice that $f'(a)$ is the slope of $f(a)$ at $x = a,$ and $f(a) - f'(a)·a$ is just $b = y - mx,$ the y-intercept. Substituting the quantities in blue into the linear equation $y = mx + b$ and rearranging, we get:
$$ \begin{align} L(x) &= f'(x)\cdot x + f(a) - f'(a)\cdot a \[5pt] &= f(a) + f'(a)(x - a) \end{align}$$
There are two features of linear approximation that we will use in developing even better polynomial approximations of functions. They are
The first derivative of the function and its approximation must be the same at $x = a$: $L(a) = f(a),$ and
The value of the function and the value of the approximation must be the same at $x = a$, where $a$ is the point around which we build our approximation: $L'(a) = f'(a).$
In what follows, we'll further insist that all derivatives of the function and its approximation must be the same at $x = a.$ It's all in the green box below.
Linear approximations
A linear approximation of a function near some point in its domain is just the equation of the line tangent to that function at that point.
The linear approximation of a function $f(x)$ near $x = a$ (left), and for $x = 0$ (right) follows from the well-known formula of a line from its slope, m, and a point, $(x_o, \; y_o)$: $y - y_o = m(x - x_o).$
$$f(x) \approx f(a) + f'(x)(x - a)$$
centered at $x = a$
$$f(x) \approx f(0) + f'(x) \cdot x$$
centered at $x = 0$
Example: Linear approximation of $f(x) = \sin(x)$ near $x = 0$
If we want to approximate $f(x) = \sin(x)$ with a line passing through $(0, f(0))$, we need a point and a slope. Our point is $(0, 0)$. The derivative of $\sin(x)$ at $x = 0$ will give us our slope: $f'(0) = \cos(0) = 1.$ So our approximation formula is
$$ \begin{align} y - y_o &= m(x - x_o) \[5pt] y - 0 &= 1(x - 0) \[5pt] y &= x \end{align}$$
That's a pretty simple approximation. It says that the value of $\sin(x)$ is approximately x for values of x near zero. That's pretty remarkable. If you can tolerate a small amount of error, this approximation involves no multiplication, the thing that really eats up computer time over many iterations. Direct calculation of sines using something like the Taylor-series representation we'll develop below, is much more time-consuming. Here is a graph of $\sin(x)$ vs. $x$, with $x$ in radians.
The table below compares this approximation with calculated values of the function for several values of $x$.
Table of approximations and exact values of sin(x)(x in radians)
| $x$ (rad) | approx. | exact | error |
--- --- |
| 0.001 | 0.001 | 0.0010 | 0% |
| 0.01 | 0.01 | 0.010 | 0.002% |
| 0.1 | 0.1 | 0.099 | 0.167% |
| 0.5 | 0.5 | 0.479 | 4.3% |
| 1.0 | 1.0 | 0.842 | 19% |
error $= \frac{exact - approx}{exact} \times 100%$
You can see that while this approximation does ultimately break down, the farther we get from x = 0, but it is very good for values of x near zero. So if someone asks you, "Hey, what's the sine of 0.1?," you can confidently answer, "About 0.1."
Next we'll ask, what about approximating these curved functions with curved functions instead of lines? We'll see how that will lead to Taylor series.
Approximating $f(x) = \cos(x)$
Can we do better than a linear approximation?
Take a look at the plot of $f(x) = \cos(x)$ below (thick black curve). Superimposed on it are the graphs of three successively better approximations, each centered around $x = 0.$ They are linear, quadratic and quartic approximations.
Notice that we haven't included any odd polynomial terms in this approximation because $\text{cos(x)}$ is an even function (symmetric across the $x$-axis). As you learn how to generate Taylor series, you'll find that those missing terms will naturally just drop out. If our example was the sine function, only polynomial terms of odd degree would appear.
The red line is our familiar linear approximation. It has the same value and slope as $cos(x)$ at $x = 0.$ It's also not a very good approximation of the cosine function in this region because $cos(x)$ is very curvy there.
The magenta approximation includes a quadratic term or "correction" to the linear approximation with the same curvature as $\cos(x).$ It bends the linear approximation downward like the cosine function. It's better, but it isn't perfect; you can see how it bends away from the cosine curve as we move away from $x = 0$ in either direction.
Finally, the dashed green curve includes a quartic term, and clearly it matches $cos(x)$ quite well over the region shown, even relatively far from $x = 0.$
Don't worry about how we got those numerical coefficients just now. That will become clear in a bit.
Remember that odd means $f(-x) = -f(x)$ and even means $f(-x) = f(x)$, which is just saying that there's symmetry across the $y$-axis.
How it's done
In the linear approximation, we made the assumptions that the value of the approximation and its slope should be the same at the point $x = a$. That just makes sense.
Well, we can generalize that kind of thinking to higher derivatives. It's also a reasonable goal to expect the curvature of a better approximation to match the curvature of the function we're trying to approximate.
If the curvature of $f(x)$ is concave-downward, then our approximation should be, too. And that means that both the function and the approximation should have the same second derivative.
We might also insist that the change in curvature (the third derivative) be the same ... and so on.
Requirements for a good polynomial approximation near x = a
In order for a polynomial function $P(x)$ to be a good approximation of a function $f(x)$,
The value of the approximation must be the same as the value of the function at $x = a$: $P(a) = f(a)$
The slope of the approximation must be the same as the slope of the function at $x = a$: $P'(a) = f'(a)$
The curvature of the approximation must be the same as the curvature of the function at $x = a$: $P''(a) = f''(a)$
... and so on. Each successive derivative of the approximation and the function at $x$ = a must be equal: $P^{(n)}(a) = f^{(n)}(a)$
Example 1
Finding a series approximation of $f(x) = e^x$
Let's explore the derivation of a Taylor-series approximation of a function by making a polynomial approximation of the exponential function,
$$f(x) = e^x$$
Our approximation will take the form of a 5th degree polynomial with unknown coefficients. I've chosen five terms because it's enough to show that important patterns emerge, but there's nothing special about 5. It looks like this:
$$f(x) \approx g(x) = Ax^5 + Bx^4 + Cx^3 +Dx^2 + Ex + F$$
Now we said above that the function and all of its derivatives have to be the equal at the point in which we're interested. We'll center this approximation about $x=0$ for convenience. Here are the derivatives of the function, those derivatives evaluated at $x=0$, and the corresponding derivatives of our approximation, $g(x)$.
A note on derivative notation
We often write the first, second and third derivatives of a function $f(x)$ as $f'(x), \, f''(x)$ and $f'''(x),$ but writing all those primes gets tedious for higher derivatives, so we write $f^{(4)}(x)$ for the fourth derivative, $f^{(5)}(x)$ for the fifth, and so on.
$$ \begin{align} f(x) = e^x \phantom{00000} f(0) &= 1 \phantom{000} g(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F \[5pt] f'(x) = e^x \phantom{0000} f'(0) &= 1 \phantom{000} g'(x) = 5 Ax^4 + 4Bx^3 + 3Cx^2 + 2Dx + E \[5pt] f''(x) = e^x \phantom{0000} f''(0) &= 1 \phantom{000} g''(x) = 20 Ax^3 + 12 Bx^2 + 6Cx + 2D \[5pt] f^{(3)}(x) = e^x \phantom{000} f^{(3)}(0) &= 1 \phantom{000} g^{(3)}(x) = 60 Ax^2 + 24 Bx + 6C \[5pt] f^{(4)}(x) = e^x \phantom{000} f^{(4)}(0) &= 1 \phantom{000} g^{(4)}(x) = 120 Ax + 24B \[5pt] f^{(5)}(x) = e^x \phantom{000} f^{(5)}(0) &= 1 \phantom{000} g^{(5)}(x) = 120 A \end{align}$$
Now if we equate $g(0) = f(0), \: g'(0) = f'(0),$ and so on, we can solve for the coefficients, $A, \; B, \; C, \; D$ & $E$, of our polynomial, $g(x).$
$$A = \frac{1}{120} \phantom{000} B = \frac{1}{24} \phantom{000} C = \frac{1}{6} \phantom{000} D = \frac{1}{2} \phantom{000} E = 1$$
The resulting polynomial, just a sum of these terms, looks like this:
$$f(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}$$
Now if we recognize some patterns, including that the factorial function is hidden in those denominators [recall that $n! = n(n-1)(n-2) \dots 3·2·1, \text{ and } 0! = 1$].
$$f(x) \approx \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \dots $$
This pattern will continue indefinitely, and we can write the series approximation in shorthand like this:
$$e^x = \sum_{n = 0}^{\infty} \, \frac{x^n}{n!}$$
We can use this sum with $x = 1$ to estimate the value of $e = e^1$. Here's a table from a spreadsheet. Each row represents one more term added to the series. Notice that each successive term adds a smaller number onto the sum. This series is said to converge to a limit of $e^x$.
The table shows successive values of n, n! and 1/n! used in the series approximation of $e^x$. Notice that each new term is smaller than the previous one. That is, the size of the next term is decreasing.
As terms are added to our series approximation of $e^x$, terms nearest the decimal point begin to be fixed and no longer change (green). This is called convergence, and we say that the series is converging to the number $e$. If more precision is required, we just add more terms to the series.
For more on the transcendental number $e$, the base of all continuously-growing exponential functions, you might want to check out the exponential functions section.
Animation: Here is an animation of the first few terms of the series we just derived, centered at $x = 0$. This animation may not show up too well on a smaller device like a phone. I'm working on that, so stand by.
A general formula for Taylor series centered on x = 0
If we go back to our derivation of the series approximation of $f(x) = e^x$, we can see that a general formula for the series approximation of any differentiable function centered around $x = 0$ is:
$$f(x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
Here $f^{(n)}(0)$ is the $n^{th}$ derivative of $f(x)$ evaluated at $x = 0$. In the special case where $x = 0$, the Taylor series is called a MacLaurin series.
The MacLaurin Series
The MacLaurin series is a Taylor series approximation of a function $f(x)$ centered at $x = 0$. $f^{(n)}(0)$ are the $n^{th}$ derivatives of $f(x)$ evaluated at $x=0$.
$$f(x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
Taylor series not centered at x = 0
If we choose to center our approximation at some other point, $x = a$, in the domain of $f(x)$, then any value we calculate from the approximation will be at $(x - a)$, and we just evaluate the derivatives at $x = a$. The generalized Taylor series looks like this:
$$f(x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$$
Notice that if $a = 0$, we just end up with the MacLaurin series formula.
The generalized Taylor series
$$f(x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$$
Example 2
Derive the Taylor series approximation of $f(x) = \sin(x)$ near $x = 0$.
Because this approximation will be centered at $x = 0$, it's a MacLaurin series.
To stay organized, try making yourself a table of derivatives, derivatives evaluated at $x=0$, and terms of the series.
| $f^{(n)}(x)$ | $f^{(n)}(0)$ | $\frac{f^{(n)}(0)}{n!} x^n$ |
---
| $f(x) = \sin(x)$ | $f(0) = 0$ | $\require{cancel} \cancel{\frac{0}{0!} x^0}$ |
| $f'(x) = \cos(x)$ | $f'(0) = 1$ | $\frac{1}{1!} x^1$ |
| $f''(x) = -\sin(x)$ | $f''(0) = 0$ | $\cancel{\frac{0}{2!} x^2}$ |
| $f^{(3)}(x) = -\cos(x)$ | $f^{(3)} = -1$ | $\frac{-1}{3!} x^3$ |
| $f^{(4)}(x) = \sin(x)$ | $f^{(4)}(0) = 0$ | $\cancel{\frac{0}{4!} x^4}$ |
| $f^{(5)}(x) = \cos(x)$ | $f^{(5)}(0) = 1$ | $\frac{1}{5!} x^5$ |
Be on the lookout for patterns as you calculate the terms of the series. Looking at the right column of the table it's pretty obvious that terms with even powers of x drop out because the derivative is zero. It also appears that the sign of each term alternates between positive and negative.
If we write the remaining terms out, we can speculate (intelligently) on further terms, such as the last term (red) below:
$$sin(x) \approx \frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots $$
Finally, we should try to capture that series of terms in compact summation notation. We try always to start the index, $n$, at zero, but sometimes it just isn't convenient. It works fine here, though. The alternating sign is represented by $(-1)^n$, giving us $1$ for $n = 0$, $-1$ for $n = 1$, and so on. The exponent and factorial terms are $2n+1$ and $(2n+1)!$ to ensure only odd numbers.
$$sin(x) \approx \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{(2n + 1)!}$$
Example 3
Find the Taylor series approximation of $f(x) = ln(x)$ near $x = 1$
Notice that this is a general Taylor series, not a MacLaurin series.
First, organize and set up a table of $f^{(n)}(x)$, $f^{(n)}(1)$ and the terms of the series (right).
Notice that in this example we've centered the approximation on $x = 1$ because $ln(x)$ is not defined at $x=0$.
You might need to take a minute to work out those derivatives for yourself, but they're correct. The first term of the series vanishes but the successive terms are quite predictable and alternate sign, + - + -.
The terms of the series are summed below. Try to notice the patterns and ask how you might extend the series by one or two more terms.
| $f^{(n)}(x)$ | $f^{(n)}(0)$ | $\frac{f^{(n)}(0)}{n!} x^n$ |
---
| $f(x) = ln(x)$ | $f(0) = 0$ | $\require{cancel} \cancel{\frac{0}{0!} x^0}$ |
| $f'(x) = \frac{1}{x}$ | $f'(0) = 1$ | $\frac{1}{1!} x^1$ |
| $f''(x) = \frac{-1}{x}$ | $f''(0) = -1$ | $\frac{-1}{2!} x^2$ |
| $f^{(3)}(x) = \frac{2}{x^3}$ | $f^{(3)} = 2$ | $\frac{2}{3!} x^3$ |
| $f^{(4)}(x) = \frac{-6}{x^4}$ | $f^{(4)}(0) = -6$ | $\frac{-6}{4!} x^4$ |
| $f^{(5)}(x) = \frac{24}{x^5}$ | $f^{(5)}(0) = 24$ | $\frac{24}{5!} x^5$ |
$$ln(x) = \frac{1}{1!}(x - 1)^1 - \frac{1}{2!} (x - 1)^2 + \frac{2}{3!} (x - 1)^3 - \frac{6}{4!} (x - 1)^4 + \frac{24}{5!} (x - 1)^5$$
Now notice that the numerators of the fractions of each term are the factorials $0!, \; 1!, \; 2!, \dots$:
$$ln(x) = \frac{0!}{1!}(x - 1)^1 - \frac{1!}{2!} (x - 1)^2 + \frac{2!}{3!} (x - 1)^3 - \frac{3!}{4!} (x - 1)^4 + \frac{4!}{5!} (x - 1)^5$$
If we use the definition of factorials, e.g. $3! = 3 \cdot 2 \cdot 1$, it's easy to see how the factorial ratios above can be reduced, like this:
$$\frac{2!}{3!} = \frac{2\cdot 1}{3\cdot 2\cdot 1} = \frac{1}{3}$$
So we can rewrite our series in its simplest form:
$$ln(x) = \frac{(x - 1)^1}{1} - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \frac{(x - 1)^5}{5} - \dots$$
Finally, write the summation notation. It's easier, in this case, to start with $n = 1$ in the sum, and the $(-1)^{(n+1)}$ term provides the right alternation of sign. You'll have to do set up these summations with a fair bit of trial and error before you develop some intuition for it. Don't worry – it'll happen with some practice.
$$ln(x) \approx \sum_{n = 1}^{\infty} (-1)^{n + 1} \frac{(x - 1)^n}{n}$$
Example 4
The MacLaurin series expansion of $f(x) = \tan(x)$
The expansion is:
$$f(x) \approx \sum_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
First we'll calculate the derivatives. These derivatives get more complicated with each round. Here are the first five:
$$ \begin{align} f(x) &= \tan(x) \[5pt] f'(x) &= \bf \sec^2(x) \[5pt] f''(x) &= 2 \sec^2(x) \tan(x) \[5pt] f'''(x) &= 4 \sec^2(x) \tan^2 (x) + \bf 2 \sec^4(x) \[5pt] f^{(4)}(x) &= 8 \sec^3(x) \tan^3(x) + 16 \sec^4(x) \tan^4(x) \[5pt] f^{(5)}(x) &= 8[2 \sec^2(x) \tan^4(x) + 3 \sec^4(x) \tan^2(x) \[5pt] &\; + 4 \sec^4(x) \tan(x) + \bf 2 \sec^6(x) \[5pt] &\; + 4 \sec^4(x) \tan(x)] \end{align}$$
The bold terms are the only parts of these derivatives that survive setting $x=0$. All of the tangent terms vanish because $\tan(0) = 0$, whereas $\sec(0) = 1$.
The derivatives, evaluated at $x = 0$, are:
$$ \begin{matrix} f(0) = 0 && f'''(0) = 2 \[5pt] f'(0) = 1 && f^{(4)}(0) = 0 \[5pt] f''(0) = 0 && f^{(5)}(0) = 16 \end{matrix}$$
Now we can construct the MacLaurin series. Notice that only odd powers of $x$ are present.
Note: We could find other representations of this function, for example, by using the trig. identity $\sec^2(x) = \tan^2(x) + 1$ in the derivatives above.
$$\tan(x) \approx \frac{0x^0}{0!} + \frac{1x^1}{1!} + \frac{0x^2}{2!} + \frac{2x^3}{3!} + \frac{0x^4}{4!} + \frac{16x^5}{5!} + \dots$$
Getting rid of the zero terms gives us the series expansion. Here I've included an $x^7$ term. As far as I know, there's no convenient formula for those coefficients: 1, 2, 4, 16, 272, ..., so I'll leave them as they are.
$$\tan(x) \approx \frac{x^1}{1!} + \frac{2x^3}{3!} + \frac{4x^5}{5!} + \frac{16x^5}{5!} + \frac{272x^7}{7!} + \dots$$
Finally, let's calculate some values of $\tan(x)$ using this expansion and a calculator, and we can compare the results. In the table below, $\tan(x)$ is calculated using the first four terms of our expansion (to the $x^7$ term) and a calculator, which presumably uses a more efficient algorithm for calculating the tangent.
The expansion is very accurate up to about a unit away from $x = 0$, where it falls apart a bit. Because of the large exponents and factorials involved in this expansion, it is an inefficient algorithm for calculating tangents to high precision far from the center of the expansion.
Example 5
Two ways to get to the MacLaurin series for $f(x) = \sin^2(x)$
In this example, we'll find the MacLaurin series for $f(x) = \sin^2(x)$ by the direct method — finding derivatives, evaluating them at x = 0, and so on, and then by squaring part of the MacLaurin series for $f(x) = \sin(x)$.
Direct method
First we need some derivatives. Many will vanish when evaluated at $x = 0$, so let's go through the painstaking process of calculating a bunch of them:
$$ \begin{align} f(x) &= \sin^2(x) \[5pt] f'(x) &= 2 \sin(x) \, \cos(x) \[5pt] f''(x) &= 2 \cos^2(x) - 2 \sin^2(x) \[5pt] f'''(x) &= -4 \cos(x) \, \sin(x) - 4 \sin(x) \, \cos(x) \[5pt] &= -8 \cos(x) \, \sin(x) \[5pt] f^{(4)}(x) &= 8 \sin^2(x) - 8 \cos^2(x) \[5pt] f^{(5)}(x) &= 16\sin(x) \cos(x) + 16 \sin(x) \cos(x) \[5pt] &= 32 \sin(x) \, \cos(x) \[5pt] f^{(6)}(x) &= 32 \cos^2(x) - 32 \sin^2(x) \[5pt] f^{(7)}(x) &= -64 \cos(x)\sin(x) - 64 \sin(x)\cos(x) \[5pt] &= -128 \sin(x) \, \cos(x) \[5pt] f^{(8)}(x) &= -128 \cos^2(x) + 128 \sin^2(x) \end{align}$$
Evaluating these at x = 0 gives:
$$ \begin{matrix} f(0) = 0 && f^{(5)}(0) = 0 \[5pt] f'(0) = 0 && f^{(6)}(0) = 32 \[5pt] f''(0) = 2 && f^{(7)}(0) = 0 \[5pt] f'''(0) = 0 && f^{(8)}(0) = -128 \[5pt] f^{(4)}(0) = -8 && \end{matrix}$$
Collecting the nonzero terms in our series formula gives:
$$\sin^2(x) \approx \frac{2x^2}{2!} - \frac{8x^4}{4!} + \frac{32x^6}{6!} - \frac{128x^8}{8!} + \dots$$
We can reduce the fractions by expanding the factorials to get this series:
$$\approx x^2 - \frac{x^4}{3} + \frac{2x^6}{45} - \frac{x^8}{315} + \dots$$
Now from the $f(x) = \sin(x)$ series
Now let's do this by squaring the first three terms of the MacLaurin series (in example 1 above) for $\sin(x)$:
$$\sin(x) \approx x^1 - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 + \dots$$
Squaring the first three terms means squaring a trinomial:
$$\sin^2(x) \approx \left( x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 \right)^2$$
That's a little tedious, but the result, after collecting terms of like powers of $x$, is:
$$\approx x^2 - \frac{2}{3!} x^4 + \frac{2}{5!} x^6 + \frac{1}{3! 3!} x^8 + \frac{1}{5! 5!} x^{10}$$
Here again, we can reduce the numerical parts to get a pretty good approximation of our series. The last two terms don't match the one we got from the direct method, but that's because we only squared the first three terms of the sine series.
Still, this method represents a powerful feature of series, namely that we can do many operations with them, some that would be more difficult with the original function.
Practice problems
Find the first four nonzero terms of the MacLaurin series for $f(x) = \sin(x^2)$
Solution
This isn't as difficult as it may at first seem, because we already know the MacLaurin series for \sin(x):
$$\sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$
To find the expansion for $\sin(x^2),$ simply plug in x2 for x in the sine series:
$$\sin(x) \approx x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \dots$$
As long as we can tolerate whatever error is introduced by using a series instead of the function it represents (and that error can usually be made to be as small as we need), series can help us to do a lot of tasks more easily.
2. Find the first two nonzero terms of the MacLaurin series for $f(x) = \tan^{-1}(x)$.
Solution
First we need some derivatives. Unfortunately, these are lousy ones. There are the first three derivatives of $tan^{-1}(x):$
$$ \begin{align} f(x) &= tan^-1(x) \[5pt] f'(x) &= \frac{1}{1 + x^2} \[5pt] f''(x) &= -(1 + x^2)^{-2} (2x) = \frac{-2x}{(1 + x^2)^2} \[5pt] f'''(x) &= \frac{-2(1 + x^2)^2 + 2x(2)(1 + x^2)(2x)}{(1 + x^2)^3} \[5pt] &= - \frac{2}{1 + x^2} + \frac{8x^2}{(1 + x^2)^2} \end{align}$$
Now evaluate these derivatives at x = 0 (MacLaurin series):
$$ \begin{matrix} f(0) = 0 && f''(0) = 0 \[5pt] f'(0) = 1 && f'''(0) = -2 \end{matrix}$$
I like to plug everything into the MacLaurin series formula as it is, just to keep things organized:
$$tan^{-1}(x) = \frac{0x^0}{0!} + \frac{1x^1}{1!} + \frac{0x^2}{2!} - \frac{2x^3}{3!} + \dots$$
Then simplify:
$$\tan^{-1}(x) \approx x - \frac{2x^3}{3!}$$
3. The indefinite integral of $f(x) = \cos(x^3)$ is tricky to do. Use the first four nonzero terms of the MacLaurin series of $\cos(x)$ to estimate the integral.
Solution
We already know the MacLaurin series for \cos(x):
$$\cos(x) \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots $$
We can approximate $f(x) = \cos(x^3)$ by inserting an $x^3$ for every x in that expansion, like this:
$$\cos(x^3) \approx 1 - \frac{x^6}{2!} + \frac{x^{12}}{4!} - \frac{x^{18}}{6!} + \dots $$
Now the integral of a sum is the sum of integrals, so we simply integrate this polynomial function term-by-term:
$$\int \cos(x^3) \, dx = x - \frac{x^7}{7 \cdot 2!} + \frac{x^{13}}{13 \cdot 4!} - \frac{x^{19}}{19 \cdot 6!} + \dots$$
Notice that it wouldn't be difficult at all to add more terms to this series – or even predict the pattern to do so. With series, we're able to approximate, with more-or-less arbitrary precision, a difficult integral.
4. Find the MacLaurin series for $\int_0^x \cos(t^3) \, dt$.
Solution
It might be tempting to start taking derivatives of this integral, recalling that the fundamental theorem of calculus gives us the first derivatives basically for free, but that gets cumbersome quickly.
We know that the series for $\cos(x)$ is:
$$\cos(x) \approx 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \dots$$
If we substitute x3 for x in the series, we get
$$\cos(t^3) \approx 1 - \frac{t^6}{2!} + \frac{t^{12}}{4!} - \frac{t^{18}}{6!} + \dots$$
Now the integral of a sum is the sum of integrals, so we simply integrate this polynomial seris term-by-term:
$$\int \cos(t^3) \, dt = t - \frac{t^7}{7\cdot 2!} + \frac{t^{13}}{13\cdot 4!} - \frac{t^{19}}{19\cdot 6!} + \dots$$
Evaluating at t = 0 (the lower limit of integration) gives all zeros, so the integral is just
$$\int \cos(x^3) \, dx = t - \frac{x^7}{7\cdot 2!} + \frac{x^{13}}{13\cdot 4!} - \frac{x^{19}}{19\cdot 6!} + \dots$$
We can write the series in summation notation like this:
$$\int \cos(x^3)\, dx = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{6n+1}}{(6n + 1)(2n)!}$$
5. Find the MacLaurin series for $f(x) = \frac{1}{1 + x^2}$ and express it with summation notation.
Solution
First we find some derivatives:
$$ \begin{align} f(x) &= \frac{1}{1 + x^2} \[5pt] f'(x) &= \frac{-2x}{(1 + x^2)^2} \[5pt] f''(x) &= - \frac{2}{(1 + x^2)^2} + \frac{8x^2}{(1 + x^2)^3} \[5pt] f^{(3)}(x) &= \frac{24x}{(1 + x^2)^3} - \frac{48x^3}{(1 + x^2)^4} \[5pt] f^{(4)}(x) &= \frac{24}{(1 + x^2)^3} - \frac{288x^2}{(1 + x^2)^4} + \frac{384x^3}{(1 + x^2)^5} \end{align}$$
Now evaluate each of those at x = 0:
$$ \begin{matrix} f(0) = 1 && f^{(3)} = 0 \[5pt] f'(0) = 0 && f^{(4)} = 24 \[5pt] f''(0) = -2 \end{matrix}$$
Next plug those values into the MacLauring series formula:
$$f(x) \approx \frac{1x^0}{0!} + \frac{0x^1}{1!} - \frac{2x^2}{2!} + \frac{0x^3}{3!} + \frac{24x^4}{4!} + \dots$$
So the series is
$$f(x) \approx 1 - x^2 + x^4 - x^6 + \dots$$
In summation notation:
$$= \sum_{n = 0}^{\infty} (-1)^n \, x^{2n}$$
6. Find the 3rd-degree Taylor series approximation of the function $f(x) = x^{1/2}$ centered around $x = 2$.
Solution
First we find the derivatives and evaluate them at x = 2:
$$ \begin{matrix} f(x) = x^{\frac{1}{2}} && f(2) = \sqrt{2} \[5pt] f'(x) = \frac{1}{2}x^{-\frac{1}{2}} && f'(2) = \frac{1}{2 \sqrt{2}} \[5pt] f''(x) = -\frac{1}{4} x^{-\frac{3}{2}} && f''(2) = \frac{1}{4 \sqrt{2^3}} \[5pt] f^{(3)}(x) = \frac{3}{8} x^{-\frac{5}{2}} && f^{(3)}(2) = \frac{3}{8 \sqrt{2^5}} \[5pt] f^{(4)}(x) = -\frac{15}{16} x^{-\frac{7}{2}} && f^{(4)}(2) = -\frac{15}{16 \sqrt{2^7}} \end{matrix}$$
Next, plug those into the Taylor series formula:( ← wide equations: scroll left & right → )
$$f(x) \approx \frac{(x - 2)^0 \cdot \sqrt{2}}{1! \cdot 2 \sqrt{2}} + \frac{(x - 2)^1}{2! \cdot 4 \sqrt{2^3}} + \frac{(x - 2)^2}{2! \cdot 4 \sqrt{2^3}} + \frac{(x - 2)^3}{3! \cdot 8 \sqrt{2^5}} + \dots$$
and simplify as much as possible:
$$f(x) \approx \sqrt{2} + \frac{x - 2}{2 \sqrt{2}} + \frac{(x - 2)^2}{8 \sqrt{2^3}} + \frac{(x - 2)^3}{48 \sqrt{2^5}}$$
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8133 | https://clu-in.org/download/techdrct/emerging_contaminant_tnt.pdf | United States Office of Solid Waste and EPA 505-F-10-010 Environmental Protection Agency Emergency Response (5106P) February 2011 1 At a Glance Highly explosive, yellow, odorless solid. Synthetic product that does not occur naturally in the environment. Has been used extensively in the manufacture of explosives and accounts for a large part of the explosives contamination at active and former U.S. military installations. Sorbed by most soils limiting its migration to water. Not expected to persist for a long period of time in surface waters or the atmosphere because of transformation processes such as photolysis. Classified as a Group C contaminant (possible human carcinogens). Primarily damages the liver and blood systems, if inhaled or ingested. SW8515 is a field screening method used detect TNT in soil by colorimetric screening. The primary laboratory methods include liquid and gas chromatography. Treatment technologies include in situ chemical and biological remediation, composting, phytoremediation, ultrafiltration, and low-temperature thermal desorption. Emerging Contaminants – 2,4,6-Trinitrotoluene (TNT) February 2011 Introduction An “emerging contaminant” is a chemical or material that is characterized by a perceived, potential, or real threat to human health or the environment or a lack of published health standards. A contaminant may also be “emerging” because a new source or a new pathway to humans has been discovered or a new detection method or treatment technology has been developed (DoD 2010). This fact sheet, developed by the U.S. Environmental Protection Agency (EPA) Federal Facilities Restoration and Reuse Office (FFRRO), provides a brief summary of 2,4,6-trinitrotoluene (TNT), including its physical and chemical properties; environmental and health impacts; existing federal and state guidelines; detection and treatment methods; and additional sources of information. While TNT is not identified as an emerging contaminant by the Department of Defense (DoD), this compound accounts for a large part of the explosives contamination at active and former U.S. military installations. With its manufacturing impurities and environmental transformation products, TNT presents various health and environmental concerns. This fact sheet is intended for use by site managers and field personnel who may address TNT contamination at cleanup sites or in drinking water supplies. What is TNT? TNT is a yellow, odorless solid that does not occur naturally in the environment. It is made by combining toluene with a mixture of nitric and sulfuric acids (ATSDR 1995). It is a highly explosive, single-ring nitroaromatic compound that is a crystalline solid at room temperature (CREEL 2006). TNT is one of the most widely used military high explosives partly because of its insensitivity to shock and friction. It has been used extensively in the manufacture of explosives since the beginning of this century and is used in military cartridge casings, bombs, and grenades. (ATSDR 1995). It has been used either as a pure explosive or in binary mixtures. The most common binary mixtures of TNT are cyclotols (mixtures with RDX) and octols (mixtures with High Melting Explosive [HMX]) (ATSDR 1995; MMR 2001). In addition to military use, small amounts of TNT may be used for industrial explosive applications, such as deep well and underwater blasting. Other industrial uses include chemical manufacturing as an intermediate in the production of dyestuffs and photographic chemicals (MMR 2001). FFRRO LogoEPA Logo FACT SHEET 2 What is TNT? (continued) Red water, the effluent from TNT manufacturing, was a major source of munitions constituent contamination in soils, groundwater, and occasionally surrounding surface water/sediment at Army ammunition plants. TNT production ended in the mid-1980s in the United States; however, contamination of soils and groundwater from red water remains in some areas (MMR 2001; EPA 2005). TNT is commonly found at hand grenade ranges, antitank rocket ranges, artillery ranges, bombing ranges, munitions testing sites, and Open Burn/Open Detonation (OB/OD) sites (CREEL 2006). Exhibit 1: Physical and Chemical Properties of TNT (ATSDR 1995; EPA 1999) Property Value CAS Number 118-96-7 Physical Description (physical state at room temperature) yellow, odorless solid Molecular weight (g/mol) 227 Water solubility (mg/L at 25oC) 130 Octanol-water partition coefficient (KOW) 1.6 Soil organic carbon-water coefficient (KOC) 300 Boiling point (oC) 240 (Explodes) Melting point (oC) 80.1 Vapor pressure at 25oC (mm Hg) 1.99 x10-4 Specific gravity 1.654 Henry’s Law Constant (atm-m3/mol at 20oC) 4.57 x10-7 Abbreviations: g/mol – gram per mole; mg/L – milligrams per liter; oC – degrees Celsius; mm Hg – millimeters of mercury; atm-m3/mol – atmosphere time cubic meter per mole.
What are the environmental impacts of TNT? TNT can be released to the environment through spills, firing of munitions, disposal of ordnance, and open incineration and detonation of ordnance, leeching from inadequately sealed impoundments, and demilitarization of munitions. The compound can also be released from manufacturing and munitions processing facilities (ATSDR 1995). TNT has been identified in at least 20 of the 1,338 hazardous waste sites that have been proposed for inclusion on the National Priorities List (NPL) (ASTDR 1995). Partition coefficients reported by most investigators indicate soils have a high capacity for rapid sorption of TNT. TNT not sorbed into soil is usually transformed rapidly under anaerobic conditions (CREEL 2006; USACE 1997). In the case of impact areas, the majority of the TNT may be degraded in the surface soil, but small quantities can reach shallow groundwater (CREEL 2006). TNT has a high aqueous solubility and is mobile in surface water and groundwater. However, if TNT reaches the water table, it continues to be sorbed by the aquifer material and undergoes transformation processes that can limit its mobility (EPA 2005; CREEL 2006). Once released to surface water, TNT undergoes rapid photolysis to a number of degradation products. 1,3,5-TNB (1,3,5-Trinitrobenzene) is the primary photodegradation product of TNT in environmental systems (ATSDR 1995; CREEL 2006). Products of photolysis of TNT have been observed as a coating on TNT particles and as a fine powdered residue surrounding TNT particles on ranges receiving limited rainfall (CREEL 2007a). TNT is broken down by biodegradation in water and surface soils but at rates much slower than photolysis. TNT photolysis is reportedly faster than biodegradation by a factor of 1,000 (ATSDR 1995; CREEL 2006). Biological degradation products of TNT in water, soil, or sediments include 2-amino-4,6-dinitrotoluene, 2,6-diamino-4-nitrotoluene, and 2,4-diamino-6-nitrotoluene (EPA 1999; CREEL 2007b). It is expected that TNT released to the atmosphere would undergo direct photolysis, as it does in surface water (ATSDR 1995). TNT does not seem to bioaccumulate in animals, but can be taken up by plants (CREEL 2006). 3 What are the health effects of TNT? For the general population, exposure to TNT is limited to areas around Army ammunition plants where these explosives are manufactured, packed, loaded, or released through the demilitarization of munitions (ATSDR 1995). Potential exposure to TNT could occur by dermal contact or inhalation exposure; however, the most likely route of exposure at or near hazardous waste sites is ingestion of contaminated drinking water (MMR 2001). EPA has assigned TNT a weight-of-evidence carcinogenic classification of C (possible human carcinogen) (IRIS 1993; OSHA 1999). A Minimal Risk Level (MRL) of 0.0005 milligrams per kilograms per day (mg/kg/day) has been derived for intermediate oral exposure to TNT (ATSDR 1995). Animal study results indicated that inhalation or ingestion of high levels of TNT may cause liver, blood, immune system, and reproductive damage (MMR 2001; EPA 2005). When TNT reaches the liver, it breaks down into several different substances. Not all of these substances have been identified (ATSDR 1995). At high air levels, workers involved in the production of TNT experienced anemia and abnormal liver tests. After long term exposure to skin and eyes, some people developed skin irritation and cataracts, respectively (MMR 2001). There is no information indicating that TNT causes birth defects in humans. However, male animals treated with high doses of TNT have developed serious reproductive system effects (ATSDR 1995b; MMR 2001). Limited information is available regarding oral, respiratory, cardiovascular, dermal, and neurological effects in humans after exposure to TNT (ATSDR 1995). Are there any federal and s tate guidelines and health s tandards for TNT? The EPA has established a lifetime Health Advisory guidance level of 2 parts per billion (ppb) for TNT in drinking water. The EPA has not established an ambient air level or a cleanup standard for TNT in soil (MMR 2001). Since TNT is explosive, flammable, and toxic, EPA has designated it as a hazardous waste and EPA regulations for disposal must be followed (ATSDR 1995). EPA assigned TNT an RfD of 5.00 x10-4 mg/kg/day (IRIS 1993). The Drinking Water Equivalent Level (DWEL), a lifetime exposure at which adverse health effects would not be expected to occur, is 20 micrograms per liter (μg/L) for TNT. The Lifetime Health Advisory is 2 μg/L (ATSDR 1995). OSHA set a general industry permissible exposure limit (PEL) of 1.5 mg/m3 (mg/m3) of workplace air for an 8-hour workday for a 40-hour workweek (OSHA 2010). The National Institute for Occupational Safety and Health (NIOSH) recommended exposure limit (REL) for TNT during an 8-hour workday, 40-hour workweek is 0.5 mg/m3 (OSHA 2010). The American Conference of Governmental Industrial Hygienists (ACGIH) has set a threshold limit value (TLV) of 0.1 mg/m3 (OSHA 2010). The Department of Transportation (DOT) specifies that when TNT is shipped, it must be wet with at least 10% water (by weight) and it must be clearly labeled as a flammable solid (ATSDR 1995). What detection and s ite characterization methods are available for TNT? High performance liquid chromatography (HLPC) and high-resolution gas chromatography (HRGC) have been paired with several types of detectors, including mass spectrometry (MS), electrochemical detection (ED), electron capture detector (ECD), and ultraviolet detector (UV) (ATSDR 1995). Laboratory Method 8330 is the most widely used analytical approach for detecting TNT in soil. The method specifies using HLPC with a UV. It has been used to detect TNT and some of its breakdown products at levels in the low ppb range (EPA 2006). Another method commonly used is Method 8095 employs the same sample-processing steps as Method 8330 but uses HRGC with an ECD for determination (EPA 2005). SW8515 is a specific field screening method used to detect TNT in soil by a colorimetric screening method (USACE 2005; Army 2009). Tested field-screening instruments for TNT include GC-IONSCAN, which uses ion mobility spectrometry (IMS), and the Spreeta Sensor, which uses surface plasma resonance (SPR) (EPA 2000; 2001). 4 What technologies are being us ed to treat TNT? Biological treatment methods such as bioreactors, bioslurry treatment, and passive subsurface biobarriers have been proven successful in reducing TNT concentrations in soil (EPA 2005; CREEL 2006; ESTCP 2010). Composting has been proven in achieving cleanup goals for TNT at field demonstrations (EPA 2005). Low-temperature thermal desorption (LTTD) can be used on soil containing low concentrations of TNT (EPA 2005). Disposal of TNT has been accomplished by burning in an incinerator equipped with an afterburner and a scrubber (ATSDR 1995). Fenton oxidation and treatment with iron metal (Fe0) has been used to remediate TNT-contaminated soil and water (NCER 2010; EPA 2005). Other methods of treating waste waters include ultrafiltration, activated carbon, and resin adsorption (ATSDR 1995). Where can I find more information about TNT? Agency for Toxic Substances and Disease Registry (ATSDR). 1995. Toxicological Profile for TNT. www.atsdr.cdc.gov/toxprofiles/TP.asp?id=677&tid =125 Cold Regions Research and Engineering Laboratory (CREEL). 2006. Conceptual Model for the Transport of Energetic Residues from Surface Soil to Groundwater by Range Activities. ERDC/CRREL TR-06-18. www.dtic.mil/cgi-bin/GetTRDoc?Location=U2&doc=GetTRDoc.pdf& AD=ADA472270 CREEL. 2007a. Photochemical Degradation of Composition B and Its Components. ERDC/EL TR-07-16. AD=ADA472238 CREEL. 2007b. Protocols for Collection of Surface Soil Samples at Military Training and Testing Ranges for the Characterization of Energetic Munitions Constituents. ERDC/CRREL TR-07-10. www.dtic.mil/cgi-bin/GetTRDoc?Location= U2&doc=GetTRDoc.pdf&AD=ADA471045. Environmental Security Technology Certification Program (ESTCP). 2010. Passive Biobarrier for Treating Comingled Perchlorate and RDX in Groundwater at an Active Range (ER-1028). Massachusetts Military Reservation (MMR) 2001. Impact Area Groundwater Study Program. Chemical Fact Sheet – TNT. Fact Sheet 2001-05. cts/tnt.pdf Occupational Safety & Health Administration (OSHA). 1999. 2,4,6-Trinitrotoluene. H_274100.html U.S. Army. 2009. Military Munitions Response Program. Munitions Response Remedial Investigation/Feasibility Study Guidance. U.S. Army Corps of Engineers (USACE). 1997. Review of Fate and Transport Processes of Explosives. Installation Restoration Research Program. Technical Report IRRP-92-2. March 1997. trirrp97-2.pdf USACE. 2005. Military Munitions Center of Expertise. Technical Update. Munitions Constituent (MC) Sampling. U.S. Department of Defense (DoD). 2010. Emerging Chemical & Material Risks. MD/ECMR U.S. Environmental Protection Agency (EPA). 1999. Office of Research and Development. Federal Facilities Forum Issue. Field Sampling and Selecting On-site Analytical Methods for Explosives in Water. EPA-600-S-99-002. EPA. 2000. Office of Research and Development. Barringer Instruments. GC-IONSCAN. Environmental Technology Verification Report. EPA/600/R-00/046. EPA. 2001. Office of Research and Development. Research International, Inc. TNT Detection Technology. Texas Instruments Spreeta Sensor. Environmental Technology Verification Report. EPA/600/R-01/064. August 2001. EPA. 2005. EPA Handbook on the Management of Munitions Response Actions. EPA 505-B-01-001 www.epa.gov/fedfac/pdf/mra_hbook_5_05.pdf EPA. 2006. 8330b. Nitroaromatics, Nitramines, and Nitrate esters by High Performance Liquid Chromatography (HLPC) Revision 2. EPA. 2009. 2009 Edition of the Drinking Water Standards and Health Advisories. EPA. IRIS. 1993. 2,4,6-Trinitrotoluene (TNT) (CASRN 118-96-7). Last Revised 1993. EPA. National Center for Environmental Research (NCER). 2010. Final Report: Fate and Transport of Munitions Residues in Contaminated Soil. Website accessed on July 2, 2010. Contact Information If you have any questions or comments on this fact sheet, please contact: Mary Cooke, FFRRO, by phone at (703) 603-8712 or by e-mail at cooke.maryt@epa.gov. |
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8135 | https://www.wikidoc.org/index.php/Cardiac_tamponade_physical_examination | Cardiac tamponade physical examination - wikidoc
Cardiac tamponade physical examination
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Overview
Physical examination may vary depending on the type of cardiac tamponade. Physical examination may show the classic Beck's triad (hypotension, muffled heart sound, and elevated jugular venous distension), tachycardia, pulsus paradoxus, and pericardial rub depending on the type, and severity of tamponade. Initial diagnosis can be challenging, as there are a number of differential diagnoses, including tension pneumothorax, and acute heart failure.
Physical Examination
Appearance of the Patient
Beck's triad
The American thoracic surgeon, Beck, has described two clinical triad in 1930. The clinical triads are for both cardiac tamponade and pericardial constriction.
| Beck's Triad for Tamponade | Beck's Triad Pericardial Constriction |
:--- |
| ▸ Hypotension (due to decreased stroke volume) | ▸ Ascites |
| ▸ Jugular venous distension (due to impaired venous return to the heart) | ▸ High venous pressure |
| ▸ Muffled heart sounds (due to fluid inside the pericardium) | ▸ Small quiet heart |
Distension of veins in the forehead and scalp
Altered sensorium (decreasing Glasgow coma scale) due to hypotension
Peripheral edema due to high venous pressure
It should be noted that Beck's triad was descriped for the tamponade casued by acute intrapericardial hemorrhage, eg, traumatic causes of tamponade. This may not apply to other more slowly progressive forms of tamponade associated with a variety of medical conditions as opposed to surgical causes.
Vital Signs
Sinus tachycardia - common finding on physical examination and seen as an adaptive response of the body towards hypotension.
Pulsus paradoxus (a drop of > 10 mmHg in arterial blood pressure on inspiration)
Diminished peripheral pulses
Skin
Skin examination of patients with cardiac tamponade is usually normal.
HEENT
HEENT examination of patients with cardiac tamponade is usually normal.
Neck
Jugular venous distension
Carotid bruits may be auscultated unilaterally/bilaterally using the bell/diaphragm of the otoscope.
Lungs
Pulmonary crackles may be heared due to acute pulmonary edema.
Cardiovascular
Auscultation
In addition to the Beck's triad and pulsus paradoxus the following can be found on cardiovascular examination:
Pericardial rub
Clicks - As Ventricular volume shrinks disproportionately, there may be psuedoprolapse/true prolapse of mitral and/or tricuspid valvular structures that result in clicks.
Kussmaul's sign - Decrease in jugular venous pressure with inspiration is uncommon.
Pulsus Paradoxus
The following video depicts and explains the concept of pulsus paradoxus. {{#ev:youtube|jTsjCZ9QxW8}}
Abdomen
Abdominal examination of patients with cardiac tamponade is usually normal.
Back
Back examination of patients with cardiac tamponade is usually normal.
Genitourinary
Genitourinary examination of patients with cardiac tamponade is usually normal.
Neuromuscular
Neuromuscular examination of patients with cardiac tampnoade is usually normal.
Extremities
Extremities examination of patients with cardiac tamponade is usually normal.
References
↑Gwinnutt, C., Driscoll, P. (Eds) (2003) (2nd Ed.) Trauma Resuscitation: The Team Approach. Oxford: BIOS Scientific Publishers Ltd. ISBN 978-1859960097
↑Gwinnutt, C., Driscoll, P. (Eds) (2003) (2nd Ed.) Trauma Resuscitation: The Team Approach. Oxford: BIOS Scientific Publishers Ltd. ISBN 978-1859960097
↑Dolan, B., Holt, L. (2000). Accident & Emergency: Theory into practice. London: Bailliere Tindall ISBN 978-0702022395
↑Guberman, BA.; Fowler, NO.; Engel, PJ.; Gueron, M.; Allen, JM. (1981). "Cardiac tamponade in medical patients". Circulation. 64 (3): 633–40. PMID6455217.Unknown parameter |month= ignored (help)
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Acetylene gas C2H2 can be produced by the reaction of calcium carbide CaC2 with water according to the following balanced equation: CaC2( s)+2H2O(l)→Ca(OH)2( s)+C2H2(g) Determine the value of ΔrH∘ for this reaction using the thermochemical equations provided below:
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Question: Acetylene gas C2H2 can be produced by the reaction of calcium carbide CaC2 with water according to the following balanced equation: CaC2( s)+2H2O(l)→Ca(OH)2( s)+C2H2(g) Determine the value of ΔrH∘ for this reaction using the thermochemical equations provided below:
Show transcribed image text
There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 Required reaction:-
CaC A 2(s)+2 H A 2 O(l)⟶Ca(OH)A 2(s)+C A 2 H A 2(g)
Given elementary reactions:- View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question
Transcribed image text:
Acetylene gas C 2H 2 can be produced by the reaction of calcium carbide CaC 2 with water according to the following balanced equation: CaC 2(s)+2 H 2O(l)→Ca(OH)2(s)+C 2H 2(g) Determine the value of Δ rH∘ for this reaction using the thermochemical equations provided below: Ca(s)+2 C(g r)→CaC 2(s)Ca(s)+2 1O 2(g)→CaO(s)CaO(s)+H 2O(l)→Ca(OH)2(s)2 C 2H 2(g)+5 O 2(g)→4 CO 2(g)+2 H 2O(l)C(g r)+O 2(g)→CO 2(g)Δ rH∘=−62.8 kJ mol−1 Δ rH∘=−635.5 kJ mol−1 Δ rH∘=−65.2 kJ mol−1 Δ rH∘=−2600 kJ mol−1 Δ rH∘=−393.5 kJ mol−1
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8137 | https://www.youtube.com/watch?v=NrQbaHCKElk | How to Rotate and Scale Images (Linear Transformation)
Mr. STEM EDU TV
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Posted: 3 Apr 2020
Tutorial for rotating and scaling images, focusing on working principle. This includes the linear transformation in linear algebra.
Video content with timestamps.
0:00 - Intro
0:30 - Linear transformation matrix
0:57 - Linear transformation applications
1:15 - How to rotate images
2:49 - How to scale images
4:01 - Outro
Linear Transformation #Tutorial
Hope you enjoy the video, and please ask any questions/comments.
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3 comments
Transcript:
Intro In this tutorial video, I will explain a linear
transformation and its applications to geometric transformations through simple image rotation
and scale. First of all, the linear transformation principle
will be explained, and I will show you how to use it for the geometric transformations
through image rotation and scale. Linear transformation matrix The linear transformation is a linear operator,
transforming from one vector space to another. It is also known as a linear mapping. If you have a column vector x, you can transform
your column vector x to another column vector x prime through the linear operator A. The
linear operator A can be transformation matrix. The linear transformation is very useful mathematical
tools in many applications including image Linear transformation applications analysis and control theories so on. In this tutorial video, I will show you how
to use the transformation matrix in 2D image transformations. Let’s go over two examples of linear transformation
in 2D geometric transformations including rotation and scaling. How to rotate images If you have a column vector x is given in
2D plane y vs x, the coordinate of vector x can be rotated in a counter-clock direction
with the theta degree through the linear transformation. The transformation matrix A for the counter-clock-wise
rotation needs cos theta in diagonal elements and sin theta in off-diagonal elements. Once you multiply A by x, you can predict
the new position for the counter-clock-wise rotated coordinate, x prime. Let’s say we have a simple triangle geometry
in y vs x plane as blue points and lines. The coordinates for three corners are given
as (0,0), (1,2), and (3,1). These coordinates are my original column vector
x. The first row will be the x coordinates and
the second row will be the y coordinates. If you want to rotate the 90 degree in the
counter-clock direction, you can simply plug 90 degree into the theta, and your transformation
matrix A becomes 0, -1, 1, and 0. Once you multiply A by x, you can calculate
the new coordinates for the rotated triangle as (0, 0), (-2, 1), and (-1, 3), which are
shown as resulting x prime. This is graphically shown in red color. Let’s go over another example, which is
the 2D image scaling. How to scale images Again, we have a point x in 2D domain. For the image scale, the transformation matrix
A is given as the scaling parameter k in the diagonal elements, and 0 in the off-diagonal
elements. If k is larger than 1, the point x will stretch,
and if k is smaller than 1, it will shrink. Once you multiply matrix A by column vector
x, you get the x prime, which represents the scaled image. I have the same triangle here as the column
vector x in blue color. I use scaling factor k = 0.5 to shrink my
triangle, so my transformation matrix A becomes 0.5, 0, 0, and 0.5. Once you multiple A by x, your new coordinate
for the shrunken triangle becomes (0, 0), (0.5, 1), and (1.5, 0.5) which are shown in
the figure as red points and lines. Outro Hope this tutorial video is useful for you. If you like, please give thumbs up, and please
ask any questions/comments in the comment section below. I will see you in the next video. |
8138 | https://math.stackexchange.com/questions/4081004/perimiter-and-area-of-minkowski-sums | calculus - Perimiter and area of Minkowski sums - Mathematics Stack Exchange
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Perimiter and area of Minkowski sums
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Prove that perimiter of A+B is equal to the sum of perimiters of A and B, where + denotes Minkowski sum(or vector sum) and A, B are convex figures in a plane.
I was originally trying to prove this where one figure was a circle, but I figured that it shouldn't really make a difference. What I actually need is to find the area of A+B in terms of perimiters of A,B and their areas. I found the result, but I think i can prove the special case(where one figure is a circle) when I know the perimiter of their sum.
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asked Mar 28, 2021 at 22:13
Огњен ПетровОгњен Петров
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Without loss of generality, we can assume that the shapes have smooth boundaries and no flat edges. Polygons and other convex shapes that don't match this description can be approximated arbitrarily well by ones that do.
If a convex shape has a smooth boundary and no flat edges, we can define a differentiable function v⃗(θ)v→(θ) that maps directions in the plane to the point on the shape furthest in that direction. Further, for any two such shapes with maps v⃗1(θ),v⃗2(θ)v→1(θ),v→2(θ), their Minkowski sum has the map v⃗1+2(θ)=v⃗1(θ)+v⃗2(θ)v→1+2(θ)=v→1(θ)+v→2(θ). (Logically, to get the point farthest in direction θ θ in the Minkowski sum, we want to add together the points farthest in θ θ from shapes 1 and 2. This is true in general, we only require convexity and smoothness and no flat edges to make sure that v⃗1(θ)v→1(θ) and v⃗2(θ)v→2(θ) are differentiable.) We can compute the perimeter of a shape s s to be:
P s=∫2 π 0∣∣∣d v⃗s(θ)d θ∣∣∣d θ P s=∫0 2 π|d v→s(θ)d θ|d θ
So:
P 1+2=∫2 π 0∣∣∣d v⃗1(θ)d θ+d v⃗2(θ)d θ∣∣∣d θ P 1+2=∫0 2 π|d v→1(θ)d θ+d v→2(θ)d θ|d θ
Now, because v⃗1(θ)v→1(θ) is the point farthest in the θ θ direction, and the function v⃗1 v→1 is differentiable, we must have that d v⃗1(θ)d θ d v→1(θ)d θ is perpendicular to the direction θ θ. (If it weren't, we could walk along the edge of the shape a small distance, and get even farther in the direction θ θ, contradicting the definition that v⃗1(θ)v→1(θ) is the farthest point in that direction.) A similar argument applies for v⃗2 v→2.
So for a given value of θ θ, d v⃗1(θ)d θ d v→1(θ)d θ and d v⃗2(θ)d θ d v→2(θ)d θ are both perpendicular to the direction given by θ θ, and so must be pointing in the same direction as each other. Thus:
∣∣∣d v⃗1(θ)d θ+d v⃗2(θ)d θ∣∣∣=∣∣∣d v⃗1(θ)d θ∣∣∣+∣∣∣d v⃗2(θ)d θ∣∣∣|d v→1(θ)d θ+d v→2(θ)d θ|=|d v→1(θ)d θ|+|d v→2(θ)d θ|
So:
P 1+2=∫2 π 0(∣∣∣d v⃗1(θ)d θ∣∣∣+∣∣∣d v⃗2(θ)d θ∣∣∣)d θ=∫2 π 0∣∣∣d v⃗1(θ)d θ∣∣∣d θ+∫2 π 0∣∣∣d v⃗2(θ)d θ∣∣∣d θ=P 1+P 2 P 1+2=∫0 2 π(|d v→1(θ)d θ|+|d v→2(θ)d θ|)d θ=∫0 2 π|d v→1(θ)d θ|d θ+∫0 2 π|d v→2(θ)d θ|d θ=P 1+P 2
EDIT: I see you're also looking for an area formula. It's not possible to compute the area of the Minkowski sum only in terms of the area and perimeters of the two shapes. Consider the shapes being identical long thin rectangles as an example. If the rectangles are pointed the same way, the sum will have a small area, while if one is rotated by a right angle relative to the other, the sum will have a much larger area. But the component shapes have the same perimeter and area in both cases.
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edited Mar 28, 2021 at 23:05
answered Mar 28, 2021 at 22:58
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8139 | https://ohsonline.com/-/media/56a19b253139413eb0296a3058ad4dfa.pdf | Red Wing for Business redwingsafety.com White Paper STAND YOUR GROUND A guide to slip resistance in industrial safety footwear Navigating the standards Slip resistance requirements for safety footwear don’t exist in the same way they do for other personal protective equipment (PPE). There are a variety of methods for testing the slip resistance of footwear, and there are guidelines that manufacturers can choose to follow (or not) during the testing phase of product development. The current testing method for slip resistance in safety footwear is relatively unknown compared with its withdrawn predecessor. In addition, there is no standard which defines “slip resistant” backed by a regulatory body with enforcement power, such as OSHA or HSE.
The relevant testing standard set forth by the American Society of Testing Materials (ASTM) is ASTM F2913, “Standard Test Method for Measuring the Coefficient of Friction for Evaluation of Slip Performance of Footwear and Test Surfaces / Flooring Using a Whole Shoe Tester” (“Whole Shoe test”). The Whole Shoe test addresses the entire shoe and tests a variety of combinations of footwear materials and surfaces, including those that are regularly contaminated with slippery substances. As such, it offers the most comprehensive picture of slip resistance. The following text lays out the scope of the standard: This test method determines the dynamic coefficient of friction between footwear and floorings under reproducible laboratory conditions for evaluating relative slip performance. The method is applicable to all types of footwear, outsole units, heel top-pieces (top-lifts) and sheet soling materials.
In the Whole Shoe test, the footwear sole and underfoot surface are brought into contact, subjected to a specified vertical force for a short period of static contact and then moved horizontally relative to one another at a constant speed. The horizontal frictional force is measured after movement starts and the dynamic coefficient of friction is calculated.
Again, the Whole Shoe test is notable because it allows for reproducible testing of footwear and footwear-related soling materials with respect to slip resistance. Other ASTM test methods in this arena typically address flooring materials.
Another slip resistance method employers may be familiar with is ASTM F1677, known as the “Mark II test” (and also known as the “Brungraber test”) because it was performed with a portable inclined articulated strut slip tester (PIAST) developed by Dr. Robert Brungraber of Bucknell University. For many years, the Mark II test was an accepted standard for measuring slip resistance.
It is critical to note that the Mark II test was withdrawn by ASTM in 2006 and replaced with the Whole Shoe test in 2011, meaning the Mark II test is no longer supported by ASTM. The Whole Shoe test is not only more versatile, but also employs test equipment capable of precise calibration. It measures the coefficient of friction of the sole of the shoe mimicking more real world conditions than the Mark II test which was created to test flooring, not footwear. Employers accustomed to requesting Mark II test data from safety footwear providers will need to switch over to asking for Whole Shoe test data. A guide to slip resistance in industrial safety footwear When it comes to footwear, the topic of slip resistance is almost as treacherous as the slick floors, wet surfaces and uneven terrain employees encounter every day. Confusion exists on the current testing standard for slip resistance of safety footwear as well as how to evaluate slip resistant footwear. This puts employers in a difficult position because slips, trips and falls are among the most reported—and most costly—injuries in many industries.
Red Wing for Business redwingsafety.com | 2 Stand Your Ground: A Guide to Slip Resistance in Industrial Safety Footwear ASTM F2913, the Whole Shoe test, replaced ASTM F1677, the Mark II test, in 2011. Employers should request Whole Shoe test data when evaluating slip resistance. Exploring a slippery issue While slip resistance seems like a straightforward concept, measuring it for safety footwear is actually quite challenging. Additional language in the Whole Shoe test documentation illustrates some of the reasons why this is the case.
1. Surface contaminants include, but are not limited to, water, ice, oil, grease and other chemicals. These contaminants are among the most prevalent but employers have no real way of knowing how slip resistance changes if the contaminant is food, gravel, cleanser, mud, or construction material.
2. The test does not account for the risk of tripping due to footwear/ground interlock. This language refers to special-purpose footwear with spikes or metal studs. The key point is that when footwear is too slip resistant for the task, it can raise additional risk of injury.
3. The standard does not address all safety concerns. A variety of other factors may potentially affect the risk of slipping beyond footwear and flooring, including ambient temperature, lighting, surface irregularities, stairs and handrails, floor mats and runners, and human factors such as overall health, inattentive behavior, taking shortcuts or carrying objects that obstruct your view.
OSHA guidelines do not address slip resistance standards for safety footwear. Its safety footwear standards focus on impact and puncture resistance, emphasizing toe and metatarsal guards. There is only a small note that foot protection may be required when working on slippery surfaces, but slipperiness is not quantified.
In general, OSHA requires slip-resistant surfaces (not footwear) in certain work environments, but its standards do not clearly define slip resistance. One proposed standard specified a 0.5 coefficient of friction (CoF), but the requirement was never adopted.
The takeaway here is that, while friction can be measured between any two surfaces, the risk of an employee slipping involves many variables that are specific to the job and the quality of the footwear. For example, one type of safety footwear may provide adequate slip resistance under optimal conditions but fail if the surface is cold or wet. Another type might initially offer substantial slip resistance but quickly wear down due to poor manufacturing quality.
Evaluating slip resistance Even if there is no way to “check the box” on a well-known slip resistance rating because it doesn’t exist, employers understand that slip resistance delivers a long list of benefits for the company and its workforce. Preventing slips and falls reduces the risk of accidents and injuries, which has a carryover effect on overhead, insurance premiums, worker’s compensation and profitability, not to mention morale.
So how can employers make a solid assessment of slip resistance? It helps to start with an understanding of the tradeoffs involved.
1. Safety vs. cost. Bottom line, it’s unlikely that the least expensive safety footwear will meet the slip resistance requirements of the job. The slip resistance of any shoe or boot comes down to the integration of several anti-slip features, all of which may affect purchase price. Shortcuts, sloppy manufacturing techniques or low-quality materials could compromise any of these Red Wing for Business redwingsafety.com | 3 Stand Your Ground: A Guide to Slip Resistance in Industrial Safety Footwear features, as well as create durability issues.
2. Safety vs. comfort. Safety features have to exist in proper balance with comfort considerations. If safety footwear does not fit properly or is uncomfortable, employees may not wear them consistently, find ways to avoid wearing them, or wear a less-safe boot or shoe instead.
3. Safety vs. overdesign. Not every job requires maximum slip resistance. In fact, it is often the case that safety footwear can be too slip resistant for a specific environment. If the shoe or boot sticks to the work surface too much, employees run the risk of injuries similar to those experienced by athletes who hurt their knees when their cleats become stuck in the turf.
It is ideal to work with a safety footwear provider who understands these tradeoffs and can help advise your organization on the right product for the work environment. This can help individuals avoid choosing footwear that is slip resistant but unsuited to a specific job, such as a boot that works well in wet conditions but fails when coming into contact with material spills that clog the tread.
To help you ask the right questions, here are some general guidelines to consider when selecting the right slip resistance features in safety footwear: Outsole The outsole compound is critical for slip resistance. In general, soft rubber compounds offer the greatest slip resistance for environments contaminated with oil and grease. However, it is important to keep in mind that small changes in the compound can result in large variations in overall slip resistance. Tread Look for outsoles that channel oil and grease away so the outsole can reach the work surface more fully, as well as outsoles with split and solid lugs specifically designed to shed debris. Treads should allow a maximum amount of material to grip the floor. Softer soles and more tightly spaced treads are better suited to fluid contaminants and indoor environments. More widely spaced treads are generally better for handling solid contaminants in outdoor areas. It is vital to avoid clogging the tread. A wider or deeper tread pattern may be necessary if footwear needs to be cleaned too often. Over time, watch for worn or flattened soles.
Insoles Insoles do not play a direct role in slip resistance, but they should offer additional cushioning and impact padding for employees who are on their feet for extended periods or work predominantly on hard surfaces.
Uppers Uppers can be made of a long list of materials, including various leathers, suede, mesh and combinations of these. Like the insole, the upper does not directly determine slip resistance but plays a key role in fit and comfort. A shafted boot (e.g., a 6” or 8” tall boot) offers more ankle stability than a shorter boot (e.g., an oxford).
Midsole The midsole should provide ample support and stability. Again, midsole technologies do not have a direct bearing on slip resistance but are important for the overall structural integrity of footwear and the amount of comfort employees can expect throughout the day.
Durability Slip resistance performance may change over time as footwear is subjected to normal wear and tear. Employers should prioritize highly durable footwear that provides the most slip resistance for the longest period of time so the workforce can operate with confidence.
Red Wing for Business redwingsafety.com | 4 Stand Your Ground: A Guide to Slip Resistance in Industrial Safety Footwear Based in Red Wing, Minnesota, Red Wing Shoe Company has a long tradition of offering premium-quality safety footwear that incorporates the best features available to protect people on the job. Right now, our boots are hard at work in thousands of applications in more than 110 countries.
To learn more about all the ways Red Wing for Business can benefit your workers and your business, please visit redwingsafety.com.
Convenient for workers With more than 1,200 U.S. retail locations, 170 Mobile Shoe Stores and over 250 purpose-built designs, your workers won’t have to travel far to find the Red Wing footwear they need to do their jobs safely and productively. Simple for you Red Wing for Business makes it easy to honor your commitment to your workers’ well-being. From initial setup through ongoing support, our team of experts will be there to ensure a smooth experience and successful program for you.
RED WING FOR BUSINESS © 2019 Red Wing Shoe Company, Inc., Rev 01/19 Setting the standard Any safety footwear provider can claim its products are slip resistant, precisely because there is no uniform global standard or rating for slip resistance. For this reason, Red Wing Safety Footwear recommends three simple principles for evaluating slip resistance: 1. Follow the ASTM standard. ASTM F2913, or the Whole Shoe test, provides the best and most complete approach to evaluating how different footwear materials will interact with various work surfaces. All slip resistant footwear should be tested to this standard to establish a common basis for comparison. Again, the Mark II test has been withdrawn by ASTM and should not be used to evaluate safety footwear.
2. Understand the details. Whether or not a specific shoe or boot will provide adequate, long-term slip resistance depends on much more than one test standard. It requires a deep understanding of what the job actually involves: the surfaces, contaminants and physical requirements of the role. Look for a provider with experience developing purpose-built safety footwear worn in a variety of industries.
3. Share knowledge. Every work environment is unique. Ensuring slip resistance depends on a strong, open and transparent collaboration between employer and safety footwear provider. The footwear manufacturer should not only provide as much detail as possible about its products, but should also seek to understand every nuance of the work environment before making a recommendation.
Stand Your Ground: A Guide to Slip Resistance in Industrial Safety Footwear |
8140 | https://simple.wikipedia.org/wiki/Solar_mass | Solar mass - Simple English Wikipedia, the free encyclopedia
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From Simple English Wikipedia, the free encyclopedia
Solar mass is a unit of measurement of mass. It is equal to the mass of the Sun, about 332,950 times the mass of the Earth, or 1,048 times the mass of Jupiter. Masses of other stars and groups of stars are listed in terms of solar masses. Its mathematical symbol and value are:
M⊙=1.98892×10 30 kg{\displaystyle M_{\odot }=1.98892\times 10^{30}{\hbox{ kg}}} (kg being kilograms)
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8141 | https://www.quora.com/If-the-discriminant-of-a-quadratic-equation-is-a-perfect-square-the-coefficient-of-X2-is-1-and-the-coefficient-of-X-and-the-constant-are-integers-will-the-roots-always-be-in-integer-form | Something went wrong. Wait a moment and try again.
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5
If the discriminant of a quadratic equation is a perfect square, the coefficient of X2 is 1 and the coefficient of X and the constant are integers, will the roots always be in integer form?
Ashish Singh
Worked at Infosys
·
Author has 383 answers and 1.4M answer views
·
10y
I think yes, roots will be integers.
Lets see how .
let the equation be x^2 - bx + c , b and c are integers
now discriminant = b^2 - 4c
let square root of discriminant be Q
so roots are (-b+Q)/2 and (-b-Q)/2
// now watch closely
b^2 - 4c will be even or odd if b is even or odd respectively as 4c is even .
hence Q will be even or odd if b is even or odd respectively
in both of those cases -b + Q and - b - Q are divisible by 2
so roots will be integers .
// Hope it helps
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, What would be the quadratic equation with integer coefficients?
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Manisha Parmar
Studied at Shreemati Nathibai Damodar Thackersey Women's University
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10y
Originally Answered: If the discrimant of a quadratic equation is a perfect square and the coefficient of x2 is 1 and the coefficient of x and the constant are integers will the roots always be in integer form?
·
NO actually, because even if the Discriminant is a perfect square and all the cofficients are integers you need to guarantee that -b +/- D^0.5 is divisible by 2a . ONLY then the equation would have integral roots.
Discriminant being a perfect square would only ensure that the roots are rational.
Daniel Claydon
Learning mathematics
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Author has 779 answers and 4.3M answer views
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6y
Related
Can all numbers be expressed as the root of a polynomial with integer coefficients?
No.
First, let's dig in to some terminology. We call any such number that is the root of a finite, integer polynomial an algebraic number, sometimes denoted by A or ¯¯¯¯Q. Any number that is not algebraic is said to be transcendental.
How many integer polynomials are there?
This sounds like a dumb question — obviously there are infinitely many. But are there countably or uncountably many of them? To answer this, we start be recalling the following theorems:
The set Z is countable.
If A is any countable set, and B is the set of all n-tuples (a1,a2,…an) of
No.
First, let's dig in to some terminology. We call any such number that is the root of a finite, integer polynomial an algebraic number, sometimes denoted by A or ¯¯¯¯Q. Any number that is not algebraic is said to be transcendental.
How many integer polynomials are there?
This sounds like a dumb question — obviously there are infinitely many. But are there countably or uncountably many of them? To answer this, we start be recalling the following theorems:
The set Z is countable.
If A is any countable set, and B is the set of all n-tuples (a1,a2,…an) of elements in A, then B is also countable.
The union of countably many countable sets is countable.
How does this help?
Well, any integer polynomial of degree n can be represented as an (n+1)-tuple of integers, corresponding to its coefficients. Thus, by theorems 1 and 2, the number of these polynomials is countable. Since each polynomial contributes a finite number of roots, the number of algebraic numbers generated by degree n polynomials is countable. Then the total number of algebraic numbers is at most the union of the number generated by degree 1,2,3… polynomials, which by theorem 3 is countable.
This immediately answers your question. The set C of complex numbers is uncountable, therefore not all its elements can be algebraic, and the same applies for R. In fact, almost no numbers are algebraic.
Ok great, you might think, we've shown the existence of transcendental numbers, but do we have any actual examples?
Yes, we do. But there are no known simple proofs for any of them, so I'll just state some of what we know. The first class of numbers whose transcendence was proven are the Liouville Numbers, including Liouville's constant:
L=∞∑k=1110k!
Later on, it emerged that e≈2.72 is also transcendental. I just want to stress: this result is not at all surprising — in fact it would have been astonishing if it wasn't true — it's just that it's hard to prove. Moreover, we found that ea is transcendental whenever a is a nonzero algebraic number. Here's a cool piece of logic: since eiπ=−1 is algebraic, it follows that iπ is transcendental. If π was the root of some integer polynomial P(x), then π must be the root of some equation
a1xb1+⋯+anxbn=c1xd1+⋯cmxdm
where the exponents on the left side are all even and on the right side all odd. Squaring both sides, rearranging to sort the exponents that are 0 and 2 modulo 4, and squaring again shows π also satisfies some integer polynomial whose exponents are all multiples of 4. But then iπ satisfies this polynomial as well, so iπ is algebraic; a contradiction. Thus we conclude π must be transcendental.
We know sin(a),cos(a), and tan(a) are transcendental for any nonzero algebraic number a, as well as their reciprocals. Also Wn(a) is transcendental for any branch of the Lambert W Function and if we further require a≠1 then ln(a) is too. Also Γ(1/3),Γ(1/4), and Γ(1/6) are all known to be transcendental, as well as the Champernowne constant, 0.1234567891011….
Perhaps the “biggest” class of transcendental numbers comes from the Gelfond-Schneider Theorem, that tells us ab is transcendental whenever a≠0,1 is algebraic and b is an irrational algebraic number. This includes the Gelfond-Schneider constant, √2√2.
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Bot
·
1y
For a quadratic equation of the form:
x2+bx+c=0
the discriminant (D) is given by:
D=b2−4ac
In this case, since the coefficient of x2 is 1, we have:
D=b2−4c
If the discriminant is a perfect square, say k2, then we can express this as:
b2−4c=k2
Rearranging gives:
b2−k2=4c
This can be factored using the difference of squares:
(b−k)(b+k)=4c
Since c is an integer, 4c is also an integer, and thus both factors b−k and b+k must be even (because their product is divisible by 4). Therefore, we can write:
b−k=2mandb+k=2n
for some integers m and n. S
For a quadratic equation of the form:
x2+bx+c=0
the discriminant (D) is given by:
D=b2−4ac
In this case, since the coefficient of x2 is 1, we have:
D=b2−4c
If the discriminant is a perfect square, say k2, then we can express this as:
b2−4c=k2
Rearranging gives:
b2−k2=4c
This can be factored using the difference of squares:
(b−k)(b+k)=4c
Since c is an integer, 4c is also an integer, and thus both factors b−k and b+k must be even (because their product is divisible by 4). Therefore, we can write:
b−k=2mandb+k=2n
for some integers m and n. Substituting these into the equations gives:
2m⋅2n=4c⟹mn=c
Now, the roots of the quadratic equation can be found using the quadratic formula:
x=−b±√D2
Since D=k2 is a perfect square, we have:
x=−b±k2
For x to be an integer, −b±k must be even. Since b and k are both integers, −b and k will both be either odd or even together (since b and k both have the same parity). Hence, −b±k will indeed be even, and thus the roots will be integers.
Conclusion
Yes, if the discriminant of a quadratic equation is a perfect square, the coefficient of x2 is 1, and the coefficients of x and the constant term are integers, the roots will always be in integer form.
Related questions
If the coefficients of x^2 and the constant term have the same sign, and the coefficient of the x term is 0, then why does the quadratic equation have no real roots?
What is the definition of a quadratic equation? Are all quadratic equations with real coefficients always solvable?
The roots of the equation x2−2px+4q=0 are consecutive integers. What will the discriminant of this equation be?
If
1
+
√
2
is a root of quadratic equation with rational coefficients, what is the value of other root?
How do you find the zeroes and nature of roots of a quadratic equation if they exist using only its coefficients?
Robert Colburn
Studied Mathematics at Cabrillo College
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4y
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If a quadratic expression is positive for all real numbers, then its discriminant must be negative. Is it always true? if yes then give the source.
If a quadratic expression is positive for all real numbers, then its discriminant must be negative. Is it always true? if yes then give the source.
PLEASE UPVOTE if you find this helpful. THANK YOU.
The discriminant can tell you how many x-intercepts a quadratic function has.
If the function is always positive, then it has no x-intercepts (since the y-value on the x-axis is zero)
The discriminant plays a role in solving ax^2 + bx + c = 0
If the discriminant is greater than zero, than there are two distinct real solutions to ax^2 + bx + c = 0 and there are two distinct x-intercepts for the graph
If a quadratic expression is positive for all real numbers, then its discriminant must be negative. Is it always true? if yes then give the source.
PLEASE UPVOTE if you find this helpful. THANK YOU.
The discriminant can tell you how many x-intercepts a quadratic function has.
If the function is always positive, then it has no x-intercepts (since the y-value on the x-axis is zero)
The discriminant plays a role in solving ax^2 + bx + c = 0
If the discriminant is greater than zero, than there are two distinct real solutions to ax^2 + bx + c = 0 and there are two distinct x-intercepts for the graph.
If the discriminant equals zero, than there is one real solution to ax^2 + bx + c = 0 and there is one x-intercept for the graph.
If the discriminant is less than zero, than there are two distinct complex solutions to ax^2 + bx + c = 0 and there are no x-intercepts for the graph.
So if the discriminant is negative, then the quadratic is either always positive or always negative.
Therefore it is true that “If a quadratic expression is positive for all real numbers, then its discriminant must be negative.”
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
1y
Related
What is the process for finding a quadratic polynomial with an integer root?
A quadratic polynomial has 2 roots. Shall I just assume that you meant “
finding a quadratic polynomial with 2 integer roots”.
Do you get the idea yet?
A quadratic polynomial has 2 roots. Shall I just assume that you meant “
finding a quadratic polynomial with 2 integer roots”.
Do you get the idea yet?
Ramnath Takiar
Former Scientists G, NCRP at Indian Council of Medical Research (1978–2013)
·
Author has 1.9K answers and 2.7M answer views
·
6y
Related
Is it possible for a quadratic equation to have 2 integer roots but cannot be directly factored (I mean “just by factoring;” solving without using the quadratic formula)?
If there are two integer roots, then it is must that the equation can be factored. Consider that a and b are two solutions. This implies that (x - a) and (x - b) are factors of the given quadartic equation. In my opinion, it is always possible to factorize any quadartic equation if it has two integer roots. The difficulty may arise if the solutions are in fractions and if the equation has imaginery roots. Not otherwise.
We know that the general form of any quadartic equation is
In case of difficulty in factorizing, it is better to go for solution given by the equation:
In all possibilities, you g
If there are two integer roots, then it is must that the equation can be factored. Consider that a and b are two solutions. This implies that (x - a) and (x - b) are factors of the given quadartic equation. In my opinion, it is always possible to factorize any quadartic equation if it has two integer roots. The difficulty may arise if the solutions are in fractions and if the equation has imaginery roots. Not otherwise.
We know that the general form of any quadartic equation is
In case of difficulty in factorizing, it is better to go for solution given by the equation:
In all possibilities, you get the required solution. In conclusion, it is always possible to factorize if there are integer solutions. If you know any such equation where the solution is in integers and you are not able to factorize, please let us know.
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Boris Sinaga
Founder of SP Learning Center
·
Author has 105 answers and 222.5K answer views
·
5y
Related
In a quadratic equation, where all the coefficients are integers, what are all the values its Discriminant CANNOT assume, despite the roots or any relation among the coefficients?(23 is one example). Where do I find references about it?
If I understand your question correctly, then we need to find all integers that can't be or impossible to be the determinant of a quadratic function when all of the coefficients are integer.
So the formula for Determinant is
D=b2−4ac
In this special case a,b and c are integers. We can also write the equation like this
b2−D=4ac
From this we can see that if we substract the value of a possible D from b2 the result should be divisible by 4. So for any impossible D, b2−D should not be divisible by 4. For now let's just assume that D is a positive integer.
We can use the unique property of the square
If I understand your question correctly, then we need to find all integers that can't be or impossible to be the determinant of a quadratic function when all of the coefficients are integer.
So the formula for Determinant is
D=b2−4ac
In this special case a,b and c are integers. We can also write the equation like this
b2−D=4ac
From this we can see that if we substract the value of a possible D from b2 the result should be divisible by 4. So for any impossible D, b2−D should not be divisible by 4. For now let's just assume that D is a positive integer.
We can use the unique property of the square numbers and of multiple of 4s to determine wether b2−D is divisible by 4 or not.
As you might have known any square numbers can be constructed as a sum of series of odd numbers.
x2=x∑i=1(2i−1)
For example
12=1
22=1+3
32=1+3+5
42=1+3+5+7
52=1+3+5+7+9
…
And any multiple of 4 can be written as the sum of a pair of consecutive odd numbers.
4=1+3
8=3+5
12=5+7
16=7+9
20=9+11
We can use these properties to determine wether b2−D is divisible by 4 or not. Suppose that D=5 and b=5. So we want to substract 5 from 25 and see wether the result is a multiple of 4 or not. Since 52 is
1+3+5+7+9
then if we substract 5 from the sequence, we'll be left with
1+3+7+9
And we can say for certain that it is divisible by 4 because we are left with 2 pairs of consecutive odd numbers that is 1+3 and 7+9.
In fact we can say for sure that for any odd square number when substracted by 5 the remainder will be divisible by 4.
So if we want the result not to be divisible by 4 then we just have to substract numbers that will leave the remaining sequence with some numbers that don't have a consecutive pair.
So from previous example if we instead substract 3 then we'll be left with
1+5+7+9
Here we have the number 1 without its consecutive odd number 3, so we can say for sure that the result is not divisible by 4.
From this can be shown that if we take every other odd number from the sequence we can be sure that we'll have at least one number without a consecutive odd number pair.
1+(5+7)+(9+…
(1+3)+5+(9+…
So we find our first batch of impossible D that is
3,7,11,15,19,23,27,…
If we substract any of these numbers from any square numbers, the results will not be divisible by 4.
Now, another property of multiple of 4 is that it can be written as sum of every other odd numbers and 1.
4=1+3
8=1+7
12=1+11
16=1+15
20=1+19
…
We can use this property to determine even D.
Notice that between 1 and the other odd number that make up a multiples of 4, there's always a gap of an even number of odd numbers. Between 1 and 3 there is 0 pair of consecutive odd numbers (But 1 and 3 themselves are consecutive odd number pair). Between 1 and 7, 1 pair. Between 1 and 11, 2 pairs. And so on.
It means that if we substract a multiple of 4 from any square numbers we can be sure that for every other square number the results will be divisible by 4. So we can rule these out.
Instead we want to have an odd number of odd numbers between 1 and the odd numbers that make up our even numbers, so we have
6=1+5(3 in between)
10=1+9(3,5,7 in between)
14=1+13(3,5,7,9,11 in between)
…
so that if we substract any of this number from the sequence, we can be sure that at least 1 odd number doesn't have a consecutive pair.
So we have our next batch of impossible D, this time even numbers:
6,10,14,18,22,26,…
While when D is 2 we can see if we substract this from any square number greater than 1, we'll have
2+(5+7)+(9+…
This is obviously not divisible by 4, because beside of any other consecutive odd number in the sequence, we also have a single 2 in the sequence that is not divisible by 4. So D=2 is impossible.
So there we have it. For any quadratic functions where all the coefficients are integers, the following odd numbers cannot be the determinant
3,7,11,15,19,23,27,…
and the following even numbers also cannot be the determinant
2,6,10,14,18,22,26,…
With the same procedure (or by noticing the pattern) we can also say that these negative numbers cannot be the determinant:
−1,−5,−9,−13,−17,−21,−25…
−2,−6,−10,−14,−18,−22,−26…
Peter Butcher
Former Professional Engineer, Teacher, Tutor
·
Author has 1.4K answers and 4.6M answer views
·
2y
Related
What is the definition of a quadratic equation? Are there any quadratic equations whose coefficients are all integers and whose solutions are both positive and negative integers?
Question
What is the definition of a quadratic equation? Are there any quadratic equations whose coefficients are all integers and whose solutions are both positive and negative integers?
Answer
Definition:
In maths a quadratic polynomial is a polynomial of degree two in one or more variables.
A quadratic function is the polynomial function defined by a quadratic polynomial.
Before the 20th century, the distinction was unclear between a polynomial and its associated polynomial function; so "quadratic polynomial" and "quadratic function" were almost synonymous.
This is still the case in many elementar
Question
What is the definition of a quadratic equation? Are there any quadratic equations whose coefficients are all integers and whose solutions are both positive and negative integers?
Answer
Definition:
In maths a quadratic polynomial is a polynomial of degree two in one or more variables.
A quadratic function is the polynomial function defined by a quadratic polynomial.
Before the 20th century, the distinction was unclear between a polynomial and its associated polynomial function; so "quadratic polynomial" and "quadratic function" were almost synonymous.
This is still the case in many elementary courses, where both terms are often abbreviated as "quadratic".
The graph above shows a quadratic polynomial with two real roots (crossings of the x axis) and hence no Complex Roots. In your question you use the term ,”Solution ” I've used the term “Root” which has the same meaning as “Solution” here.
Some other quadratic polynomials have their minimum above the x axis, in which case there are no real roots and two complex roots.
For example, a univariate (single-variable) quadratic function has the form:
f(x) = ax² + bx + c where a ≠ 0 and b and c can be any real number (which includes being an integer). x is its variable.
The graph of a univariate quadratic function is a parabola which is a curve that has an axis of symmetry parallel to the y-axis.
If a quadratic function is equal to zero, then the result is a Quadratic Equation
The solutions of a quadratic equation are the zeroes of the corresponding quadratic function.
Quadratic equations whose coefficients are all integers and whose slutions are both positive and negative integers
Let's consider one form of a quadratic equation
f(x) = ( x + A)(x - B) = 0, where (in this case) A and B are integers and A and B are the roots of the quadratic
Specifically one root is x = - A and the second root is x = + B.
This means that if we substitute x = B into the function or x = - A the function equals zero.
Let's multiply f(x) = ( x + A)(x - B) = 0 out.
It becomes (f(x) = x² + Ax - Bx - AB = 0 or
f(x) = x² + (A - B)x - AB = 0
IF we started with A and B as integers, then (A - B) and -AB are still integers with one solution negative (- A) and one positive (+B)
Some real examples
Consider f(x) = (x - 1)(x + 2) = 0
It has 2 solutions, being x = +1 and x = -2.
Let's expand f(x)
,f(x) = x² - x + 2x - 2 = 0 or
f(x) = x² + x - 2 = 0
Note:
All the coefficients are integers
There is one positive solution and one negative solution.
We could also have had both solutions positive or both negative if we wished.
The graph I used above had one positive and one negative solution (+2 and -1)
Finally, what about our equation f(x) = x² + x - 2 = 0
Consider the graph below:
It has 2 solutions, being x = +1 and x = -2 and the equation (upper left hand corner of the graph) has integer coefficients.
Conclusion
There are thousands of examples where quadratics have integer coefficients and both positive and negative solutions.
I chose a simple example here.
If unsure, try some examples yourself and create your own quadratic equations and solutions.
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Kermit Rose
Studied Mathematics & Statistics (academic discipline) at Florida State University School
·
Author has 963 answers and 651K answer views
·
2y
Related
How do you determine if a quadratic equation has two solutions or no solution in integers?
When does the quadratic equation ax^2 + bx + c have integer solutions?
Assume that a, b, and c are integers. If they were rational numbers, then the equation could be converted to integer coefficients by multiplying by the least common multiple.
If the equation is not equivalent to one equation with integer coefficients, then there are no integer solutions.
If b^2 - 4ac is < 0, then there are no real solutions, and therefore, no integer solutions.
If b^2 - 4ac = 0, then the solutions are -(b/a)/(c/a) = -b/c
(x+b/c)^2 = x^2 + 2b/c x + b^2/c^2 = 0
(c/2)x^2 + bx + (b^2)/(2c) = 0
If b^2 - 4a
When does the quadratic equation ax^2 + bx + c have integer solutions?
Assume that a, b, and c are integers. If they were rational numbers, then the equation could be converted to integer coefficients by multiplying by the least common multiple.
If the equation is not equivalent to one equation with integer coefficients, then there are no integer solutions.
If b^2 - 4ac is < 0, then there are no real solutions, and therefore, no integer solutions.
If b^2 - 4ac = 0, then the solutions are -(b/a)/(c/a) = -b/c
(x+b/c)^2 = x^2 + 2b/c x + b^2/c^2 = 0
(c/2)x^2 + bx + (b^2)/(2c) = 0
If b^2 - 4ac = w^2 is the square of an integer, and a and c are integers,
and both (w+b) and (w-b) are divisible by (2a) then there are two integer solutions.
If exactly one of (w+b) and (w-b) is divisible by (2a) then there is one integer solution and one rational number solution.
b^2 - w^2 = 4ac,
and (b-w) (b+w) = 4ac.
set x1 = (-b-w)/(2a) and x2 = (-b+w)/(2a)
(x - (-b-w)/(2a) ) (x - (-b+w)/(2a) ) =
x^2 -((-b)/(a))x + (b^2-w^2)/(2a)^2 =
x^2 + ((b)/(a))x + (4ac)/(2a)^2 = 0
x^2 + (b/a) x + (c/a) = 0
Multiply by the value of a.
ax^2 + b x + c = 0
Max Gretinski
Studied Mathematics
·
Author has 6.5K answers and 2.5M answer views
·
2y
Related
What are the real values of a for which the quadratic equation x^2 + ax + 6a = 0 has only distinct integer roots ?
Begin with x2 + ax + 6a = 0
The discriminant of the quadratic is Δ = a2 - 4(1)(6a) = a2 - 24a = a(a - 24).
Now, in order for the equation to have distinct rational roots, the discriminant must be a positive perfect square: 1; 4; 9; 16; etc..
If we wish, we may also write… a2 - 24a = (a−12)2 - 144 or (a−12)2 -(12)2
Since Δ > 0 is necessary, we observe that either a < 0 or a > 24. Any value of a between those two numbers yields a negative discriminant.
Since the question does not limit a to the integers or rational numbers, there are infinitely many solutions.
For example, set
a2
Begin with x2 + ax + 6a = 0
The discriminant of the quadratic is Δ = a2 - 4(1)(6a) = a2 - 24a = a(a - 24).
Now, in order for the equation to have distinct rational roots, the discriminant must be a positive perfect square: 1; 4; 9; 16; etc..
If we wish, we may also write… a2 - 24a = (a−12)2 - 144 or (a−12)2 -(12)2
Since Δ > 0 is necessary, we observe that either a < 0 or a > 24. Any value of a between those two numbers yields a negative discriminant.
Since the question does not limit a to the integers or rational numbers, there are infinitely many solutions.
For example, set
a2 - 24a = 1
and we learn
a2 - 24a - 1 = 0
when a = 12 ±√145
In that case, a - 24 = -12 ±√145
so that a(a - 24) = 145 - 144 = 1
We can likewise set
a2 - 24a = 4 (the next perfect square)
and obtain
a = 12 ±√148
so that a(a - 24) = 148 - 144 = 4
We may continue in this fashion forever.
a = 12 ±√153a = 12 ±√160a = 12±√169 → a = 25 or a = -1 → p(x) = x2 + 25x + 150, or p(x) = x2 - x - 6
a = 12 ±√180a = 12±√193a = 12 ±√208a = 12 ±√225 → a = 27 or a = -3
a = 12 ±√244
a = 12 ±√265
a = 12 ±√288
a = 12 ±√313
a = 12 ±√340
a = 12 ±√369a = 12 ±√400 → a = 32 or a = -8
We later discover
a = 49 or a = -25
also yields integer solutions.
Any time two numbers that differ by 24 are both squares (25 and 1, 49 and 25), and in certain other situations (27 and 3, 32 and 8), the product of the integers will be a square.
In the case of
p(x) = x2 + 25x + 150 = (x + 10)(x + 15)
p(x) = x2 - x - 6 = (x - 3)(x + 2)
p(x) = x2 + 27x + 162 = (x + 9)(x + 18)
p(x) = x2 - 3x - 18 = (x - 6)(x + 3)
p(x) = x2 + 32x + 192 = (x + 8)(x + 24)
p(x) = x2 - 8x - 48 = (x - 12)(x + 4)
p(x) = x2 + 49x + 294 = (x + 7)(x + 42)
p(x) = x2 - 25x - 150 = (x - 30)(x + 5)
Notice that the factors x + 2 to x + 5 and x + 7 through x + 10 appear.
Anurag Roy
B Tech in Metallurgy and Materials Engineering, Jadavpur University (Graduated 2025)
·
3y
Related
One root of a quadratic equation is square of the other root & the discriminant is
900
. What is the quadratic equation with integer coefficients?
Here according to the question we get two such kinds of quadratic equation satisfying the above condition..
P.S:The question in particular can be solved in 1 page(maximum) but for better clarity and sake of understanding has been solved elaborately…
Thanks a lot…
Here according to the question we get two such kinds of quadratic equation satisfying the above condition..
P.S:The question in particular can be solved in 1 page(maximum) but for better clarity and sake of understanding has been solved elaborately…
Thanks a lot…
Edward M Young
Dermatologist at Veteran's Administration Hospital (2015–present)
·
Author has 1.3K answers and 268K answer views
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1y
Related
How many complex values can be generated from the roots of a quadratic equation with integer coefficients?
At most, two complex conjugate roots of the form (p + qi) and (p - qi), where i = square root of -1, can be generated from the roots of a quadratic equation with integer coefficients. This occurs whenever the Discriminant (D = b^2 - 4ac) is negative.
Example: Quadratic Q: y = x^2 - 2x + 13 = 0.
The two roots of Q, x1 and x2, are easily found by using the quadratic equation formula with (a, b, c) = (1, -2, 13):
x1 = (1/2)(2 + sqrt [4 - 52]) = (1/2)(2 + sqrt [-48]) =
(1/2)(2 + 4i sqrt3) = 1 + 2i sqrt3 = 1 + 3.464(i)
x2 = 1 - 2i(sqrt3) = 1 - 3.464(i)
where i is defined as the square root of -1
At most, two complex conjugate roots of the form (p + qi) and (p - qi), where i = square root of -1, can be generated from the roots of a quadratic equation with integer coefficients. This occurs whenever the Discriminant (D = b^2 - 4ac) is negative.
Example: Quadratic Q: y = x^2 - 2x + 13 = 0.
Solution:
The two roots of Q, x1 and x2, are easily found by using the quadratic equation formula with (a, b, c) = (1, -2, 13):
x1 = (1/2)(2 + sqrt [4 - 52]) = (1/2)(2 + sqrt [-48]) =
(1/2)(2 + 4i sqrt3) = 1 + 2i sqrt3 = 1 + 3.464(i)
x2 = 1 - 2i(sqrt3) = 1 - 3.464(i)
where i is defined as the square root of -1 or as i^2 = -1.
Proof: plug roots x1 then x2 into Q:
x1: 1 + 4i sqrt3 -12 -2 -4i sqrt3 + 13 = 1 - 14 + 13 = 0.
x2: 1 - 4i sqrt3 - 12 - 2 + 4i sqrt3 + 13 = 1 - 14 + 13 = 0.
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8142 | https://math.stackexchange.com/questions/1859652/where-do-you-see-cyclic-quadrilaterals-in-real-life | Skip to main content
Where do you see cyclic quadrilaterals in real life?
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I've just been studying cyclic quads in geometry at school and I'm thinking see seems pretty interesting, but where would I actually find these in the real world? They seem pretty useless to me...
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edited Jul 14, 2016 at 20:17
mr-matt
asked Jul 14, 2016 at 20:03
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You may or may not know this... but your school is a part of the real world. When you learn in school, you learn for real
– Ben Grossmann
Commented
Jul 14, 2016 at 20:09
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I realise that, but right now it seems as though the only use for learning cyclic quadrilaterals is so that you can come back later and teach it, just a pointless loop...
– mr-matt
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Jul 14, 2016 at 20:10
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Well, at one time they were pretty useful, hardcore applied math even. In the Almagest, Ptolemy uses the result we now know as Ptolemy's Theorem as an aid for computing his table of chords, an important tool for astronomical calculations.
– André Nicolas
Commented
Jul 14, 2016 at 20:12
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I am going to ask, when do plan to use what you learned about Shakespeare's Tragedies, Ancient History, or Chemistry "in the real world." Only in math class to people ever ask, "When am I going to use this." And you are missing the point. Yes, frequently math is useful. But you learn math to learn how to think logically -- to apply reason and rationality.
– Doug M
Commented
Jul 14, 2016 at 20:13
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Um... you don't? They occur every time you inscribe something in a circle! So constructing efficient habitats in space, beehives and tectonic pressure plates, efficiency in resource allocations within city bounderies, hubcaps, steam railroad wheel ratios, architecture, trigonometry and surveying, the list is endless.
– fleablood
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I can't think of any applications, and I doubt any satisfactory ones exist - for example, as noted in the comments there may well have been connections to astronomy, but I think it's fair to suggest that almost no-one who is being taught circle theorems is going to use them in their life at any point.
Thus, I'm going to interpret this question as:
Why would you bother learning a theorem that has no application in real life?
And I think there are two good answers to this question:
1. It is interesting
This is, really, the only reason you're taught anything in your life other than how to pay taxes. Geometry is something that lots of people over a long period of time have found to be intrinsically interesting. The reasons for this are complicated - it's a good intellectual exercise, and for many people intellectual exercises are something they enjoy doing.
2. It forces you to think logically
The patterns of thought people generally use in mathematics are valuable. Logical arguments are important in all walks of life, and being able to understand and interpret them is an extremely valuable life skill which you really should want to have.
I have a lot of sympathy with this question, for the following reason: you are probably taught mathematics very badly. The arguments I give above really rely on the idea that you are taught how to prove theorems (and Euclidean geometry is a fantastic exercise in proof). Without that, I would claim that learning geometry really has no value. I would even go so far as to say you shouldn't bother going so far as to learn basic trigonometry (unless you need it to be an engineer or something), unless you study its proof. That really is where all the value, and all the fun, is.
This is not your fault. But there is something you can do about it. Look up a proof, try to understand it, and if you're lucky you'll get a little intellectual buzz from the 'aha!' moment of it all coming together. But, I'm sorry to say, you'll probably have to do this yourself. Mathematics teaching is woeful in the vast majority of schools, and statistically speaking you are unlikely to even have a teacher capable of explaining to you why these results are true, let alone interesting or useful.
So, on the off chance that this answer has spiked your curiosity, I recommend writing another question, called "How do you prove interesting facts about cyclic quadrilaterals?", and you might get a more satisfying answer.
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answered Jul 14, 2016 at 20:32
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When I was working in math fields a while ago and my students approached me with the same question I would answer that math science is 200 years ahead from physical world. Who know what we may come across in nature but by the time discover new phenomena, we might try mathematical apparatus already available to us to describe it. Not to mention, math is abstract, it flexes your mind, teaches you critical thinking. Things like Sufficient vs. Necessary conditions are all over our lives. I find that a lot of people do not understand this.
– AstroSharp
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Jul 14, 2016 at 22:25
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@AstroSharp I see what you're saying, but I don't think I agree. I don't think anyone seriously does mathematics with the hope of "setting up the future". I prefer to emphasise the very real fact that maths is a good exercise in logical thinking, which is good for enriching one as a human being.
– preferred_anon
Commented
Jul 14, 2016 at 23:15
@AstroSharp: all of these circle-theorems are a lot more than 200 years old though, so while your first argument might help to justify academic study of number theory (used for crypto) or higher-dimensional manifolds (used for physics), it's not relevant for 2000-ish-year-old geometry. There should be (and are) some applications of that visible by now, but the trouble comes in convincing a high-school student that those applications will occur in their personal life :-)
– Steve Jessop
Commented
Jul 15, 2016 at 9:28
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That is to say, you have a great argument for persuading a funding body not to entirely abandon pure mathematics on principle, but schoolkids probably don't expect to be physicists in 200 years time!
– Steve Jessop
Commented
Jul 15, 2016 at 9:44
I'm pretty sure back when my mother taught me to look left and right before crossing the street, she neither thought it was an interesting thing to learn, nor did she think about the taxes I might not be able to pay later if I didn't do it.
– celtschk
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Jul 15, 2016 at 10:28
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Theorem 3 of the Bern-Eppstein paper cited below proves that any polygon of n vertices
may be partitioned into O(n) cyclic quadrilaterals.
A hint of how this might be achieved can be glimpsed in the figure below, where
all the white quadrilaterals are cyclic.
Fig.5 from cited paper.
Quadrilateral meshing is important in many applications. The cyclic
quads produced by their algorithm have desirable "quality" characteristics.
Bern, Marshall, and David Eppstein. "Quadrilateral meshing by circle packing." International Journal of Computational Geometry & Applications 10.04 (2000): 347-360. (Pre-journal arXiv abstract.)
(Journal link.)
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Joseph O'RourkeJoseph O'Rourke
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Ptolemy's theorem says that if a,b,c,d are the lengths of the sides of a cyclic quadrilateral, with a opposite c and b opposite d, and e,f are the diagonals, then ac+bd=ef. In the second century AD, Ptolemy used that to prove identities that today we would express as
sin(a+b)cos(a+b)=sinacosb+cosasinb,=cosacosb−sinasinb.
As to where these come up in Reality, you can start with this:
PS: A bit more on what Ptolemy did:
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answered Jul 14, 2016 at 21:18
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8143 | https://www.youtube.com/watch?v=8ykEIVYdAGM | OpenStax: Algebra and Trigonometry - Chapter 2, Section 5 | Quadratic Equations
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Posted: 12 Jul 2022
Welcome to Huzefa’s explanation video of OpenStax Algebra and Trigonometry textbook. This is a full walkthrough of Chapter 2, Equations and Inequalities, Section 5, Quadratic Equations. Watch Huzefa as he reviews exercises 1-57 odd.
Credit:
Link to Exercises:
To skip to a particular question, use the chapters below:
00:00 Introduction
00:15 Exercise 1
00:30 Exercise 3
01:23 Exercise 5
02:36 Exercise 7
03:07 Exercise 9
04:35 Exercise 11
05:14 Exercise 13
06:08 Exercise 15
06:49 Exercise 17
07:17 Exercise 19
07:36 Exercise 21
08:05 Exercise 23
08:46 Exercise 25
10:51 Exercise 27
11:53 Exercise 29
14:05 Exercise 31
15:37 Exercise 33
16:21 Exercise 35
16:59 Exercise 37
17:38 Exercise 39
18:48 Exercise 41
19:39 Exercise 43
20:55 Exercise 45
21:34 Exercise 47
22:05 Exercise 49
24:42 Exercise 51
26:56 Exercise 53
28:03 Exercise 55
29:53 Exercise 57
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Transcript:
Introduction [Music] what's up everybody and welcome back to open stacks algebra and trigonometry chapter 2 section 5 quadratic equations let's do it how do we recognize when an equation is quadratic that's when the highest degree Exercise 1 is 2. so for example if we had x squared minus 4 equals 10 this would be a quadratic equation because the highest degree is two when we solve a quadratic equation by factoring why do we move all the Exercise 3 terms to one side having zero on the other side so the reason why we do that is because when we're solving a quadratic equation we're factoring we're breaking it up into two different binomials and the zero product rule comes in really handy here because let's say i have i factored into x minus seven and x plus three it becomes a much simpler equation because if it's two things multiplying each other and we're trying to get a result of zero as long as this one piece equals zero the entire function goes to zero likewise if this one piece equals zero the entire thing goes to zero because anything times zero is zero so what makes this binomial go to zero seven minus seven so when x equals seven what makes this one go to zero when x equals negative three negative three plus three is zero and that's how we get our solutions describe two scenarios we're Exercise 5 using the square root property to solve a quadratic equation would be the most efficient method so the first would be if we simply had x squared equals 64. in this case i can just take the square root of both sides and get x equals plus or minus eight don't forget when we're solving in an equation context we need to acknowledge that the square root of 64 is both positive and negative eight because both when multiplied by themselves gives 64. another example is if we had a perfect binomial square like this isolated and then equal to let's say 16. so again here instead of foiling out and setting equal to zero and then factoring we could just take the square root of both sides and we get x plus eight equals plus or minus four then by subtracting eight from both sides for both the negative four and four we get two solutions right negative four minus eight would give us negative 12 and then negative 8 plus 4 would give us negative 4. and then if you plug these back into the original you'll see it works you get negative 4 plus 8 which is 4 4 squared is 16 negative 12 plus 8 is negative 4 negative 4 squared is 16 boom done all right to solve this equation by factoring first we're going to break it into two binomials and we're going to Exercise 7 figure out what the binomials are as follows what two numbers multiply to 18 and add to negative nine that is going to be negative three and negative six right multiplies to eighteen adds to negative nine now what are the solutions what makes this go to zero positive three what makes this go to zero positive six those are your solutions done in this one we have a leading coefficient which means we have to use Exercise 9 the star method to factor so again this is my a b and c i put the a up here and here that would be 6 and 6. i put the b down here which is 17 and then i put a times c 6 times 5 up here which is 30. what your numbers multiply to 30 and add to 17 that would be 2 and 15. then we're going to reduce these like fractions 6 over 15 reduces to 2 over 5 divide both by 3 and then 6 over 2 divide both by 2 and we get 3 over 1. so this is how it factors into three x plus one that's these parts times two x plus five and this still equals zero so now the question is what makes this go to zero and what makes this go to zero so if you're not sure how to do this mentally you can always set it equal to zero like so and then solve for x so you can just say like okay i'm going to subtract one from both sides and then i'm going to divide it by three so we got negative one third but the way to do it mentally if you want is we're trying to get this to turn into negative five so if i multiply two times negative five over two that will give me negative five so it's kind of like you take this number over this and then negate it if it's a plus so the other one would be negative five halves these are the solutions done the first thing Exercise 11 i'm gonna do is i'm gonna factor out a gcf which is three both are divisible by three so then i got x squared minus 25 right because 75 divided by 3 is 25. then i've got a difference of squares so check this out this is really nice i can do x minus 5 times x plus 5 equals zero okay now we got a simple task what makes this go to zero positive five what makes this go to zero negative five those are the two solutions done now by the way we could also solve by adding 75 to both sides dividing by 3 and then just taking the square root it's an alternative method whatever you prefer so for this one once again we're asked to solve by factoring so we're going to follow the same protocol we're going to Exercise 13 subtract at 9 from both sides and set everything equal to 0. now what you'll notice is this is a difference of squares which means this is a perfect square and this is a perfect square and it's subtraction meaning difference as such we're going to factor it as follows we're going to factor the square root of 2x 4x squared which is 2x plus that here and here i'm going to take the square root of 9 which is 3 place that here and here and then alternate with plus minus and then we're going to say what makes this go to zero remember my little trick it's going to be this value over that value but then negated and then it's going to be this value which is actually negative 3 but i'm going to write positive 3. so it's going to be negative 3 over 2 but then negated which is going to make it positive 3 halves so here are your solutions boom done all right here i'm going to set everything equal to 0 first so i'm going Exercise 15 to subtract 5x from both sides i'm going to subtract 30 from both sides and i'm going to set it equal to 0. then i'm going to divide everything by five i don't even need to factor i can just divide because i can divide both sides and that zero divided by five is zero so it just goes away so now i have x squared minus x minus six because 30 divided by five is six now i can factor so again we use that idea what multiplies to negative six adds to negative one that would be negative three and positive two so now my solutions are what makes this go to zero which is three what makes this go to zero which is negative two boom done Exercise 17 so in this case what we're gonna do is we're gonna factor out the x so this is when there's no c term so i'm gonna factor out the x and then i've got seven x plus three and we've basically got it split up into two pieces which is what we want so let me say what makes this go to zero that's just zero and then what makes this whole thing go to zero that's negative three sevenths again i take this divided by this and flip it since it's a plus so those are my two answers Exercise 19 done so on this one we're using the square root property which is pretty straightforward we just take the square root of both sides boom boom and just don't forget we're taking plus or minus so square root of 36 is six but it's plus or minus six those are the two two solutions positive six negative six done so in this case i've Exercise 21 got a square and a square i'm going to square both sides or square root both sides excuse me square root of x minus 1 squared is x minus 1. square root of 25 squared is plus or minus 5. so that gives me two equations x minus 1 equals 5 x minus 1 equals negative 5. then i'm going to add 1 to both sides and i get 6 add 1 to both sides and negative 5 plus 1 is negative 4 those are my two solutions done here i'm going to square root both sides Exercise 23 and i get 2x plus 1 equals plus or minus three because once again square root of nine we have to take the positive and the negative root so i'll break this up into two separate equations equals positive three and negative three and then we'll solve subtract one from both sides two x equals two divide both sides by two and we get x equals one over here we're gonna subtract one we get two x equals negative four then we divide both sides by two and we get x equals negative two those are the two solutions done here we're going to solve by completing the square so i'm going to show you the steps for completing the Exercise 25 square so first i'm going to add 22 to both sides so i got x squared minus 9x equals 22. then what i'm going to do is i'm going to take half of this middle term this x term the b term and half of that is 9 over two and i know it's negative but it doesn't really matter because then we're going to square it and that squared is positive anyway so it's 81 over four and i'm going to add it to both sides like so then what i'm going to do on the left side is i'm going to factor it now you might be like well this is difficult to factor but i've actually created a perfect square so it factors to x minus this term negative 9 over 2 squared that's what we're doing we're creating a perfect square by adding 81 over 4. if we foil this out it would equal this and then on this side we just got to add these two together 22 is the same as 88 over four right if i put it over one i just multiply top and bottom by four now i can add these together and i get 169 over four now i got a nice situation where i can take the square root of both sides and i get x minus 9 halves equals square root of 169 is 13 square root of 4 is 2 but again remember this is going to be plus or minus so now i got x minus 9 halves equals positive 13 halves x minus 9 halves equals negative 13 halves so i'm going to add 9 halves in both sides so then here i got x equals 22 over 2 right nine halves plus thirteen halves is twenty two halves which is eleven and here i would add nine halves to both sides nine halves plus negative thirteen halves is negative four over two which reduces to negative 2. so my two solutions are 11 and negative 2 done so again we're going to solve by Exercise 27 completing the square so here we have x squared minus 6x we've already got the constant on this side which is nice and what's what we need to start with let me take half of negative 6 which is negative 3 and i'm going to square it which is 9 and i'm going to add that to both sides now i've got a perfect square on this side to factor so it becomes x minus 3 times x minus 3 or x minus 3 squared again the shortcut for that is going to be x and then this value plus that value so x plus negative 3 or x minus 3 squared equals 13 plus 9 which is 22. now i'm going to take the square root of both sides boom boom and i get x minus 3 equals plus or minus square root of 22. now there's this is not a perfect square so we kind of have to leave it as plus or minus square root of 22. last step is we add 3 to both sides and we get 3 plus or minus square root of 22 that encapsulates both solutions done first i'm going to start by putting the Exercise 29 2 alone so i'm going to make it 6z squared i'm going to minus z and look here the 2 is on the left side the 6z squareds on the right side it doesn't matter so i just kind of flipped them around to make it easier for myself now we want to complete the square but this one's a little tricky because we don't want to have 6 in front of the z squared when we're completing the square so what we're going to do is we're going to divide everything by 6 so we've got z squared minus 1 6 z equals 2 divided by 6 is actually 1 3. okay now we're going to still use completing the square as as what they want so we're going to take half of this b term half of negative 1 6 is negative 1 12 and then i'm going to square it which is going to be 1 over 144 so i'm going to add 1 over 144 to both sides like so so of course this left side is going to factor into z minus this value plus that value which is minus 1 12 and that's going to be squared equals and then to get these to have common denominators i'm going to multiply by 48 and by 48 so then it's going to be 48 over 144 plus 1 over 44 which is going to be 49 over 144 like so then i'm going to take the square root of both sides like so and we've got z minus 1 12 equals square root of 49 is 7 square root of 144 is 12 plus or minus so now i'm going to break it up into two equations z minus 1 12 equals 7 12 and z minus 1 12 equals negative 7 12. so last but not least i'm going to add 1 12 to both sides and i'm going to get z equals 8 12 which is the same as two-thirds if we divide the top and bottom by four and this one if i add 1 12 to negative 7 12 i'm going to get negative 6 12 which is the same as negative one-half so these are my two solutions two-thirds and negative one-half boom Exercise 31 done here we're going to start by adding one to both sides and then so the minus one becomes a plus one over here then i'm going to divide everything by 2 and make it x squared minus 3 over 2 x equals 1 half now i'm going to take half of this which is negative 3 4 and i'm going to square it which is going to be 9 16 and i'm going to add that to both sides now i've created a perfect square trinomial and i'm going to factor it like so again the shortcut is its x plus this value which is minus 3 4. without the square of course and that's what this whole side factors to and that of course equals these added together so i'm going to turn that into 8 over 16 common denominators 8 plus 9 is 17 16. oh and this is squared by the way then i'm going to square root both sides and i get x minus 3 4 equals square root of 17 is the square root of 17. square root of 16 is 4. oh and i also sorry i forgot when you take the square root it's meant to be plus or minus so this is going to be plus or minus red 17 over 4. last but not least i'm going to add 3 4 to both sides and i get 3 4. plus or minus square root of 17 over 4. since they have common denominators i can also rewrite it but this is optional you can also rewrite it in a slightly more simplified format as 3 plus or minus square root of 17 over 4 Exercise 33 boom done here we are using the discriminant to determine the number of solutions to discriminate is of course b squared minus 4ac where this is a b and c so b squared is 4 squared which is 16 minus 4 a is 1 that's the coefficient there it's an invisible 1 times c which is 7. so 4 times 1 times 7 is 28 so 16 minus 28 is negative 12. so the fact is it's negative so what does that mean if it's positive it would have two real solutions zero it would have one real solution here it's negative it has no real solutions and there's your winner done Exercise 35 so once again the discriminant is b squared minus 4 ac here it would be negative 30 squared minus 4 times a times c and negative 30 squared is of course 900. here i'm going to take a little shortcut 4 times 25 is 100 times 9 is 900 900 minus 900 is zero when it's zero it means we have one real and rational we know it's not only real but it's rational because there's going to be nothing under the square root one real rational solution boom done again the discriminant is b squared minus 4ac where we've got a b and c so Exercise 37 it's negative 1 squared which is 1 minus 4 times a which is 6 times c which is negative 2. so this becomes 4 times 6 is 24 times 2 is negative 48 but minus a negative becomes a plus so we got 49. so it's positive so since this is positive we know we have two real solutions but also since this is a perfect square it's going to take the square root of that is seven so it's going to be two real rational solutions boom done for these problems we're going to use the quadratic formula and the first thing we need to do is set everything Exercise 39 equal to 0 so i'm going to subtract 4 from both sides and then we've got our a which is in front of the x squared our b which is in front of the x meaning it's 1 and 1 for a let's see right here the quadratic formula is negative b plus or minus the square root of b squared minus 4 ac all over 2 a so we're going to plug and chug so we got negative b which is 1 so it's negative 1 plus or minus the square root of b squared 1 squared is just 1 minus 4 times a which is 1 times c which is negative 4 all over 2 times a 2 a 2 times 1 which is 2. and then to simplify we have 4 times 1 times 4 which is 16 negative with another minus in front that becomes a plus 16 so it's negative 1 plus or minus square root of 1 plus 16 over 2 1 plus 16 is 17 so my solutions here are negative 1 plus or minus the square root of 17 over 2. that's two solutions because of the plus or minus boom done once again using the quadratic formula i'm going to mark this as a that's b is negative 5 and c is 1. Exercise 41 the formula is negative b plus or minus the square root of b squared minus 4 a c all over 2 a let's go so we got negative negative 5 which is 5 plus or minus the square root of negative 5 squared which is 25 that's the b squared minus 4 times a which is 3 times c which is 1 all over 2 times 3. so 4 times 3 times 1 is 12 so 25 minus 12 5 plus or minus square root of 25 minus 12 over 2 times 3 which is 6. 25 minus 12 is of course 13. so we got 5 plus or minus square root of 13 over 6 are the solutions done so first we have Exercise 43 to get this into our nice quadratic format and you might be like well how do i do that if you just multiply everything by x squared it gets really easy so then that becomes 4 x squared and then x squared times 1 over x it'll become x squared over x which just cancels into a singular x and then x squared times 1 over x squared simply x squared over x squared or 1 with the minus there and now we've got it in the right format so now we've got a b b of course is one invisible one there and then c so we got our quadratic formula negative b plus or minus the square root of b squared minus 4ac all over 2a let's go negative b is negative 1 plus or minus square root of b squared 1 squared is 1 minus 4 a c all over 2 times a as we simplify on the inside 4 times 4 is 16 that's negative 16 minus negative 16 becomes plus 16 so it's like a 1 plus 16 becomes 17 over eight and there are your two solutions negative one plus or minus square root of 17 over eight boom done Exercise 45 for this question we're going to use a calculator to solve i'm not going to use the ti calculators the texas instruments instead i'm going to show you how to do this on desmos so i've simply written in the function like this and these are what we're looking for we're looking for the x intercept so normally we're setting it equal to 0 and finding the solutions for x but when we're given the quadratic what we're doing is we're basically it's like we're setting y equal to zero by virtue of looking for the actual x-intercepts so these are your two solutions rounded to the nearest thousandth point one three one whoops and two point five three five so here are your answers 0.131 and 2.535 boom done to find the answer to this they're recommending we graph both Exercise 47 of these individually and then find the points of intersection so i'm going to show you how to do this on desmos so now we can see the two equations this is y equals four this one is the quadratic we look at the intersection points and when we find the solutions we're looking for the x values so it's negative 6.653 and 1.653 so this one has to round to the nearest tenth so we say negative 6.7 and 1.7 boom done beginning with the general form of a quadratic equation solve for x by using the completing the square method thus Exercise 49 deriving the quadratic formula okay check this out so we're first going to start by following our normal protocol we're going to move that c value to the other side so we got subtracting c from both sides then we want to get rid of that a we don't want to have a leading coefficient so i'm going to divide everything by a so now i've got x squared right the a goes away a over a is just 1 plus b over a x equals negative c over a now as we're completing the square we're going to take half of this which is b over 2 a and we're going to square it and i'm going to add that to both sides that's going to be b squared over 4 a squared plus b squared over 4 a squared now i'm going to factor the left side so again i've created a nice perfect square trinomial and again the key is it's going to be x plus b over 2a it's going to be whatever this value is right there equals and i want to combine these guys ideally so if i multiply the top and bottom by 4a i get 4 a and then here i get 4 a squared a times a is a squared and now i can kind of consolidate that so i'm going to put it as since that's negative i'm going to put that second just for fun so i'm going to make it b squared minus 4ac all over 4a squared so you can kind of see this is coming together as we might expect so now i'm going to take the square root of both sides and since the bottom is a perfect square that can just become 2a and on top we can't really do anything with that so that's going to just stay trapped inside as b squared minus 4ac but once again we're going to leave that plus or minus out there because we know that's a plus or minus radical and then on the left side we get x plus b over 2a last but not least i'm going to subtract b over 2a from both sides so x equals i'll just make it a negative b over 2a and then that's plus or minus the square root b squared minus 4ac all over 2a now here's the deal they have the same denominator so let's merge that and it's going to become negative b plus or minus the square root of b squared minus 4 a c all over 2a and there is a quadratic formula boom done a person has a garden that has Exercise 51 length 10 feet longer than the width so i can say this is the width this is the length that is 10 feet long instead of a quadratic equation to find the dimensions the area is 119. so check it out i know that w with times length which is w plus 10 equals 119 feet and they want us to use the quadratic equation so so what we're going to do first we're going to distribute we got w squared plus 10 w and then i'm going to subtract 119 from both sides in the same step because use the quadratic i want to set everything equal to zero okay now to get the solution we're gonna do negative b plus or minus the square root of b squared minus 4ac all over 2a so i'll write it out up here okay so b is 10 so we got negative 10 plus or minus the square root of 10 squared which is 100 minus 4 a is 1 c is negative 119 that's a huge number all over 2 times a a is 1 so that's just 2. so negative 4 times 1 is negative 4 times 119 is 476 positive because the negatives cancel out so we got 476 all over two that becomes 576 which i believe is a perfect square i believe that's 24 squared so square root of 576 is 24 so now we got negative 10 plus or minus 24 over 2. now here's the deal we're going to get two solutions but we're talking about width and length so talking about positive values so i'm not going to worry about the negative 24. so let's see we don't have a ton of room here let's bring it up here so i'm just going to worry about the positive value negative 10 plus 24 because minus 24 would give us negative 34 divided by 2 that doesn't make sense divided by 2 so negative 10 plus 24 is 14 divided by 2 is 7 and since we were talking about feet squared the width is 7 feet and the length would be 7 plus 10 which is 17 feet and those are winners done Exercise 53 suppose that an equation is given here where x represents the number of items sold in an auction and p is the profit how many items sold would you make or would make this profit a maximum so we're going to do this by graphing now when we're talking about a maximum profit this is of course going to look like a parabola like this because that a term is negative the maximum profit because this whole thing represents profit is going to be that max point and when they say how many items sold would make a maximum profit we're not looking for the y value because that would that would actually give us the value of the max profit we're looking at the number of items sold to get this amount that's the x value so i'm going to graph it i'm going to show you what it looks like okay so you see how i adjusted the window as the instruction zero to 200 on the x zero to ten thousand on the y what we're looking for is we're looking for that vertex and it's really nice on desmos because it kind of auto finds it for you and as we mentioned before we're trying to find that x value so that means 70 items will give us our max profit the max profit happens to be 8 800 they're not asking for that so the answer is 70 boom done Exercise 55 the cost function for a certain company is this and the revenue is given by this recall that profit is revenue minus cost set up a quadratic and find two values of x that will create a profit of 300. so profit is revenue minus cost therefore the profit will be 100 x minus 0.5 x squared minus 60x minus 300 because i'm subtracting this entire cost we got to distribute that like so so we're going to combine like terms and we're going to get profit equals negative 0.5 x squared plus 40x right 100 minus 60 is 40 minus 300. and we're trying to get a profit of 300 so i'm going to plug this 300 in for profit and then we're going to solve so that's going to be 300 equals all of this now i need to subtract 300 from both sides and that's going to cancel that out and i'm going to get 0 equals negative 0.5 x squared plus 40x minus 600 this is what we want set it equal to zero now i'm going to try to factor this but i don't like this negative point five x squared here but i can remedy that by multiplying everything by negative two so i'm gonna do it up here so i have more room so we're gonna do a little arrow like this and boom so multiply everything by negative 2 i get x squared minus 80x plus 1200. now i'm going to factor this by asking what multiplies to 1200 adds to negative 80. that's going to be negative 60 and negative 20. now to solve what makes this go to 0 60 what makes this go to 0 20 so the two values of production are 60 and 20 done a vacant lot is being converted into a community garden the garden and the walkway and its perimeter Exercise 57 have an area of 378 that means everything total find the width of the walkway if the garden is 12 feet wide by 15 feet long so we've kind of got this whole thing in the mix and they're saying the entire thing you know this entire width times this entire length is going to be 378 so let's set that up now what is this entire width well area called width length whatever it's 12 feet plus x plus x which means it's 12 plus two x's what about here it's 15 plus two x's now let's create an area equation to relate everything so we're going to say 15 plus 2x times 12 plus 2x gaussian not a ton of room here equals that 378. now we gotta solve for x so we're going to do it as follows first i'm going to foil everything out and we're going to simplify 15 times 12 is 180 15 times 2x is 30x 2x times 12 is 24x and then 2x times 2x is 4x squared and this all equals 378 still now we'll combine like terms i'm going to write it in the right order so we got 4x squared 30 and 24 make 54x plus 180 equals 378. now we want to set everything equal to zero let's box this off so i'm going to subtract both sides by 378 then i have 4x squared plus 54x minus 198 and that now equals zero beautiful now i'm gonna divide everything by two to simplify a little bit more we got two x squared plus 27 x minus 99 equals zero now i'm going to factor using the star method so we've got a b and c so i'm gonna set it up down here so we've got two two uh we've got uh b which is 27 and then a times c that's negative 198 up top so what multiplies to negative 198 and adds to 27 that's gonna be 33 and negative six now we're going to reduce these like fractions that becomes 1 and negative 3 and this can't be reduced which is nice so it becomes 2x plus 33 times x minus 3 equals 0. what's gonna send this to zero it's gonna be a negative number so i don't care because we can't have negative length or width but this one's going to be positive what sends that to zero three which means the width of this walkway is three feet boom done i hope you guys enjoyed this video and if you did please click that like button and if you want to see more from the scala learning channel make sure to click subscribe thank you guys so much for joining and i'll see you in the next video take it easy you |
8144 | https://www.youtube.com/watch?v=mH6JRjE7Egk | Solving Two - Step Linear Equations
Patrick J
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Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) !! Solving Two - Step Linear Equations. Here we look at solving linear equations where we have to perform addition / subtraction as well as division / multiplication to solve.
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Transcript:
okay here we're going to look at solving a a few different equations more specifically what are known as linear equations and we'll have to do a couple a couple steps to solve them um so in part A here we have -4x - 14 = 10 typically uh the way I always thought about it is first I want to get uh for an equation where there's an x to the first power or variable to the first Power I want to get the term involving just X all by itself so if there are other X's I'll I'll try to combine them so on the left side I see two terms I see one involving X and then I see just a number and then on the other side I see just a number so to get the uh the X's basically by itself since I'm subtracting 14 I'm going to add 14 to both sides so on the left side that's going to leave me with -4x = let's see 10 + 14 is going to be 24 and now what I need to do at this point since I'm multiplying by -4 uh to get the X by itself I'm just simply going to divide both sides by -4 so -4 / -4 is 1x let's see uh a positive over a negative is a negative 24 / 4 will give us six so the solution to our first equation will be x = -6 |
8145 | https://www.youtube.com/watch?v=V-2OTlRH65g | Enantiomers
The Organic Chemistry Tutor
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Posted: 25 Apr 2018
This organic chemistry video tutorial explains how to draw the enantiomer of a molecule by drawing its reflection across a vertical line.
Finding Chirality Centers:
Organic Chemistry - Video Lessons:
Final Exam and Test Prep Videos:
Chiral and Achiral Molecules:
Stereochemistry:
Enantiomers:
Diastereomers:
Meso Compounds:
Stereoisomers:
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Enantiomeric Excess Test Question:
SN2 SN1 E1 E2 Reaction Mechanisms:
Alkene Reactions Review:
Alkyne Reactions Review:
Organic Chemistry PDF Worksheets:
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Transcript:
in this video we're gonna focus on enantiomers inators are isomers of the same chemical formula but there are non-superimposable mirror images of each other think of your left hand and your right hand they're similar but they're not exactly the same your left hand is a mirror image of your right hand so this molecule has an inertia for any type of molecule that has one chiral Center it's gonna have one in Ashman so how can we draw the enantiomer of this molecule what we need to do is draw the mirror image of that molecule so let's say the red line represents a mirror so here's the carbon atom we have a ch3 that should be faced in this mirror because we have one here and then we have a hydrogen atom in the back I'll chlorine atom at the top and a bromine atom at the wedge so that's how you can draw an enantiomer of a molecule you simply need to draw the mirror image of that molecule now let's identify the configuration of each chiral carbon atom so the group with the highest priority is the bromine atom because it has the highest atomic number and then it's the chlorine atom and then the methyl group then the hydrogen now the hydrogen is on it's on the dash so it's where it needs to be so it's going into the page which is what we want so all we need to do is count it from 1 to 2 to 3 ignore and 4 and so what we have is the our configuration since we're rotating it clockwise now the enantiomer needs to have the opposite configuration so bromine is still going to be number one chlorine is 2 methyl is 3 H is 4 so we're going to count it from 1 2 2 2 3 and so we're going down or clockwise and so we have the S configuration so whenever you have one chiral Center you automatically have an enantiomer and the enantiomers will always have opposite configurations if you have a molecule with two chiral centers then to draw the enantiomer simply reverse both Carle centers their configurations and make sure that the molecule doesn't have a line of symmetry other wives it could be a meso compound here's another example you could try so go ahead and draw the announcement for this molecule so let's try the mirror first and this time the chlorine is going to be faced on the left side as opposed to the right side and then we have the bromine at the top and then the hydrogen is facing away from the mirror and then we have the methyl group which we can write like this now let's assign the configuration to each chiral Center so this is going to be grouping one two three and four so this time the age is on the wedge so because it's coming out of a page what we need to do is we need to reverse it so if we count it this way 1 2 2 2 3 it appears to be our but because H is in the front not in the back we need to reverse it so what we have is the S isomer now for the chiral Center at the right this is number 1 number 2 3 & 4 so we need to count it from 1 2 to 3 and so it appears to be s but age is in the front so we need to reverse it and so we're gonna get the our isomer so as you can see the configurations are opposite from each other now go ahead and draw the enantiomer for these two Fischer projections so feel free to pause the video as you try these examples so let's start with the first example so let's draw the mirror all we need to do is basically reverse hydrogen and fluorine to draw the enantiomer and so it's going to look like this and that's all we need to do for the first example now for the second one we need to reverse the bromine atom with the hydrogen and the hydrogen with the chlorine and so the enantiomer is going to look like this so bromine is going to be facing towards the mirror in this case the blue line and so as the hydrogen atom so here bromine was on the right side and now it's on the left side and here the hydrogen was on the right and now it's on the left so you just got to reverse the groups in the Fischer projection to draw the unnatural so how would you draw the enantiomer of this molecule so the bromine is on the wedge all you need to do is put it on the dash and so that's how you can draw the enantiomer of that molecule try this one so let's say we have two substituents a bromine atom and a chlorine atom go ahead and draw the nature of that molecule so the bromine is on the wedge just reverse and put it on a dash the chlorine is on a dash now we need to place it on a wedge and so that's a simple way in which we could draw the national file drawn or using a mirror try this one draw the enantiomer of this molecule so one way is we can draw the mirror image of it and so that's one way we can draw the image for the other way is I'm going to redraw the original molecule and if we're gonna keep the bromine atoms in the same position all I need to do is switch the wedge with the dash and vice versa so here we have a wedge let's replace it with a dash and here there's a wedge I mean there's a dash so let's replace that with a wedge and so that's how we can draw then answer this molecule that's another way we can do it just by changing the configuration of the chiral centers |
8146 | https://courses.dcs.wisc.edu/wp/grammar/spanish-28/ | English Grammar for Second Language Learners
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Tanto en inglés como en español hay artículos definidos o determinados y artículos indefinidos o indeterminados.
Los artículos definidos en español son el, la, los, las.
¿Cuál es el coche que venden?
Estas son las cartas que quiero enviar hoy.
Los artículos indefinidos son un/uno, una, unos, unos.
Tengo un secreto.
Nunca han visitado una ciudad tan grande.
Aunque el significado de estos artículos es el mismo en inglés y en español, los usos son distintos muchas veces. Por ejemplo, cuando en inglés se usa un adjetivo posesivo para hablar de una parte del cuerpo, el español emplea el artículo definido. Nota la traducción de la siguiente frase:
How did she break her ankle?
¿Cómo rompió el tobillo?
A modo de contraste, aquí hay una breve lista de cuando se usa el artículo definido en español y no en inglés:
El artículo definido con conceptos generales:
la libertad (liberty, freedom)
el socialismo (socialism)
No se debe hablar de la religión en una fiesta. (You should not talk about religion at a party.)
El artículo definido con sustantivos que representan a todos los miembros de un grupo:
Se dice que los elefantes recuerdan todo. (They say that elephants remember everything.)
Los alemanes tienen fama de ser metódicos. (Germans are famous for being methodical.)
El artículo definido con los días de la semana:
Todos los viernes nos reunimos con amigos para relajarnos. (Every Friday we get together with friends to relax.)
¿Dónde vas el sábado? (Where are you going on Saturday?)
Hay distintos usos del artículo indefinido que hay que notar también. Se omite el artículo indefinido en los siguientes contextos (pero sí se usa en inglés):
Con profesiones:
Soy profesora de español. ¿Y Ud.? (I am a Spanish professor. And you?)
¿Cuándo se hizo médico? (When did you become a doctor?)
muchas veces después del verbo ser, especialmente en frases negativas:
No es problema llegar tarde. (It’s not a problem to arrive late.)
Es (una) cuestión de tiempo. (It’s a question of time.)
Recomendamos consultar la lista de recursos para más reglas y ejemplos porque esta lista no es exhaustiva (¿Notó la omisión del artículo después de ser?).
Recursos:
Vídeo completo sobre los usos de los artículos (9:17)
Actividades para practicar el uso de los artículos |
8147 | https://www.universoformulas.com/fisica/cinematica/caida-libre/ | Caída libre: caso particular de MRUA
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Caída libre
11 comentarios / Cinemática / Por Bernat Requena Serra
Un caso particular de movimiento rectilíneo uniformemente variado (MRUV) es la caída libre de un cuerpo o su lanzamiento vertical con una velocidad v 0.
En este caso, al espacio lo llamaremos h (altura), la aceleración será g (la gravedad, normalmente se considera g = 9.81 m/s 2), que será positiva en el caso de la caída y negativa en el caso del lanzamiento vertical hacia arriba. En caída libre, la velocidad inicial V 0, aquí y para facilitar la comprensión siempre la consideraremos 0 m/s. Sin embargo, en el lanzamiento vertical hacia arriba siempre será mayor que 0 m/s.
Las dos fórmulas a aplicar serán:
Podemos calcular la velocidad a la que cae un cuerpo sabiendo a la altura que es lanzado o el tiempo en el aire.
En el caso de caída libre desde una altura h (recordemos que la velocidad inicial la consideramos = 0 m/s). La velocidad final, en el punto de contacto con el suelo, si conocemos la altura h, será:
El tiempo de caída, con estas premisas y desde una altura h, es, despejando t en la segunda fórmula:
Si conocemos el tiempo de caída, t:
En el caso contrario, de lanzamiento vertical, deberíamos tener en cuenta la dirección contraria de la aceleración de la gravedad y de la velocidad inicial V 0.
Recordemos que para el caso del movimiento de caída libre ideal, que estamos exponiendo, no tenemos en cuenta el rozamiento con el aire. Y que el sentido de la gravedad g en la caída es el mismo que el sentido de la velocidad, por lo tanto en las fórmulas g toma valor positivo, mientras que en el lanzamiento vertical el sentido de la gravedad g es contrario al del vector velocidad, por lo que introduciremos en las fórmulas la gravedad con signo negativo (-g).
11 comentarios en “Caída libre”
Juan 28 mayo, 2019 a las 03:51
Los dos vectores g de módulo distintos en el dibujo rompe los ojos! Qué quisieron expresar?
Responder
1. Respuestas 28 mayo, 2019 a las 14:11
En efecto, Juan.
El dibujo actual ilustra una caída libre con el vector de la aceleración de la gravedad y el de la velocidad de caída, ambos en este caso del mismo sentido.
En lanzamiento hacia arriba tendrían sentido contrario.
Gracias.
2. Image 10 eli 26 agosto, 2025 a las 15:40
Juan el vector g, no cambia a lo largo de la caida libre, pero el vector velocidad si crece, es mayor a medida que pasa el tiempo v=gxt g es constante y t varia
jesus garcia 4 noviembre, 2017 a las 22:14
es buena la explicacion
Responder
3. Nacho 5 junio, 2017 a las 18:52
Hola, tengo una duda, ¿la aceleración no tendría que ser negativa cuando el cuerpo cae y positiva cuando sube?
La dirección del eje Y es positiva hacia arriba.
Gracias de antemano.
Responder
1. Respuestas 5 junio, 2017 a las 23:26
En el movimiento rectilinieo uniformemte acelerado hay que referenciarlo, como en los demás, a un sistema de coordenadas en donde se refleje el sentido positivo o negativo de la velocidad, aceleración o posición.
En la caida libre (caso particular de MRUA) es por comodidad que tomamos como referencia el punto donde se inicia la caida libre.
En todo caso, si en la caida libre se utilizasen los sistemas de referncia generales del MRUA, como planteas, sería totalmente correcto y arrojaría idéntico resultado.
Gracias.
María Elena Juárez 4 diciembre, 2015 a las 04:48
Gracias ya que me ha servido de mucha ayuda
Responder
5. María Elena Juárez 4 diciembre, 2015 a las 04:47
Felicidades por todo el trabajo que han realizado y muchas gracias ya que me ha servido de mucho.
Pero seria bueno que le anexaran ejercicios como ejemplos y ejercicios para realizar.
Responder
6. María Elena Juárez 4 diciembre, 2015 a las 04:44
la teoría esta muy buena y las imagenes excelentes, pero les quedaria aun mejor si le agregan ejercicios resueltos y para resolver.
Gracias por todo el trabajo y bendiciones a todos
Responder
7. BACETE 14 octubre, 2015 a las 20:09
¿Podrían poner algún ejemplo?
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8. Roxana 6 abril, 2015 a las 02:55
Muy bueno
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8148 | https://www.youtube.com/watch?v=bA6aJ0bNGh8 | Union and intersection of sets with examples
The Engineering Tutoress
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Posted: 3 May 2021
This video tutorial provides introduction to universal set and compliment of a set. It explains how to find a compliment of a set. It also explains through two examples of how to find intersection and union of two sets.
Transcript:
[Music] welcome everyone in this video we will work on two examples of union and intersection problem statement number one set a and set b are subsets of set q the values within each set are so set a contains the elements 4 7 9 set b contains the elements 4 5 9 and 10 and set q contains the elements 4 5 6 7 8 9 and 10. what is the union of the complement of set a and set b so q in this example is the universal set meaning that a and b are subsets of q when we are talking about union we mean combining the sets in one larger set or in our case we will combine the complement of set a and set b now what is complement of set a complement of a set a is set that contains all elements in the universal set queue that are not in the set a or simply said the universal set q minus the subset a now let's write this down so we have the universal set q with the elements 4 5 6 7 8 9 and 10. and we have the subset a with the elements 4 7 and 9. and in this case we want the complement of set a so we're looking at the numbers and we see that 4 is taken away from the universal set next is 5. so we have five in the complement of set a then six seven is taken away eight nine is taken away from the universal set and then ten so this is the complement of set a and we have the set b so we'll write it down here so we have the elements 4 5 9 and 10. in this case we want to find the union of these two sets so as i said earlier when we talk about union we think of combining the two sets so the larger set or the union will contain all the elements from both sets but keep in mind that both of these sets contains the elements 5 and 10 and we don't write them twice we write them only once so we have union of complement of set a with the set b is equal to so we have 4 then 5 6 8 9 and 10 once so this is the union between the complement of set a and set b now let's work on our second example set a consists of elements 1 3 and 6 and set b consists of elements 1 2 6 and 7. both sets come from the universe of 1 2 3 4 5 6 7 and 8. what is the intersection between the complement of set a and b so here again we are looking for the complement of set a again complement of set a consists of all elements in the universal set in this example you here that are not in the set a so we have the universal set u with the elements 1 2 3 4 five six seven and eight we have the subset a with the elements one three and six and we want the complement of set a we are looking at the numbers and we see that one is taken away from the universal set and we have two and three is taken away we have four we have five here six is taken away and we're left with 7 and 8. so this is the complement of set a we have the set b with the elements 1 2 6 and 7. from this example we can see that we are looking for the intersection between the complement of set a and set b so we need to find out which elements are common for both sets so both sets contain the element two and both sets contain the element seven so complement of set a intersection with set b is equal to we have 2 and 7 as a result so this is it for this video and thank you for watching |
8149 | https://www.savemyexams.com/a-level/chemistry/edexcel/17/revision-notes/5-advanced-physical-chemistry/5-5-kinetics-ii/5-5-6-reaction-order-graphs/ | A LevelChemistryEdexcelRevision Notes5. Advanced Physical ChemistryKinetics IIReaction Order Graphs
Reaction Order Graphs (Edexcel A Level Chemistry): Revision Note
Exam code: 9CHO
Author
Richard Boole
Last updated
Reaction Order Graphs
Reaction Order Using Concentration-Time Graphs
In a zero-order reaction, the concentration of the reactant is inversely proportional to time
This means that the reactant concentration decreases as time increases
The graph is a straight line going down as shown:
Concentration-time graph of a zero-order reaction
The gradient of the line is the rate of reaction
Calculating the gradient at different points on the graph, will give a constant value for the rate of reaction
When the order with respect to a reactant is 0, a change in the concentration of the reactant has no effect on the rate of the reaction
Therefore:
Rate = k
This equation means that the gradient of the graph is the rate of reaction as well as the rate constant, k
In a first-order reaction, the concentration of the reactant decreases with time
The graph is a curve going downwards and eventually plateaus:
Concentration-time graph of a first-order reaction
In a second-order reaction, the concentration of the reactant decreases more steeply with time
The concentration of reactant decreases more with increasing time compared to a first-order reaction
The graph is a steeper curve going downwards:
Concentration-time graph of a second-order reaction
Order of reaction from half-life
The order of a reaction can also be deduced from its half-life (t1/2 )
For a zero-order reaction the successive half-lives decrease with time
This means that it would take less time for the concentration of reactant to halve as the reaction progresses
The half-life of a first-order reaction remains constant throughout the reaction
The amount of time required for the concentration of reactants to halve will be the same during the entire reaction
For a second-order reaction, the half-life increases with time
This means that as the reaction is taking place, it takes more time for the concentration of reactants to halve
Half-lives of zero, first and second-order reactions
Examiner Tips and Tricks
Make sure that you know the correct shapes for the concentration-time graphs. It can be easy to confuse some concentration-time graphs with the following rate-concentration graphs, particularly;
The straight line of a zero-order concentration-time graph with the straight line of a first-order rate-concentration graph.
The curve of a first-order concentration-time graph with the curve of a second-order rate-concentration graph.
Reaction order using rate-concentration graphs
In a zero-order reaction, the rate doesn’t depend on the concentration of the reactant
The rate of the reaction therefore remains constant throughout the reaction
The graph is a horizontal line
The rate equation is rate = k
Rate-concentration graph of a zero-order reaction
In a first-order reaction, the rate is directly proportional to the concentration of a reactant
The rate of the reaction increases as the concentration of the reactant increases
This means that the rate of the reaction decreases as the concentration of the reactant decreases when it gets used up during the reaction
The graph is a straight line
The rate equation is rate = k[A]
Rate-concentration graph of a first-order reaction
In a second-order reaction, the rate is directly proportional to the square of concentration of a reactant
The rate of the reaction increases more as the concentration of the reactant increases
This means that the rate of the reaction decreases more as the concentration of the reactant decreases when it gets used up during the reaction
The graph is a curved line
The rate equation is rate = k[A]2
Rate-concentration graphs of a second-order reaction
Examiner Tips and Tricks
Careful - sometimes when asked to complete calculations for the rate constant, k, the exam question will give you a graph as well as tabulated data. Do not ignore the graph as this demonstrates the order of one of the reactants, while the tabulated data allows you to determine the order for the other reactants.
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Author:Richard Boole
Expertise: Chemistry Content Creator
Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME. |
8150 | https://en.wikipedia.org/wiki/Cameron%E2%80%93Martin_theorem | Published Time: 2006-08-20T13:57:18Z
Cameron–Martin theorem - Wikipedia
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1 Motivation
2 Statement of the theoremToggle Statement of the theorem subsection
2.1 For abstract wiener spaces
2.2 Version for locally convex vector spaces
3 Integration by parts
4 An application
5 See also
6 References
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From Wikipedia, the free encyclopedia
Theorem defining translation of Gaussian measures (Wiener measures) on Hilbert spaces.
In mathematics, the Cameron–Martin theorem or Cameron–Martin formula (named after Robert Horton Cameron and W. T. Martin) is a theorem of measure theory that describes how abstract Wiener measure changes under translation by certain elements of the Cameron–Martin Hilbert space.
Motivation
[edit]
The standard Gaussian measureγ n{\displaystyle \gamma ^{n}} on n{\displaystyle n}-dimensional Euclidean spaceR n{\displaystyle \mathbf {R} ^{n}} is not translation-invariant. (In fact, there is a unique translation invariant Radon measure up to scale by Haar's theorem: the n{\displaystyle n}-dimensional Lebesgue measure, denoted here d x{\displaystyle dx}.) Instead, a measurable subset A{\displaystyle A} has Gaussian measure
γ n(A)=1(2 π)n/2∫A exp(−1 2⟨x,x⟩R n)d x.{\displaystyle \gamma {n}(A)={\frac {1}{(2\pi )^{n/2}}}\int {A}\exp \left(-{\tfrac {1}{2}}\langle x,x\rangle {\mathbf {R} ^{n}}\right)\,dx.}
Here ⟨x,x⟩R n{\displaystyle \langle x,x\rangle {\mathbf {R} ^{n}}} refers to the standard Euclidean dot product in R n{\displaystyle \mathbf {R} ^{n}}. The Gaussian measure of the translation of A{\displaystyle A} by a vector h∈R n{\displaystyle h\in \mathbf {R} ^{n}} is
γ n(A−h)=1(2 π)n/2∫A exp(−1 2⟨x−h,x−h⟩R n)d x=1(2 π)n/2∫A exp(2⟨x,h⟩R n−⟨h,h⟩R n 2)exp(−1 2⟨x,x⟩R n)d x.{\displaystyle {\begin{aligned}\gamma {n}(A-h)&={\frac {1}{(2\pi )^{n/2}}}\int {A}\exp \left(-{\tfrac {1}{2}}\langle x-h,x-h\rangle {\mathbf {R} ^{n}}\right)\,dx\[4pt]&={\frac {1}{(2\pi )^{n/2}}}\int {A}\exp \left({\frac {2\langle x,h\rangle {\mathbf {R} ^{n}}-\langle h,h\rangle {\mathbf {R} ^{n}}}{2}}\right)\exp \left(-{\tfrac {1}{2}}\langle x,x\rangle _{\mathbf {R} ^{n}}\right)\,dx.\end{aligned}}}
So under translation through h{\displaystyle h}, the Gaussian measure scales by the distribution function appearing in the last display:
exp(2⟨x,h⟩R n−⟨h,h⟩R n 2)=exp(⟨x,h⟩R n−1 2‖h‖R n 2).{\displaystyle \exp \left({\frac {2\langle x,h\rangle {\mathbf {R} ^{n}}-\langle h,h\rangle {\mathbf {R} ^{n}}}{2}}\right)=\exp \left(\langle x,h\rangle {\mathbf {R} ^{n}}-{\tfrac {1}{2}}\|h\|{\mathbf {R} ^{n}}^{2}\right).}
The measure that associates to the set A{\displaystyle A} the number γ n(A−h){\displaystyle \gamma {n}(A-h)} is the pushforward measure, denoted (T h)∗(γ n).{\displaystyle (T{h}){}(\gamma ^{n}).} Here T h:R n→R n{\displaystyle T{h}:\mathbf {R} ^{n}\to \mathbf {R} ^{n}} refers to the translation map: T h(x)=x+h{\displaystyle T_{h}(x)=x+h}. The above calculation shows that the Radon–Nikodym derivative of the pushforward measure with respect to the original Gaussian measure is given by
d(T h)∗(γ n)d γ n(x)=exp(⟨h,x⟩R n−1 2‖h‖R n 2).{\displaystyle {\frac {\mathrm {d} (T_{h}){}(\gamma ^{n})}{\mathrm {d} \gamma ^{n}}}(x)=\exp \left(\left\langle h,x\right\rangle {\mathbf {R} ^{n}}-{\tfrac {1}{2}}\|h\|_{\mathbf {R} ^{n}}^{2}\right).}
The abstract Wiener measure γ{\displaystyle \gamma } on a separableBanach spaceE{\displaystyle E}, where i:H→E{\displaystyle i:H\to E} is an abstract Wiener space, is also a "Gaussian measure" in a suitable sense. How does it change under translation? It turns out that a similar formula to the one above holds if we consider only translations by elements of the densesubspacei(H)⊆E{\displaystyle i(H)\subseteq E}.
Statement of the theorem
[edit]
For abstract wiener spaces
[edit]
Let i:H→E{\displaystyle i:H\to E} be an abstract Wiener space with abstract Wiener measure γ:Borel(E)→[0,1]{\displaystyle \gamma :\operatorname {Borel} (E)\to [0,1]}. For h∈H{\displaystyle h\in H}, define T h:E→E{\displaystyle T_{h}:E\to E} by T h(x)=x+i(h){\displaystyle T_{h}(x)=x+i(h)}. Then (T h)∗(γ){\displaystyle (T_{h})_{}(\gamma )} is equivalent to γ{\displaystyle \gamma } with Radon–Nikodym derivative
d(T h)∗(γ)d γ(x)=exp(⟨h,x⟩∼−1 2‖h‖H 2),{\displaystyle {\frac {\mathrm {d} (T_{h}){}(\gamma )}{\mathrm {d} \gamma }}(x)=\exp \left(\langle h,x\rangle ^{\sim }-{\tfrac {1}{2}}\|h\|{H}^{2}\right),}
where
⟨h,x⟩∼=i(h)(x){\displaystyle \langle h,x\rangle ^{\sim }=i(h)(x)}
denotes the Paley–Wiener integral.
The Cameron–Martin formula is valid only for translations by elements of the dense subspace i(H)⊆E{\displaystyle i(H)\subseteq E}, called Cameron–Martin space, and not by arbitrary elements of E{\displaystyle E}. If the Cameron–Martin formula did hold for arbitrary translations, it would contradict the following result:
If E{\displaystyle E} is a separable Banach space and μ{\displaystyle \mu } is a locally finite Borel measure on E{\displaystyle E} that is equivalent to its own push forward under any translation, then either E{\displaystyle E} has finite dimension or μ{\displaystyle \mu } is the trivial (zero) measure. (See quasi-invariant measure.)
In fact, γ{\displaystyle \gamma } is quasi-invariant under translation by an element v{\displaystyle v}if and only ifv∈i(H){\displaystyle v\in i(H)}. Vectors in i(H){\displaystyle i(H)} are sometimes known as Cameron–Martin directions.
Version for locally convex vector spaces
[edit]
Consider a locally convex vector spaceE{\displaystyle E}, with a Gaussian measure γ{\displaystyle \gamma } on the cylindrical σ-algebraσ(Cyl(E,E′)){\displaystyle \sigma (\operatorname {Cyl} (E,E'))} and let γ m:=γ(⋅−m){\displaystyle \gamma {m}:=\gamma (\cdot -m)} denote the translation by m∈E{\displaystyle m\in E}. For an element in the topological dual f∈E′{\displaystyle f\in E'} define the distance to the mean t γ(f):=f−E γ[f],{\displaystyle t{\gamma }(f):=f-\mathbb {E} {\gamma }[f],} and denote the closure in L 2(E,γ){\displaystyle L^{2}(E,\gamma )} as E a γ:=clos{(t γ(f n))n:f∈E′}{\displaystyle E{a}^{\gamma }:=\operatorname {clos} \left{(t_{\gamma }(f_{n})){n}\colon \ f\in E'\right}}. Define the covariance operator R γ¯:E a γ→(E′)∗{\displaystyle {\overline {R{\gamma }}}:E_{a}^{\gamma }\to (E')^{}} extended to the closure as
R γ¯(f)(g)=⟨f,g−E γ[g]⟩L 2(γ){\displaystyle {\overline {R_{\gamma }}}(f)(g)=\langle f,g-\mathbb {E} {\gamma }[g]\rangle {L^{2}(\gamma )}}.
Define the norm
‖h‖H γ:=sup{f(h):f∈E′,R γ¯(f)(f)≤1},{\displaystyle \|h\|{H{\gamma }}:=\sup{f(h)\colon f\in E',\;{\overline {R_{\gamma }}}(f)(f)\leq 1},}
then the Cameron-Martin spaceH γ{\displaystyle H_{\gamma }} of γ{\displaystyle \gamma } in E{\displaystyle E} is
H γ={h∈E:‖h‖H γ<∞}{\displaystyle H_{\gamma }={h\in E\colon \|h\|{H{\gamma }}<\infty }}.
If for h∈E{\displaystyle h\in E} there exists an g∈E a γ{\displaystyle g\in E_{a}^{\gamma }} such that h=R γ¯(g){\displaystyle h={\overline {R_{\gamma }}}(g)} then h∈H γ{\displaystyle h\in H_{\gamma }} and ‖h‖H γ=‖g‖L 2(γ){\displaystyle \|h\|{H{\gamma }}=\|g\|{L^{2}(\gamma )}}. Further there is equivalenceγ h∼γ{\displaystyle \gamma {h}\sim \gamma } with Radon-Nikodým density
d γ h d γ=exp(g(x)−1 2‖h‖H γ 2).{\displaystyle {\frac {d\gamma {h}}{d\gamma }}=\exp \left(g(x)-{\frac {1}{2}}\|h\|{H_{\gamma }}^{2}\right).}
If h∉H γ{\displaystyle h\not \in H_{\gamma }} the two measures are singular.
Integration by parts
[edit]
The Cameron–Martin formula gives rise to an integration by parts formula on E{\displaystyle E}: if F:E→R{\displaystyle F:E\to \mathbf {R} } has boundedFréchet derivativeD F:E→Lin(E;R)=E∗{\displaystyle \mathrm {D} F:E\to \operatorname {Lin} (E;\mathbf {R} )=E^{}}, integrating the Cameron–Martin formula with respect to Wiener measure on both sides gives
∫E F(x+t i(h))d γ(x)=∫E F(x)exp(t⟨h,x⟩∼−1 2 t 2‖h‖H 2)d γ(x){\displaystyle \int {E}F(x+ti(h))\,\mathrm {d} \gamma (x)=\int {E}F(x)\exp \left(t\langle h,x\rangle ^{\sim }-{\tfrac {1}{2}}t^{2}\|h\|_{H}^{2}\right)\,\mathrm {d} \gamma (x)}
for any t∈R{\displaystyle t\in \mathbf {R} }. Formally differentiating with respect to t{\displaystyle t} and evaluating at t=0{\displaystyle t=0} gives the integration by parts formula
∫E D F(x)(i(h))d γ(x)=∫E F(x)⟨h,x⟩∼d γ(x).{\displaystyle \int {E}\mathrm {D} F(x)(i(h))\,\mathrm {d} \gamma (x)=\int {E}F(x)\langle h,x\rangle ^{\sim }\,\mathrm {d} \gamma (x).}
Comparison with the divergence theorem of vector calculus suggests
d i vV h=−⟨h,x⟩∼,{\displaystyle \mathop {\mathrm {div} } V_{h}=-\langle h,x\rangle ^{\sim },}
where V h:E→E{\displaystyle V_{h}:E\to E} is the constant "vector field" V h(x)=i(h){\displaystyle V_{h}(x)=i(h)} for all x∈E{\displaystyle x\in E}. The wish to consider more general vector fields and to think of stochastic integrals as "divergences" leads to the study of stochastic processes and the Malliavin calculus, and, in particular, the Clark–Ocone theorem and its associated integration by parts formula.
An application
[edit]
Using Cameron–Martin theorem one may establish (See Liptser and Shiryayev 1977, p.280) that for a q×q{\displaystyle q\times q} symmetric non-negative definite matrixH(t){\displaystyle H(t)} whose elements H j,k(t){\displaystyle H_{j,k}(t)} are continuous and satisfy the condition
∫0 T∑j,k=1 q|H j,k(t)|d t<∞,{\displaystyle \int {0}^{T}\sum {j,k=1}^{q}|H_{j,k}(t)|\,dt<\infty ,}
it holds for a q{\displaystyle q}−dimensional Wiener processw(t){\displaystyle w(t)} that
E[exp(−∫0 T w(t)∗H(t)w(t)d t)]=exp[1 2∫0 T tr(G(t))d t],{\displaystyle E\left[\exp \left(-\int {0}^{T}w(t)^{}H(t)w(t)\,dt\right)\right]=\exp \left[{\tfrac {1}{2}}\int {0}^{T}\operatorname {tr} (G(t))\,dt\right],}
where G(t){\displaystyle G(t)} is a q×q{\displaystyle q\times q} nonpositive definite matrix which is a unique solution of the matrix-valued Riccati differential equation
d G(t)d t=2 H(t)−G 2(t){\displaystyle {\frac {dG(t)}{dt}}=2H(t)-G^{2}(t)}
with the boundary condition G(T)=0{\displaystyle G(T)=0}.
In the special case of a one-dimensional Brownian motion where H(t)=1/2{\displaystyle H(t)=1/2}, the unique solution is G(t)=tanh(t−T){\displaystyle G(t)=\tanh(t-T)}, and we have the original formula as established by Cameron and Martin: E[exp(−1 2∫0 T w(t)2 d t)]=1 coshT.{\displaystyle E\left[\exp \left(-{\tfrac {1}{2}}\int _{0}^{T}w(t)^{2}\,dt\right)\right]={\frac {1}{\sqrt {\cosh T}}}.}
See also
[edit]
Girsanov theorem– Theorem on changes in stochastic processes
Sazonov's theorem
References
[edit]
^Bogachev, Vladimir (1998). Gaussian Measures. Rhode Island: American Mathematical Society.
Cameron, R. H.; Martin, W. T. (1944). "Transformations of Wiener Integrals under Translations". Annals of Mathematics. 45 (2): 386–396. doi:10.2307/1969276. JSTOR1969276.
Liptser, R. S.; Shiryayev, A. N. (1977). Statistics of Random Processes I: General Theory. Springer-Verlag. ISBN3-540-90226-0.
Elworthy, David (2008), MA482 Stochastic Analysis(PDF), Lecture Notes, University of Warwick
Lunardi, Alessandra; Miranda, Michele; Pallara, Diego (2016), Infinite Dimensional Analysis, Lecture Notes, 19th Internet Seminar, Dipartimento di Matematica e Informatica Università degli Studi di Ferrara
| v t e Measure theory |
| Basic concepts | Absolute continuityof measures Lebesgue integration L p spaces Measure Measure space Probability space Measurable space/function |
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| Types of measures | Atomic Baire Banach Besov Borel Brown Complex Complete Content (Logarithmically)Convex Decomposable Discrete Equivalent Finite Inner (Quasi-)Invariant Locally finite Maximising Metric outer Outer Perfect Pre-measure (Sub-)Probability Projection-valued Radon Random Regular Borel regular Inner regular Outer regular Saturated Set function σ-finite s-finite Signed Singular Spectral Strictly positive Tight Vector |
| Particular measures | Counting Dirac Euler Gaussian Haar Harmonic Hausdorff Intensity Lebesgue Infinite-dimensional Logarithmic Product Projections Pushforward Spherical measure Tangent Trivial Young |
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| Main results | Carathéodory's extension theorem Convergence theorems Dominated Monotone Vitali Decomposition theorems Hahn Jordan Maharam's Egorov's Fatou's lemma Fubini's Fubini–Tonelli Hölder's inequality Minkowski inequality Radon–Nikodym Riesz–Markov–Kakutani representation theorem |
| Other results | Disintegration theorem Lifting theory Lebesgue's density theorem Lebesgue differentiation theorem Sard's theorem Vitali–Hahn–Saks theorem For Lebesgue measure Isoperimetric inequality Brunn–Minkowski theorem Milman's reverse Minkowski–Steiner formula Prékopa–Leindler inequality Vitale's random Brunn–Minkowski inequality |
| Applications&related | Convex analysis Descriptive set theory Probability theory Real analysis Spectral theory |
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| Results | Cameron–Martin theorem Inverse function theorem Nash–Moser theorem Feldman–Hájek theorem No infinite-dimensional Lebesgue measure Sazonov's theorem Structure theorem for Gaussian measures |
| Related | Crinkled arc Covariance operator |
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| v t e Hilbert spaces |
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| Examples | C n(K) with K compact &n<∞ Segal–Bargmann F |
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Cameron–Martin theorem
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8151 | https://planetmath.org/convergenceofthesequence11nn | convergence of the sequence (1+1/n)^n
Theorem 1.
The following sequence:
| | |
--- |
| | (1) |
is convergent.
Proof.
The proof will be given by demonstrating that the sequence (1) is:
1.
monotonic (increasing), that is
2. 2.
bounded above, that is for some
In order to prove part 1, consider the binomial expansion for :
| |
| |
Since , and since the sum has one term more than , it is demonstrated that the sequence (1) is monotonic. In order to prove part 2, consider again the binomial expansion:
| |
| |
Since and :
| |
| |
where the formula giving the sum of the geometric progression with ratio has been used. ∎
In conclusion, we can say that the sequence (1) is convergent and its limit corresponds to the supremum of the set , denoted by , that is:
| |
| |
which is the definition of the Napier’s constant.
| | |
--- |
| Title | convergence of the sequence (1+1/n)^n |
| Canonical name | ConvergenceOfTheSequence11nn |
| Date of creation | 2013-03-22 17:43:26 |
| Last modified on | 2013-03-22 17:43:26 |
| Owner | kfgauss70 (18761) |
| Last modified by | kfgauss70 (18761) |
| Numerical id | 7 |
| Author | kfgauss70 (18761) |
| Entry type | Theorem |
| Classification | msc 33B99 |
| Related topic | NondecreasingSequenceWithUpperBound | |
8152 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8198809/ | Design and Calibration of a Hall Effect System for Measurement of Six-Degree-of-Freedom Motion within a Stacked Column - PMC
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Sensors (Basel)
. 2021 May 27;21(11):3740. doi: 10.3390/s21113740
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Design and Calibration of a Hall Effect System for Measurement of Six-Degree-of-Freedom Motion within a Stacked Column
Olafur Oddbjornsson
Olafur Oddbjornsson
1 EFLA Consulting Engineers, Lynghals 4, 110 Reykjavik, Iceland; Olafur.oddbjornsson@efla.is
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1, Panos Kloukinas
Panos Kloukinas
2 Department of Civil Engineering, School of Engineering, University of Greenwich, Central Avenue, Chatham Maritime, Kent ME4 4TB, UK; P.Kloukinas@greenwich.ac.uk
Find articles by Panos Kloukinas
2, Tansu Gokce
Tansu Gokce
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Kate Bourne
Kate Bourne
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Tony Horseman
Tony Horseman
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Luiza Dihoru
Luiza Dihoru
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Matt Dietz
Matt Dietz
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Rory E White
Rory E White
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3, Adam J Crewe
Adam J Crewe
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3,, Colin A Taylor
Colin A Taylor
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
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3
Editor: Arcady Zhukov
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1 EFLA Consulting Engineers, Lynghals 4, 110 Reykjavik, Iceland; Olafur.oddbjornsson@efla.is
2 Department of Civil Engineering, School of Engineering, University of Greenwich, Central Avenue, Chatham Maritime, Kent ME4 4TB, UK; P.Kloukinas@greenwich.ac.uk
3 Earthquake and Geotechnical Engineering Group, Faculty of Engineering, University of Bristol, University Walk, Bristol BS8 1TR, UK; tansu.gokce@bristol.ac.uk (T.G.); k.bourne@bristol.ac.uk (K.B.); tony.horseman@bristol.ac.uk (T.H.); luiza.dihoru@bristol.ac.uk (L.D.); m.dietz@bristol.ac.uk (M.D.); rory.white@bristol.ac.uk (R.E.W.); colin.taylor@bristol.ac.uk (C.A.T.)
Correspondence: A.J.Crewe@bristol.ac.uk
Roles
Arcady Zhukov: Academic Editor
Received 2021 May 6; Accepted 2021 May 26; Collection date 2021 Jun.
© 2021 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC8198809 PMID: 34072278
Abstract
This paper presents the design, development and evaluation of a unique non-contact instrumentation system that can accurately measure the interface displacement between two rigid components in six degrees of freedom. The system was developed to allow measurement of the relative displacements between interfaces within a stacked column of brick-like components, with an accuracy of 0.05 mm and 0.1 degrees. The columns comprised up to 14 components, with each component being a scale model of a graphite brick within an Advanced Gas-cooled Reactor core. A set of 585 of these columns makes up the Multi Layer Array, which was designed to investigate the response of the reactor core to seismic inputs, with excitation levels up to 1 g from 0 to 100 Hz. The nature of the application required a compact and robust design capable of accurately recording fully coupled motion in all six degrees of freedom during dynamic testing. The novel design implemented 12 Hall effect sensors with a calibration procedure based on system identification techniques. The measurement uncertainty was ±0.050 mm for displacement and ±0.052 degrees for rotation, and the system can tolerate loss of data from two sensors with the uncertainly increasing to only 0.061 mm in translation and 0.088 degrees in rotation. The system has been deployed in a research programme that has enabled EDF to present seismic safety cases to the Office for Nuclear Regulation, resulting in life extension approvals for several reactors. The measurement system developed could be readily applied to other situations where the imposed level of stress at the interface causes negligible material strain, and accurate non-contact six-degree-of-freedom interface measurement is required.
Keywords: Hall effect sensors, displacement measurement, 6DoF, stacked column, seismic testing, nonlinear calibration
1. Introduction
Within the UK, there are currently seven nuclear power plants which have an Advanced Gas-cooled Reactor (AGR) core . The AGR cores are constructed using graphite bricks of varying geometry. A full description of the function of each brick type and detail of the key–keyway system is provided in the second paper of Neighbour , ‘AGR Core Design, Operation and Safety Functions’. The AGR cores are now approaching the end of their design life [3,4], and in order to maintain safe operation, software tools are being used to assist in detailed analysis of the core components [5,6]. The safety case for each AGR facility includes consideration of the response of the nuclear reactor core to seismic events . This includes requirements to maintain the ability to safely shut down the reactor in the case of a seismic event. To this end, the University of Bristol developed a Multi Layer Array (MLA), which is a quarter-sized model of an AGR core. The rig has a platform size of 2.497 m × 2.497 m and a height of 1.731 m, and weighs approximately 9T. Full detail of the MLA design and construction is provided in , and an introduction to the interface instrumentation is provided in . Experiment results generated by the MLA are used to verify and validate software tools that subsequently provide analysis of the graphite bricks within AGR cores. The MLA is placed on a shaking table and shaken to allow investigation of the dynamic response of the array to inputs representative of different seismic events. Typical excitations include sine dwells and artificial earthquake motions. The peak accelerations of the input motions vary from 0.4 g to 1 g and the motions have broad band frequency content from 0 Hz to 100 Hz. To record the motion of individual components within the array, selected columns required instrumentation capable of measuring fully coupled six-degree-of-freedom (6DoF) motion. The columns that form the basis of this study comprise scale models of graphite moderator bricks within an AGR core. The graphite moderator brick types include fuel bricks, unkeyed interstitial bricks and keyed interstitial bricks. Within the context of the MLA, the model fuel bricks are referred to as lattice bricks, the model unkeyed interstitial bricks are referred to as filler bricks and the keyed interstitial bricks are referred to as interstitial bricks.
The MLA is composed of eight layers representing the inner 10 octagonal rings of an AGR core. The outer ring and bottom layer of the MLA are made up of restraint components which are fastened to the outer structure of the MLA rig. As such, the active array comprises the nine central rings and the top seven layers of the MLA which represent the dynamic behaviour of the lower eight layers of radially keyed bricks in AGR cores. The bottom layer of bricks in the MLA is restrained by a set of interlocking base plates, and the detail of one is provided in Figure 1. The lattice bricks are installed on rocking features on the base plate, so whilst they are not mechanically fastened to the MLA frame, and are allowed to rock, the horizontal motion of the bottom layer of bricks is constrained to the motion of the base plate. In the interstitial columns, the bottom layer of the filler bricks is rigidly connected to the base plate so the input motion is applied at the interface below the lowest interstitial brick. This interface can accommodate motion in all six degrees of freedom. The top layer of the array is made up of lattice bricks that are two-thirds the full brick height, which ensures that the lattice columns do not protrude above the interstitial columns. The reduction in height of the top layer lattice bricks effectively provides a symmetry constraint and avoids the overturning moments that would have existed had standard bricks been used, protruding above the installed interstitial bricks.
Figure 1.
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(a) Detail of the base plate. (b) Installation of fuel and interstitial bricks on the base plate, side view. (c) Installation of fuel and interstitial bricks on the base plate, isometric view.
A plan view of the MLA is shown in Figure 2 along with detail of the lattice and interstitial columns. The lattice bricks are stacked to form the lattice columns, each comprising six full-sized bricks plus the top layer brick two-thirds the full brick height. Between the lattice columns, filler bricks are stacked alternately with interstitial bricks to form the interstitial columns, each comprising seven full-size interstitial bricks and six full-size filler bricks. Due to the geometry of the base plate, the filler brick installed at the bottom layer is slightly shorter than full-size and is referred to as a hybrid brick. Each hybrid brick is fastened to the base plate using two 5 mm screws. Geometric detail of the interstitial and filler bricks is provided in Figure 3.
Figure 2.
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(a) MLA dimensions. (b) Layout of interstitial and lattice columns within the MLA.
Figure 3.
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(a) Interstitial brick dimensions. (b) Filler brick dimensions.
In total, the active section of the MLA comprises 284 lattice columns and 301 interstitial columns, equating to 4095 bricks. Of these 585 columns, up to eight interstitial and six lattice columns are instrumented per experiment. The six-degree-of-freedom motion at each interface is fully coupled on each axis and presents a significant challenge from both an instrumentation and sensor processing perspective. These interface measurements are particularly important as they are one of main outputs of the dynamic testing of the MLA physical model.
This paper presents a novel solution and framework for the design and subsequent calibration of the fully coupled six-degree-of-freedom motion of one of these interfaces in a column of stacked interstitial bricks. The ranges of motions to be measured are identified and the selection of the measurement devices is described in Section 2. The optimum design of sensor locations within the interface is discussed in Section 3. A detailed description of the calibration technique, using a FARO coordinate measurement arm as a reference, is presented in Section 4 and Section 5. The final design is detailed in Section 6 along with calibration results. The final calibration matrix for the interface employs an 8th order polynomial fit to the data and the uncertainties associated with the calibration are presented in Section 7. Due to the redundancy of the sensors in the interface, the calibration procedure can also be modified to cope with the loss of some sensor channels. Section 8 outlines the changes to the calibration procedure which can account for the loss of sensor channels without significant degradation in the accuracy of results. The system presented here also meets experiment requirements of the MLA tests for a compact and robust design. This measurement system could be readily applied to other situations where accurate non-contact six-degree-of-freedom interface measurement is required.
2. Sensor Selection
During each MLA experiment, the interstitial and filler bricks are positioned as per Figure 2, and as such brick-to-brick interface measurement ranges are physically constrained. To ensure that the sensor design was robust and capable of measuring displacement beyond the anticipated range of motion, the values presented in Table 1 were adopted as the design measurement ranges. The relatively low weight of the bricks, approximately 130 g, excludes the use of conventional spring-loaded contact instruments such as potentiometric and inductive sensors. Since any instruments requiring physical contact between the bricks would excessively influence the dynamic behaviour of the brick interfaces during testing, the range of measurement instruments investigated was limited to non-contact sensors.
Table 1.
Interface measurement ranges for design.
| Measurement Axis | Range | Measurement Angle | Range |
:---: :---: |
| X | ±6 mm | Roll | ±6° |
| Y | ±6 mm | Pitch | ±6° |
| Z | +4 mm | Yaw | ±6° |
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A wide range of non-contact measurement sensors were considered for the interface measurement system. These included Capacitive sensors, Eddy Current sensors, Optical sensors, and Hall effect sensors. The MLA design required not only that the sensors be small enough to be installed within the bricks, but that additional processing such as signal conditioning and data acquisition also be accommodated within each individual brick. To ensure that the dynamic response of the column was preserved, the wiring associated with the instrumentation had to be minimised to make certain that the componentry and associated connections did not impede or influence the movement of individual bricks.
Neither the Capacitive nor Eddy Current sensors were small enough to be accommodated within the bricks, which would have necessitated an excessive amount of instrumentation to be fitted within the column and would have interfered with the dynamic behaviour of the bricks. Of the range of optical sensors available, confocal sensors were sufficiently small to be installed within a brick. By mounting three confocal sensors at the brick interface, it would be possible to accurately measure the roll and pitch movement at the brick-to-brick interface. When considering the practicalities of the confocal sensor installation, the extensive signal conditioning required by the sensors would lead to a prohibitive number of cables passing through the column, which would likely interfere with the dynamic response. Additionally, the confocal sensors were an expensive option, and given the quantity required to appropriately instrument the array for an experiment, the full cost was beyond the project budget.
Finally, the Hall effect sensors were assessed. While these sensors are often used as current sensors , using them in conjunction with permanent magnets allows them to be used as displacement sensors. Hall effect sensors are available in small, Integrated Circuit (IC) based formats and can be easily installed within a brick, with magnets of appropriate strength mounted in the opposing brick. By fitting multiple Hall effect sensors in a combination of orientations, it is possible to track the brick-to-brick interface in six degrees of freedom. The Hall effect sensors also presented an affordable option, and as such were chosen as the most suitable non-contact sensor.
A Hall effect sensor is a transducer whose output voltage changes in the presence of an applied magnetic field. The sensor consists of a thin metal strip with a constant electric current flowing along its length. In the presence of a magnetic field, the charge carriers, electrons and holes move to the opposite non-energised edges of the metal strip, creating a measurable voltage differential between the edges of the strip that is proportional to the applied external magnetic field [11,12,13].
The advantages of using Hall effect sensors to measure position are numerous. Hall effect sensors are available ICs with inbuilt Hall sensing elements and linear amplifiers. These ICs, requiring only a stable supply voltage, produce an output voltage that varies with the applied magnetic field. Hall effect ICs are cost-effective, compact, easy to handle and are available in varying layouts and for different ranges of magnetic strength. For the application at hand, minimisation of the required cabling within the array was also a critical consideration and another reason that the Hall effect sensors presented a practical solution to non-contact displacement measurement at the brick interfaces.
The use of Hall effect sensors for positional measurement is, however, not without disadvantages. While, for most Hall effect sensors, the relationship between sensor output and magnetic field strength is linear, when used as a displacement sensor the strength of the magnetic field does not change linearly as the distance between the magnetic source and the sensor is increased. Instead, the magnetic field dissipates as a function of the inverse square of the separation distance. For a 10 mm displacement between the sensor and the magnet, the strength of the magnetic field is reduced to only 1% of the strength exhibited at a separation distance of 1 mm. It was therefore concluded that measuring displacements greater than 10 mm was not feasible. As a result of this non-linearity in the sensor response [11,12], the best measurement accuracy is achieved over short separation distances, with the measurement accuracy deteriorating with increasing separation.
There are many examples of Hall effect sensors being applied for the purposes of displacement measurement. For example, Giovanola et al. designed a Hall effect displacement transducer for the measurement of crack propagation using two single-axis Hall effect sensors, Schott et al. designed a magnetic displacement sensor which made use of a two-axis Hall effect sensor to measure displacement on two axes, and Jones et al. used Hall sensors to measure normal and shear displacements in an elastomer to create Hall effect-based tactile sensors. Kawato and Kim and Zhang et al. implemented Hall effect sensors for accurate 2D positioning above a magnet matrix, Yang and Huang designed a miniature system comprising six single-axis Hall effect sensors for 3D displacement measurement, and Cole et al. used multiple Hall sensors to measure the 3D position of a test foot impacting onto artificial grass. Yu and Kim implemented three two-axis Hall effect sensors to measure lateral motion about the x- and y-axis, and rotation about the z-axis. Fontana at al. made use of Hall effect sensors to measure the rotation angle of a shaft, and Paul and Chang determined the position of a linear motor using a Hall effect sensor. Zhao at al. achieved measurements in three degrees of freedom using three single-axis Hall effect sensors, and Nie and Sup used two three-axis Hall effect sensors to successfully demonstrate a four-degree-of-freedom load cell. The application of Hall effect sensors for the direct measurement of fully coupled six-degree-of-freedom motion, as is required in this unusual application, is a more complex measurement task requiring a completely different approach.
3. Interface Instrumentation Concept
To enable measurement of the brick interface motion in six degrees of freedom, it was determined that a set of two bi-axial Hall effect transducers should be installed in each of three corners of a filler brick, with opposing magnets installed in the corresponding three corners of an interstitial brick. Such a configuration, consisting of six bi-axial Hall effect sensors and three magnets, allows the interface movement to be tracked in six degrees of freedom. The use of six bi-axial sensors, yielding 12 channels of data, also provides redundancy in an application where the risk of sensor damage mid-experiment is significant. The effect of the loss of working data channels is presented in Section 8.
With the configuration of the Hall effect sensors specified, the most appropriate magnet type for the application needed to be identified. This required consideration of both physical dimensions and constituent materials. Key requirements included an appropriate strength for the stipulated dimensions and a stable magnetic field which did not show variation with changing temperature. Three candidate materials were identified , the most suitable being AlNiCo (Aluminium–Nickel–Cobalt), followed by SmCo (Samarium–Cobalt) and Neodymium. Although the Neodymium magnets provided the strongest magnetic field, they were significantly more sensitive to temperature variations than the other two magnet types. The AlNiCo magnets offered the most stable solution with low sensitivity to thermal variation and a strong magnetic field; however, they can become demagnetised if handled incorrectly. Given the nature of the MLA experiments, it was preferable to prioritise a robust solution, and as such SmCo was chosen for the application, even though the magnetic field had some sensitivity to temperature variations. The physical dimensions of the required magnet were defined by the geometry of the test specimen, and the strength of the magnet was matched to the maximum range of the Hall effect sensor.
To achieve the best performance from an instrumented interface employing Hall effect sensors and magnets, the placement of sensors and magnets needs to be assessed to optimise the layout and maximise measurement accuracy. The optimal location of the sensors is dependent on the performance required. When choosing a layout optimised for the measurement of rotation, the sensor–magnet pairs should be in close proximity. Conversely, for the measurement of displacement which predominantly consists of shear and normal translation, with minimal rotations, the sensor–magnet pairs yield acceptable results when placed a greater distance apart. For the application at hand, both displacements and rotations need to be recorded with appropriate accuracy. There is also a practical limit to how close the sensor–magnet pairs may be positioned to prevent interference between each sensor pair. A separation distance 2.5 times the absolute maximum measurement range should be maintained to ensure interference is minimised and measurement accuracy is retained. In this case, the minimum separation distance was set at 25 mm between sensor–magnet pairs.
In general, the sensor–magnet pairs should be placed in an orientation as near to a square grid as is possible, with the centre of the grid at the centre of the interface. Placing the sensor–magnet pairs at the minimum separation distance will deliver the maximum measurement range for rotation, thus spreading the measurement accuracy over a wider range. In contrast, placing the sensor–magnet pairs further apart will decrease the measurement range for rotation and provide a better measurement accuracy over a narrow range. As such, it is important that the layout of the sensor–magnet pairs be considered in the context of each application.
It should be noted that the proposed instrumented interface, with embedded Hall effect sensors and magnets, is limited in its measurement to blocks which remain essentially rigid and un-deformed. If the housing of the sensors or the magnets becomes deformed, the calibration is no longer valid. The calibration procedure, which is discussed in detail in the following section, is an integral part of the measurement process. To obtain meaningful results from the raw data acquired from the Hall effect sensors, the sensor array in its entirety is calibrated against the magnet array in its entirety. The application presented here did not involve appreciable component deformation, and as such the relationship between the level of deformation and what effect it has on the measurement error was not studied.
4. Interface Calibration Concept
Each interface, comprising three sets of Hall effect sensors and corresponding magnets, required calibration to convert the raw sensor data to six-degree-of-freedom displacement measurements in the appropriate engineering units. Owing to the measurement complexity, the calibration process was non-trivial. For a given degree of freedom, the relationship between the sensor output and the movement at the component interface is highly non-linear and fully coupled to all other degrees of freedom and the brick sensor geometry itself. Standard calibration procedures which are typically implemented for more conventional instruments, such as single axis measurements or multi-axis applications where axes are normally orthogonal and cross-coupling is minimal, cannot be applied.
Examples of non-traditional approaches to calibration include the approach taken by Zhao et al. , who placed particular emphasis on decoupling the raw sensor data. By analytically fitting an elliptic function to the output of the three Hall effect sensors, the x and y displacements, along with the rotation about the z axis, could be determined. The work presented by Zhao et al. was subsequently extended to include analysis of a similar system comprising six Hall effect sensors yielding results in six degrees of freedom . Nie and Sup used experimental measurements to formulate analytical expressions to relate the magnetic field strength to the position of the magnet. The distribution of the magnetic flux density was analytically described using an expression derived from the Biot–Savart law . The approach of Northey et al. , who utilised four Hall effect sensors to measure displacement in three degrees of freedom (x, y and z), was to employ a feed forward neural network for calibration. This overcame difficulties associated with the non-linear behaviour of the sensors and crosstalk effects. To calibrate a system of three Hall effect sensors measuring three degrees of motion (x, y and rotation about z), Chen et al. made use of a so-called ‘soft sensor’ method to account for the non-linearity of the sensors. Key parameters were obtained numerically using optimal approximation theory, making use of the mathematical relationship between the parameters which could be directly measured and those which must be inferred.
The calibration of a six-degree-of-freedom measurement system is not only complicated by the inherent non-linearity of the Hall effect sensors, but in this case also by the fact that the movement of the interface is fully coupled in all degrees of freedom. Unlike other examples in the literature, the motion being measured at each brick interface is a response to an external input and constrained only by the presence of surrounding components. The calibration procedure must therefore be able to appropriately characterise this coupling in an efficient and robust manner.
To account for both the non-linearity of the sensors and the coupling between axes, the interface calibration was carried out in all six axes simultaneously. To achieve this, a high precision six-degree-of-freedom reference measurement device was required. The only appropriate devices available were instrumented hexapods, six axis precision stages and coordinate measurement arms.
The reference measurement device chosen for this application was a FARO Prime 6ft, six axis coordinate measurement arm made by FARO Technologies Inc. , with a measurement accuracy of ±0.015 mm. The rationale for this selection was the fact that the arm could record its position with high precision in six degrees of freedom at timing intervals dictated by an external trigger, which enabled time synchronisation between the data from the interface sensors, and that of the reference measurement. Following on from the choice of apparatus for the reference measurements was the identification of a motion sequence that should be applied at each interface to appropriately capture the dynamic response expected during experiments. The motion sequences applied during calibration should capture and characterise the system response for every combination of axis motion. As a fully coupled six-degree-of-freedom system, the so-called ‘calibration volume’, can be difficult to visualise as it is a sixth-order problem space. The calibration volume is essentially the parameter space of the sensors and is a 6DoF hypersphere.
To ensure that the motion sequences were suitable, an empirical approach was adopted whereby preliminary testing and evaluation was used to derive a predefined set of movement patterns covering all practical combinations of axial movements. To assess the quality of the resulting calibration volume, the response on each axis was plotted against all others. The movement patterns could then be revised to fill any unexpected gaps in the calibration volume. It is acknowledged that a more robust method for evaluating the calibration space could be devised in future, such that there can be high confidence that the calibration volume sufficiently covers the range of interface motion captured during each experiment. Several attempts were made to automate the calibration process, but the nature of the contacts between the bricks required a system that would guarantee there would be no damage to any component. At the time of development, no solution could be identified within the time and resource constraints.
Dataset Reduction Prior to Calibration
For a manually operated calibration using a coordinate measurement arm, it is not possible to generate a perfectly well-defined calibration volume that exactly matches the limits of the experiment data. As such, the interfaces should be calibrated beyond the anticipated measurement range of the experiment to ensure that the calibration volume covers all motion captured during the experiment. Noting that practical measurement limits exist based on the constrained nature of the bricks within the MLA, any data pairs within the calibration dataset which exceed these limits should be discarded based on the values recorded by the FARO arm. When an interface does not reach the projected movement limits during an experiment, the calibration volume can be retrospectively reduced to increase the measurement accuracy. In this way, it is possible to increase the accuracy of the calibration, and by extension the accuracy of the experiment results. Shown in Figure 4 and Figure 5 are examples of the reference measurements obtained from the FARO arm on each axis, before and after dataset reduction. The dataset comprises 12 individual motion sequences that characterise the fully coupled nature of the interface motion. Presented as image (b) in Figure 4 and Figure 5 are the reduced datasets as implemented during the calibration process described in the following section. Please refer to Section 6 for full discussion of the interface measurement limits for data reduction.
Figure 4.
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Reference displacement data measured by the FARO arm. (a) Full reference dataset for measured displacements at the interface. (b) Reduced reference dataset for use during calibration.
Figure 5.
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Reference rotation data measured by the FARO arm. (a) Full reference dataset for measured rotations at the interface. (b) Reduced reference dataset for use during calibration.
5. Calibration Procedure
Nonlinear system identification based on a standard linear least squares method is a technique commonly used to identify the equations of a motion of a system where the underlying physics cannot be easily derived. The method is often applied to empirical datasets measured from complex systems. In this case, the system identification approach is used to correlate the output from the Hall effect sensors to the reference data, which are the data collected by the coordinate measurement arm. By doing so, it is then possible to determine the relative motion at each interface, in six degrees of freedom, from the data recorded by the Hall effect sensors.
A polynomial fit was chosen to correlate the output from the Hall effect sensors to the six-degree-of-freedom position recorded by the coordinate measurement arm. Additional terms were then added to determine if the accuracy of the fit could be improved. Coupled terms, log and exponential terms were assessed and it was found that the best fit was achieved when a linearly independent polynomial was implemented to model each sensor independently. Assuming 12 Hall effect sensors monitoring the position of three magnets on the opposite side of an interface, the polynomial fit for each degree of freedom is as follows:
(1)
where n is the order of the polynomial, r(t) is the reference measurement at time t for the degree of freedom to be fitted, v j(t) is the voltage output from Hall effect sensor j at time t, and c i,j is the fitted polynomial coefficient for the output from Hall effect sensor j and for the i-th term of the polynomial.
Now the system of equations to be solved in the linear least square sense can be presented in a matrix-vector form as:
(2)
where matrix V incorporates the Hall effect output data as:
(3)
and where matrix R holds the reference data from the FARO coordinate measurement arm as:
(4)
where x and y are horizontal translations, z is vertical translation, φ is rotation about the x-axis (roll), θ is rotation about the y-axis (pitch) and ψ is rotation about the z-axis (yaw). Finally, the calibration coefficient matrix C is of the following form:
(5)
The calibration coefficient matrix C is evaluated from the calibration data contained within matrices V and R by solving Equation (2) for C in the least square sense. The Hall effect output data matrix V is the size of the number of samples m by the number of sensors, 12 for the case at hand, by the number of polynomial terms (n + 1). The size of the reference data matrix R is given by the number of samples (m) by the number of degrees of freedom, six in this case. The size of the calibration matrix C is given by the number of sensors, the number of polynomial terms (n + 1) and the number of degrees of freedom.
To generate the calibration matrix C, and thus the calibration dataset, the only variable left to be determined is the order of the polynomial required to generate an appropriate fit between the Hall effect sensor data and the reference data. To assess the required polynomial order, the most pragmatic approach in this instance was to generate calibrations for increasing orders of polynomial. Residuals between the ‘fit’ and reference data were then used to calculate the Sum of Squared Residuals (SSR). It is acknowledged that increasing the order of the polynomial will inevitably yield decreasing values for SSR and can lead to over-fitting of the data . As such, the SSR was used qualitatively as an indicative metric to threshold the lowest acceptable order of the polynomial. More detailed analysis of the resulting calibration volumes for each order of the polynomial were then assessed to determine the most suitable polynomial order, as presented in the following section.
6. Final Design and Calibration Results
To validate the capabilities and performance of the proposed interface measurement technique, a prototype interface was produced. Two bricks were fabricated, each a quarter-size model of the interstitial bricks that compose an AGR core. The filler brick, modelling an unkeyed interstitial brick, had a 47.2 mm by 47.2 mm square section, a 39.5 mm diameter by 3.4 mm high spigot at its top face and a 31.7 mm diameter bore centred on the middle of its cross section. As shown in Figure 6a, each of the three corners of the filler brick were fitted with four Hall effect sensors. A non-magnetic peg was fitted to the corner of the filler brick that did not house Hall effect sensors. During installation in the MLA, the peg on the filler brick aligned with a small cavity on the corresponding face of the interstitial brick. This constrained the motion of the filler bricks in yaw (rotations about the z-axis), ensuring that the behaviour of the filler bricks was representative of the unkeyed interstitial graphite bricks of an AGR core.
Figure 6.
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(a) Instrumented filler brick with Hall effect sensors and custom data acquisition board installed. (b) Instrumented interstitial brick with magnets installed.
The second brick, modelling a keyed interstitial brick with keys on each face, had a 47.2 mm by 47.2 mm square section, a 31.7 mm diameter hole at the middle of its cross section and a 40.0 mm diameter by 3.16 mm deep recess at its bottom face. As presented in Figure 6b, the lower face of the brick was fitted with magnets corresponding to the locations fitted with Hall effect sensors on the filler brick. The geometry of the prototype interface restricted horizontal travel to ±0.5 mm as a result of the clearance between the spigot and socket joint. The vertical translation and rotations about all three axes were not limited by the geometry of the interface, as the peg on the filler brick was removed during calibration.
To undertake the calibration process, as described in Section 5, customised benchtop equipment was designed and manufactured to accommodate the FARO coordinate measurement arm. The calibration setup is shown in Figure 7. A clamping frame held the bottom brick, housing the Hall effect sensors, below the interface. The top brick, housing the magnets, was attached directly to the measurement arm using a custom-made connector. The clamping frame was designed and manufactured to a high degree of precision to ensure that the longitudinal axis of the brick was parallel with the vertical axis of the coordinate measurement arm. Custom probes were also fabricated to enable rigid mounting of the top brick, such that the centre of the measurement arm probe was coincident with the centre of the brick. The custom probes were made from titanium to minimise magnetic interference and manufactured with a range of fractionally different diameters to accommodate the manufacturing tolerances of the keyed brick bores.
Figure 7.
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Photo showing calibration of the prototype interface as installed in the calibration rig.
A custom designed data acquisition system was used to collect the data from both the FARO arm and the hall sensors in the interface being calibrated. More details about the design of the miniaturised 32 channel simultaneous sample and hold data acquisition units that fitted inside the bricks, and of the end-of-column hardware that distributed the sampling clock, can be found in Crewe et al. .
Following the calibration procedure outlined in Section 5, the prototype interface was calibrated concentrating on rotations in roll and pitch of ±1.2°, and ±2.4° in yaw. The full set of limits imposed for the calibration of the prototype interface is provided in Table 2. It should be noted that the limits presented in Table 1 were for the purposes of instrumentation design, whereas the limits presented in Table 2 reflect the maximum brick displacements anticipated during testing. Horizontal translation was limited by the clearance between the recess and the spigot. The vertical translation was limited by the fact that the two bricks were required to always maintain at least one point of contact to appropriately replicate the anticipated conditions of the MLA. The calibration session consisted of 12 individual tests, each approximately 90 s in length. Each individual test consisted of a pre-defined combination of movements to maximise the response on a single axis. For example, the first test recorded involved horizontal translation combined with rotation in yaw while minimising motion in roll and pitch. This aimed to ensure that the full range of motion was captured for the calibration parameter space. At the start of each test, the operator was required to align the interface such that there was zero translation or rotation. This position was then maintained while recording commenced such that each test point had a zero-position logged to quantify the DC offset. Once all calibration test points were completed, the full calibration dataset could be generated by combining and concatenating data from the reference arm and the Hall effect sensors. The zero-position of the interface was defined as the average of the zero-point positions recorded for each test point. Finally, any data points having one or more degrees of freedom beyond the limits imposed for the test were removed from the dataset. All post-processing of the calibration data was undertaken using MATLAB .
Table 2.
Interface measurement limits.
| Measurement Axis | Range | Measurement Angle | Range |
:---: :---: |
| X | ±1.2 mm | Roll | ±1.2° |
| Y | ±1.2 mm | Pitch | ±1.2° |
| Z | ±1.2 mm | Yaw | ±2.4° |
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Once acquired, the raw calibration dataset was reduced by applying the limits in Table 2, as discussed in section “Dataset Reduction Prior to Calibration” of Section 4. The calibration procedure could then be applied. A polynomial fit between the sensor data and the reference data was produced from second to tenth order. The quality of the fit was assessed using both the SSR and analysis of the resulting calibration volume, for each order of the polynomial in each degree of freedom. Section 7 discusses the statistical characterisation of the measurement uncertainty, which takes into account the uncertainty associated with the independent polynomial fitting method.
Presented in Figure 8 is the SSR associated with increasing order of polynomial for each degree of freedom. From these results, a 6th order polynomial was deemed the lowest order of polynomial suitable for the application at hand. Calibration volumes were then generated from each polynomial fit from 6th to 10th order. The coverage associated with each calibration volume was then individually assessed based on the interface measurement limits presented in Table 2. Within the context of this application, an eighth-order polynomial was determined the most appropriate for the purposes of defining a calibration volume. The calibration volumes generated by an 8th order polynomial fit are presented in Figure 9a,b for displacement and rotation, respectively.
Figure 8.
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Square Sum of Residual (SSR) errors with increasing polynomial order. (a) Displacement in x, y and z. (b) Rotation in pitch, roll and yaw.
Figure 9.
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Calibration volume generated by an eighth-order polynomial fit to the reference data. Data points (blue) are overlaid with surfaces (red) for visualisation of the full volume to be applied to the raw sensor data. (a) Displacement calibration volume. (b) Rotation calibration volume.
As an alternative to using a system identification technique for the Hall effect sensor calibration, an approach using Neural Networks (NN) was also investigated. The NN consisted of 12 inputs which were the Hall Effect transducer measurements, 6 outputs which were the FARO Arm recordings, 3 hidden layers of 8 neurons each and an output layer. Full details of the method are given in Dihoru et al. . While the machine learning method proved to be viable, the training of the NN required long computational times and the subsequent calibration was not as accurate as that achieved using a polynomial fit. The NN was able to measure the movements of the interface with accuracies of ±0.069 mm for displacement and ±0.074 degrees for rotation or better, while the 8th order polynomial fit achieved accuracies of ±0.050 mm for displacement and ±0.052 degrees for rotation or better. In addition, for this application, it was deemed preferable to utilise a deterministic analysis procedure based on the physical relationship between the input and output measurements. Consequently, the system identification calibration process was adopted for the experimental test program.
7. Measurement Uncertainty
Given the nature of the calibration method, it was not possible to achieve a perfect fit for all spatial coordinates. As shown in Figure 8, there are residuals between the reference measurements and those generated by polynomial fit. Such a difference in results is to be expected and stems from the error inherent in the instrumentation and calibration process. This includes environmental effects such as electrical noise and mechanical vibrations along with thermal effects. As previously noted, the magnetic field of SmCo magnets is sensitive to operating temperature. During calibration, air cooling is applied to the filler brick to stabilise and control the operating temperature of the installed electronics, which results in some flow of hot air over the magnets on the keyed brick. This setup aimed to replicate the temperature distribution within the MLA during experiments but could not fully account for temperature effects on the sensors. As it was not possible to directly quantify the measurement uncertainty from such error sources, a statistical approach was adopted.
It is possible to statistically define a confidence interval for the measurement uncertainty whereby a 95% confidence interval can be determined. For each axis, the mean and standard deviation of the residual between the reference data and that generated by the polynomial fit can be calculated to generate a confidence interval. This analysis assumes that the measurement errors are independent of the spatial location within the calibration volume. Within the context of the MLA, every interface will have an individual confidence interval associated with the measured data, as each interface is individually calibrated. By way of example, the following analysis is presented for a case study that involves an external input oriented at 45 degrees to the sensors, to ensure that there is excitation on both the x- and y-axes. The magnitude of the input is indicative of the input magnitude applied during experiments.
Shown below in Table 3 and Table 4 are representative examples of the results from the top and bottom interface of an interstitial brick. From these tables, the measurement uncertainty associated with translation motion is 0.050 mm for the top interface and 0.032 mm for the bottom interface. The measurement uncertainty in rotation is 0.050 degrees for the top interface and 0.052 degrees for the bottom interface. From these results, it can be surmised that despite the errors identified with the 8th order polynomial fit, as presented in Figure 8, very good performance can be achieved from the instrumentation and calibration procedure.
Table 3.
Statistical characteristics of a 95% confidence interval for the top interface of an interstitial brick.
| Axis | Mean (μ) | Standard Deviation (σ) | 95% Confidence Interval (μ ± 2σ) |
:---: :---: |
| X (mm) | 0.000 | 0.016 | ±0.032 |
| Y (mm) | 0.000 | 0.017 | ±0.034 |
| Z (mm) | 0.000 | 0.025 | ±0.050 |
| Roll (degrees) | 0.000 | 0.023 | ±0.046 |
| Pitch (degrees) | 0.000 | 0.025 | ±0.050 |
| Yaw (degrees) | 0.000 | 0.022 | ±0.044 |
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Table 4.
Statistical characteristics of a 95% confidence interval for the bottom interface of an interstitial brick.
| Axis | Mean (μ) | Standard Deviation (σ) | 95% Confidence Interval (μ ± 2σ) |
:---: :---: |
| X (mm) | 0.000 | 0.016 | ±0.032 |
| Y (mm) | 0.000 | 0.015 | ±0.030 |
| Z (mm) | 0.000 | 0.009 | ±0.018 |
| Roll (degrees) | 0.000 | 0.015 | ±0.030 |
| Pitch (degrees) | 0.000 | 0.026 | ±0.052 |
| Yaw (degrees) | 0.000 | 0.016 | ±0.032 |
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8. Faulty Channel Handling
Given the nature of the MLA experiments, which can comprise up to 500 individual tests, the electronics are subject to a range of environmental influences and it is not uncommon for sensors to fail partway through a test programme. Some of the testing, where the interface measurement system was deployed, involved excitation of the whole MLA at levels up to 1 g, resulting in accelerations in excess of 2 g in some bricks, particularly during impact events between bricks. These high forces occasionally lead to failures in the connectors between the sensors and the local acquisition system in the bricks, resulting in partial data loss for an interface. When sensor failures do occur, it is not possible to restore or repair the faulty channels midway through an experiment. As a result, erroneous data are recorded at that interface for every test after the sensor failure. If faulty channels are not removed, they cause anomalous results when applying the calibration and subsequent calculation of interface position. At the end of each experiment, the data undergo preliminary post-processing to calibrate the raw data and to assess the results for potential hardware failures. Given that the calibration procedure applies nonlinear system identification, it is possible to account for sensor failures at each individual interface using a method that makes use of the redundancy inherent in using 12 Hall effect sensor channels to capture six-degree-of-freedom motion.
Faulty channels are typically easy to identify, yielding either data that resemble electrical noise or a flatline response. Once identified, the faulty channel is removed from the raw dataset for all tests after the point of failure. The original calibration dataset for that interface is then reduced to remove all input from the faulty channel, as per the application of Equations (3)–(5), and a new calibration volume is generated. This new reduced calibration volume is then applied to the reduced raw dataset to generate good quality displacement data. It should be noted that the reduced calibration volume still includes the full range of motion sequences used to create the original calibration, and it simply reduces the number of data channels used to generate the calibration volume. This process provides a convenient and efficient way to overcome sensor failures, ensuring that data quality and integrity are maintained without the requirement for onerous post-processing.
A sensitivity study was undertaken to investigate the effect of multiple sensor failures on the measurement accuracy. For consistency, the same case study as Section 6 was investigated. Individual sensor channels were systematically removed from both the calibration and measurement datasets until only a single data channel remained at each of the three instrumented corners. Presented in Figure 10 is a drawing of the layout of the Hall effect data channels. Each bi-axial Hall effect sensor is aligned at 45 degrees to the global MLA axes, denoted by subscript ‘G’, and each instrumented corner comprises four data channels.
Figure 10.
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Schematic of data channel layout in a filler brick showing the orientation of local sensor axes and the global MLA axes.
The measurement uncertainty, as detailed in the previous section, was then calculated to determine the change in accuracy. Presented in Figure 11 is the effect of the number of sensors on the measurement uncertainty for translation and rotation, showing a non-linear increase in measurement uncertainty as the number of sensors is reduced. These results can be considered representative for an interface between interstitial and filler bricks noting that the exact value of measurement uncertainty varies for each individual interface. The increase in measurement uncertainty with a decreasing number of sensor channels is more pronounced for rotation than for translation. This is because the rotations are not directly measured and are instead inferred from the translational displacement. As a result, the error propagation is higher for pitch, roll and yaw as measurement uncertainty increases.
Figure 11.
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Effect of the number of sensors on the measurement uncertainty in translation and rotation.
With all sensors operating, the measurement uncertainty for the case study was 0.05 mm for translation and 0.049 degrees for rotation. Reducing the number of available sensors to 10 yielded a measurement uncertainty of 0.061 mm in translation and 0.088 degrees in rotation. With only eight channels available, the measurement uncertainty increased to 0.099 mm in translation and 0.148 degrees in rotation. In practical terms, the loss of two sensor channels from two different locations was deemed acceptable for the purposes of the MLA experiments. A loss of a further two channels could be tolerated if the data were restricted to qualitative analysis, and if there remained a minimum of two sensor channels per corner. The accuracy of the data was not considered fit for purpose if fewer than eight sensor channels were available.
The interface measurement system had to be miniaturised to enable installation within each instrumented brick in the MLA to meet the requirements of the seismic testing application. It was therefore important that the sensor system could tolerate faulty sensors as there was not enough space to ruggedize the connectors to ensure they could cope with high vibration levels (>2 g) at frequencies from 0 to 100 Hz. For other applications where space is not such an issue, there is no reason that the whole system could not be encapsulated or vibration proof connectors implemented, to ensure that the risk of data loss from a sensor is minimised.
9. Conclusions
This paper presents a compact and robust method of utilising an array of 12 Hall effect sensors and magnets to measure fully coupled interface movements in six degrees of freedom within a column of rigid stacked bricks. The selection of the measurement devices and the optimum design of sensor locations within the interface are discussed together with a detailed description of the calibration technique, using a FARO coordinate measurement arm as a reference. Novel application of a linearly independent polynomial fitting approach, based on nonlinear system identification techniques, is presented as a method by which the acquired Hall effect data can be accurately calibrated. The measurement uncertainty associated with the calibration procedure was ±0.050 mm for displacement and ±0.052 degrees for rotation. The system was accurate enough to be used subsequently to instrument 13 interfaces in several columns of bricks in a large research programme looking at the performance of AGR cores subjected to earthquake motions with peak accelerations up to 1 g with broad band frequency content from 0 Hz to 100 Hz. The data from that work have been used to verify and validate software tools that subsequently provide analysis of the graphite bricks within AGR cores.
The calibration procedure implemented can capitalise on the redundancy inherent in the interface instrumentation when sensor channels fail partway through an experiment. The analysis of faulty channels and the associated sensor redundancy indicates that the loss of two channels still provides acceptable performance for the calibration of interface measurements within the context of the MLA experiments. The loss of data from two sensors only increases the uncertainly of the interface measurements to 0.061 mm in translation and 0.088 degrees in rotation. With minimal effort, the calibration datasets can be modified to ensure that good quality results can still be generated from the incomplete raw data.
The system has been deployed in an experimental research programme where the interface measurements were particularly important as they were used to determine the dynamic behaviour of the MLA model reactor core. This research programme has enabled EDF to present seismic safety cases for nuclear reactors to the Office for Nuclear Regulation, resulting in life extension approvals for several reactors.
The measurement system presented could be readily applied to other situations where the imposed level of stress at the interface causes negligible material strain, and accurate non-contact six-degree-of-freedom interface measurement is required.
Acknowledgments
The authors would like to thank EDF for financial and technical support for the larger project that this work contributed towards.
Author Contributions
Conceptualization: A.J.C. and C.A.T.; Data curation: T.G., R.E.W. and T.H.; Formal analysis: O.O., P.K., T.G., T.H., K.B. and M.D.; Funding acquisition: A.J.C. and C.A.T.; Investigation: O.O., P.K., T.G., T.H. and R.E.W.; Methodology: O.O., P.K. and T.G.; Project administration: A.J.C. and C.A.T.; Software programming: O.O., P.K., T.G., T.H., K.B., L.D., M.D., R.E.W. and A.J.C.; Supervision: A.J.C. and C.A.T.; Validation: O.O., P.K., T.G., T.H., K.B., L.D., M.D., R.E.W. and A.J.C.; Visualization: T.G. and K.B.; Writing—original draft: O.O., K.B. and T.G.; Writing—review and editing: O.O., P.K., T.G., K.B., T.H., L.D., M.D., R.E.W., A.J.C. and C.A.T. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded by EDF under contract number 4840522977.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
The data presented in this paper is specific to the particular interface being calibrated, so data sharing is not applicable to this article.
Conflicts of Interest
The authors declare no conflict of interest. The sponsors provided funding to allow the development of this new sensor technique by the authors, but had no role in the design, implementation, or validation of the technique. The views expressed in this paper are those of the authors and do not necessarily represent those of EDF.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Data Availability Statement
The data presented in this paper is specific to the particular interface being calibrated, so data sharing is not applicable to this article.
Articles from Sensors (Basel, Switzerland) are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI)
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Abstract
1. Introduction
2. Sensor Selection
3. Interface Instrumentation Concept
4. Interface Calibration Concept
5. Calibration Procedure
6. Final Design and Calibration Results
7. Measurement Uncertainty
8. Faulty Channel Handling
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8153 | https://pubmed.ncbi.nlm.nih.gov/26936942/ | An introduction to the Endocrine Society Clinical Practice Guideline on treatment of symptoms of the menopause - PubMed
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. 2016 Mar;22(1):6-8.
doi: 10.1177/2053369115626029.
An introduction to the Endocrine Society Clinical Practice Guideline on treatment of symptoms of the menopause
Cynthia A Stuenkel1,Richard J Santen2
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1 University of California, San Diego, School of Medicine, Department of Medicine, La Jolla, CA, USA castuenkel@ucsd.edu.
2 University of Virginia Health System, Division of Endocrinology and Metabolism, Charlottesville, VA, USA.
PMID: 26936942
DOI: 10.1177/2053369115626029
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An introduction to the Endocrine Society Clinical Practice Guideline on treatment of symptoms of the menopause
Cynthia A Stuenkel et al. Post Reprod Health.2016 Mar.
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. 2016 Mar;22(1):6-8.
doi: 10.1177/2053369115626029.
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Cynthia A Stuenkel1,Richard J Santen2
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1 University of California, San Diego, School of Medicine, Department of Medicine, La Jolla, CA, USA castuenkel@ucsd.edu.
2 University of Virginia Health System, Division of Endocrinology and Metabolism, Charlottesville, VA, USA.
PMID: 26936942
DOI: 10.1177/2053369115626029
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Treatment of symptoms of menopause remains a challenge for many health care practitioners. In an effort to facilitate this process, the Endocrine Society convened an international Task Force of menopause experts to review the relevant clinical evidence and formulate practical recommendations for relieving the most common menopausal symptoms. The result is a comprehensive evidence-based guideline, which emphasizes an individualized approach to alleviate bothersome vasomotor symptoms and those related to postmenopausal changes of the vagina and urinary tract. Therapies including estrogen, either alone or in combination with progestogen or bazedoxifene, tibolone, antidepressants, gabapentin, as well as complementary approaches are discussed. In this commentary, the chairs of the Task Force highlight the organization and content of the guideline and the processes involved in its development.
Keywords: Estrogen therapy; guideline; menopause symptoms.
© The Author(s) 2016.
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8154 | https://www.sciencedirect.com/science/article/pii/S0093641323001076 | Independent friction-restitution description of billiard ball collisions - ScienceDirect
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1. Introduction
2. Experimental
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Appendix A1. Extended IFR modeling.
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Mechanics Research Communications
Volume 131, August 2023, 104149
Independent friction-restitution description of billiard ball collisions
Author links open overlay panelAntonio Doménech-Carbó
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The oblique impact of a cue ball moving on a horizontal surface with an arbitrary spin on an object ball at rest is described using the independent friction-restitution modeling of impact. The model yields theoretical expressions for the post-collision linear and angular velocities in terms of ‘constant’ coefficients of normal restitution, tangential restitution, and friction for both stick and slip regimes of impact. The model also predicts the post-impact velocities reached by the balls when, as a result of friction with the supporting surface, the pure rolling motion is established. Theoretical predictions are in satisfactory agreement with experimental data for collisions between regulation billiard balls, steel, brass, and rubber spheres.
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Keywords
Billiard balls
Oblique impact
Restitution
Friction
1. Introduction
Billiards and pools have received considerable attention within the frame of impact physics, , . A part of this research was developed in the educational context, using simplified models limited to normal impulses acting throughout the impact [4,5]. The accurate description of impact events, however, involves the consideration of tangential impulses acting between the colliding balls. This is complicated by the fact that collisions may occur in two impact regimes, stick and slip, depending on a series of factors [6,7].
This matter was faced by Brach describing the tangential effects in terms of frictional impulses mimicking the classical Amontons-Coulomb laws. Walton further expanded this model introducing the concept of tangential restitution by analogy to the normal restitution previously -and widely- used to describe the inelasticity of the impact. This model, as well as the ‘continuous’ model developed by Maw et al. , have been extensively used in literature to interpret abundant experimental data concerning, among other fields, fluids and granular matter, , , , , , , , , , , , , , , , , , .
The above formulations, however, lead to two problematic issues from the conceptual point of view: i) the accommodation to the requirement of kinetic energy loss, and ii) the variation of the coefficients of tangential restitution (and friction) with the impact angle, in sharp contrast with the constancy of the normal coefficient of restitution. The former will be treated in the general theoretical approach. With regard to the second item, the Walton's modeling, where friction and restitution appear as ‘mixed’, is inconsistent with the widely extended idea, clearly represented via spring plus dashpot or spring plus compliance , , schemes used to describe impact dynamics, that these are different ‘mechanisms’ depending on different factors at the so-called mesoscopic scale. In short, restitution depends on the mechanical properties (Youngs modulus, Poisson coefficient) of the bulk materials whereas friction depends on surface properties (asperity, roughness), a separation implicit in much literature on collisions based on Hertz-type contact force models , , , , , , , , .
In this context, it was developed an alternative formulation, the so-called independent friction-restitution (IFR) approach, based on the idea that friction and restitution operate simultaneously but independently during collisions. A simple redefinition of friction and restitution permitted to develop a model of impact using constant coefficients of restitution and friction [44,45]. The closure was subsequently refined by introducing a definition of sticking friction in terms of the classical theory of percussion center [46,47] and considering rolling friction and normal friction effects. This formulation permits to avoid the previously mentioned conceptual problem and the justification of a series of experimental results considered as ‘anomalous’ in literature [14,17], repeatedly appearing in other reports [15,16,, , ,, , , , ].
In previous studies we described billiard ball collisions based on the Brach's model and a semiempirical approximation to frictional impulses . Here, the IFR modeling is applied to describe billiard ball-type collisions. A generalized view is proposed, assuming that the cue ball is projected with an arbitrary spin on an object ball at rest. This situation is of interest not only for billiard and pool games but also as approaching to the general impact conditions existing in granular flows. The current description expands the description of sphere rebound and two-disk collisions already reported accounting for the influence of friction with the horizontal supporting surface in the post-collision paths. The theoretical model is tested using experimental data for regulation billiard balls and steel, brass, and rubber balls.
2. Experimental
Experiments on two-sphere collisions were performed by means of the arrangement already reported [47,48]. Regulation billiard balls (diameter 6.1 cm; mass 205.0 g) and steel (diameter 2.50 cm, mass 70.30 g), brass (diameter 2.50 cm, mass 68.20 g), and rubber (diameter 4.60 cm, mass 46.40 g) balls were used moving on the laboratory bench with a PVC surface. The cue ball was projected at 0.80±0.05 m s−1 with the help of a slanted track whose edge was placed at 50 cm from the contact point to ensure that the cue ball moves in pure rolling regime. The angles of impact and scattering were measured from the photographs recorded with a conventional camera placed in a zenithal position 75 cm just over the point of contact between the spheres.
3. Theory
3.1. General approach
Let us consider the oblique impact of a cue, homogeneous ball of mass m 1, radius R, and inertia momentI 1 (= 2 m 1 R 2/5) on an object ball of the same radius and mass m 2 and inertia momentI 2 (= 2 m 2 R 2/5). It will be assumed that both balls move over a rough horizontal plane as schematized in Fig.1. The cue ball is projected along a horizontal plane with center of mass velocityv o and a rotation rateω o while the object ball is initially at rest. To describe the dynamics of the system, two reference systems will be combined. In the first one, the x-axis is defined by the center of mass velocity of the cue ball and the z-axis is perpendicular to the supporting plane. The second reference system uses normal-tangential coordinates; the normal axis is defined by the line joining the centers of the balls at the collision and the tangential vertical and horizontal axes are perpendicularly directed to the above in vertical and horizontal planes. It will be assumed that arbitrary x-, y- and z-components of the angular rate (ω ox,ω oy, ω oz) are imparted to the cue ball, as depicted in Fig.2a,b. For simplicity, the case where Rω oy/v o ≥ 1 will be that treated in the following.
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Fig. 1. Schematic representation in zenithal view of the oblique impact of a cue ball on an object ball initially at rest.
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Fig. 2. a) Zenithal and b) frontal views of the ball 1 immediately before the impact schematizing the center of mass linear velocity and the (positive) components of the angular velocity in the x-, y- and z-directions.
According to the discontinuous modeling of impact events, it will be assumed that during the collision, a series of impulses act between the spheres so that the balls will move with center of mass velocities v 1, v 2, and rotation rates ω 1, ω 2. The general problem of billiard ball collisions can be formulated as consisting on the determination of the normal, tangential and vertical components of linear and angular postimpact velocities of the balls from the initial values of the corresponding quantities of the cue ball. This implies the determination of twelve unknowns with the concomitant requirement of twelve constitutive equations. Solving this problem departs from the three equations expressing the conservation of linear momentum in the normal, tangential and vertical directions and the six equations relating linear and angular velocities of the balls by applying the angular momentum law. Additionally, the inelasticity of the impact is expressed by the following relationship between the preimpact (E Ko) and postimpact (E K1, E K2) kinetic energies,(1)ξ=1−E K 1+E K 2 E K 0>0 that expresses that the energy efficiency ξ has to be positive. In the widely used impulse-based models, the set of problem-solving equations is completed by those introducing the coefficients of friction (μ) and the coefficients of normal (e n) and tangential (e t) restitution.
The differences among the different formulations [8,9,41,42] are focused on the definitions of these coefficients regardless the kinetic energy inequality (Eq.(1)). As a result, these models apply only for a set of values of the coefficients of restitution and friction, as discussed by Louge and Adams . To overcome this difficulty, Stronge introduced the energetic coefficient of restitution in the context of ‘continuous’, force-based modeling [7,, , ] of impact events, in principle involving computational methods .
3.2. The IFR formalism
According to the IFR formalism, the net tangential (P t) and normal (P n) impulses will be expressed as the sum of independent tangential and normal impulses due to restitution (P et, P en) and friction (P ft, P fn).
Fig.3 depicts a zenithal (a) and lateral (b) view of the net impulses acting between the spheres during the impact event. These will be defined independently; i.e., friction (restitution) impulses will be taken as those operating in the absence of restitution (friction) effects as already described , , , , , . Given the normal-tangential configuration of the impulses, the post-collision velocities will be expressed in an equivalent reference system; i.e., v jn, v jt, v jz, and ω 1t, ω 1n, ω jz (j=1,2), as schematized in Fig.3c,d.
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Fig. 3. a) Representation of the net horizontal impulses acting between the spheres and b) frontal view of the sphere 2 showing the vertical and horizontal components of the net tangential impulse acting on it. c) Zenithal and d) frontal views of the ball 2 just after the collision illustrating the corresponding linear and angular velocities referred to the normal-tangential coordinate system.
The system will satisfy the Newtonian laws of linear and angular momentum. These can be expressed taking separately the normal impulse, P n, and the horizontal and vertical components (P th, P tv, respectively) of the tangential impulse, in terms of the impact angle, ψ, and the scattering angles δ 1 (= α 1+ψ) and δ 2 depicted in Fig.1. First of all, one can write for the horizontal impulses,(2)P n=m 2 v 2 cos δ 2=−(m 1 v 1 cos δ 1−m 1 v o cos ψ)(3)P th=m 2 v 2 sin δ 2=−(m 1 v 1 sin δ 1−m 1 v o sin ψ)(4)P th=2 5 m 2 R 2 ω 2 z=−2 5 m 1 R 2(ω 1 z−ω oz)
With regard to the vertical impulses, it has to be taken into account the appearance of a normal impulse (N) exerted by the infinitely massive supporting plane on the sphere 2, as schematized in Fig.4a. This will be accompanied by a tangential force (F), as discussed in . For simplicity, it will be assumed that F=0; a refined treatment of this matter will be treated in the Appendix A1. Then, the vertical component of the net tangential impulse will be,(5)P tv=2 5 m 2 R 2 ω 2 y=−2 5 m 1 R 2(ω 1 y−ω 1 y)=m 1 v 1 z whereas N+P tv=0.
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Fig. 4. Lateral representation along the normal axis of: a) normal and tangential vertical impulses acting on the spheres for the oblique impact when Rω oy/v o ≥ 1; b) variation of the linear and angular velocities for the ball 2 from just after the impact to the postransition regime where the ball reaches the pure rolling motion.
Adopting the usual kinematic definition of normal coefficient of restitution (e n) as the negative ratio of the post-impact to preimpact relative normal velocities at the contact point (identical in this case to the center of mass velocities), one obtains,(6)v 2 cos δ 2−v 1 cos δ 1=e n v o cos ψ
Combining Eq.(6) with Eqs.(2) and (3), and introducing the mass ratio M (= m 1/m 2) yields the normal components of the post-impact center of mass velocities as,(7)v 1 n=v 1 cos δ 1=(M−e n 1+M)v o cos ψ(8)v 2 n=v 2 cos δ 2=M(1+e n 1+M)v o cos ψ
Then, the normal impulse due to restitution, P en, is given by,(9)P en=m 1(1+e n 1+M)v o cos ψ
According to the generalized IFR closure , this will be accompanied by a normal friction impulse, P fn. This is only relevant in cases where significant adhesive of viscous effects appear so that it can, in principle, be neglected to describe billiard ball impacts. The corresponding generalized treatment will be summarized in the Appendix A1. In turn, the impulse due to tangential restitution will be expressed in terms of the tangential coefficient of restitution, e t. This is defined as the negative ratio of the post-impact to preimpact relative velocities of the contact point. To obtain the analytical expressions for the tangential restitution impulse it is convenient to define the angle σ between the horizontal and vertical components of the initial velocity of the ball 1 at the contact point, illustrated in Fig.3b,(10)tan σ=v o sin ψ+R ω oz R ω oy cos ψ
The absolute value of the velocity of the contact point of the ball 1 just before the impact, V ot, is,(11)V ot=(v o sin ψ+R ω oz)2+(R ω oy cos ψ)2=R ω oy(cos ψ cos σ)
Notice that when the cue ball is launched with pure rolling motion, v o=Rω oy, Rω ox=0, Rω oy=0, and σ=ψ.
Applying a treatment similar to that described for two-disk collisions , the impulse due to tangential restitution will be,(12)P et=2 7 m 1(1+e t 1−M)V ot=2 7 m 1 R ω oy(1+e t 1+M)(cos ψ cos σ)
In turn, the friction impulse will be defined based on the well-known Amontons-Coulomb approach now discriminating between two impact regimes. In sliding regime, the friction impulse will be the coefficient of friction (μ) times the normal impulse,(13)P f=μ P n=μ m 1(1+e n 1+M)v o cos ψ while in sticking regime, the friction impulse will be defined using the percussion point approach already described , , , ,(14)P f=2 5 m 1(1 1+M)V ot=2 5 m 1 R ω oy(1 1+M)(cos ψ cos σ)
It will be assumed that the ratio between the horizontal and vertical components of the tangential impulse as well as the corresponding components of the friction impulse equal to tan σ. Accordingly, the net normal impulse acting throughout the collision will be given by Eq.(9) and the horizontal and vertical components of the net tangential impulse will be given by the following expressions.
Sliding regime:(15)P th=μ m 1(1+e n 1+M)v o cos ψ sin σ+2 7 m 1 R ω oy(1+e t 1+M)cos ψ tan σ(16)P tv=μ m 1(1+e n 1+M)v o cos ψ cos σ+2 7 m 1 R ω oy(1+e t 1+M)cos ψ
Sticking regime:(17)P th=[2 5+2 7(1+e t 1+M)]m 1 R ω oy cos ψ tan σ(18)P tv=[2 5+2 7(1+e t 1+M)]m 1 R ω oy cos ψ
The transition between the sticking and sliding regimes can be obtained by equalizing Eqs.(13) and (14). This transition occurs for a threshold value of σ given by,(19)cos σ T=2 5 μ(1+e n)
When the cue ball is projected with pure rolling motion, the transition occurs for an impact angle ψ T given by the same relationship; i.e., cos ψ T=2/5 μ(1+e n).
3.3. Post-impact angles
The post-impact linear and angular velocities of the balls can be obtained combining Eqs.(9) and (15)-(18) with the general equations of motionEqs.(2)-((5)). It will be assumed that there are no rolling friction and pivoting friction effects so that the x-component of the angular velocity does not influence the post-impact velocities. The corresponding expressions for impacts in sticking and sliding regime are summarized in the Appendix A2. Here, the equations for the post-collision angles, which are those able to be directly compared with experimental data, are presented. First of all, the post-impact angles δ 1, δ 2 in sliding regime will be,(20)tan δ 1=(1+M M−e n)tan ψ−μ(1+e n M−e n)sin σ−2 7(1+e t M−e n)(R ω oy v o)tan σ(21)tan δ 2=μ sin σ−2 7 M(1+e t 1+M)(R ω oy v o)tan σ
These equations become notably simplified for collisions of spheres of the same mass when e t=−1, a situation which applies (see Section 3) reasonably for billiard balls,(22)tan δ 1=(2 1−e n)tan ψ−μ(1+e n 1−e n)sin σ(23)tan δ 2=μ sin σ
Additionally, if the cue ball is projected with pure rolling motion, Rω y o/v o=1 and then σ=ψ so that for collisions in sticking regime, the postcollision angles are,(24)tan δ 1=(1+M M−e n)tan ψ−2 5+2 7(1+e t M−e n)tan σ(25)tan δ 2=2 5+2 7(1+e t)(R ω oy v o)tan σ whose simplified expressions for M=1 and e t=−1, are,(26)tan δ 1=(2 1−e n)tan ψ−2 5(R ω oy v o)tan σ(27)tan δ 2=2 5(1 1+e n)(R ω oy v o)tan σ
3.4. Post-transition angles
As illustrated in Fig.1, after the impact the spheres move in directions that define post-impact angles δ 1, δ 2, relative to the normal axis (or α 1, α 2, relative to the x-axis) with a combination of rolling and sliding. Then, the friction with the horizontal supporting surface determines the appearance of continuous forces. As a result, as schematized in Fig.4b, the spheres change continuously their trajectories and velocities until pure rolling motion is reached [, , ,5]. This corresponds to center of mass velocities u 1, u 2, and rectilinear trajectories defining post-transition angles λ 1, λ 2, relative to the normal axis (or β 1, β 2, relative to the x-axis). The rotation rates Ω 1, Ω 2 (respectively equal to u 1/R, u 2/R) will satisfy the law of angular momentum so that,(28)u jn=5 7 v jn+2 7 R ω jt;u jt=5 7 v jt+2 7 R ω jn where j=1,2 and the subscripts n and t refer, as before, to the normal and tangential directions. Notice that the x-component of the initial rotation rate of the cue ball, ω ox, that does not influence the post-impact angles, here influences the motion of the ball 1. The values of the post-transition velocities, u 1n, u 2n, u 1t, u 2t, calculated from the post-impact angles are summarized in the Appendix A2. Combining these values, one obtains the following post-transition angles. The post-transition angle λ 2 in sliding and sticking regimes are,(29)tan λ 2=μ sin σ+2 7(1+e t 1+e n)(R ω oy v o)tan σ 1−μ cos σ−2 7(1+e t 1+e n)(R ω oy v o)and(30)tan λ 2=2 5+2 7(1+e t)tan σ 1+e n−2 5+2 7(1+e t)respectively. As before, these equations become simplified taking e t=−1, being in the case of sliding,(31)tan λ 2=μ sin σ 1−μ cos σ and, for collisions in sticking regime,(32)tan λ 2=2 5(R ω oy v o)tan σ 1+e n−2 5(R ω oy v o)
The corresponding expressions for the angle λ 1 are much more complicated. For testing purposes, it is convenient to limiting it to those when Rω ox=0, Rω oz=0, Rω oy=v o, and e t=−1. The angle λ 1 in sliding regime is,(33)tan λ 1=7 5(1+M)tan ψ−μ(1+e n)sin σ M−e n+2 5(1+M)−μ(1+e n)cos σ
The corresponding expression for collisions in sticking regime is,(34)tan λ 1=(1+M)tan ψ−2 5 tan σ M−e n−2 5+2 5(1+M)
3.5. Energetic considerations
The above formulation has to be balanced with the general condition expressed by Eq.(1), the energy inequality. For simplicity, it will be considered the frontal impact between two identical billiard balls where the cue ball is initially in pure rolling motion. The above condition can be expressed in terms of energy efficiency by means of Eq.(1). This can be seen as imposing a general condition to be accomplished by the possible impulse functions. For our particular case, E K0=(7/10)mu 2 and introducing the impulses in the normal and tangential vertical directions, P n, P tv, Eq.(1) becomes,(35)(P n m v 0)[1+3(P t v P n)2]<[1+P t v P n]
If the impact occurs in sliding regime, Eqs.(9), (12), and (13) lead to,(36)(1+e n 2)[1+3(μ+2 7(1+e t 1+e n))2]<[1+μ+2 7(1+e t 1+e n)]
This expression is simplified taking e t=−1. Then,(37)(1+e n 2)<1+μ 1+3 μ 2
Eqs.(36) and (37) limit the possible values of the coefficients of restitution and friction; i.e., the IFR closure is valid only for several sets of values of these coefficients. Experimental data (see the next section) indicate that the above condition is clearly accomplished under ordinary conditions. Generically, Eqs.(36) and (37) can be interpreted as denoting that these parameters must be related through any subtle, ‘hidden’ relationship, as has been discussed in the context of the Walton's model . This is reasonable taking into account that the above parameters can be linked to the ‘mesoscopic’ mechanical properties (there is abundant literature on the relationship between en and the Young's modulus and the Poisson's coefficient [32-54]) and, ultimately, on their ‘atomic’ properties. Despite these limitations, the IFR modeling presented here can be considered as a reasonable approach to provide analytical solutions of the motion equations and interpret experimental data under a wide set of conditions.
3.6. Comparison with experimental data
As described in Section 2, experimental data were obtained for the oblique impact of different homogeneous balls launching the cue ball with pure rolling motion on the object ball initially at rest on a horizontal surface. Under these conditions, the angle δ 2 was particularly sensitive to the impact regime. This can be seen in Fig.5 where experimental values of δ 2 recorded at different impact angles ψ for billiard ball collisions are accompanied by theoretical lines from Eqs.(23) and (27) taking e n=0.98 and a) μ=0.10, b) μ=0.20, and c) μ=0.40. The value of the normal coefficient of restitution was that calculated for the impact of billiard balls using the experimental variation of the angles α 1 and β 1 with the impact angle (vide infra). In agreement with Eq.(19), the impact occurs in sliding regime at all incidence angles for μ=0.10 and μ=0.20 while for μ=0.40 there is transition between the sticking regime at low impact angles, and the sliding regime at high impact angles. According to Eq.(19), the transition takes place at an impact angle of 58 deg.
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Fig. 5. Plots of δ 2 vs. the impact angle ψ. Experimental data for billiard ball collisions (circles) and theoretical lines from Eqs.(23) and (27) taking Rω oy/v o=1, e n=0.98 and a) μ=0.10, b) μ=0.20, and c) μ=0.40. In the cases (a) and (b) the impact occurs in sliding regime at all impact angles and only Eq.(23) applies. In the case (c) there is transition between the sticking (low impact angles) and sliding (high impact angles) regimes at 58 deg (Eq.(19)).
As can be seen in Fig.5, experimental δ 2 vs. ψ data for billiard ball collisions agree with theory from Eq.(23) taking μ=0.10. Consistently, experimental data for the variation of the angle α 1 on the impact angle agrees with theoretical curve from Eq.(20) taking e n=0.98, e t=−1.00 and μ=0.10, as can be seen in Fig.6. This variation is compared in the same figure with the corresponding representation for the oblique impact of brass balls. Again, experimental data can satisfactorily be reproduced now inserting e n=0.66, e t=−0.95 and μ=0.14 into Eq.(20). As in the case of billiard balls the collisions take place in sliding regime at all impact angles.
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Fig. 6. Variation of α 1 on the impact angle for the collisions of billiard balls (circles) and brass balls (solid circles) with Rω oy/v o=1. Experimental data are superimposed to the theoretical lines from Eq.(20) taking e n=0.98, e t=−1.00 and μ=0.10 (billiard balls) and e n=0.66, e t=−0.95 and μ=0.14 (brass balls). In both cases the collisions occur in sliding regime at all impact angles.
The agreement between experimental data and the theoretical IFR modeling is extended to post-transition angles. An example is provided in Fig.7 where the experimental variation of the angles α 1 and β 1 with the incident angle for the oblique impact of steel balls is depicted. Experimental data agree with theoretical expectances from Eqs.(20) and (29) (continuous lines) taking e n=0.98, e t=−0.95, and μ=0.10, corresponding to impact in sliding regime at all incidence angles.
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Fig. 7. Plots of α 1 (circles) and β 1 (solid circles) vs. the impact angle experimentally determined for the impact of steel balls in the conditions described in the experimental section. Continuous lines represent the theoretical lines from Eqs.(20) and (29) taking Rω oy/v o=1, e n=0.98, e t=−0.95, μ=0.10.
The opposite behavior was obtained for the impact of rubber balls for which experimental data agree with theory for collisions in sticking regime. Fig.8 shows the δ 2 vs. impact angle plot superimposing the experimental data and the theoretical lines from Eqs.(23) and (27) taking e n=0.98, e t=−0.80, and μ=1.0. Experimental data at low impact angles fit well with the theoretical sticking curve while clearly diverge from the expected behavior in sliding regime (compare with Fig.5). This regime is only attained at large impact angles, the transition occurring at ψ=78 deg.
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Fig. 8. Variation of δ 2 on the impact angle for rubber ball collisions. Experimental data (circles) and theoretical lines inserting Rω oy/v o=1, e n=0.98, e t=−0.80, and μ=1.0 into a) Eq.(23), sliding regime, and b) Eq.(27), sticking regime.
Similar agreement between theory and experiment appears for the post-transition angle λ 2 measured for the collisions between rubber balls. Again, as illustrated in Fig.9, experimental data at impact angles below ca. 70 deg. can be fitted to theory for impact in sticking regime and only at larger incidence angles there is transition to the sliding regime.
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Fig. 9. Variation of λ 2 on the impact angle for rubber ball collisions. Experimental data (circles) and theoretical lines inserting Rω oy/v o=1, e n=0.98, e t=−0.80, and μ=1.0 into Eqs.(29) (a, sliding regime), and Eq.(30) (b, sticking regime).
Interestingly, the values of the e n and μ determined using the IFR model for impacts occurring in sliding regime of billiard balls, steel and brass spheres fall within the ‘allowed region’ defined by Eq.(37) in Fig.10 for e t=−1. In principle, the unique constraint is imposed by the condition μ> 0.
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Fig. 10. Two-dimensional diagram corresponding to the allowed and prohibited values of e n and μ for the frontal impact of two identical spheres assuming sliding regime and initial pure rolling motion of the cue ball for e t=−1.
4. Conclusions
Billiard ball collisions corresponding to the impact of a homogeneous sphere on an identical sphere initially at rest can be described using the independent friction-restitution (IFR) closure. Application of the model for the case of the impact of a cue ball projected with an arbitrary spin permits to predict the post-collision linear and angular velocities, as well as the scattering angles in terms of the mass ratio and the coefficients of normal and tangential restitution and sliding friction. The model accounts for the occurrence of two impact regimes, sticking and sliding whose transition depends on several impact parameters.
Consideration of frictional effects accompanying the motion of the spheres over the horizontal supporting surface after the impact permits to describe the linear and angular velocities and the scattering angles when the balls reach pure rolling motion, of practical application in billiards and pools. Experimental data on post-impact and post-transition angles for regulation billiard balls, steel, brass and rubber spheres projected with pure rolling motion satisfactorily agree with theoretical model, evidencing the suitability of the IFR formalism for describing relatively complex impact events.
Declaration of Competing Interest
The author declares no existing conflicts of interest.
Acknowledgments
Project PID2020–113022GB-I00 supported by MCIN/AEI/ 10.13039/501100011033, Fondo Europeo de Desarrollo Regional (ERDF) and Agencia Estatal de Investigación (AEI), is gratefully acknowledged.
Appendix A1. Extended IFR modeling.
The description of impact events where significant adhesive and/or viscous effects occur within the proposed IFR modeling can be made, as previously described , incorporating a normal frictional impulse (P fn) which adds to the normal restitution impulse given by Eq.(9). For practical purposes, only the sliding regime is operative so that the friction impulse can be taken as proportional to the tangential restitution impulse, introducing a normal friction coefficient, μ n. This formalism permitted to interpret a series of experimental results considered as ‘anomalous’ in literature [15,17,19]. Then, the net normal impulse will be,(A1.1)P n=P en+P fn=m 1(1+e n 1+M)v o cos ψ+2 7 μ n m 1(1+e t 1+M)(cos ψ cos σ)R ω oy
The application of this approach to the motion of the ball 2 requires, however, the consideration of the impulses associated to the ball-supporting surface interaction depicted in Fig.3a. The value of the normal impulse N will be equal to the net tangential impulse in the vertical direction. For impacts in sliding regime:(A1.2)N=μ m 1(1+e n 1+M)v o cos ψ cos σ+2 7 m 1(1+e t 1+M)R ω oy cos ψ
Assuming, as already discussed that N is accompanied by a friction impulse F also operating in sliding regime, one can write,(A1.3)F=μN=μμ m 1(1+e n 1+M)v o cos ψ cos σ+2 7 μm 1(1+e t 1+M)R ω oy cos ψ where μ represents the coefficient of friction between the sphere and the supporting massive plane. As a result, the post-impact angle δ 2 will be given by the relationship,(A1.4)tan δ 2=μ(1+e n)sin σ+2 7(1+e t)(R ω yo v o)tan σ(1+e n)(1−μμ)+2 7(1+e t)(R ω yo v o)(μ n cos σ−μ)
When e t=−1, the resulting equation slightly modifies Eq.(22):(A1.5)tan δ 2=μ sin σ 1−μμ
Appendix A2. Post-impact velocities.
The linear and angular velocities of the balls after the impact can be obtained combining the general equations of motion Eqs.(2)-((5)) with the constitutive equations for the impulses expressed by Eqs.(15)-(18). As previously noted, rolling and pivotment frictional effects are neglected. Now, the normal (n)-tangential (t)-vertical (z) reference system is used. For collisions in sliding regime,(A2.1)v 1 t=v o sin ψ−μ(1+e n 1+M)v o cos ψ sin σ−2 7(1+e t 1+M)R ω oy cos ψ tan σ(A2.2)v 2 t=μ M(1+e n 1+M)v o cos ψ sin σ+2 7 M(1+e t 1+M)R ω oy cos ψ tan σ(A2.3)R ω 1 t=R ω oy sin ψ+5 2 μ(1+e n 1+M)v o cos ψ cos σ+5 7(1+e t 1+M)R ω oy cos ψ(A2.4)R ω 2 t=−5 2 μ M(1+e n 1+M)v o cos ψ cos σ−5 7 M(1+e t 1+M)R ω oy cos ψ(A2.5)R ω 1 z=R ω oz+5 2 μ(1+e n 1+M)v o cos ψ sin σ+5 7(1+e t 1+M)R ω oy cos ψ tan σ(A2.6)R ω 2 z=−5 2 μ M(1+e n 1+M)v o cos ψ sin σ−5 7 M(1+e t 1+M)R ω oy cos ψ tan σ
For collisions in sticking regime,(A2.7)v 1 t=v o sin ψ−(1 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ tan σ(A2.8)v 2 t=(M 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ tan σ(A2.9)R ω 1 t=R ω oy sin ψ+5 2(1 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ(A2.10)R ω 2 t=−5 2(M 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ(A2.11)R ω 1 z=R ω oz+5 2(1 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ tan σ(A2.12)R ω 2 z=−5 2(M 1+M)[2 5+2 7(1+e t)]R ω oy cos ψ tan σ
In both impact regimes,(A2.13)R ω 2 n=0(A2.14)R ω 1 n=R ω ox cos ψ+R ω oy sin ψ
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8155 | https://askfilo.com/physics-question-answers/three-masses-are-connected-by-light-inextensible-strings-which-pass-over-smooth | Solving time: 4 mins
Three masses are connected by light, inextensible strings which pass over smooth light, fixed pulleys. The masses m1 and m3 are 4 kg and 6 kg respectively and the coefficient of friction for all surfaces in contact is 0.3. The minimum mass of m2 and the tension in the horizontal part of the string (θ=sin−1(54)) for no motion (g=10 m s−2) are respectively
Views: 5,525 students
Text SolutionText solutionverified iconVerified
T1 T1T2 m1 gm2===m1 g=T2+m2 gsinθ+μm2 gcosθ=μm3 g=0.3×6×10=18 N=μm3 g+m2 gsinθ+μm2 gcosθ=sinθ+μcosθm1−μm3sin53∘+0.3cos53∘4.0−6×0.3=0.8+0.3×0.64−1.80.982.2≃2.2 kg
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| Question Text | Three masses are connected by light, inextensible strings which pass over smooth light, fixed pulleys. The masses m1 and m3 are 4 kg and 6 kg respectively and the coefficient of friction for all surfaces in contact is 0.3. The minimum mass of m2 and the tension in the horizontal part of the string (θ=sin−1(54)) for no motion (g=10 m s−2) are respectively |
| Topic | Laws of motion |
| Subject | Physics |
| Class | Class 11 |
| Answer Type | Text solution:1 |
| Upvotes | 119 |
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8156 | https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOorvq_JfzEaa3-NjRx0f_Y38WPUYZKW4poXgQ-MP8ARAKAJPGlYP | Olympiad Inequalities
Thomas J. Mildorf December 22, 2005
It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay...
1
and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities.
2
Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation.
3
The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.)
4
where a seemingly inexhaustible supply of Olympiads is available.
5
Such as the source of the last problem in this document.
11 The Standard Dozen
Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if
λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I
if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If
x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence
{xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers
t,
|t − x1| + |t − x2| + · · · + |t − xn| ≥ | t − y1| + |t − y2| + · · · + |t − yn|
We use these definitions to introduce some famous inequalities.
Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I
and any nonnegative reals ω1, . . . , ω n,
ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f
(ω1x1 + · · · + ωnxn
ω1 + · · · + ωn
)
If f is concave, then the inequality is flipped.
Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n
are nonnegative reals with a postive sum, then
f (r) :=
(ω1xr
1
· · · + ωnxrn
ω1 + · · · + ωn
)1
r
is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.
Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then
(a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa
1
bλb
1
· · · zλz
1
· · · + aλz
n
bλb
n
· · · zλz
n
This theorem is customarily identified as Cauchy when there are just two sequences.
Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have
a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1
with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively.
2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then
a1b1 + a2b2 + · · · + anbn
n ≥ a1 + a2 + · · · + an
n ·b1 + b2 + · · · + bn
n ≥ a1bn + a2bn−1 + · · · + anb1
n
Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then
ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0
with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0.
Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest.
Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni
). Then
d2
i
≥ di+1 di−1
Theorem 8 (Maclaurin) Let di be defined as above. Then
d1 ≥ √d2 ≥ 3
√d3 ≥ · · · ≥ n
√dn
Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence
x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then
f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn)
Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r ,
pf (x) + qf (y) + rf (z) + (p + q + r)f
(px + qy + rz p + q + r
)
≥ (p + q)f
(px + qy p + q
)
( q + r)f
(qy + rz q + r
)
( r + p)f
(rz + px r + p
)
Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1,
(1 + x)r ≥ 1 + xr
3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n,
∑
sym
xa1
1
xa2
2
· · · xan
n
≥ ∑
sym
xb1
1
xb2
2
· · · xbn
n
where the sums are taken over all permutations of n variables.
Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter.
We now present an array of problems and solutions based primarily on these inequalities and ideas.
2 Examples
When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2
Solution 1. Simply use AM-GM on the terms within each factor, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥
(
3 3
√a3b3c3
) (
3 3
√a3b3c3
)
= 9 a2b2c2
6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases.
4Solution 2. Rearrange the terms of each factor and apply Cauchy,
(a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥
(√a3b3c3 + √a3b3c3 + √a3b3c3
)2
= 9 a2b2c2
Solution 3. Expand the left hand side, then apply AM-GM, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc
ab 4c + b3c3 + a2b2c2
a2b2c2 + abc 4 + a3c3
≥ 9 9
√a18 b18 c18 = 9 a2b2c2
We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that
a + b + c ≤ a2 + b2 + c2
Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree
d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero
k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3
√abc , obtaining
a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2
As abc = 1 is not homogenous, the above inequality must be true for all nonnegative
a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2
3 + b2
6 + c2
6 = a2 + a2 + a2 + a2 + b2 + c2
6 ≥ 6
√a8b2c2 = a43 b13 c13
Let P (x) be a polynomial with positive coefficients. Prove that if
P
( 1
x
)
≥ 1
P (x)5holds for x = 1, then it holds for all x > 0.
Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1
x
)
into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives
P (x)P
( 1
x
)
= (anxn + · · · + a1x + a0)
(
an
1
xn + · · · + a1
1
x + a0
)
≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that
a + b + c + d + e = 8
a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e?
Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2)
− 2r(a + b + c + d + e) + 5 r2
= 16 − 16 r + 5 r2
Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when
f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and
r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when
a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write
a = b = c = d = 8−e
4
since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval
I together (that is, replacing xi1 < x i2 with x′
i1
= xi1 + ≤ < x i2 − ≤ = x′
i2
for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading
xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1
a + 1
b + 4
c + 16
d ≥ 64
a + b + c + d
Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by
a + b + c + d and applying Cauchy, we have (a + b + c + d)
(12
a + 12
b + 22
c + 42
d
)
≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1,
ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1
Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2
∂a 2
f =
2b
(c+a+1) 3
2c
(a+b+1) 3
≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e)
( 1
a + 1
b + 1
c + 1
d + 1
e
)
≤ 25 + 6
(√ pq −
√ qp
)2
and determine when equality holds.
Solution. As a function f of five variables, the left hand side is convex in each of
a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other,
f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that
a3 + b3 + c3 + 6 abc ≥ 14
Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3
= a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc
⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution.
Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13
or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c
2
)3 + c3 + 6 (1−c
2
)2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of
aa + b + d + bb + c + a + cb + c + d + da + c + d
Solution. We can obtain any real value in (1 , 2). The lower bound is approached by
a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ ,
b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because
aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and
aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) ≥ 32
Solution 1. Let x = 1
a
, y = 1
b
, and z = 1
c
. We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) = x2
y + z + y2
x + z + z2
x + y
Now, multiplying through by ( x + y)( y + z)( z + x), we seek
x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥
3
√xyz ·
(
3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2))
which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3
√x4y4z4, x3y+xy 3+x3z
3
≥
3
√x7y4z and 7x4+4 y4+z4
12
≥ 3
√x7y4z - as guaranteed by Muirhead’s inequality.
Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1
for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen,
x2
y + z + y2
z + x + z2
x + y = xf
(y + zx
)
yf
(z + xy
)
zf
(x + yz
)
≥ (x + y + z)f
((y + z) + ( z + x) + ( x + y)
x + y + z
)
= x + y + z
2But x + y + z ≥ 3 3
√xyz = 3, as desired.
Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y))
( x2
y + z + y2
z + x + z2
x + y
)
≥ 12(x + y + z)2
Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc.
Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously,
xy+z
≥ yz+x ≥ zx+y . Hence, by Chebyshev,
x ·
( xy + z
)
y ·
( yz + x
)
z ·
( zx + y
)
≥ x + y + z
3
( xy + z + yx + z + zx + y
)
Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1
Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2
a2 + b2 + 2 a + 2 ≤ 1
ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1
ab +a+1
= 1
xz+xy+1
= yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π
2
) such that tan
(
a0 − π
4
)
tan
(
a1 − π
4
)
· · · + tan
(
an − π
4
)
≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1
Solution 1. Let yi = tan (x − π
4
). We have tan( xi) = tan ((xi − π
4
) + π
4
) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi
1−yi
≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression
( 1 + a
1 − a
)
·
( 1 + b
1 − b
)
= 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′
i
= sn+1 and
y′
j
= yi + yj − sn+1 , decreasing the product. Now we have
n
∏
i=0
1 + yi
1 − yi
≥
( 1 + sn+1
1 − sn+1
)n+1
≥
( 2nn+1 2
n+1
)n+1
= nn+1
Where the last inequality follows from the fact that 1+ x
1−x
is an increasing function of x.
Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥
∑
j6=i
(1 − yj ), which by AM-GM gives 1+ yn
n
≥ ∏
j6=i
(1 − yj ) 1
n
. Now we have
n
∏
i=0
1 + yi
n ≥
n
∏
i=0
∏
j6=i
(1 − yj ) 1
n
=
n
∏
i=0
(1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1
a(1 + b) + 1
b(1 + c) + 1
c(1 + a) ≥ 31 + abc
with equality if and only if a = b = c = 1.
Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c)
a(1 + b) + (1 + b) + bc (1 + a)
b(1 + c) + (1 + c) + ac (1 + b)
c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a)
a(1+ b)
and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a)
b(1+ c)
so that 1 + b = b + bc = b + 1
a
. It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that
xay + bz + yaz + bx + zax + by ≥ 3
a + b
Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by ))
( xay + bz + yaz + bx + zax + by
)
≥
(x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3
n
= 0. Prove that
x1 + x2 + · · · + xn ≤ n
3
Solution 1. Substitute yi = x3
i
so that y1 + · · · + yn = 0. In maximizing 3
√y1 +
· · · + 3
√yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with
|f ′(y1)| ≥ | f ′(y2)| ⇐⇒ 0 < |y1| ≤ | y2|. Hence, we may put y1 = · · · = yk = −1;
−1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1
n−k−1
. We first show that yk+1 leads to a maximal sum of 3
√yi if it is -1 or can be made positive. If |yk+1 | < |yk+2 |, we set 11 y′
k+1
= y′
k+2
= yk+1 +yk+2
2
, increasing the sum while making yk+1 positive. Otherwise, set y′
k+1
= −1 and y′
k+2
= 1 − yk+1 − yk+2 , again increasing the sum of the 3
√yi. Now we may apply Jensen to equate all positive variables, so that we need only show
k 3
√−1 + ( n − k) 3
√ kn − k ≤ n
3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 |n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8.
Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3
n
= sin 3(α1) + · · · + sin 3(αn) =
14
((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n
3
. The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 +
· · · + x2
n
= 1. Determine the smallest possible value of
x51
x2 + x3 + · · · + xn
x52
x3 + · · · + xn + x1
· · · + x5
n
x1 + · · · + xn−1
Solution. Observe that ∑ni=1 xi
∑
j6=i
xj ≤ n − 1, so that
( n∑
i=1
xi
(∑
j6=i
xj
)) ( n∑
i=1
x5
i
∑
j6=i
xi
)
≥ (x31 + · · · + x3
n
)2
= n2
(x31 + · · · + x3
n
n
)2
≥ n2
(x21 + · · · + x2
n
n
)3
= 1
n
Leads to n∑
i=1
x5
i
∑
j6=i
xi
≥ 1
n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum
x1 + x22
2 + x33
3 + · · · + xnn
n
12 where x1, x 2, . . . , x n are positive reals such that 1
x1
1
x2
· · · + 1
xn
= n
Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM
x1 + x22
2 + x33
3 + · · · + xnn
n ≥
(
1 + 12 + 13 + · · · + 1
n
)
1+ 12 +··· + 1
n
√x1x2 · · · xn
= 1 + 12 + 13 + · · · + 1
n
Prove that for all positive reals a, b, c, d ,
a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d)
Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4
51 ≥ 51
√a102 b51 c51 d51 = a2bcd
from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights
x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system
x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that
a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2
Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c >
Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac +
bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1
x1 + 1998 + 1
x2 + 1998 + · · · + 1
xn + 1998 = 11998 Prove that n
√x1x2 · · · xn
n − 1 ≥ 1998
Solution. Let yi = 1
xi+1998
so that y1 + · · · + yn = 11998 and xi = 1
yi
− 1998. Now
n
∏
i=1
xi =
n
∏
i=1
( 1
yi
− 1998
)
= e
Pni=1 ln
“1
yi−1998
”
Hence, to minimize the product of the xi, we equivalently minimize the sum of ln
( 1
yi
− 1998
)
.In particular,
ddy
(
ln
( 1
y − 1998
))
= 1
( 1
y
− 1998
)2 · −1
y2
= −1
y − 1998 y2
d2
dy 2
(
ln
( 1
y − 1998
))
= 1 − 3996 y
(y − 1998 y2)2
So ln
( 1
y
− 1998
)
is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider
( 1
a − 1998
) ( 1
b − 1998
)
≥
( 2
a + b − 1998
)2
⇐⇒ 1
ab − 1998
( 1
a + 1
b
)
≥ 4(a + b)2 − 4 · 1998
a + b
⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998
⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2
which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1
n − 1 + x1
· · · + 1
n − 1 + xn
≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1
k+ey
for an arbitrary nonnegative constant k. We have f ′(y) = −ey
(k+ey)2
and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence,
f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and
n
∑
i=1
1
n − 1 + xi
=
n
∑
i=1
f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥
ym+1 ≥ · · · xn for some positive m. Then by, Majorization,
f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f
((m − 1) k0 + ym+1 + · · · + yn
n − 1
)
Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N
Applying the lemma, it would suffice to show
kk + x + 1
k + 1
xk
≤ 1Clearing the denominators,
(
k2 + kxk
)
( k + x) ≤ k2 + k
(
x + 1
xk
)
x1−k
−xk + x + k ≤ x1−k
But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥
1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c ,
√b + ca +
√c + ab +
√a + bc ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal.
15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now,
∑
cyc
√b + ca ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
⇐⇒ ∑
cyc
b + ca
√(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show
∑
cyc
(b + c)
√bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM,
∑
cyc
(b + c)
√bc a ≥
(2( a + b + c))
(√bc a + √ca b + √ab c
)
3
≥ 2( a + b + c)as desired.
Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to
∑
cyc
xy2 + z2 − x2 ≥ x2 + y2 + z2
xyz
Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent
∑
cyc
x2
cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n,
x31
x21 + x1x2 + x22
x32
x22 + x2x3 + x23
· · · + x3
n
x2
n
xnx1 + x21
≥ x1 + · · · + xn
3
Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1
x3
i−x3
i+1
x2
i+xixi+1 +x2
i+1
.Hence, (where xn+1 = x1)
n
∑
i=1
x3
i
x2
i
xixi+1 x2
i+1
= 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
≥ 12
n
∑
i=1
xi + xi+3
3 = 13
n
∑
i=1
xi
as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc
Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z +
x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that
a
√2b2 + 2 c2 − a2 + b
√2c2 + 2 a2 − b2 + c
√2a2 + 2 b2 − c2 ≥ √3
Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have
∑
cyc
a
√2b2 + 2 c2 − a2 = ∑
cyc
y + z
√4x2 + 4 xy + 4 xz + y2 + z2 − 2yz
17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have
∑
cyc
(y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥
(( y + z) + ( z + x) + ( x + y)) f
(∑
cyc
(y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y)
)
= 2√2
√∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2
Noting that ∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 =
∑
cyc
2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑
cyc
2x3 + 7 x2(y + z) + 8 xyz
⇐⇒ ∑
sym
4x3 + 24 x2y + 8 xyz ≥ ∑
sym
3x3 + 21 x2y + 12 xyz
⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz
which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality)
(∑
i
αiβi
) ( ∑
i
αi
1
√βi
)2
≥
(∑
i
αi
)3
Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c
in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C
respectively. Recall from triangle goemetry that ma = 12
√2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a
4
, 3b
4
, and 3c
4
. The desired inequality is therefore equivalent to 43 ma
a
43 mb
b
43 mc
c
≥ 2√3 where we refer to the new triangle ABC . Recalling that
23
ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8
8For a complete proof of this last inequality, see post #14.
18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n,
∑
1≤i<j ≤n
xixj (x2
i
x2
j
) ≤ C
( n∑
i=1
xi
)4
Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 =
( n∑
i=1
x2
i
2 ∑
1≤i<j ≤n
xixj
)2
≥ 4
( n∑
i=1
x2
i
) (
2 ∑
1≤i<j ≤n
xixj
)
= 8 ∑
1≤i<j ≤n
xixjn∑
k=1
x2
k
But x21 + · · · + x2
n
≥ x2
i
x2
j
with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑
1≤i<j ≤n
xixj
(x2
i
x2
j
)
as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2)
Solution 1. Consider
∑
cyc
(a − b)6 = ∑
cyc
a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑
cyc
ab (a − b)4 = ∑
cyc
a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result.
Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have
a6 − 2a3b3 + b6 = (a3 − b3)2
= ((a − b)( a2 + ab + b2))2
≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4
19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of
a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that
a3 + b3 + c3 + 4 abc ≤ 932
Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because
f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As
∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have
λ = 3a2 + 4 bc
= 3b2 + 4 ca
= 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve
h(t) = f
(12, t, 12 − t
)
= 18 + t3 +
(12 − t
)3
2 t
(12 − t
)
= 14 + t
4 − t2
2
h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking,
h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9
(Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals
a, b, c ,
3
√
a3 + pabc
1 + p + 3
√
b3 + pabc
1 + p + 3
√
c3 + pabc
1 + p ≤ a + b + c
Solution. By H¨ older,
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤
(∑
cyc
11 + p
) ( ∑
cyc
a
) ( ∑
cyc
a2 + pbc
)
But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that
∑
cyc
a2 + pbc ≤ (p + 1) ∑
cyc
a2 + 2 bc
3 = p + 1 3 (a + b + c)2
From which we conclude
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤ (a + b + c)3
as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that
a4 + b4 + c4 + 3( a + b + c) ≥ a2
b + a2
c + b2
c + b2
a + c2
a + c2
b
Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a +
b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c
9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution.
21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c ,
a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc )
≥ 2a3b3 + 2 b3c3 + 2 c3a3
Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c −
a)|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial
P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x −
y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that
a
√b + c + b
√c + a + c
√a + b ≥√22 · (a + b + c)
Solution. By Cauchy
(∑
cyc
a√b + c
) ( ∑
cyc
a
√b + c
)
≥ (a + b + c)2
But, also by Cauchy,
√(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑
cyc
a√b + c
Hence, ∑
cyc
a
√b + c ≥√22 · (a + b + c) ·
√ a + b + cab + bc + ca
22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that
ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2
(a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
≥ 94
Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
=
∑
sym
4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2
on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 =
∑
sym
9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2
Canceling like terms, we seek
∑
sym
4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2
Sure enough, this is true, since 3a5b+ab 5
4
≥ a4b2 and a4b2+a2b4
2
≥ a3b3 by AM-GM, and
abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2
(a + b)2 + c2 + (b + c − a)2
(b + c)2 + a2 + (c + a − b)2
(c + a)2 + b2 ≥ 35
10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force.
23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2
(b + c)2 + a2 = (3 − 2a)2
(3 − a)2 + a2 ≥ 15 + ka − k
for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where
a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 −
(15 + ka − k
) ((3 − a)2 + a2) = 18 25 − 54 a2
25 + 36 a3
25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that
( 2ab + c
)23
+
( 2bc + a
)23
+
( 2ca + b
)23
≥ 3
Solution. Suppose that there exists some r such that
( 2ab + c
)23
≥ 3ar
ar + br + cr
We could sum the inequality cyclicly to deduce what we want. Since equality holds at
a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a
3 3
√
a · (b+c
2
)2
11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0,
∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative.
24 = a23
(b+c
2
)23
=
( 2ab + c
)23
by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals
p, k ≥ 1, ( a1 + a2 + · · · + an
ap
1
ap
2
· · · + apn
)k
≥ ak
1
ak
2
· · · + akn
apk
1
apk
2
· · · + apk n
where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1.
Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑
i=1
k
√
aki
ak
1
ak
2
· · · + akn
≥
n
∑
i=1
k
√
apk i
apk
1
apk
2
· · · apk n
WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api
ap
1+··· +apn
and
Ti = aqi
aq
1+··· +aqn
for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n
majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for
m ≤ n,
m
∑
i=1
Si ≥
m
∑
i=1
Ti
⇐⇒ (ap
1
· · · + apm) ( aq
1
· · · + aqn) ≥ (aq
1
· · · + aqm) ( ap
1
· · · + apn)
⇐⇒ (ap
1
· · · + apm) (aqm+1 + · · · + aqn
) ≥ (aq
1
· · · + aqm) (apm+1 + · · · + apn
)
⇐⇒ ∑
(i,j )| { 1≤i≤m<j ≤n}
api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k
√x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a)
25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12
∑
cyc
(a2 − 2ab + bc − c2 + ca )2
Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π
7
) : sin 2 (2π
7
) : sin 2 (π
7
).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c ,
a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2
Solution. Upon observing that this inequality is stronger than Schur’s inequality for
r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2
2( x + y) ≥ √2x2 + 2 y2
This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now,
∑
cyc
ab √2a2 + 2 b2 ≤ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b)So it would suffice to show
∑
cyc
a(a − b)( a − c) = ∑
cyc
(a3 + abc − ab (a + b))
≥ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b) − ab (a + b)= ∑
cyc
3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3
2( a + b)= ∑
cyc
ab (a − b)2
2( a + b)But ∑
cyc
(b + c − a)( b − c)2 = 2 ∑
cyc
a(a − b)( a − c)26 so that the desired is
∑
cyc
(
b + c − a − bc b + c
)
(b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find
(
a + b − c − ab a + b
)
(a − b)2 ≥ 0
(
c + a − b − ac a + c
) ((a − c)2 − (b − c)2) ≥ 0
(
b + c − a − bc b + c
)
(b − c)2 +
(
c + a − b − ac a + c
)
(b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved.
Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b,
(1 + k)( a − b)2 + 8 ab
4( a + b) ≥ k
√ak + bk
2
with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b.
Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as
(a − b)4 · P (a, b ), etc.
For x ≥ y ≥ 1, prove that
x
√x + y + y
√y + 1 + 1
√x + 1 ≥ y
√x + y + x
√x + 1 + 1
√y + 1
12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤
,≥respectively, with equality iff a=bor k=±1.
27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write
x − y
√x + y + y − 1
√y + 1 + 1 − x
√1 + x ≥ 0
⇐⇒ a
√2 + a + 2 b + b
√2 + b ≥ a + b
√2 + a + b
But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since
af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f
(a(2 + a + 2 b) + b(2 + b)
a + b
)
= (a + b)f
((a + b)2 + 2( a + b)
a + b
)
= a + b
√2 + a + b
as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that
x1 + x2 + · · · + xn = 1
x1
1
x2
· · · + 1
xn
Prove that 1
n − 1 + x1
1
n − 1 + x2
· · · + 1
n − 1 + xn
≤ 1
Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take
yi = 1
n−1+ xi
, and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi
are too large, the xi are too small and we shall prove 1
x1
· · · + 1
xn
x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1)
(
yi + 1 −
n
∑
j=1
yj
)
= (n − 1) yi − 1 +
n
∑
j=1
(1 − (n − 1) yj )= −xiyi +
n
∑
j=1
xj yj (∗)=⇒ n − 1
xi
−1 +
n
∑
j=1
xj yj
xiyi
(∗∗ )28 Summing () over i,(n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(( n∑
j=1
1
xj yj
)
− 1
xiyi
)
But by Cauchy and (), we have
( n∑
j=1
1
xj yj
)
− 1
xiyi
≥ (n − 1) 2
(∑nj=1 xj yj
)
− xiyi
(n − 1) 2
(n − 1) yi
= n − 1
yi
Hence, (n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(n − 1
yi
)
= ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2)
Solution. Upon expansion, we see that it is equivalent to
∑
sym
56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities:
∑
sym
a6 − 2a5b + a4bc ≥ 0 (1)
∑
sym
a5b − 4a4b2 + 3 a3b3 ≥ 0 (2)
∑
sym
a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3)
∑
sym
a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑
cyc
ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑
sym
a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2
i+j
≥
xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that
(M + z2 + z3 + · · · + z2n
2n
)2
≥
(x1 + · · · + xn
n
) ( y1 + · · · + yn
n
)
Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j
without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥
x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take
r < 1. Let A and B denote the set of i for which xi > r and the set of j for which
yj > r respectively, and write a = |A|, b = |B|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have |C| ≥ | A + B|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, |A + B| ≥ | A| + |B| − 1, since if A and B consist of the values p1 < · · · < p a and
q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb
in A + B. Hence, |C| ≥ a + b − 1. Since |C| ≥ 1, there is some zk > r , and hence,
M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. •
30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be.
3 Problems
(MOP 04) Show that for all positive reals a, b, c ,
( a + 2 ba + 2 c
)3
+
( b + 2 cb + 2 a
)3
+
(c + 2 ac + 2 b
)3
≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏
i=1
1 − aki
aki
≥ (nk − 1)n
Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that
a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that
ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c
Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1
· · · + 11 + an
≤ a1 + · · · + an + n
(Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either
x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that
f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0
13 This improvement is more widely known than the other one in this packet, and is published in his book,
Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”)
31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that
a
√a2 + 8 bc + b
√b2 + 8 ca + c
√c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that
(a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3
(Titu Andreescu) Show that for all nonzero reals a, b, c ,
a2
b2 + b2
c2 + c2
a2 ≥ ac + cb + ba
(IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that
ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that
√ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca
(APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that
a2
√(a3 + 1) ( b3 + 1) + b2
√(b3 + 1) ( c3 + 1) + c2
√(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c ,
a3
b2 − bc + c2 + b3
c2 − ca + a2 + c3
a2 − ab + b2 ≥ a + b + c
(USAMO 97/5) Prove that for all positive reals a, b, c ,1
a3 + b3 + abc + 1
b3 + c3 + abc + 1
c3 + a3 + abc ≤ 1
abc
(Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k
(IMO 05/3) Prove that for all positive a, b, c with product at least 1,
a5 − a2
a5 + b2 + c2 + b5 − b2
b5 + c2 + a2 + c5 − c2
c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that
aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c ,
3
√4a3 + 4 b3 + 3
√4b3 + 4 c3 + 3
√4c3 + 4 a3 ≤ 4a2
a + b + 4b2
b + c + 4c2
c + a
Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that
ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1
a2 − 1 + 1
b2 − 1 + 1
c2 − 1 = 1 Prove that 1
a + 1 + 1
b + 1 + 1
c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x
for all real numbers 0 < x < π
4
.24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a
√a2 + b2 + b
√b2 + c2 + c
√c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals
a1, a 2, . . . , a n with a1a2 . . . a n = 1,
√a21 + 1 2 + · · · +
√a2
n
1 2 ≤ a1 + · · · + an
33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1
n
be a real number. Show that for all positive reals a1, a 2, . . . , a n,
( (n − 1) a1
a2 + · · · + an
)k
+
( (n − 1) a2
a3 + · · · + an + a1
)k
· · · +
( (n − 1) an
a1 + · · · + an−1
)k
≥ n
(Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that
x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n,
∑
i,j ∈S
min {aiaj , b ibj } ≤ ∑
i,j ∈S
min {aibj , a j bi}
(Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n,
a1 + √a1a2 + · · · + n
√a1 · · · an
n ≤ n
√
a1 · a1 + a2
2 · · · a1 + · · · + an
n
(Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3,
aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+| xy + yz + zx = 3,
a(y + z)
b + c + b(z + x)
c + a + c(x + y)
a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k,
∑
cyc
max( ak, b k)( a − b)2
2 ≥ ∑
cyc
ak(a − b)( a − c) ≥ ∑
cyc
min( ak, b k)( a − b)2
2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and
k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that
ab + ( k − 3) bc + ca
(b − c)2 + kbc + bc + ( k − 3) ca + ab
(c − a)2 + kca + ca + ( k − 3) ab + bc
(a − b)2 + kab ≥ 3( k − 1)
k
(Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have
abcd
(2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4
(2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4
34 |
8157 | http://joelcorbo.com/docs/notes/kleppner-kolenkow-errata.pdf | Kleppner & Kolenkow Errata Updated September 12, 2008 This is a list of typos from An Introduction to Mechanics by Daniel Kleppner and Robert Kolenkow compiled for Physics H7A at UC Berkeley. To reach the compiler, please email Joel Corbo at jcorbo@berkeley.edu. Thanks to all those who have contributed to this list.
Chapter 1: Vectors and Kinematics • Page 17, Example 1.7 In the figure at the top of the page, the labels r(0) and r(t) on the position vectors have been inverted.
• Page 26, Section 1.9 In the line “Using the angle ∆θ defined in the sketch,” there is no such sketch. This is my best guess as to what was intended: A(t1) ∆A A(t2) ∆θ • Page 42, Notes 1.1 The first equation in the Taylor’s Series section should read: f(x) = a0 + a1x + a2x2 + · · · = ∞ X k=0 akxk.
• Page 48, Problem 1.13 Replace the first sentence of the problem with “At t = 0, an elevator departs from the ground with uniform speed.” • Page 49, Problem 1.19 Replace the first sentence of the problem with “A tire of radius R rolls in a straight line without slipping.” 1 Chapter 2: Newton’s Laws • Page 103, Problem 2.1 Replace the second sentence of the problem with, “It starts at rest from the origin at t = 0.” • Page 105, Problem 2.17b The second sentence should begin, “Assuming that tan θ > µ,” not “Assuming that tan θ < µ.” Chapter 3: Momentum • Page 129, Example 3.8 The very first sentence on the page should be replaced with, “The instantaneous length of the spring is ra −rb = r′ a −r′ b. The instantaneous departure of the spring from its equilibrium length is ra −rb −l = r′ a −r′ b −l, where l is the unstretched length of the spring.” • Page 131, Example 3.9 In the last sentence of the second paragraph, replace “If the ball hits a resilient surface” with “If the ball hits a softer surface”.
• Page 137, Section 3.5 Towards the bottom of the page, replace the section beginning with “The change in momentum is” and ending with equation 3.18 to: The change in momentum is ∆P = P(t + ∆t) −P(t) = M∆v + (∆m)u + ∆m∆u Therefore, dP dt = lim ∆t→0 ∆P ∆t = lim ∆t→0 M ∆v ∆t + u∆m ∆t + ∆m∆u ∆t = M dv dt + udm dt .
3.18 • Page 149, Problem 3.17 In the answer clue, the weight W should be equal to 8.2 N, not 10 kg.
2 Chapter 4: Work and Energy • Page 157, Example 4.3 The left-hand side of the second equation on the page should read “K(r) −K(Re)”, not “K(r) −K(re)”.
• Page 161, Section 4.5 The line after equation 4.15 should read, “where V = ˙ R is the velocity of the center of mass.” • Page 164, Example 4.6 About half way down the page, replace each instance of ϕ with φ in the equations after “The work done by gravity is”. (Note that these two symbols are different ways to write the Greek letter phi.) Chapter 5: Some Mathematical Aspects of Force and Energy • Page 205, Example 5.2 In the equation below “The differential of f is,” replace the first dy with dx, producing d f = ∂(xy) ∂x dx + ∂(xy) ∂y dy.
Chapter 6: Angular Momentum and Fixed Axis Rotation • Page 237, Example 6.2 The last equation on the page should read LA = Mr2ωˆ k.
• Page 242, Example 6.4 The second sentence in the second-to-last paragraph on the page should begin “The total energy E = K + U is”.
• Page 261, Section 6.7 The second line of the set of equations after “Eliminating rj from Eq. (6.9) gives” should read Σ(R + r′ j) × mj( ˙ R + ˙ r′ j).
• Page 276, Note 6.2 In the equation following E = K + U, the term 1 2l2 ˙ φ2 should read 1 2ml2 ˙ φ2.
• Page 282, Problem 6.23 The answer clue should read, “If A = 2a, then α = 3a R .” 3 Chapter 7: Rigid Body Motion • Page 294, Example 7.5 About a third of the way down the page, the equations after “The torque is” should read: τ = dL dt = Lω sin α(−ˆ ı sin ωt +ˆ cos ωt).
• Page 336, Problem 7.8b The inequality should read F ≪2Mv2 b , not F ≪Mv2 b .
Chapter 8: Noninertial Systems and Fictitious Forces • Page 362, Example 8.8 Delete the first instance of the equation “Fθ = −2m ˙ rΩ” that appears on this page.
• Page 371, Note 8.2 The equation between equations 3 and 4 should read dB dt = d dt(Bxˆ ı + Byˆ + Bzˆ k).
Chapter 9: Central Force Motion • Page 381, Section 9.3 In the paragraph after equation 9.11, change the second line to, “to simply as the effective potential. Ueff differs from the true.” • Page 383, Section 9.4 The sentence just before the end of the section should refer to Eqs. (9.17a and b), not Eqs. (9.16a and b), which do not exist.
• Page 392, Equation 9.22 In equation 9.22, the term (1 −ϵ2)x should read (1 −ϵ2)x2 • Page 394, Example 9.4 The equation at the top of the page should be replaced with cos θa = 1 ϵ.
• Page 396, Example 9.5 In the line after “Knowing A, we can find E from Eq. (9.26):”, change the equation after the word “or” to E = −C A.
• Page 406, Problem 9.3 Replace the last sentence of the problem with, “Find θ as a function of r for motion with uniform radial velocity v.” 4 Chapter 10: The Harmonic Oscillator • Page 419, Example 10.2 At the end of the first set of calculations, the result should be approximately 7000, not approximately 700.
• Page 427, Section 10.3 The last sentence before the plot should read “γ is given in units of s−1.” • Page 438, Problem 10.3 In the figure accompanying this problem, replace the symbol ϕ with φ.
5 |
8158 | https://openstax.org/books/calculus-volume-2/pages/5-1-sequences | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Calculus Volume 2
5.1 Sequences
Calculus Volume 25.1 Sequences
Search for key terms or text.
Learning Objectives
5.1.1 Find the formula for the general term of a sequence.
5.1.2 Calculate the limit of a sequence if it exists.
5.1.3 Determine the convergence or divergence of a given sequence.
In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.
Terminology of Sequences
To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form
Each of the numbers in the sequence is called a term. The symbol is called the index variable for the sequence. We use the notation
to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number exists for each positive integer we can also define a sequence as a function whose domain is the set of positive integers.
Let’s consider the infinite, ordered list
This is a sequence in which the first, second, and third terms are given by and You can probably see that the terms in this sequence have the following pattern:
Assuming this pattern continues, we can write the term in the sequence by the explicit formula Using this notation, we can write this sequence as
Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the term in terms of the previous term In particular, we can define this sequence as the sequence where and for all each term is defined by the recurrence relation
Definition
An infinite sequence is an ordered list of numbers of the form
The subscript is called the index variable of the sequence. Each number is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case for some function defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.
Note that the index does not have to start at but could start with other integers. For example, a sequence given by the explicit formula could start at in which case the sequence would be
Similarly, for a sequence defined by a recurrence relation, the term may be given explicitly, and the terms for may be defined in terms of Since a sequence has exactly one value for each positive integer it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence consists of all points for all positive integers Figure 5.2 shows the graph of
Figure 5.2 The plotted points are a graph of the sequence
Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence
You can see that the difference between every consecutive pair of terms is Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation
Note that
Thus the sequence can also be described using the explicit formula
In general, an arithmetic sequence is any sequence of the form
In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence
We see that the ratio of any term to the preceding term is Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as
Alternatively, since
we see that the sequence can be described by using the explicit formula
The sequence that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is In general, a geometric sequence is any sequence of the form
Example 5.1
Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the term of the sequence.
Solution
First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the term includes a factor of Next, consider the sequence of numerators and the sequence of denominators We can see that both of these sequences are arithmetic sequences. The term in the sequence of numerators is and the term in the sequence of denominators is Therefore, the sequence can be described by the explicit formula
The sequence of numerators is a geometric sequence. The numerator of the term is The sequence of denominators is an arithmetic sequence. The denominator of the term is Therefore, we can describe the sequence by the explicit formula
Checkpoint 5.1
Find an explicit formula for the term of the sequence
Example 5.2
Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence.
for
for
Solution
Writing out the first few terms, we have
In general,
Write out the first few terms:
From this pattern, we derive the explicit formula
Checkpoint 5.2
Find an explicit formula for the sequence defined recursively such that and
Limit of a Sequence
A fundamental question that arises regarding infinite sequences is the behavior of the terms as gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as For example, consider the following four sequences and their different behaviors as (see Figure 5.3):
The terms become arbitrarily large as In this case, we say that as
The terms as
The terms alternate but do not approach one single value as
The terms alternate for this sequence as well, but as
Figure 5.3 (a) The terms in the sequence become arbitrarily large as (b) The terms in the sequence approach as (c) The terms in the sequence alternate between and as (d) The terms in the sequence alternate between positive and negative values but approach as
From these examples, we see several possibilities for the behavior of the terms of a sequence as In two of the sequences, the terms approach a finite number as In the other two sequences, the terms do not. If the terms of a sequence approach a finite number as we say that the sequence is a convergent sequence and the real number is the limit of the sequence. We can give an informal definition here.
Definition
Given a sequence if the terms become arbitrarily close to a finite number as becomes sufficiently large, we say is a convergent sequence and is the limit of the sequence. In this case, we write
If a sequence is not convergent, we say it is a divergent sequence.
From Figure 5.3, we see that the terms in the sequence are becoming arbitrarily close to as becomes very large. We conclude that is a convergent sequence and its limit is In contrast, from Figure 5.3, we see that the terms in the sequence are not approaching a finite number as becomes larger. We say that is a divergent sequence.
In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.
Definition
A sequence converges to a real number if for all there exists an integer such that if The number is the limit of the sequence and we write
In this case, we say the sequence is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.
We remark that the convergence or divergence of a sequence depends only on what happens to the terms as Therefore, if a finite number of terms are placed before to create a new sequence
this new sequence will converge if converges and diverge if diverges. Further, if the sequence converges to this new sequence will also converge to
Figure 5.4 As increases, the terms become closer to For values of the distance between each point and the line is less than
As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences and shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence diverges because the terms alternate between and but do not approach one value as On the other hand, the sequence diverges because the terms as We say the sequence diverges to infinity and write It is important to recognize that this notation does not imply the limit of the sequence exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence diverges to negative infinity because as We write this as
Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence and a related function defined on all positive real numbers such that for all integers Since the domain of the sequence is a subset of the domain of if exists, then the sequence converges and has the same limit. For example, consider the sequence and the related function Since the function defined on all real numbers satisfies as the sequence must satisfy as
Theorem 5.1
Limit of a Sequence Defined by a Function
Consider a sequence such that for all If there exists a real number such that
then converges and
We can use this theorem to evaluate for For example, consider the sequence and the related exponential function Since we conclude that the sequence converges and its limit is Similarly, for any real number such that and therefore the sequence converges. On the other hand, if then and therefore the limit of the sequence is If and therefore we cannot apply this theorem. However, in this case, just as the function grows without bound as the terms in the sequence become arbitrarily large as and we conclude that the sequence diverges to infinity if
We summarize these results regarding the geometric sequence
Later in this section we consider the case when
We now consider slightly more complicated sequences. For example, consider the sequence The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences and As we describe in the following algebraic limit laws, since and both converge to the sequence converges to Just as we were able to evaluate a limit involving an algebraic combination of functions and by looking at the limits of and (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of and by evaluating the limits of and
Theorem 5.2
Algebraic Limit Laws
Given sequences and and any real number if there exist constants and such that and then
provided and each
Proof
We prove part iii.
Let Since there exists a constant positive integer such that for all Since there exists a constant such that for all Let be the larger of and Therefore, for all
□
The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence As shown earlier, Similarly, for any positive integer we can conclude that
In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.
Example 5.3
Determining Convergence and Finding Limits
For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.
Solution
We know that Using this fact, we conclude that
Therefore,
The sequence converges and its limit is
By factoring out of the numerator and denominator and using the limit laws above, we have
The sequence converges and its limit is
Consider the related function defined on all real numbers Since and as apply L’Hôpital’s rule and write
We conclude that the sequence diverges.
Consider the function defined on all real numbers This function has the indeterminate form as Let
Now taking the natural logarithm of both sides of the equation, we obtain
Since the function is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,
Using properties of logarithms, we write
Since the right-hand side of this equation has the indeterminate form rewrite it as a fraction to apply L’Hôpital’s rule. Write
Since the right-hand side is now in the indeterminate form we are able to apply L’Hôpital’s rule. We conclude that
Therefore, and Therefore, since we can conclude that the sequence converges to
Checkpoint 5.3
Consider the sequence Determine whether or not the sequence converges. If it converges, find its limit.
Recall that if is a continuous function at a value then as This idea applies to sequences as well. Suppose a sequence and a function is continuous at Then This property often enables us to find limits for complicated sequences. For example, consider the sequence From Example 5.3a. we know the sequence Since is a continuous function at
Theorem 5.3
Continuous Functions Defined on Convergent Sequences
Consider a sequence and suppose there exists a real number such that the sequence converges to Suppose is a continuous function at Then there exists an integer such that is defined at all values for and the sequence converges to (Figure 5.5).
Proof
Let Since is continuous at there exists such that if Since the sequence converges to there exists such that for all Therefore, for all which implies We conclude that the sequence converges to
□
Figure 5.5 Because is a continuous function as the inputs approach the outputs approach
Example 5.4
Limits Involving Continuous Functions Defined on Convergent Sequences
Determine whether the sequence converges. If it converges, find its limit.
Solution
Since the sequence converges to and is continuous at we can conclude that the sequence converges and
Checkpoint 5.4
Determine if the sequence converges. If it converges, find its limit.
Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.
Theorem 5.4
Squeeze Theorem for Sequences
Consider sequences and Suppose there exists an integer such that
If there exists a real number such that
then converges and (Figure 5.6).
Proof
Let Since the sequence converges to there exists an integer such that for all Similarly, since converges to there exists an integer such that for all By assumption, there exists an integer such that for all Let be the largest of and We must show that for all For all
Therefore, and we conclude that for all and we conclude that the sequence converges to
□
Figure 5.6 Each term satisfies and the sequences and converge to the same limit, so the sequence must converge to the same limit as well.
Example 5.5
Using the Squeeze Theorem
Use the Squeeze Theorem to find the limit of each of the following sequences.
Solution
Since for all integers we have
Since and we conclude that as well.
Since
for all positive integers and we can conclude that
Checkpoint 5.5
Find
Using the idea from Example 5.5b. we conclude that for any real number such that If the sequence diverges because the terms oscillate and become arbitrarily large in magnitude. If the sequence diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.
(5.1)
(5.2)
(5.3)
(5.4)
Bounded Sequences
We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.
Definition
A sequence is bounded above if there exists a real number such that
for all positive integers
A sequence is bounded below if there exists a real number such that
for all positive integers
A sequence is a bounded sequence if it is bounded above and bounded below.
If a sequence is not bounded, it is an unbounded sequence.
For example, the sequence is bounded above because for all positive integers It is also bounded below because for all positive integers n. Therefore, is a bounded sequence. On the other hand, consider the sequence Because for all the sequence is bounded below. However, the sequence is not bounded above. Therefore, is an unbounded sequence.
We now discuss the relationship between boundedness and convergence. Suppose a sequence is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms that are arbitrarily large in magnitude as gets larger. As a result, the sequence cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.
Theorem 5.5
Convergent Sequences Are Bounded
If a sequence converges, then it is bounded.
Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence is bounded, but the sequence diverges because the sequence oscillates between and and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.
Consider a bounded sequence Suppose the sequence is increasing. That is, Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that converges. For example, consider the sequence
Since this sequence is increasing and bounded above, it converges. Next, consider the sequence
Even though the sequence is not increasing for all values of we see that Therefore, starting with the eighth term, the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.
Definition
A sequence is increasing for all if
A sequence is decreasing for all if
A sequence is a monotone sequence for all if it is increasing for all or decreasing for all
We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.
Theorem 5.6
Monotone Convergence Theorem
If is a bounded sequence and there exists a positive integer such that is monotone for all then converges.
The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).
Figure 5.7 Since the sequence is increasing and bounded above, it must converge.
In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.
Example 5.6
Using the Monotone Convergence Theorem
For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.
defined recursively such that
Solution
Writing out the first few terms, we see that
At first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all We can show this as follows.
Therefore, the sequence is decreasing for all Further, the sequence is bounded below by because for all positive integers Therefore, by the Monotone Convergence Theorem, the sequence converges. To find the limit, we use the fact that the sequence converges and let Now note this important observation. Consider Since
the only difference between the sequences and is that omits the first term. Since a finite number of terms does not affect the convergence of a sequence,
Combining this fact with the equation
and taking the limit of both sides of the equation
we can conclude that
Writing out the first several terms,
we can conjecture that the sequence is decreasing and bounded below by To show that the sequence is bounded below by we can show that
To show this, first rewrite
Since and is defined as a sum of positive terms, Similarly, all terms Therefore,
if and only if
Rewriting the inequality as and using the fact that
because the square of any real number is nonnegative, we can conclude that
To show that the sequence is decreasing, we must show that for all Since it follows that
Dividing both sides by we obtain
Using the definition of we conclude that
Since is bounded below and decreasing, by the Monotone Convergence Theorem, it converges. To find the limit, let Then using the recurrence relation and the fact that we have
and therefore
Multiplying both sides of this equation by we arrive at the equation
Solving this equation for we conclude that which implies Since all the terms are positive, the limit
Checkpoint 5.6
Consider the sequence defined recursively such that Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.
Student Project
Fibonacci Numbers
The Fibonacci numbers are defined recursively by the sequence where and for
Here we look at properties of the Fibonacci numbers.
Write out the first twenty Fibonacci numbers.
Find a closed formula for the Fibonacci sequence by using the following steps.
Consider the recursively defined sequence where and Show that this sequence can be described by the closed formula for all
Using the result from part a. as motivation, look for a solution of the equation
of the form Determine what two values for will allow to satisfy this equation.
Consider the two solutions from part b.: and Let Use the initial conditions and to determine the values for the constants and and write the closed formula
Use the answer in 2 c. to show that
The number is known as the golden ratio (Figure 5.8 and Figure 5.9).
Figure 5.8 The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)
Figure 5.9 The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon appears to have these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)
Section 5.1 Exercises
Find the first six terms of each of the following sequences, starting with
1.
for
for
3.
and for
and for
5.
Find an explicit formula for where and for
Find a formula for the term of the arithmetic sequence whose first term is such that for
7.
Find a formula for the term of the arithmetic sequence whose first term is such that for
Find a formula for the term of the geometric sequence whose first term is such that for
9.
Find a formula for the term of the geometric sequence whose first term is such that for
Find an explicit formula for the term of the sequence whose first several terms are (Hint: First add one to each term.)
11.
Find an explicit formula for the term of the sequence satisfying and for
Find a formula for the general term of each of the following sequences.
(Hint: Find where takes these values)
13.
Find a function that identifies the term of the following recursively defined sequences, as
and for
15.
and for
and for
17.
and for
and for
Plot the first terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.
19.
[T] and for
[T] and for
21.
[T] and for
[T] and for
Suppose that and for all Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.
23.
24.
25.
26.
Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.
27.
28.
29.
(Hint:
For each of the following sequences, whose terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.
31.
32.
33.
34.
35.
36.
37.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
etc.
39.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
Use the Squeeze Theorem to find the limit of each of the following sequences.
40.
41.
42.
43.
For the following sequences, plot the first terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.
[T]
45.
[T]
Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.
46.
47.
48.
49.
50.
51.
52.
53.
Newton’s method seeks to approximate a solution that starts with an initial approximation and successively defines a sequence For the given choice of and write out the formula for If the sequence appears to converge, give an exact formula for the solution then identify the limit accurate to four decimal places and the smallest such that agrees with up to four decimal places.
[T]
55.
[T]
[T]
57.
[T]
[T] Suppose you start with one liter of vinegar and repeatedly remove replace with water, mix, and repeat.
Find a formula for the concentration after steps.
After how many steps does the mixture contain less than vinegar?
59.
[T] A lake initially contains fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by each month. However, factoring in all causes, fish are lost each month.
Explain why the fish population after months is modeled by with
How many fish will be in the pond after one year?
[T] A bank account earns interest compounded monthly. Suppose that is initially deposited into the account, but that is withdrawn each month.
Show that the amount in the account after months is
How much money will be in the account after year?
Is the amount increasing or decreasing?
Suppose that instead of a fixed amount dollars is withdrawn each month. Find a value of such that the amount in the account after each month remains
What happens if is greater than this amount?
61.
[T] A student takes out a college loan of at an annual percentage rate of compounded monthly.
If the student makes payments of per month, how much does the student owe after months?
After how many months will the loan be paid off?
[T] Consider a series combining geometric growth and arithmetic decrease. Let Fix and Set Find a formula for in terms of and and a relationship between and such that converges.
63.
[T] The binary representation of a number between and can be defined as follows. Let if and if Let Let if and if Let and in general, and if and if Find the binary expansion of
[T] To find an approximation for set and, in general, Finally, set Find the first ten terms of and compare the values to
For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of integers by fixing two special integers and and letting be the remainder after dividing into then creates a bit sequence of zeros and ones whose term is equal to one if is odd and equal to zero if is even. If the bits are pseudorandom, then the behavior of their average should be similar to behavior of averages of truly randomly generated bits.
65.
[T] Starting with and using ten different starting values of compute sequences of bits up to and compare their averages to ten such sequences generated by a random bit generator.
[T] Find the first digits of using either a computer program or Internet resource. Create a bit sequence by letting if the digit of is odd and if the digit of is even. Compute the average value of and the average value of Does the sequence appear random? Do the differences between successive elements of appear random?
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Authors: Gilbert Strang, Edwin “Jed” Herman
Publisher/website: OpenStax
Book title: Calculus Volume 2
Publication date: Mar 30, 2016
Location: Houston, Texas
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© Jul 24, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. |
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Hematology: Basic Principles and Practice, Expert Consult - Online and Print
This leading text reflects both the new direction and explosive growth of the field of hematology. Edited and written by practitioners who are the leaders in the field, the book covers basic scientific foundations of hematology while focusing on its clinical aspects. This edition has been thoroughly updated and includes ten new chapters on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more. The table of contents itself has been thoroughly revised to reflect the rapidly changing nature of the molecular and cellular areas of the specialty. Over 1,000 vivid images, now all presented in full color for the first time, include a collection of detailed photomicrographs in every chapter, selected by a hematopathology image consultant. What's more, this Expert Consult Edition includes access to the complete contents of the book online, fully searchable.
A multidisciplinary approach from a unique team of specialists delivers well-rounded guidance on every topic.
"Red boxes" present the authors' unique personal strategies for diagnosis and treatment.
An intuitively re-designed table of contents makes information easier than ever to find.
Online access allows you to rapidly search the entire contents of the book.
Ten cutting-edge new chapters - on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more - keep you completely current.
More than 1,000 vibrant images, all presented in full color for the first time, include a collection of detailed photomicrographs in several chapters selected by a hematopathology image consultant.
An enhanced clinical focus makes actionable information easier to find and apply.
Enhanced coverage of pediatric hematology reflects the supervision of new section editor Helen Heslop, MD, FRCPA, FRACP.
1147752736
Hematology: Basic Principles and Practice, Expert Consult - Online and Print
This leading text reflects both the new direction and explosive growth of the field of hematology. Edited and written by practitioners who are the leaders in the field, the book covers basic scientific foundations of hematology while focusing on its clinical aspects. This edition has been thoroughly updated and includes ten new chapters on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more. The table of contents itself has been thoroughly revised to reflect the rapidly changing nature of the molecular and cellular areas of the specialty. Over 1,000 vivid images, now all presented in full color for the first time, include a collection of detailed photomicrographs in every chapter, selected by a hematopathology image consultant. What's more, this Expert Consult Edition includes access to the complete contents of the book online, fully searchable.
A multidisciplinary approach from a unique team of specialists delivers well-rounded guidance on every topic.
"Red boxes" present the authors' unique personal strategies for diagnosis and treatment.
An intuitively re-designed table of contents makes information easier than ever to find.
Online access allows you to rapidly search the entire contents of the book.
Ten cutting-edge new chapters - on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more - keep you completely current.
More than 1,000 vibrant images, all presented in full color for the first time, include a collection of detailed photomicrographs in several chapters selected by a hematopathology image consultant.
An enhanced clinical focus makes actionable information easier to find and apply.
Enhanced coverage of pediatric hematology reflects the supervision of new section editor Helen Heslop, MD, FRCPA, FRACP.
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Overview
This leading text reflects both the new direction and explosive growth of the field of hematology. Edited and written by practitioners who are the leaders in the field, the book covers basic scientific foundations of hematology while focusing on its clinical aspects. This edition has been thoroughly updated and includes ten new chapters on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more. The table of contents itself has been thoroughly revised to reflect the rapidly changing nature of the molecular and cellular areas of the specialty. Over 1,000 vivid images, now all presented in full color for the first time, include a collection of detailed photomicrographs in every chapter, selected by a hematopathology image consultant. What's more, this Expert Consult Edition includes access to the complete contents of the book online, fully searchable.
A multidisciplinary approach from a unique team of specialists delivers well-rounded guidance on every topic.
"Red boxes" present the authors' unique personal strategies for diagnosis and treatment.
An intuitively re-designed table of contents makes information easier than ever to find.
Online access allows you to rapidly search the entire contents of the book.
Ten cutting-edge new chapters - on cellular biology, haploidentical transplantation, hematologic manifestations of parasitic diseases, and more - keep you completely current.
More than 1,000 vibrant images, all presented in full color for the first time, include a collection of detailed photomicrographs in several chapters selected by a hematopathology image consultant.
An enhanced clinical focus makes actionable information easier to find and apply.
Enhanced coverage of pediatric hematology reflects the supervision of new section editor Helen Heslop, MD, FRCPA, FRACP.
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| ISBN-13: | 9780443067150 |
| Publisher: | Elsevier Health Sciences |
| Publication date: | 10/30/2008 |
| Edition description: | Older Edition |
| Pages: | 2560 |
| Product dimensions: | 8.80(w) x 11.10(h) x 2.70(d) |
About the Author
Table of Contents
Part I. MOLECULAR AND CELLULAR BASIS OF HEMATOLOGY
Chapter 1 Anatomy and Physiology of the Gene
Edward J. Benz, Jr., Andrew Wagner, Nancy Berliner
Chapter 2 Genomic Approaches to the Study of Hematologic Science
Todd Golub and Scott Armstrong
Chapter 3 Protein Synthesis, Processing, and Degradation
Garabet G. Toby and David L. Wiest
Chapter 4 Protein Architecture: Relationship of Form and Function
Alan C. Rigby and Barbara C. Furie
Chapter 5 Regulation of Cellular Response
Benjamin G. Neel and Christopher Carpenter
Chapter 6 Cell Adhesion
Rodger P. McEver
Chapter 7 Control of Cell Division
William M.F. Lee and Chi V. Dang
Chapter 8 Cell Death
David M. Hockenbery and Stanley J. Korsmeyer
Chapter 9 Pharmacogenomics and Hematological Diseases
Loe Kager and William E. Evans
Part II. IMMUNOLOGIC BASIS OF HEMATOLOGY
Chapter 10 Overview and Compartmentalization of the Immune System
Leland D. Powell and Linda G. Baum
Chapter 11 B-Cell Development
Kenneth Dorshkind and David J. Rawlings
Chapter 12 T-Cell and NK-Cell Immunity
Carol Clayberger and Alan M. Krensky
Chapter 13 Dendritic Cell Biology
Zbigniew M. Szczepiorkowski and Nina Bhardwhaj
Chapter 14 Regulation of Activation of B and I Lymphocytes
John C. Monroe and Lawrence Turka
Chapter 15 Leukocyte Trafficking
James J. Campbell, Steven Sloan, and Leslie E.Silberstein
Chapter 16 Tolerance and Autoimmunity
Mark J. Shlomchik
Part III. BIOLOGY OF STEM CELLS AND DISORDERS OF HEMATOPOIESIS
Chapter 17 Stem Cell Model of Hematopoiesis
Anskar YH Leung and Catherine M. Verafaillie
Chapter 18 Bone Marrow Microenvironment
Mo Dao and Catherine M. Verfaillie
Chapter 19 The Humeral Regulation of Hematopoiesis
Montaser Shaheen and Hal Broxmeyer
Chapter 20 Biology of Erythropoiesis, Erythroid Differentiation, and Maturation
Thalia Papayannopoulou, Alan D'Andrea, Janis Abkowitz, and Anna Rita Migliaccio
Chapter 21 Granulopoiesis and Monocytopoiesis
Arati Khanna-Gupta and Nancy Berliner
Chapter 22 Thrombocytopoiesis
Michael W. Long and Ronald Hoffman
Chapter 23 Thrombocytopenia Due to Decreased Platelet Production
J. B. Opalinska and Alan M. Gewirtz
Chapter 24 Inherited Forms of Bone Marrow Failure
Melvin H. Freedman
Chapter 25 Aplastic Anemia
Neal S. Young and Jaroslaw P. Maciejewski
Chapter 26 Paroxysmal Nocturnal Hemoglobinuria
Rob A. Brodsky
Chapter 27 Pure Red Cell Aplasia
Emmanuel N. Dessypris
Part IV. RED BLOOD CELLS
Chapter 28 Pathobiology of the Human Erythrocyte and Its Hemoglobins
Martin H. Steinberg, Edward J. Benz, Jr., Henry Adewoye, and Ben Ebert
Chapter 29 Approach to Anemia in the Adult and Child
Peter Marks and Bertil Glader
Chapter 30 Anemia of Chronic Diseases
Lawrence B. Gardner and Edward J. Benz, Jr.
Chapter 31 Pathology of Iron Metabolism
Nancy C. Andrews
Chapter 32 Disorders of Iron Metabolism
: Iron Deficiency and Overload
Gary M. Brittenham
Chapter 33 Heme Biosynthesis and Its Disorders: Porphyrias and Sideroblastic Anemias
James S. Wiley and Michael R. Moore
Chapter 34 Megablastic Anemias
AÅ“ok C. Anthony
Chapter 35 Thalassemia Syndromes
Bernard G. Forget and Alan R. Cohen
Chapter 36 Pathology of Sickle Cell Disease
Robert P. Hebbel
Chapter 37 Sickle Cell Disease
Stephen H. Embury, Elliot Vichinsky, and Yogen Saunthararajah
Chapter 38 Hemoglobin Variants Associated with Hemolytic Anemia, Altered Oxygen Affinity, and
Methemoglobinemias
Edward J. Benz, Jr.
Chapter 39 Red Cell Enzymopathies
Xylina T. Gregg and Josef T. Prchal
Chapter 40 Red Cell Membrane Disorders
Patrick G. Gallagher and Petr Jarolim
Chapter 41 Autoimmune Hemolytic Anemias
Melody J. Cunningham and Leslie E. Silberstein
Chapter 42 Extrinsic Nonimmune Hemolytic Anemias
Stanley L. Schrier and Erin Gourley Reid
Part V. HOST DEFENSE AND ITS DISORDERS
Chapter 43 Complement Biology and Immunoglobulin
Gregory Kline, Robert Barrington, and Michael C. Carroll
Chapter 44 Normal Phagocyte Structure and Function
Robert L. Baehner
Chapter 45 Eosinophilia, Eosinophil-Associated Diseases, Chronic Eosinophil Leukemia, and the Hypereosinophilic Syndromes
Steven J. Ackerman and Joseph H. Butterfield
Chapter 46 Disorders of Phagocyte Function and Number
Mary C. Dinauer and Thomas D. Coates
Chapter 47 Disorders of Lymphocyte Function
Aileen M. Cleary, Richard A. Insel, and David B. Lewis
Chapter 48 Histiocytic Disorders
Jeffrey M. Lipton and Robert J. Arceci
Chapter 49 Lysosomal Storage Diseases: Perspective and Principles
Gregory A. Grabowski and Nancy Leslie
Chapter 50 Infectious Mononucleosis and Other Epstein-Barr Virus-Associated Diseases
Stephen Gottschalk, Cliona M. Rooney, and Helen E. Heslop
Chapter 51 The Spleen and Its Disorders
Susan B. Shurin
Chapter 52 Basophils, Mast Cells, and Systemic Mastocytosis
Robert I. Parker and Dean D. Metcalfe
Part VI. HEMATOLOGIC MALIGNANCIES
Chapter 53 Conventional Cytogenetics and Molecular Cytogenetics in Hematological
Malignancies
Gordon Dewald and Rhett P. Ketterling
Chapter 54 The Molecular Basis of Neoplasia
Carol Westbrook
Chapter 55 Pharmacology and Molecular Mechanisms of Antineoplastic Agents for Hematologic Malignancies
Stanton Gerson, Kapil N. Bhalla, Steven Grant, Richard Creeger, Nizar Bahlis
Chapter 56 Radiation Therapy in the Treatment of Hematologic Malignancies
Andrea K. Ng and Peter M. Mauch
Chapter 57 Clinical Use of Hematopoietic Growth Factors
Jacob M. Rowe and Irit Avivi
Chapter 58 Radioimmunotherapy for B-Cell and Non-Hodgkin Lymphoma
Thomas E. Witzig
Chapter 59 Pathobiology of Acute Myeloid Leukemia
Brian Huntly and D. Gary Gilliland
Chapter 60 Clinical Manifestations of Acute Myeloid Leukemia
Kenneth B. Miller and Philip R. Daoust
Chapter 61 Therapy for Acute Myeloid Leukemia
Elihu H. Estey and Hagop Kantarijan
Chapter 62 Acute Myeloid Leukemia in Children
Gary V. Dahl and Howard J. Weinstein
Chapter 63 Pathobiology of Acute Lymphoblastic Leukemia
A. Thomas Look and Adolpho Ferrando
Chapter 64 Clinical Manifestations of Acute Lymphoblastic Leukemia
Stacey L. Berg, C. Philip Steuber, and David G. Poplack
Chapter 65 Treatment of Childhood Acute Lymphoblastic Leukemia
Lewis B. Silverman, Stephen E. Sallan, and Harvey J. Cohen
Chapter 66 Acute Lymphoblastic Leukemia in Adults
Dieter Hoelzer and Nicola Gokbudget
Chapter 67 Myelodysplastic Syndrome: Biology and Treatment
Richard Stone and Daniel DeAngelo
Chapter 68 Polycythemia Vera
Ronald Hoffman, Kelty R. Baker and Joe Prchal
Chapter 69 Chronic Myeloid Leukemia
Charles Sawyers and Neil P. Shah
Chapter 70 Myelofibrosis with Myeloid Metaplasia
Ronald Hoffman and Farhad Ravandi-Kashani
Chapter 71 Essential Thrombocythemia
Steven Fruchtman and Ronald Hoffman
Chapter 72 Myelodysplastic Syndromes and Myeloproliferative Syndromes in Children
David M. Loeb, Robert P. Castleberry, and Cindy L. Schwartz
Chapter 73 Pathobiology of Non-Hodgkin Lymphomas
Gianluca Gaidano and Riccardo Dalla-Favera
Chapter 74 Patho-Biology of Hodgkin's Lymphoma
Stefano A. Pileri, Brunangelo Falini and Harold Stein
Chapter 75 Hodgkin's Disease: Clinical Manifestations, Staging, and Therapy
Volker Diehl, Daniel Re, and Andreas Josting
Chapter 76 The Pathologic Basis for the Classification of Non-Hodgkin Lymphomas
Elaine Jaffe and Stefania Piltaluga
Chapter 77 Clinical Manifestations, Staging, and Treatment of Non-Hodgkin Lymphoma
John G. Gribben and Ann S. La Casce
Chapter 78 Malignant Lymphomas in Childhood
John T. Sandlund, Jr.Michael P. Link
Chapter 79 Chronic Lymphocytic Leukemia
Dilip V. Patel and Kanti R. Rai
Chapter 80 Hairy Cell Leukemia
Annaadriana Zakanja, LoAnne C. Petersen, and Martin S. Tallman
Chapter 81 Cutaneous T-Cell Lymphomas
Timothy M. Kuzel, Joan Guitart, and Steven T. Rosen
Chapter 82 AIDS-Related Lymphomas
David T. Scadden
Chapter 83 Multiple Myeloma and Other Plasma Cell Disorders
Guido Tricot
Chapter 84 Immunoglobulin Light Chain Amyloidosis (Primary Amyloidosis, AL)
Morrie A. Gertz, Martha Q. Lacy, Angela Dispenzien
Chapter 85 Atypical Immune Lymphoproliferations
Timothy C. Greiner and Thomas G. Gross
Chapter 86 Clinical Approaches to Infections in the Compromised Host
Jo-Anne H. van Burik
Chapter 87 Nutritional Support of Patients with Hematologic Malignancies
Polly Lenssen and Saundra N. Aker
Chapter 88 Psychosocial Aspects of Hematologic Disorders
Ruth McCorkle, Siew Tzuh Tang, Jeannie Pasacreta
Chapter 89 Pain Management and Antiemetic Therapy in Hematologic Disorders
Janet L. Abrahm, Lauren Dias, and Bridget Fowler
Chapter 90 Palliative Care
Janet L. Abrahm and Barbara Sourkes
Chapter 91 Indwelling Access Devices
Janet L. Abrahm and Linda M. Pellerin
Chapter 92 Late Complications of Hematologic Diseases and Their Therapies
Wendy Landier and Smita Bhatia
Part VII. TRANSPLANTATION
Chapter 93 Hematopoietic Stem Cell Transplantation for Acquired Nonmalignant Diseases and
Myelodysplastic Syndrome
William Tse and H. Joachim Deeg
Chapter 94 Hematopoietic Cell Transplant for Immune Deficiences and Genetic Diseases
Stella M. Davies
Chapter 95 Results of Allogeneic Hematopoietic Cell Transplant for Hematologic
Malignancies
Andrea Bacigalupo
Chapter 96 Autologous Transplant for Hematologic malignancies and Solio Tumors
Stephen J. Forman and Jeffery R. Schriber
Chapter 97 Unrelated Donor Stem Cell Transplantation Therapy
Effie W. Petersdorf and Claudio Anasetti
Chapter 98 Umbilical Cord Blood Transplantation
Juliet N. Barker and John E. Wagner
Chapter 99 Preparative Regimens for Hematopoietic Cell Transplantation
Marcos de Lima and Sergio Giralt
Chapter 100 Graft-versus-Host Disease and Graft-versus-Leukemia
Pavan Reddy and James L.M. Ferrara
Chapter 101 Graft Engineering to Enhance Engraftment, Reduce Graft-versus-Host Disease and
Provide a Graft-versus-Tumor Effect
Catherine M. Bollard and Adrian P. Gee
Chapter 102 Gene Transfer for Hematologic Disorders
Katherine A. High and Malcolm K. Brenner
Chapter 103 Experimental Cell Therapy
Robert S. Negrin
Chapter 104 Complications After Hematopoietic Cell Transplantation
Daniel J. Weisdorf
Part VIII. HEMOSTASIS AND THROMBOSIS
Chapter 105 Megakaryocyte and Platelet Structure
Joseph E. Italiano and John Hartwig
Chapter 106 The Molecular Basis for Platelet Function
Edward F. Plow and Charles S. Abrams
Chapter 107 The Molecular Basis for Platelet Activation
Lawrence F. Brass
Chapter 108 The Blood Vessel Wall
Aly Karsan and John M. Harlan
Chapter 109 Molecular Basis of Blood Coagulation
Bruce Furie and Barbara C. Furie
Chapter 110 Molecular and Cellular Basis of Fibrinolysis
H. Roger Lijnen and Désiré Collen
Chapter 111 Regulatory Mechanisms in Hemostasis: Natural Anticoagulants
Charles T. Esmon
Chapter 112 Clinical Evaluation of Hemorrhagic Disorders: Bleeding History and Differential
Diagnosis of Pupura
Barry S. Coller and Paul I. Schniederman
Chapter 113 Laboratory Evaluation of Hemostatic Disorders
Jacob H. Rand and Lisa Senzel
Chapter 114 Structure, Biology, and Genetics of Factor VIII
Randal J. Kaufman and Stylianos E. Antonarakis
Chapter 115 Biochemistry of Factor IX and Molecular Biology of Hemophilia B
S. Jamie Freedman and Bruce Furie
Chapter 116 Clinical Aspects and Therapy of Hemophilia
Jay Nelson Lozier and Craig M. Kessler
Chapter 117 Inhibitors in Hemophilia
Donald I. Feinstein
Chapter 118 Other Clotting Factor Deficiencies
Harold R. Roberts and Miguel A. Escobar
Chapter 119 Disorders of Fibrinogen
José Martinez and Andres Ferber
Chapter 120 Structure, Biology, and Genetics of von Willebrand Factor
David Ginsburg and Denisa D. Wagner
Chapter 121 von Willebrand Disease: Clinical Aspects and Therapy
Gilbert C. White, II, and J. Evan Sadler
Chapter 122 Vitamin K: Metabolism and Disorders
Barbara C. Furie and Bruce Furie
Chapter 123 Inhibitors of Blood Coagulation
Donald I. Feinstein
Chapter 124 Disseminated Intrvascular Coagulation
Howard A. Liebman and Ilene C. Weitz
Chapter 125 Hemostatic Defects Associated with Dysproteinemias
Howard A. Liebman
Chapter 126 Disorders of Coagulation in the Neonate
Leonardo R. Brandao and Donna DiMichele
Chapter 127 Hypercoagulable States
Kenneth A. Bauer
Chapter 128 Venous Thromboembolism
Mark A. Crowther and Jeffrey S. Ginsberg
Chapter 129 Arterial Thromboembolism
Mark A. Crowther and Jeffrey S. Ginsberg
Chapter 130 Anticoagulation and Thrombolytic Therapy
Jeffrey I. Weitz
Chapter 131 Immune Thrombocytopenic Purpura, Neonatal Alloimmune
Thrombocytopenia, and Posttransfusion Purpura
James Bussel and Douglas Cines
Chapter 132 Thrombotic Thrombocytopenic Purpura and the Hemolytic Uremic Syndrome
Keith R. McCrae, J Evan Sadler, and Douglas Cines
Chapter 133 Thrombocytopenia Due to Platelet Destruction and Hypersplenism
Theodore E. Warkentin and John G. Kelton
Chapter 134 Hereditary Disorders of Platelet Function
Joel S. Bennett
Chapter 135 Acquired Disorders of Platelet Function
Jose Lopez and Perumal Thiagarajan
Part IX. TRANSFUSION MEDICINE
Chapter 136 Human Blood Group Anitgens and Antibodies
Part 1: Carbohydrate Determinants
Part 2: Protein Determinants
Marion E. Reid and Martin L. Olsson
Chapter 137 Human Platelet Antigens
Thomas J. Kunicki and Diane J. Nugent
Chapter 138 Human Leukocyte Antigen (HLA) and Human Neutrophil Antigen (HNA) Systems
Ena Wang, Francesco M. Marincola, and David Stroncek
Chapter 139 Principles of Red Blood Cell Transfusion
Paul M. Ness and Margot S. Kruskall
Chapter 140 Principles of Platelet Transfusion Therapy
Thomas S. Kickler
Chapter 141 Principles of Neutrophil (Granulocyte) Transfusions
Ronald G. Strauss
Chapter 142 Transfusion of Plasma Derivatives: Fresh-Frozen Plasma,
Cryoprecipitate, Albumin, and Immunoglobulins
Cassandra D. Josephson and Christopher D. Hillyer
Chapter 143 Preparation of Plasma-Derived and Recombinant Human Plasma Proteins
William N. Drohan and David B. Clark
Chapter 144 Transfusion Therapy for Coagulation factor Deficiencies
Catherine S. Manno and Peter Larson
Chapter 145 Hemapheresis and Cellular Therapy
Ellen F. Lazarus and Harvey G. Klein
Chapter 146 Practical Aspects of Stem Cell Collection
Scott D. Rowley, Stuart L. Goldberg, and Jack Hsu
Chapter 147 Transfusion Reactions
Edward L. Snyder and YanYun Wu
Chapter 148 Transfusion-Transmitted Diseases
Eberhard W. Fiebig, Michael P. Busch and Jay E. Menitove
Chapter 149 Transfusion Medicine in Hematopoietic Stem Cell and Solid Organ Transplantation
Richard M. Kaufman, Jerome Ritz and Walter H. Dzik
Chapter 150 Red Blood Cell Substitutes
Robert M. Winslow
Part X. CONSULTATIVE HEMATOLOGY
Chapter 151 Hematologic Manifestations of Systemic Disease: Liver and Renal Disease
Rachel Rosovsky and Peter W. Marks
Chapter 152 Hematologic Manifestations of Systemic Disease: Infection, Chronic Inflammation, and
Cancer
Peter W. Marks and David S. Rosenthal
Chapter 153 Hematologic Manifestations of AIDS
David H. Henry and James A. Hoxie
Chapter 154 Hematologic Problems in the Surgical Patient: Bleeding and Thrombosis
Mark T. Reding and Nigel S. Key
Chapter 155 Hematologic Aspects of Pregnancy
Edward J. Benz, Jr.
Chapter 156 Hematologic Manifestations of Childhood Illness
A. Kim Ritchey and Frank G. Keller
Part XI. SPECIAL TESTS AND PROCEDURES
Chapter 157 Bone Marrow Examination
Daniel H. Ryan and Harvey J. Cohen
Chapter 158 Automated Analysis of Blood Cells
Raymond E. Felgar and Daniel H. Ryan
Chapter 159 Laboratory Detection of Hemoglobinopathies and Thalassemias
Martin H. Steinberg and David Chui
Chapter 160 Antenatal Diagnosis of Hematologic Disorders
Kim Kramer and Harvey J. Cohen
Chapter 161 Use of Molecular Techniques in the Analysis of Hematologic Diseases
Michele Paessler and Adam Bagg
Chapter 162 Molecular Imaging in Hematology
Francis G. Blankenberg
Appendix Normal Blood Values: Selected Reference Values for Neonatal, Pediatric, and Adult
Populations
Sharon M. Geaghan
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Editorial Reviews
"Reflect[s] the state of the art. This is an excellent textbook that will appeal to a broad range of readers." - JAMA, review of previous edition
From the Publisher
This third edition is a comprehensive compilation of both malignant and non-malignant hematology. The previous edition was published in 1995. The purpose, according to the editors, is to provide a resource for both beginners and experts eager to learn about the science and practice of hematology. They have achieved this laudatory goal rather well. The book is targeted for a broad audience including both clinical hematologists and bench scientists at all levels of training. The editors are all eminent authorities in their respective fields. This edition is particularly exhaustive in coverage of hematology with numerous new contributors and chapters being added. For example, the emerging area of stem cell transplantation is now extensively covered in depth. A unique feature of this textbook is the inclusion in most chapters of the contributors' preferred work-up and treatment of a clinical entity; this feature is particularly helpful to the non-expert. Also of benefit is the liberal use of charts and tables, which reinforce the main points of the text. The book could have benefited from color glossies, especially in the presentation of hematopathology slides. This third edition is a truly remarkable textbook with a thorough yet highly readable presentation of the hematology field. While the primary purpose is to provide a standard reference, it can very well be read from cover to cover. The quality of chapters is uniformly excellent with virtually all topics covered by experts in the field. This third edition compares favorably with other hematology references, including William's Hematology and Wintrobe's Clinical Hematology.
Ron Gartenhaus
This book is designed to be a comprehensive textbook for practitioners in the field of hematology. The purpose is to provide up-to-date clinical insights into the management of hematologic disorders. It is designed to provide clinicians with up-to-date knowledge about the rapid advances in molecular biology and immunology, especially as they relate to major clinical applications. Therefore, besides providing information that is state-of-the-art on the practice of hematology, this textbook succinctly and successfully bridges the interface between bench research and clinical practice. This book is aimed at individuals involved in the practice of hematology. It is written at a level that is understandable yet interesting to medical students, interns, residents, fellows, faculty, and private practitioners in the field of hematology. The book is well illustrated and contains numerous useful tables. The references are extensive and current. A nice feature of this book is the use of red highlighted boxes that emphasize clinically relevant information and clinical problems. This book is essential for individuals in the practice of hematology. It will be useful both as quick reference and as a complete authoritative source. The editors should be commended for coordinating the efforts of such a large group of outstanding contributors. Information is presented in a scholarly and clinically pertinent manner. It is difficult for any textbook to completely represent all areas of immunology; however, this textbook could have presented more information on the unique clinical and laboratory problems involved in bone marrow transplantation. Nevertheless, it is an excellent textbook without significantomissions.
Richard K. Burt
In less than a decade of existence, this textbook has swiftly established itself as an authoritative and reliable reference source in the hematology field. It appears to me that among the four major textbooks, Hematology: Basic Principles and Practice has achieved the most harmonious and balanced blend of basic science and clinical information. It is a welcome development of the past decade in the field of hematology.
Journal of the American Medical Association
This new edition of a text about the science and practice of hematology reflects the new directions and growth of the field as an academic and clinical discipline. It includes expanded information on stem cell transplantation as well as the personal strategies of many practitioners for the diagnosis and treatment of hematologic disorders. The 160 chapters discuss molecular, cellular, and immunologic bases; biology of stem cells and disorders of hematopoiesis; red blood cells, host defense and its disorders; malignancies; transplantation; hemostasis and thrombosis; transfusion medicine; consultative hematology; and special tests and procedures. Contains many b&w and color illustrations. The editors are professors of medicine associated with leading medical centers. Annotation c. Book News, Inc., Portland, OR (booknews.com)
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8160 | https://onlinelibrary.wiley.com/doi/10.1111/srt.12424 | The relationship between skin function, barrier properties, and body‐dependent factors - Dąbrowska - 2018 - Skin Research and Technology - Wiley Online Library
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Skin Research and Technology
Volume 24, Issue 2 pp. 165-174
INVITED REVIEW
Free Access
The relationship between skin function, barrier properties, and body-dependent factors
A.K. Dąbrowska,
Corresponding Author
A.K. Dąbrowska
agnieszka.dabrowska@empa.ch
orcid.org/0000-0003-3599-0625
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
Laboratory for Surface Science and Technology, Department of Materials, ETH Zürich, Zürich, Switzerland
Correspondence
A.K. Dąbrowska, Laboratory for Protection and Physiology, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland.
Email: agnieszka.dabrowska@empa.ch
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F. Spano,
F. Spano
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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S. Derler,
S. Derler
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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C. Adlhart,
C. Adlhart
Institute of Chemistry and Biotechnology, Zurich University of Applied Sciences, ZHAW, Wädenswil, Switzerland
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N.D. Spencer,
N.D. Spencer
Laboratory for Surface Science and Technology, Department of Materials, ETH Zürich, Zürich, Switzerland
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R.M. Rossi,
R.M. Rossi
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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A.K. Dąbrowska,
Corresponding Author
A.K. Dąbrowska
agnieszka.dabrowska@empa.ch
orcid.org/0000-0003-3599-0625
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
Laboratory for Surface Science and Technology, Department of Materials, ETH Zürich, Zürich, Switzerland
Correspondence
A.K. Dąbrowska, Laboratory for Protection and Physiology, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland.
Email: agnieszka.dabrowska@empa.ch
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F. Spano,
F. Spano
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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S. Derler,
S. Derler
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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C. Adlhart,
C. Adlhart
Institute of Chemistry and Biotechnology, Zurich University of Applied Sciences, ZHAW, Wädenswil, Switzerland
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N.D. Spencer,
N.D. Spencer
Laboratory for Surface Science and Technology, Department of Materials, ETH Zürich, Zürich, Switzerland
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R.M. Rossi,
R.M. Rossi
Laboratory for Biomimetic Membranes and Textiles, Empa, Swiss Federal Laboratories for Materials Science and Technology, St. Gallen, Switzerland
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First published: 23 October 2017
Citations: 340
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Abstract
Background
Skin is a multilayer interface between the body and the environment, responsible for many important functions, such as temperature regulation, water transport, sensation, and protection from external triggers.
Objectives
This paper provides an overview of principal factors that influence human skin and describes the diversity of skin characteristics, its causes and possible consequences. It also discusses limitations in the barrier function of the skin, describing mechanisms of absorption.
Methods
There are a number of in vivo investigations focusing on the diversity of human skin characteristics with reference to barrier properties and body-dependent factors.
Results
Skin properties vary among individuals of different age, gender, ethnicity, and skin types. In addition, skin characteristics differ depending on the body site and can be influenced by the body-mass index and lifestyle. Although one of the main functions of the skin is to act as a barrier, absorption of some substances remains possible.
Conclusions
Various factors can alter human skin properties, which can be reflected in skin function and the quality of everyday life. Skin properties and function are strongly interlinked.
1 Introduction
With a surface of some 2 m 2 and a mass equal to around 15% of total body mass, skin is the human body's largest single organ.1 The main functions of human skin are to regulate the temperature, both by insulation and sweating, to be involved in the functioning of the nervous system and the regulation of water content, and to protect the organism from mechanical injuries, microorganisms, substances, and radiation present in the environment.2, 3
The properties and condition of the skin vary with body site and can be influenced by various inherent body-dependent factors, such as skin type, ethnicity, gender, or even lifestyle and body-mass index (BMI). Skin can be also influenced by the penetration of various substances to which it is exposed.3, 4 The permeability of skin can be used for cosmetic purposes or drug delivery, which has been a developing technology since the 1970s, but can also lead to harmful effects.5, 6
This review article presents various major factors influencing and determining human skin properties and performance and provides an overview of skin barrier functions and their limitations, focusing on skin-penetration mechanisms and the methods used to enable or improve drug-delivery processes. Our aim is to provide readers with an understanding of the complex relationship between specific skin functions. Figure1 shows an overview of principal body-dependent factors and penetrating substances that can influence human skin.
Figure 1
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The overview of principal factors influencing human skin [Colour figure can be viewed at wileyonlinelibrary.com]
1.1 Body-dependent factors influencing human skin
1.1.1 Ethnicity
Skin tone varies among ethnic groups due to different levels of the four chromophores responsible for skin color: hemoglobin and oxyhemoglobin (pinkish tones), melanin (brownish shades), and carotenoids, responsible for yellow-orange tones.7, 8 Ethnicity also influences the natural hydration level of the skin as well as the decrease in hydration level with age.9, 10 It has been reported that Caucasians and African-Americans have slightly drier skin, compared to Chinese, due to the lower levels of SC natural moisturizing factors.7, 10 Other experiments have shown no difference in skin hydration level between ethnic groups,11 which may be due to the significant variations in skin properties among individuals, caused either by different skin types and habits, outweighing the influence of ethnicity.12
Caucasians exhibit higher dryness with age than Chinese, which may be caused by different eating habits and avoidance of sun exposure among Chinese.13 It has been shown that darker skin is more resistant to photo aging processes.8 Also, the thickness of the SC was found to be greater for darker skin.13, 14 Some studies have also indicated differences in the number of layers within comparable SC thickness, suggesting that African-Americans have more cell layers within the SC than do Caucasians (16 and 9 layers, respectively), the skin thus being more resistant to chemicals and damage.15 Most studies demonstrated higher transepidermal water loss (TEWL) for African-Americans than for Caucasians15 as well as larger gland pore sizes and a higher level of sebum secretion.7
1.1.2 Gender
Some researchers found that TEWL is higher for men, explaining this by the fact that they spend more time outdoors and their skin is more damaged and therefore more prone to transepidermal water loss.4 Sebum secretion is considered to be either independent of gender16 or slightly higher in males,4, 10, 17 due to a higher testosterone level.4, 10 The sweating rate was found to be 30%-40% higher in males than in females (taking the difference in body surface area into consideration).18 Results show no clear relationship between gender and SC thickness,14, 19, 20 but it was also found that the cellular _epidermis_ is slightly thicker in males than in females.14 Measurements have shown no clear and statistically significant difference between the elastic properties and pH of the skin in females and males for the same anatomical sites.4, 16, 21, 22 Any relationship between gender and skin hydration seems to be outweighed by the individualized factors.4, 10, 16
1.1.3 Age
Overall, the hydration level of the skin decreases significantly with age, mainly because of the decrease in the amount of natural moisturizers present in the skin,4, 10, 22 but some studies showed contrary results.23 It was also found that the influence of the skin-aging processes on skin hydration can vary between different ethnic groups, being the most significant for Caucasians.9 Furthermore, the elasticity of the skin is also known to decrease, predominantly due to the decreasing collagen production in the dermis and reduction in the resilience of existing collagen and elastin fibers.4, 17, 22, 24 The thickness of skin initially increases with age, showing a maximum value for women at around 30-40 years of age and for men at around 40-50 years, and then significantly decreases with age.25, 26 The amount of sebum secretion has been reported to be either independent or slightly decreasing with age.4, 16, 17 The sweating rate slightly decreases with age.27, 28 Skin pH has been found to be higher for older subjects, due to an age-related decrease in the amount of acidic natural moisturizing factors present in the skin.10, 29 Skin roughness, the dimensions of the primary lines and anisotropy were all found to increase with age.12, 30-32
1.1.4 Skin type
Skin type, its pigmentation, hydration, roughness, and many other parameters are very individual. Significant variations can be observed between people from the same ethnic groups, living in the same environment and sharing the same lifestyle, but having different complexions.
Table 1. Dependence of main properties of human skin on the anatomical site
| Human skin characteristics | Anatomical site | Value/range (reference) |
---
| SC thickness | Volar forearm | 18.3±4.9 μm14 (71 subjects; 37 males, 34 females, age: 20-68 years, median: 47 years) |
| 22.6±4.33 μm36 (15 subjects; 6 males, 9 females, age: 23-49 years, mean: 35.4 years) |
| Shoulder | 11±2.2 μm14 (71 subjects; 37 males, 34 females, age: 20-68 years, median: 47 years) |
| 21.8±3.63 μm36 (15 subjects; 6 males, 9 females, age: 23-49 years, mean: 35.4 years) |
| Buttock | 14.9±3.4 μm14 (71 subjects; 37 males, 34 females, age: 20-68 years, median: 47 years) |
| Cheek | 16.8±2.84 μm36 (15 subjects; 6 males, 9 females, age: 23-49 years, mean: 35.4 years) |
| Back of hand | 29.3±6.84 μm36 (15 subjects; 6 males, 9 females, age: 23-49 years, mean: 35.4 years) |
| Roughness Ra | Forehead | 12-15 μm12 (age: 20-45 years) |
| Volar forearm | 17-20 μm12 (age: 20-45 years) |
| Index finger | 19-33 μm12 (age: 20-45 years) |
| Hydration level | Forehead | 53.54±16.49 corneometer units4 (50 subjects, age: 10-60 years) |
| Forearm | 51.00±15.92 corneometer units4 (50 subjects, age: 10-60 years, corneometry) |
| 26.2±3.4%37 (single-person study, 26 years, confocal Raman spectroscopy) |
| Leg | 37.22±17.50 corneometer units4 (50 subjects, age: 10-60 years) |
| Palm | 40.47±18.47 corneometer units4 (50 subjects, age: 10-60 years, corneometry) |
| 30±5%38 (two subjects: male and female, 30 years, confocal Raman spectroscopy) |
| TEWL | Forehead | 12.27±10.05 g/m 2/h4 (50 subjects, age: 10-60 years) |
| Forearm | 10.12±9.54 g/m 2/h4 (50 subjects, age: 10-60 years) |
| Leg | 9.68±9.52 g/m 2/h4 (50 subjects, age: 10-60 years) |
| Palm | 23.47±9.67 g/m 2/h4 (50 subjects, age: 10-60 years) |
| Sweating rate | Whole body | 95±13 g/m 2 h39 (average for 10 subjects walking on a flat treadmill in 20 °C, 40% RH with a speed of 1.34 m/s) |
| Sebum secretion | Forehead | 95.65±51.38 μg/cm 24 (50 subjects, age: 10-60 years) |
| Forearm | 18.45±37.88 μg/cm 24 (50 subjects, age: 10-60 years) |
| Cheek | 73.39±64.05 μg/cm 24 (50 subjects, age: 10-60 years) |
| Leg | 2.88±6.42 μg/cm 24 (50 subjects, age: 10-60 years) |
| Palm | 9.82±10.11 μg/cm 24 (50 subjects, age: 10-60 years) |
| Elasticity | Forearm | E=8.3±2.1 kPa40 (20 women, age: 55-70 years, indentation in vivo; load: 20 mN, penetration depth: 1000 μm, mean Hertzian contact pressure: 1 kPa) |
| E=0.42-0.85 MPa41 (138 subjects, age: 3-89 years, torsion test in vivo; applied force: 28.6×10−3 Nm) |
| E=129±88 kPa42 (10 subjects, age: 20-30 years, suction test in vivo; suction pressure: 100 mbar) |
| pH | Forehead | 6.43±0.4429 (average amount for 10 subjects at the age of 24-34 years) |
| Forearm | 5.30±0.3029 (average amount for 10 subjects at the age of 24-34 years) |
| Cheek | 5.07±0.4529 (average amount for 10 subjects at the age of 24-34 years) |
1.1.5 Anatomical sites
Skin differs not only between different people but also between anatomical sites for the same person.33-35 Table1 summarizes exemplary characteristics of human skin for different anatomical sites.
For example, the SC thickness varies significantly with the investigated anatomical site.20, 43 It was found that the thickest SC layer is to be found in heels, having 86±36 cell layers, whereas the smallest number of cell layers (6±2) has been observed for genital skin.20 The thickness of the _epidermis_ depends on the body site in a similar way.14, 44 The SC thickness is related to many other phenomena, such as surface morphology, hydration level, and permeability to various substances.37, 45 Since the SC acts as a barrier layer, penetration through the skin is higher on body sites with thinner SC.46, 47 The roughness (Ra) of the index finger lies within the range of 19-33 μm, whereas it is lower (12-20 μm) for the volar forearm.12, 48, 49 Also, sebum secretion varies between different anatomical regions, not only as far as the amount is concerned, but also the chemical composition.29, 50, 51 Due to many reasons, such as exposure to harsh environmental conditions or the frequency of washing with detergents, certain body parts, eg, hands, are more prone to having a lower hydration level of the superficial _stratum corneum_ (SSC).36, 52 The highest density of sweat glands can be found on the soles of the feet (620±20 sweat glands per cm 2), whereas the lowest density of sweat glands is found on the upper lips (16 sweat glands per cm 2).53 In general, the SC thickness is directly related to the TEWL and anatomical sites characterized by the thickest SC display the lowest TEWL values.54 However, TEWL levels are also dependent on other factors, such as SC lipid content, blood flow, or skin temperature.4, 54 This explains the fact, that the TEWL of the palm (characterized by a thick SC layer but a low level of barrier lipids) is higher than that of the leg.4, 29 The elastic properties of skin depend on the collagen structure in the _dermis_, the local thickness of the skin and other parameters that depend on the anatomical site.22, 55 For example, facial skin was found to be less elastic than the skin on the arm and on the back.24 As the properties of human skin are rarely independent of each other but work as a system, a variation in one property is generally coupled with a variation in others. For example, the frictional behavior of human skin varies with body site because it depends on SC roughness, elastic properties, thickness, hydration level, sweating rate as well as on the presence of hair and sebum.12, 36, 56-59
1.1.6 Lifestyle and body-mass index
Lifestyle is considered to be one of the factors influencing the extrinsic aging process, which is related to visible aging caused by the exposure to external factors.60 Proper sun protection prevents accelerated skin aging and the risk of skin cancer and other skin damage.61 A healthy diet, containing significant amount of fruits and vegetables, as well as a calm, low-stress lifestyle, leads to a higher concentration of carotenoids and may result in a slower rate of skin aging.62, 63 Higher daily vitamin C intake has been connected to a decreased formation of wrinkles, while a higher linoleic acid dose has been associated with a lower level of dryness.64 A study has shown that smoking is associated with an altered skin condition (wrinkling, larger pores, rougher skin surface, and presence of discoloration).60 It also significantly increases the risk of skin diseases, such as necrosis after surgery, as it decreases the self-healing abilities of the skin.65, 66
Skin properties are also dependent on the body-mass index (BMI). TEWL is usually higher for obese people.7, 67 Obesity is also correlated to elevated sweat gland activity and higher skin blood flow.67, 68 Obesity can also increase the risk of various skin disorders67, 68 and impaired wound healing.68 For example, 74% of examined obese people have been found to suffer from _acanthosis nigricans_, also related to insulin resistance.68 In another study, 40% of obese children were diagnosed with striae disease.68
1.2 Cause and effect chain
It has to be pointed out that several different factors can influence the skin at the same time in everyday life. The same effect can have various origins while the same cause can result in different symptoms, depending on individuals and circumstances. Moreover, human skin characteristics are generally mutually dependent. Therefore, a variation of one parameter will influence interrelated parameters. An example, showing the cause and effect chain connected with the hydration level of skin, is summarized in Figure2.
Figure 2
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Cause and effect chain for the example of the decrease in skin hydration level. “+” and “-” symbolize positive and negative correlation [Colour figure can be viewed at wileyonlinelibrary.com]
As skin hydration is a very important parameter responsible for skin homeostasis, all deviations from a normal hydration level can result in significant changes in human skin properties and functions.69 Among the main causes of dry skin, one can list skin aging,4 the wrong or no skin care70 or malnutrition.71 Skin hydration can also be influenced by environmental factors72 or by anatomical location (eg, skin on the palms and legs is drier than on the forehead).4 Skin dryness can also be a consequence of various diseases, not only directly related to the skin, such as atopic dermatitis, but also other health problems, eg, hypothyroidism.73-76 A lower hydration level results in a lower elasticity of the skin,4 faster skin aging and wrinkle creation,49 higher surface roughness,77 and lower mechanical resistance.3 Dry skin is also more susceptible to skin diseases and more prone to redness and itchiness.45, 78 The frictional behavior of human skin also depends on hydration.12, 79-82 It was reported that moist skin shows higher friction-coefficient values than dry or completely wet skin. Drier skin is more prone to mechanical failure, flakiness, irritation, and other problems.45, 78 Irritated skin leads to difficulties in achieving and maintaining an adequate hydration level. This results in drier skin and may lead to more severe skin conditions, if untreated.
1.3 Penetration through skin
Our skin is constantly in contact with various substances that are either present in the environment or deliberately applied to the surface of the skin.83 Numerous substances have been applied to the skin surface for medical or religious reasons since the beginning of humanity, which provides a hint that the absorption properties of the skin were already known a long time ago.84 Depending on the circumstances, the barrier properties of human skin, given mainly by its horny layer (SC), may be perceived as being either an advantage or an obstacle.85 In everyday life, the skin can be exposed to various substances in the solid, liquid, or gaseous states. Some of them, such as harmful chemicals, allergens, pathogens etc. can be dangerous and lead to irritation, rashes, burns, or other health problems following the topical application or penetration of these substances into deeper layers of the skin.84, 86 Due to the skin's large surface area (around 2 m 2), the topical dosage of drugs seems to be an interesting alternative for medication, but, because of the barrier function of the skin, this method is far from straightforward.5, 87, 88
The _epidermis_ and _dermis_ are the skin layers involved in the penetration processes, but the SC composition and properties are mainly responsible for the barrier function of human skin.3 Skin protects the body from penetrating substances through various mechanisms, either mechanically blocking particles from further migration into the skin or neutralizing, attacking, or degrading them.3 Substances that penetrate through the SC barrier layer still have to overcome many other obstacles, such as the antimicrobial barrier and immunological or enzymatic systems.3, 86
There are three different pathways that can be used by substances penetrating the skin mentioned in the literature: intercellular,transcellular, and transappendageal.5 Figure3 shows a simplified scheme of skin penetration.
Figure 3
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Three penetration pathways for the skin: intercellular, transcellular, and transappendageal [Colour figure can be viewed at wileyonlinelibrary.com]
The intercellular pathway involves the transport of substances between the cells of the SC layer.5, 89 This mechanism plays a major role in skin permeability and requires the presence of component lipids, such as ceramides, that allow free lateral water diffusion by forming nanometric spaces via short range repulsive forces.89, 90 The diffusion rate depends on the properties of penetrating particles, such as volume, weight, solubility, lipophilicity, or hydrogen-bonding ability.6 It is assumed that particles with a size of 5-7 nm can be efficiently transported through the intercellular pathway.3 Although the SC is a thin layer, reaching a thickness of some 20 μm for the volar forearm,14 the intercellular pathway is much longer and reaches 400 μm, which reduces penetration rate significantly.91-93
The transcellular pathway involves keratinocytes in the transport of substances.5 Despite the seemingly short distances involved, this pathway is very selective. Penetrating particles have to overcome various barriers that are repeated many times in the skin structure; lipophilic cell membranes, hydrophilic cellular contents with keratin, and phospholipidic cell barriers.94, 95
The transappendageal pathway involves appendages, such as sweat and sebaceous glands and hair follicles and is a typical route for the penetration of water-soluble substances.87, 89 Some studies have shown that the size of particles penetrating the skin through aqueous pores can be around 36 nm, whereas trans-follicularly penetrating particles may potentially have a diameter of up to 210 μm (this being the maximum size of the follicular openings).3 However, other researchers have argued that only particles with sizes up to 40 nm88 or even as small as 20 nm6 can effectively penetrate through follicles into deeper skin layers, whereas bigger particles will only be transported deep into the hair follicle.
The transappendageal pathway used to be considered as the least significant penetration passage, as the appendages cover only 0.1% of the skin surface.5, 91 On the other hand, it is the only penetration pathway for particles larger than few nm.3 In addition, appendages may play a role as reservoirs for topically applied substances and therefore could potentially be an efficient penetration path.5
Consistent with the first part of this paper, the properties of human skin can vary significantly not only between different people but also between anatomical areas. The absorption properties of skin are subject to analogous variations, in particular, when skin barrier properties are altered due to a skin disease. Figure4 presents the main body-dependent factors influencing skin permeability.
Figure 4
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Body-dependent factors influencing skin permeability. “+” and “-” symbolize positive and negative correlation between single factors and skin permeability [Colour figure can be viewed at wileyonlinelibrary.com]
The barrier properties of human skin vary significantly among body parts, as the anatomical site strongly influences the majority of skin characteristics, such as SC thickness, density of appendages, and hydration level.3, 96 Skin permeability is facilitated when the SC is thinner, due to a smaller distance having to be covered by penetrating substances.3 The distribution, density, and diameter of follicles also have a major influence on skin barrier properties.3 The maximum density of follicles is to be found on the forehead (292 follicles/cm 2 compared to 50 follicles/cm 2 in other body parts) whereas the maximum diameter of follicles has been reported for the calf (165±45 μm).3 The water-diffusion rate varies depending on the body site. Diffusion rates of 2.1 mg/cm 2/h were reported for the SC on the sole, 0.4 mg/cm 2/h on the calf and 0.1 mg/cm 2/h on the thigh (isolated _dermis_ showed very poor barrier properties and the diffusion rate persisted at a level of 5 mg/cm 2/h).97 The hydration level of skin can vary significantly between body parts. Enhanced hydration leads to an increased penetration rate, since water can act as a penetration enhancer.98 On the other hand, dehydrated skin loses its elasticity, which can lead to surface fracture even during normal everyday activities. Penetration of substances through impaired skin is significantly enhanced.87 For example, eczema (dry and flaky skin) may cause the skin to be 8-10 times more permeable than is the case for normal, healthy skin.98
It was found that sweating and high skin hydration result in increased water absorption.96, 99 Both these parameters are reduced in older subjects.3, 27, 28
It was found that the presence of hair makes men's faces more permeable than women's.18 Men sweat to a greater extent than women (800 ml/h for men versus 450 ml/h for women during physical activity), which can also result in higher skin permeability in males.18 Sex hormones influence the SC chemical composition and may also influence skin permeability.3
All parameters contributing to personal variations in skin permeability are also dependent on the skin type, as all above-mentioned characteristics, such as SC thickness, density of appendages, sweat rate, and hydration level can vary among individuals of the same gender and at the same age.
All the above-mentioned factors influencing human skin barrier properties are interlinked. Therefore, it is impossible to explicitly determine the importance of individual parameters. Moreover, different body-dependent factors will influence skin permeability to a different extent, depending on the penetration pathway preferable for a certain type of penetrating substance.3 In addition, skin permeability can also be influenced by the climate and environmental conditions.3, 37 Increased hydration level of the skin, which can be caused by increased air humidity as well as increased temperature, for example, can act as penetration enhancers.3, 37 This leads to the conclusion, that transdermal drug delivery should be personalized, taking into account both individual skin characteristics and the living environment of patients.
The transport of substances through the structure of human skin strongly depends on the substances themselves, as well as on the accompanying excipients.3 It has been reported that elastic particles can migrate through human skin more efficiently than rigid ones, and that there are higher dimensional limits for elastic penetrating particles than for rigid ones, due to the difference in deformability.100 In addition, the size and shape of the molecule, pH of the solution, as well as other physicochemical descriptors, eg, water-octanol partition coefficients and Abraham solute descriptors, are decisive factors for the ability of the substances to migrate through the skin structure and its penetration depth, the possible pathway and the diffusion coefficient.3, 101
As the penetration of substances is quite often desired, (eg, delivery or application of active substances in cosmetics), many compounds have been investigated as potential penetration enhancers: surfactants, esters, fatty acids, alcohols, amines, terpenes, alkanes, phospholipids, sulfoxides, amides, or pyrrolidones.102 The role of such substances is to reversibly decrease the barrier resistance of human skin.103, 104 Besides chemical penetration enhancers, there are also some techniques enhancing skin penetration, such as electroporation, which leads to the creation of aqueous pores by the application of an electric pulse.105 In addition, it has been observed that massage can increase the transappendageal penetration rate.6
2 Summary
In this review, we have discussed a number of important factors that influence skin properties and function. The foregoing literature survey has disclosed how many factors can alter human skin in everyday life. The complexity of the skin cannot be neglected: the properties and performance of the skin are all interdependent and need to be considered as a system rather than as individual characteristics. Consequently, a change of one property can lead to a wide variety of effects.
Attributes influencing skin properties and functions can be divided into body-dependent factors and factors connected with the skin barrier function.
The majority of body-dependent factors influencing skin are congenital. Ethnicity and gender influence skin properties to a lesser extent than other factors. Skin type can influence UV protection and skin-aging processes. Skin aging results in, _inter alia_, a decrease in hydration level and elasticity. Generally, skin properties, such as thickness, roughness, hydration, TEWL, sweating rate, sebum secretion, elasticity or pH, vary with body site.
Healthy diet and adequate body weight may slow the skin-aging process, decrease the risk of skin disorders and accelerate the wound-healing process. Also, the avoidance of excessive UV exposure allows the maintenance of better skin health.
Skin acts as a barrier that protects the body from the environment. However, certain substances can penetrate into deeper layers of the skin and even be absorbed into the bloodstream. This imperfection in skin's barrier properties leads to the need for adequate protection upon exposure to dangerous substances, but it is also an opportunity for new modalities of drug delivery.
Because of the diversity of skin properties and the complex relationship between skin permeability and other characteristics, individualized factors have to be taken into account for skin modeling or the design of transdermal delivery-based treatment methods.
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8161 | https://www.youtube.com/watch?v=1iSpmHc2q-c | Set Theory Proof: If A n B = empty! then A is a subset of B complement
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Set Theory Proof: If A n B = empty! then A is a subset of B complement
B-Roll - Islandesque by Kevin MacLeod is licensed under a Creative Commons Attribution license (
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[Music] what's up YouTube for this problem I'm going to prove that if you have a intersection B equal to the enemy set that a is a subset of B complement this is called the absolute complement let me explain what that is before we did improve so let's protect me have a complement okay this is the set of all X such that X is not a and if you want to add an extra condition here you can say that X is in U where u is your Universal set you might say what is the universal set was it's the universe's it's not but when you're dealing with sets okay everything all your elements are in the universal set so all of your sets are subsets of the universal set it's kind of like us where people who live in a world who live on planet earth and we're part of the universe well the elements and all of your steps are in the universal set therefore every set is a subset of the universal set so you can omit this or you can leave it it's kind of understood okay let's go to the proof so let's start the proof will assume that this is true and then we have to show that this is true so we'll start by supposing a intersection B is equal to the empty set so suppose a intersect B and the claim is that a is a subset of B complement so claim plane a is a subset of B why am I writing it again just for added clarity right just just for a declaring so once that is a subset of the other to show that every element in this is also in this set so we'll start by taking any element in this set and then trying to show it to you so take a take any X and a take any exudate so now we have to show that it's a B complement so how do we do that so what I'm thinking is what we have to somehow use this well we know that a can't be in B because if it isn't the than a and B as in the intersection which is empty which is a contradiction so again we know that it can't be in both so we have to show that right so one way to do that is to do it use a contradiction so suppose for the chronic contrary suppose to the contrary just a really short contradiction contradiction statement suppose to the contrary that X's would be then that would be that X is an aien V and X is an N and X is B expert verbose eater so that means that X is in the intersection which is empty so we have a contradiction simple these contradictions like opposing arrows okay where's that comfortable have no idea I saw some guy do it once in class I'm like that's really friggin cool so I copied it I stole it from this guy really good professor really really really cool symbol contradiction okay so we took in XNA we're trying to show it's not in V suppose it is in B then that means X isn't named next to me that means is in the intersection was this empty a contradiction hence X is not in B but it is not a B that means that is in the complement of V so we started with an X and a we showed us in the complement this shows rose to it this shows that a is a subset of that's it hope that made some sense thanks for checking out my channel to mathematics take care |
8162 | https://freeessay.com/pars-pro-toto/ | Literary devices - Pars Pro Toto
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Pars Pro Toto
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Pars pro toto is Latin for "(taking) a part for the whole"; it is a kind of synecdoche. When used in a context of language it means that something is named after a part of it (or after a limited characteristic, in itself not necessarily representative for the whole). E.g. "glasses" is a "pars pro toto" name for something that consists of more than just two pieces of glass. The opposite of a pars pro toto is a totum pro parte, in which the whole is used to describe a part.
Pars pro toto
Pars pro toto, Latin for "a part for the whole", is a figure of speech where the name of a portion of an object, place or concept represents the entire object, place or concept. It is distinct from a merism, which is a reference to a whole by an enumeration of parts, metonymy, where an object, place or concept is called by something or some place associated with the object, place or concept, or synecdoche, which can refer both to this and its inverse of the whole representing a part. In the context of language, pars pro toto means that something is named after a part of it, or after a limited characteristic, in itself not necessarily representative for the whole. For example, "glasses" is a pars pro toto name for something that consists of more than just two pieces of glass. Pars pro toto is a common device in iconography, where a particular icon can stand for a complete set of characteristics. James George Frazer used "pars pro toto" to explain his concept of contagious magic.
Pars Pro Toto
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Pars pro toto
The Pars pro toto is a stylistic device of rhetoric and a special form of Synekdoche as well as metonymy and belongs to the group of the tropics. The pars pro toto replaces one expression by another, which is a part of the replaced term. The stylistic counterpart of Pars pro toto is the Totum pro parte. Here the whole stands for a part of the whole.
The word group Pars pro toto comes from the Latin and can be translated with A part [stands] for the whole. The translation thus already reveals what is at stake: namely, a word which stands as a substitute for a whole for this whole. Let us look at an example.
After all, we have a roof over our heads.
The above example is familiar to most people from the everyday language. The term “roof” is used for the entire house. A part of the meant expression (roof) thus stands for the whole (house). There are many such examples in the colloquial language: head for person, sail for ship, four eyes for two persons.
Pars pro toto and Totum per parte
The stylistic counterpart to the stylistic figure described is the Totum pro parte. This is therefore also a special form of Synekdoche. Here the whole thing describes a part.
Germany gets the gold medal in the hurdles.
In the example sentence it is said that Germany cleans a gold medal. But, of course, the individual competitor, who has overshadowed his fellow competitors, is meant here. Therefore the whole (Germany) stands for a part (competitor).
Note: Basically, both style figures are about a relationship between a part and the whole. As a special form of the Synekdoche, however, there are still a few marginal cases which then have a different name. Various types are available under Synekdoche.
Examples of the Pars pro toto
There are also many examples of the stylistic figure in the colloquial language. Some of these have even become so established in our linguistic usage that it is hardly apparent that the term is only a sub-range. This is often the case especially in geography.
The territory of Holland for the Netherlands.
England for the United Kingdom
Russia for the whole Soviet Union
A good drop instead of a good wine
Souls for people
Function and effect of Pars pro toto at a glance
The style figure is usually used to avoid word repetitions. Primarily in speeches revolving around the same theme, this can be regarded as a revaluation of the style.
Furthermore, the whole can focus the receiver on a particular detail. When we say that our blade has defeated the enemy, we will, of course, take it to the fore, in contrast to the “rest” of the weapon.
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8163 | https://math.libretexts.org/Courses/Western_Connecticut_State_University/Draft_Custom_Version_MAT_131_College_Algebra/05%3A_Polynomial_and_Rational_Functions/5.02%3A_Power_Functions_and_Polynomial_Functions | f(x)
f(x)=8x5
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5.2: Power Functions and Polynomial Functions
Last updated
: Jan 16, 2020
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5.1: Quadratic Functions
5.3: Graphs of Polynomial Functions
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Learning Objectives
Identify power functions.
Identify end behavior of power functions.
Identify polynomial functions.
Identify the degree and leading coefficient of polynomial functions.
Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 5.2.15.2.1.
Table 5.2.15.2.1
| Year | 2009 | 2010 | 2011 | 2012 | 2013 |
| Bird Population | 800 | 897 | 992 | 1,083 | 1,169 |
The population can be estimated using the function P(t)=−0.3t3+97t+800P(t)=−0.3t3+97t+800, where P(t)P(t) represents the bird population on the island tt years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes.
Identifying Power Functions
In order to better understand the bird problem, we need to understand a specific type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. (A number that multiplies a variable raised to an exponent is known as a coefficient.)
As an example, consider functions for area or volume. The function for the area of a circle with radius rr is
A(r)=πr2
A(r)=πr2
and the function for the volume of a sphere with radius rr is
V(r)=43πr3
V(r)=43πr3
Both of these are examples of power functions because they consist of a coefficient, ππ or 43π43π, multiplied by a variable rr raised to a power.
Definition: Power Function
A power function is a function that can be represented in the form
f(x)=kxp
f(x)=kxp(5.2.1)
where kk and pp are real numbers, and kk is known as the coefficient.
Q&A: Is f(x)=2xf(x)=2x a power function?
No. A power function contains a variable base raised to a fixed power (Equation 5.2.15.2.1). This function has a constant base raised to a variable power. This is called an exponential function, not a power function. This function will be discussed later.
Example 5.2.15.2.1: Identifying Power Functions
Which of the following functions are power functions?
f(x)=1Constant functionf(x)=xIdentify functionf(x)=x2Quadratic functionf(x)=x3Cubic functionf(x)=1xReciprocal functionf(x)=1x2Reciprocal squared functionf(x)=√xSquare root functionf(x)=3√xCube root function
f(x)f(x)f(x)f(x)f(x)f(x)f(x)f(x)=1=x=x2=x3=1x=1x2=x−−√=x−−√3Constant functionIdentify functionQuadratic functionCubic functionReciprocal functionReciprocal squared functionSquare root functionCube root function
Solution
All of the listed functions are power functions.
The constant and identity functions are power functions because they can be written as f(x)=x0f(x)=x0 and f(x)=x1f(x)=x1 respectively.
The quadratic and cubic functions are power functions with whole number powers f(x)=x2f(x)=x2 and f(x)=x3f(x)=x3.
The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f(x)=x−1f(x)=x−1 and f(x)=x−2f(x)=x−2.
The square and cube root functions are power functions with fractional powers because they can be written as f(x)=x1/2f(x)=x1/2 or f(x)=x1/3f(x)=x1/3.
Exercise 5.2.15.2.1
Which functions are power functions?
f(x)=2x2⋅4x3f(x)=2x2⋅4x3
g(x)=−x5+5x3−4xg(x)=−x5+5x3−4x
h(x)=2x5−13x2+4h(x)=2x5−13x2+4
Answer
: f(x) is a power function because it can be written as f(x)=8x5. The other functions are not power functions.
Identifying End Behavior of Power Functions
Figure 5.2.25.2.2 shows the graphs of f(x)=x2f(x)=x2, g(x)=x4g(x)=x4 and and h(x)=x6h(x)=x6, which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin.
To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol ∞∞ for positive infinity and −∞−∞ for negative infinity. When we say that “x approaches infinity,” which can be symbolically written as x→∞x→∞, we are describing a behavior; we are saying that xx is increasing without bound.
With the even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as xx approaches positive or negative infinity, the f(x)f(x) values increase without bound. In symbolic form, we could write
as x→±∞,f(x)→∞
as x→±∞,f(x)→∞
Figure 5.2.35.2.3 shows the graphs of f(x)=x3f(x)=x3, g(x)=x5g(x)=x5, and h(x)=x7h(x)=x7, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin.
These examples illustrate that functions of the form f(x)=xnf(x)=xn reveal symmetry of one kind or another. First, in Figure 5.2.25.2.2 we see that even functions of the form f(x)=xnf(x)=xn, nn even, are symmetric about the yy-axis. In Figure 5.2.35.2.3 we see that odd functions of the form f(x)=xnf(x)=xn, nn odd, are symmetric about the origin.
For these odd power functions, as xx approaches negative infinity, f(x)f(x) decreases without bound. As xx approaches positive infinity, f(x)f(x) increases without bound. In symbolic form we write
as x→−∞,f(x)→−∞as x→∞,f(x)→∞
as x→−∞,f(x)→−∞as x→∞,f(x)→∞
The behavior of the graph of a function as the input values get very small (x→−∞)(x→−∞) and get very large x→∞x→∞ is referred to as the end behavior of the function. We can use words or symbols to describe end behavior.
Figure 5.2.45.2.4 shows the end behavior of power functions in the form f(x)=kxnf(x)=kxn where nn is a non-negative integer depending on the power and the constant.
How To: Given a power function f(x)=kxnf(x)=kxn where nn is a non-negative integer, identify the end behavior.
Determine whether the power is even or odd.
Determine whether the constant is positive or negative.
Use Figure 5.2.45.2.4 to identify the end behavior.
Example 5.2.25.2.2: Identifying the End Behavior of a Power Function
Describe the end behavior of the graph of f(x)=x8f(x)=x8.
Solution
The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As xx approaches infinity, the output (value of f(x)f(x) ) increases without bound. We write as x→∞,x→∞, f(x)→∞.f(x)→∞. As xx approaches negative infinity, the output increases without bound. In symbolic form, as x→−∞,x→−∞, f(x)→∞.f(x)→∞. We can graphically represent the function as shown in Figure 5.2.55.2.5.
Example 5.2.35.2.3: Identifying the End Behavior of a Power Function.
Describe the end behavior of the graph of f(x)=−x9f(x)=−x9.
Solution
The exponent of the power function is 9 (an odd number). Because the coefficient is –1 (negative), the graph is the reflection about the x-axis of the graph of f(x)=x9. Figure 5.2.6 shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write
as x→−∞,f(x)→∞as x→∞,f(x)→−∞
Analysis
We can check our work by using the table feature on a graphing utility.
Table 5.2.2
| x | f(x) |
| -10 | 1,000,000,000 |
| -5 | 1,953,125 |
| 0 | 0 |
| 5 | -1,953,125 |
| 10 | -1,000,000,000 |
We can see from Table 5.2.2 that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value).
Exercise 5.2.2
Describe in words and symbols the end behavior of f(x)=−5x4.
Answer
: As x approaches positive or negative infinity, f(x) decreases without bound: as x→±∞, f(x)→−∞ because of the negative coefficient.
Identifying Polynomial Functions
An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r of the spill depends on the number of weeks w that have passed. This relationship is linear.
r(w)=24+8w
We can combine this with the formula for the area A of a circle.
A(r)=πr2
Composing these functions gives a formula for the area in terms of weeks.
A(w)=A(r(w))=A(24+8w)=π(24+8w)2
Multiplying gives the formula.
A(w)=576π+384πw+64πw2
This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power.
Definition: Polynomial Functions
Let n be a non-negative integer. A polynomial function is a function that can be written in the form
f(x)=anxn+...+a2x2+a1x+a0
This is called the general form of a polynomial function. Each ai is a coefficient and can be any real number. Each product aixi is a term of a polynomial function.
Example 5.2.4: Identifying Polynomial Functions
Which of the following are polynomial functions?
f(x)=2x3⋅3x+4
g(x)=−x(x2−4)
h(x)=5√x+2
Solution
The first two functions are examples of polynomial functions because they can be written in the form of Equation 5.2.2, where the powers are non-negative integers and the coefficients are real numbers.
f(x) can be written as f(x)=6x4+4.
g(x) can be written as g(x)=−x3+4x.
h(x) cannot be written in this form and is therefore not a polynomial function.
Identifying the Degree and Leading Coefficient of a Polynomial Function
Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term.
Terminology of Polynomial Functions
We often rearrange polynomials so that the powers are descending.
When a polynomial is written in this way, we say that it is in general form.
How To: Given a polynomial function, identify the degree and leading coefficient
Find the highest power of x to determine the degree function.
Identify the term containing the highest power of x to find the leading term.
Identify the coefficient of the leading term.
Example 5.2.5: Identifying the Degree and Leading Coefficient of a Polynomial Function
Identify the degree, leading term, and leading coefficient of the following polynomial functions.
f(x)=3+2x2−4x3
g(t)=5t5−2t3+7t
h(p)=6p−p3−2
Solution
For the function f(x), the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4.
For the function g(t), the highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5.
For the function h(p), the highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3; the leading coefficient is the coefficient of that term, −1.
Exercise 5.2.3
Identify the degree, leading term, and leading coefficient of the polynomial f(x)=4x2−x6+2x−6.
Answer
: The degree is 6. The leading term is −x6. The leading coefficient is −1.
Identifying End Behavior of Polynomial Functions
Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the term of highest degree (Table 5.2.3).
Table 5.2.3
| Polynomial Function | Leading Term | Graph of Polynomial Function |
| f(x)=5x4+2x3−x−4 | 5x4 | |
| f(x)=−2x6−x5+3x4+x3 | −2x6 | |
| f(x)=3x5−4x4+2x2+1 | 3x5 | |
| f(x)=−6x3+7x2+3x+1 | −6x3 | |
Example 5.2.6: Identifying End Behavior and Degree of a Polynomial Function
Describe the end behavior and determine a possible degree of the polynomial function in Figure 5.2.8.
Solution
As the input values x get very large, the output values f(x) increase without bound. As the input values x get very small, the output values f(x) decrease without bound. We can describe the end behavior symbolically by writing
as x→∞,f(x)→∞
as x→−∞,f(x)→−∞
In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity, the function values approach negative infinity.
We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive.
Exercise 5.2.1
Describe the end behavior, and determine a possible degree of the polynomial function in Figure 5.2.9.
Answer
: As x→∞, f(x)→−∞; as x→−∞, f(x)→−∞. It has the shape of an even degree power function with a negative coefficient.
Example 5.2.7: Identifying End Behavior and Degree of a Polynomial Function
Given the function f(x)=−3x2(x−1)(x+4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function.
Solution
Obtain the general form by expanding the given expression for f(x).
f(x)=−3x2(x−1)(x+4)=−3x2(x2+3x−4)=−3x4−9x3+12x2
The general form is f(x)=−3x4−9x3+12x2. The leading term is −3x4; therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is
as x→−∞,f(x)→−∞
as x→∞,f(x)→−∞
Exercise 5.2.7
Given the function f(x)=0.2(x−2)(x+1)(x−5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function.
Answer
: The leading term is 0.2x3, so it is a degree 3 polynomial. As x approaches positive infinity, f(x) increases without bound; as x approaches negative infinity, f(x) decreases without bound.
Identifying Local Behavior of Polynomial Functions
In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing.
We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept (0,a0). The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 5.2.10.
Defintion: Intercepts and Turning Points of Polynomial Functions
A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero.
Given a polynomial function, determine the intercepts.
Determine the y-intercept by setting x=0 and finding the corresponding output value.
Determine the x-intercepts by solving for the input values that yield an output value of zero.
Example 5.2.8: Determining the Intercepts of a Polynomial Function
Given the polynomial function f(x)=(x−2)(x+1)(x−4), written in factored form for your convenience, determine the y- and x-intercepts.
Solution
The y-intercept occurs when the input is zero, so substitute 0 for x.
f(0)=(0−2)(0+1)(0−4)=(−2)(1)(−4)=8
The y-intercept is (0,8).
The x-intercepts occur when the output is zero.
0=(x−2)(x+1)(x−4)
x−2=0orx+1=0orx−4=0x=2orx=−1orx=4
The x-intercepts are (2,0),(–1,0), and (4,0).
We can see these intercepts on the graph of the function shown in Figure 5.2.11.
Example 5.2.9: Determining the Intercepts of a Polynomial Function with Factoring
Given the polynomial function f(x)=x4−4x2−45, determine the y- and x-intercepts.
Solution
The y-intercept occurs when the input is zero.
f(0)=(0)4−4(0)2−45=−45
The y-intercept is (0,−45).
The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial.
f(x)=x4−4x2−45=(x2−9)(x2+5)=(x−3)(x+3)(x2+5)
0=(x−3)(x+3)(x2+5)
x−3=0orx+3=0orx2+5=0x=3orx=−3or(no real solution)
The x-intercepts are (3,0) and (–3,0).
We can see these intercepts on the graph of the function shown in Figure 5.2.12. We can see that the function is even because f(x)=f(−x).
Exercise 5.2.5
5.2.5: Given the polynomial function f(x)=2x3−6x2−20x, determine the y- and x-intercepts.
Solution
: y-intercept (0,0); x-intercepts (0,0),(–2,0), and (5,0)
Comparing Smooth and Continuous Graphs
The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n–1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors.
A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth.
Intercepts and Turning Points of Polynomials
A polynomial of degree n will have, at most, n x-intercepts and n−1 turning points.
Example 5.2.10: Determining the Number of Intercepts and Turning Points of a Polynomial
Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f(x)=−3x10+4x7−x4+2x3.
Solution
The polynomial has a degree of 10, so there are at most n x-intercepts and at most n−1 turning points.
Exercise 5.2.6
Without graphing the function, determine the maximum number of x-intercepts and turning points for f(x)=108−13x9−8x4+14x12+2x3
Answer
: There are at most 12 x-intercepts and at most 11 turning points.
Example 5.2.11: Drawing Conclusions about a Polynomial Function from the Graph
What can we conclude about the polynomial represented by the graph shown in Figure 5.2.12 based on its intercepts and turning points?
Solution
The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 5.2.14.
The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4.
Exercise 5.2.7
What can we conclude about the polynomial represented by the graph shown in Figure 5.2.15 based on its intercepts and turning points?
Figure 5.2.15.
Answer
: Add texts here. Do not delete this text first.
Solution
The end behavior indicates an odd-degree polynomial function; there are 3 x-intercepts and 2 turning points, so the degree is odd and at least 3. Because of the end behavior, we know that the lead coefficient must be negative.
Example 5.2.12: Drawing Conclusions about a Polynomial Function from the Factors
Given the function f(x)=−4x(x+3)(x−4), determine the local behavior.
Solution
The y-intercept is found by evaluating f(0).
f(0)=−4(0)(0+3)(0−4)=0
The y-intercept is (0,0).
The x-intercepts are found by determining the zeros of the function.
0=−4x(x+3)(x−4)x=0orx+3=0orx−4=0x=0orx=−3orx=4
The x-intercepts are (0,0),(–3,0), and (4,0).
The degree is 3 so the graph has at most 2 turning points.
Exercise 5.2.8
Given the function f(x)=0.2(x−2)(x+1)(x−5), determine the local behavior.
Answer
: The x-intercepts are (2,0), (−1,0), and (5,0), the y-intercept is (0,2), and the graph has at most 2 turning points.
Key Equations
general form of a polynomial function: f(x)=anxn+an−1xn−1...+a2x2+a1x+a0
Key Concepts
A power function is a variable base raised to a number power.
The behavior of a graph as the input decreases without bound and increases without bound is called the end behavior.
The end behavior depends on whether the power is even or odd.
A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power.
The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient.
The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function.
A polynomial of degree n will have at most n x-intercepts and at most n−1 turning points.
Glossary
a nonzero real number that is multiplied by a variable raised to an exponent (only the number factor is the coefficient)
continuous function
a function whose graph can be drawn without lifting the pen from the paper because there are no breaks in the graph
the highest power of the variable that occurs in a polynomial
end behavior
the behavior of the graph of a function as the input decreases without bound and increases without bound
leading coefficient
the coefficient of the leading term
leading term
the term containing the highest power of the variable
polynomial function
a function that consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power.
power function
a function that can be represented in the form f(x)=kxp where k is a constant, the base is a variable, and the exponent, p, is a constant
smooth curve
a graph with no sharp corners
term of a polynomial function
any aixi of a polynomial function in the form f(x)=anxn+an−1xn−1...+a2x2+a1x+a0
turning point
the location at which the graph of a function changes direction
5.1: Quadratic Functions
5.3: Graphs of Polynomial Functions |
8164 | https://www.ee.columbia.edu/~dpwe/classes/e4810-2004-05/lectures/L09-firdesign.pdf | 2003-11-18 Dan Ellis 1 ELEN E4810: Digital Signal Processing Topic 9: Filter Design: FIR 1. Windowed Impulse Response 2. Window Shapes 3. Design by Iterative Optimization 2003-11-18 Dan Ellis 2 1. FIR Filter Design FIR filters no poles (just zeros) no precedent in analog filter design Approaches windowing ideal impulse response iterative (computer-aided) design 2003-11-18 Dan Ellis 3 Least Integral-Squared Error Given desired FR Hd(ejω), what is the best finite ht[n] to approximate it?
Can try to minimize Integral Squared Error (ISE) of frequency responses: best in what sense?
€ φ = 1 2π Hd e jω ( ) −Ht e jω ( ) 2 dω −π π ∫ = DTFT{ht[n]} 2003-11-18 Dan Ellis 4 Least Integral-Squared Error Ideal IR is hd[n] = IDTFT{Hd(ejω)}, (usually infinite-extent) By Parseval, ISE But: ht[n] only exists for n = -M..M , € φ = hd n [ ] −ht n [ ] 2 n=−∞ ∞ ∑ € ⇒φ = hd n [ ] −ht n [ ] 2 n=−M M ∑ + hd n [ ] 2 nM ∑ minimized by making ht[n] = hd[n], -M ≤ n ≤ M not altered by ht[n] 2003-11-18 Dan Ellis 5 Least Integral-Squared Error Thus, minimum mean-squared error approximation in 2M+1 point FIR is truncated IDFT: Make causal by delaying by M points → h't[n] = 0 for n < 0 € ht n [ ] = 1 2π Hd e jω ( )e jωndω −π π ∫ −M ≤n ≤M 0 otherwise 2003-11-18 Dan Ellis 6 Approximating Ideal Filters From topic 6, ideal lowpass has: and: ω −π π ωc −ωc € HLP e jω ( ) = 1 ω < ωc 0 ωc < ω < π € hLP n [ ] = sinωcn πn (doubly infinite) 2003-11-18 Dan Ellis 7 Approximating Ideal Filters Thus, minimum ISE causal approximation to an ideal lowpass € ˆ h LP n [ ] = sinωc n −M ( ) π n −M ( ) 0 ≤n ≤2M 0 otherwise 2M+1 points Causal shift 2003-11-18 Dan Ellis 8 Freq. Resp. (FR) Arithmetic Ideal LPF has pure-real FR i.e.
θ(ω) = 0, H(ejω) = |H(ejω)| → Can build piecewise-constant FRs by combining ideal responses, e.g. HPF: ω −π π ωc −ωc ω ω 1 1 1 δ[n] – hLP[n] = hHP[n] i.e. H(ejω) = 1 HLP(ejω) = 1 for |ω| < ωc = δ[n] - (sinωcn)/πn wouldn’t work if phases were nonzero!
2003-11-18 Dan Ellis 9 Gibbs Phenomenon Truncated ideal filters have Gibbs’ Ears: Increasing filter length → narrower ears (reduces ISE) but height the same → not optimal by minimax criterion (11% overshoot) 2003-11-18 Dan Ellis 10 Where Gibbs comes from Truncation of hd[n] to 2M+1 points is multiplication by a rectangular window: Multiplication in time domain is convolution in frequency domain: ht n [ ] = hd n [ ]⋅wR n [ ] wR n [ ] = 1 −M ≤n ≤M 0 otherwise wR[n] g[n]⋅h[n] 1 2π G(e jθ )H(e j(ω−θ))dθ −π π ∫ 2003-11-18 Dan Ellis 11 Where Gibbs comes from Thus, FR of truncated response is convolution of ideal FR and FR of rectangual window (pdc.sinc): Hd(ejω) DTFT{wR[n]} periodic sinc...
Ht(ejω) 2003-11-18 Dan Ellis 12 Where Gibbs comes from Rectangular window: Mainlobe width (∝ 1/L) determines transition band Sidelobe height determines ripples wR n [ ] = 1 −M ≤n ≤M 0 otherwise € WR e jω ( ) = e−jωn n=−M M ∑ = sin 2M +1 [ ] ω 2 ( ) sin ω 2 “periodic sinc” ⇒ doesn’t vary with length 2003-11-18 Dan Ellis 13 2. Window Shapes for Filters Windowing (infinite) ideal response → FIR filter: Rectangular window has best ISE error Other “tapered windows” vary in: mainlobe → transition band width sidelobes → size of ripples near transition Variety of ‘classic’ windows...
€ ht n [ ] = hd n [ ]⋅w n [ ] 2003-11-18 Dan Ellis 14 0.42 + 0.46cos(2π n 2M +1) +0.08cos(2π 2n 2M +1) Window Shapes for FIR Filters Rectangular: Hann: Hamming: Blackman: w n [ ] =1 −M ≤n ≤M 0.5 + 0.5cos(2π n 2M +1) 0.54 + 0.46cos(2π n 2M +1) double width mainlobe reduced 1st sidelobe triple width mainlobe big sidelobes!
2003-11-18 Dan Ellis 15 Window Shapes for FIR Filters Comparison on dB scale: 2π 2M+1 2003-11-18 Dan Ellis 16 Adjustable Windows Have discrete main-sidelobe tradeoffs...
Kaiser window = parametric, continuous tradeoff: Empirically, for min. SB atten. of α dB: € w n [ ] = I0 β 1−( n M )2 ( ) I0(β) −M ≤n ≤M modified zero-order Bessel function β = 0.11(α −8.7) 50 < α 0.58(α −21)0.4 +0.08(α −21) 21 ≤α ≤50 0 α < 21 € N = α −8 2.3Δω required order transition width 2003-11-18 Dan Ellis 17 Windowed Filter Example Design a 25 point FIR low-pass filter with a cutoff of 600 Hz (SR = 8 kHz) No specific transition/ripple req’s → compromise: use Hamming window Convert the frequency to radians/sample: ω 2π 8 kHz π 4 kHz 0.15 π 600 Hz H(ejω) € ωc = 600 8000 × 2π = 0.15π 2003-11-18 Dan Ellis 18 Windowed Filter Example 1.
Get ideal filter impulse response: 2.
Get window: Hamming @ N = 25 → M = 12 (N = 2M+1) 3.
Apply window: € ωc = 0.15π € ⇒hd n [ ] = sin0.15πn πn € ⇒w n [ ] = 0.54 + 0.46cos 2π n 25 ( ) −12 ≤n ≤12 h n [ ] = hd n [ ]⋅w n [ ] = sin0.15πn πn 0.54 + 0.46cos 2πn 25 ( ) −12 ≤n ≤12 M 2003-11-18 Dan Ellis 19 3. Iterative FIR Filter Design Can derive filter coefficients by iterative optimization: Gradient descent / nonlinear optimiz’n Filter coefs h[n] Goodness of fit criterion → error ε Estimate derivatives ∂ε/∂h[n] Update filter to reduce ε desired response H(ejω) 2003-11-18 Dan Ellis 20 Error Criteria € ε = W ω ( )⋅D e jω ( ) −H e jω ( ) [ ] ω∈R ∫ p dω error measurement region error weighting desired response actual response exponent: 2 → least sq’s ∞ → minimax = W(ω)·[D(ejω) – H(ejω)] 2003-11-18 Dan Ellis 21 Minimax FIR Filters Iterative design of FIR filters with: equiripple (minimax criterion) linear-phase → symmetric IR h[n] = (–)h[-n] Recall, symmetric FIR filters have FR with pure-real i.e. combo of cosines of multiples of ω n € H e jω ( ) = e−jωM ˜ H ω ( ) € ˜ H ω ( ) = a k [ ]cos kω ( ) k=0 M ∑ a = h[M] a[k] = 2h[M - k] M (type I) 2003-11-18 Dan Ellis 22 Minimax FIR Filters Now, cos(kω) can be expressed as a polynomial in cos(ω)k and lower powers e.g. cos(2ω) = 2(cosω)2 - 1 Thus, we can find α’s such that Mth order polynomial in cosω α[k]s are simply related to a[k]s € ˜ H ω ( ) = α k [ ] cosω ( ) k k=0 M ∑ Mth order polynomial in cosω 2003-11-18 Dan Ellis 23 Minimax FIR Filters An Mth order polynomial has at most M - 1 maxima and minima: has at most M-1 min/max (ripples) € ˜ H ω ( ) = α k [ ] cosω ( ) k k=0 M ∑ Mth order polynomial in cosω € ˜ H ω ( ) ⇒˜ H ω ( ) 5th order polynomial in cosω 1 2 3 4 2003-11-18 Dan Ellis 24 Key ingredient to Parks-McClellan: is the unique, best, weighted-minimax order 2M approx. to D(ejω) has at least M+2 “extremal” freqs over ω subset R error magnitude is equal at each extremal: peak error alternates in sign: Alternation Theorum € ˜ H ω ( ) € ˜ H ω ( ) € ω0 < ω1 < ...< ωM < ωM +1 € ε ωi ( ) =ε ∀i ε ωi ( ) = −ε ωi+1 ( ) ⇔ 2003-11-18 Dan Ellis 25 Alternation Theorum Hence, for a frequency response: If ε(ω) reaches a peak error magnitude ε at some set of extremal frequences ωi And the sign of the peak error alternates And we have at least M+2 of them Then optimal minimax (10th order filter, M = 5) 2003-11-18 Dan Ellis 26 Alternation Theorum By Alternation Theorum, M+2 extrema of alternating signs ⇒ optimal minimax filter But has at most M-1 extrema ⇒ need at least 3 more from band edges 2 bands give 4 band edges ⇒ can afford to “miss” only one Alternation rules out transition band edges, thus have 1 or 2 outer edges € ˜ H ω ( ) 2003-11-18 Dan Ellis 27 Alternation Theorum For M = 5 (10th order): 8 extrema (M+3, 4 band edges) - great!
7 extrema (M+2, 3 band edges) - OK!
6 extrema (M+1, only 2 transition band edges) → NOT OPTIMAL 2003-11-18 Dan Ellis 28 Parks-McClellan Algorithm To recap: FIR CAD constraints D(ejω), W(ω) → ε(ω) Zero-phase FIR H(ω) = Σkαkcoskω → M-1 min/max Alternation theorum optimal → ≥M+2 pk errs, alter’ng sign Hence, can spot ‘best’ filter when we see it – but how to find it?
~ 2003-11-18 Dan Ellis 29 Parks-McClellan Algorithm Alternation → [H(ω)-D(ω)]/W(ω) must = ±ε at M+2 (unknown) frequencies {ωi}...
Iteratively update h[n] with Remez exchange algorithm: estimate/guess M+2 extremals {ωi} solve for α[n], ε ( → h[n] ) find actual min/max in ε(ω) → new {ωi} repeat until |ε(ωi)| is constant Converges rapidly!
~ ~ 2003-11-18 Dan Ellis 30 Parks-McClellan Algorithm In Matlab, >> h=remez(10, [0 0.4 0.6 1], [1 1 0 0], [1 2]); filter order (2M) band edges ÷ π desired magnitude at band edges error weights per band |
8165 | https://www.coursehero.com/file/201714400/Test-Casesdocx/ | Binary, Decimal, Hexadecimal Conversion Test Cases | Course Hero
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Emily Musgrove 2/10/23 CS-274-A These are the test cases for all possible scenarios and their outputs:
1.)Converting 101100 from binary to decimal:What is the base you are converting from?
2
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What is the base you are converting to?
10
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What is the value of the base you are converting from?
101100
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You are converting 101100 from base 2 to base 10.The result is 44 Correct!
2.)Converting 715 from decimal to binary:What is the base you are converting from?
10
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What is the base you are converting to?
2
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What is the value of the base you are converting from?
715
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You are converting 715 from base 10 to base 2.The result is 1011001011 Correct!
3.)Converting 1FA from hexadecimal to binary:What is the base you are converting from?
16
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What is the base you are converting to?2 What is the value of the base you are converting from?1FA You are converting 1FA from base 16 to base 2.The result is 111111010 Correct!4.)Converting 819 from decimal to hexadecimal:What is the base you are converting from?10 What is the base you are converting to?16 What is the value of the base you are converting from?819 You are converting 819 from base 10 to base 16.The result is 333 Correct!
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In this unit, you have read four stories/poems that were written by a diverse array of authors and poets. These writers use tone to express emotion and, thus, a message. Select one of the stories/poem
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8166 | https://homepages.cwi.nl/~lex/files/link.pdf | A Borsuk theorem for antipodal links and a spectral characterization of linklessly embeddable graphs1 L´ aszl´ o Lov´ asz2 and Alexander Schrijver3 Abstract.
For any undirected graph G, let µ(G) be the graph parameter introduced by Colin de Verdiere. In this paper we show that µ(G) ≤4 if and only if G is linklessly embeddable (in R3). This forms a spectral characterization of linklessly embeddable graphs, and was conjectured by Robertson, Seymour, and Thomas.
A key ingredient is a Borsuk-type theorem on the existence of a pair of antipodal linked (k −1)-spheres in certain mappings φ : S2k →R2k−1. This result might be of interest in its own right.
We also derive that λ(G) ≤4 for each linklessly embeddable graph G = (V, E), where λ(G) is the graph paramer introduced by van der Holst, Laurent, and Schrijver. (It is the largest dimension of any subspace L of RV such that for each nonzero x ∈L, the positive support of x induces a nonempty connected subgraph of G.) 1. Introduction Motivated by estimating the maximum multiplicity of the second eigenvalue of Schr¨ odin-ger operators, Colin de Verdi ere (cf. ) introduced an interesting new invariant µ(G) for graphs G, based on spectral properties of matrices associated with G. He showed that the invariant is monotone under taking minors (that is, if H is a minor of G then µ(H) ≤µ(G)), that µ(G) ≤1 if and only if G is a disjoint union of paths, that µ(G) ≤2 if and only if G is outerplanar, and that µ(G) ≤3 if and only if G is planar.
In this paper we show that µ(G) ≤4 if and only if G is linklessly embeddable in R3.
An embedding of a graph G in R3 is called linkless if each pair of disjoint circuits in G are unlinked closed curves in R3 (for our ourposes, the following definition of linking suffices: two disjoint curves are unlinked, if there is a mapping of the unit disc into R3 such that its boundary is mapped onto the first curve and the image of the disc is disjoint from the second). G is linklessly embeddable if G has a linkless embedding in R3. (‘Embedding’ presumes that vertices and edges have disjoint images.) Our result was conjectured by Robertson, Seymour, and Thomas , and has the fol-lowing context. Robertson, Seymour, and Thomas showed that a graph G is linklessly embeddable if and only if G does not have any of the seven graphs in the Petersen family as a minor — the Petersen family consists of all graphs obtainable from K6 by so-called ∆Y- and Y∆-operations (it includes the Petersen graph). (Y∆means deleting a vertex of degree 3, and making its three neighbours mutually adjacent. ∆Y is the reverse operation (applied to a triangle).) 11991 Mathematics Subject Classification: primary: 05C10, 05C50, 57M15, secondary: 05C50, 57N15 2Department of Computer Science, Yale University, New Haven, Connecticut 06520, U.S.A. email: lovasz@cs.yale.edu 3CWI, Kruislaan 413, 1098 SJ Amsterdam, The Netherlands and Department of Mathematics, University of Amsterdam, Plantage Muidergracht 24, 1018 TV Amsterdam, The Netherlands. Research partially done while visiting the Department of Computer Science at Yale University. email: lex@cwi.nl 1 Since any of the graphs G in the Petersen family has µ(G) ≥5 (Bacher and Colin de Verdiere ), it follows that if µ(G) ≤4 then G is linklessly embeddable. So we prove the reverse implication. Thus, next to the combinatorial characterization (in terms of minors) of the (topologically defined) class of linklessly embeddable graphs, there is a spectral characterization.
Our proof method also applies to a related parameter called λ(G), introduced by van der Holst, Laurent, and Schrijver . It is defined as follows. Let G = (V, E) be a graph. Call a linear subspace L of RV a representation of G if for each x ∈L, supp+(x) is nonempty and G|supp+(x) is a connected graph. Here we use the following notation. If U ⊆V then G|U is the subgraph of G induced by U (and G −U = G|(V \ U)). For any vector x ∈RV , supp(x) is the support of x, that is, supp(x) := {v ∈V |x(v) ̸= 0}. The positive support is supp+(x) := {v ∈V |x(v) > 0} and the negative support is supp−(x) := {v ∈V |x(v) < 0}.
Now by definition, λ(G) is the maximum dimension of any representation L of G. It is easy to see that λ(G) is monotone under taking minors. Moreover, if G is a clique sum of graphs G1 and G2 (that is, if G arises from G1 and G2 by identifying a clique in G1 and G2), then λ(G) = max{λ(G1), λ(G2)}. In it is shown that λ(G) ≤1 if and only if G is a forest, that λ(G) ≤2 if and only if G is series-parallel, and that λ(G) ≤3 if and only if G arises by taking clique sums and subgraphs from planar graphs.
In this paper we show that λ(G) ≤4 if G is linklessly embeddable — hence, more generally, if G arises by taking clique sums and subgraphs from linklessly embeddable graphs. We do not know if the reverse implication holds.
A key ingredient in our proof is a Borsuk-type theorem on the existence of antipodal links, which we formulate and prove in Section 2. We derive it from an extension of a theorem of Bajm´ oczy and B´ ar´ any establishing a polyhedral form of Borsuk’s antipodal theorem.
In Section 3 we derive that λ(G) ≤4 for linklessly embeddable graphs G. In Section 4 we give some preliminaries on the Colin de Verd ere parameter µ(G), and after that, in Section 5, we show that µ(G) ≤4 for linklessly embeddable graphs G. Finally, in Section 6 we consider a number of open questions related to µ(G) and λ(G).
2. A Borsuk-type theorem for antipodal links Let P be a convex polytope in Rn. We say that two faces F and F ′ are antipodal if there exists a nonzero vector c in Rn such that the linear function cTx is maximized by every point of F and minimized by every point of F ′ (Figure 1). So F and F ′ are antipodal if and only if F −F ′ is contained in a face of P −P.
Call a continuous map φ of a cell complex into Rm generic if the images of a k-face and an l-face intersect only if k + l ≥m, and for k + l = m they have a finite number of intersection points, and at these points they intersect transversally. (In this paper, faces are relatively open.) For any convex polytope P in Rn, let ∂P denote its boundary.
The following theorem extends a result of Bajm´ oczy and B´ ar´ any . (The difference is that their theorem concludes that φ(F) ∩φ(F ′) is nonempty. Their proof uses Borsuk’s theorem. We give an independent proof.) 2 Theorem 1. Let P be a full-dimensional convex polytope in Rn and let φ be a generic continuous map from ∂P to Rn−1. Then there exists a pair of antipodal faces F and F ′ with dim(F) + dim(F ′) = n −1 such that |φ(F) ∩φ(F ′)| is odd.
Proof. We prove a more general fact. Call two faces parallel if their projective hulls have a nonempty intersection that is contained in the hyperplane at infinity. So faces F and F ′ are parallel if and only if their affine hulls are disjoint while F −F and F ′ −F ′ have a nonzero vector in common. (Note that two antipodal faces are parallel if dim(F) + dim(F ′) ≥n.) Figure 1: Parallel and non-parallel antipodal faces Now it suffices to show Let P be a full-dimensional convex polytope in Rn having no parallel faces and let φ be a generic continuous map from ∂P to Rn−1. Then X |φ(F) ∩φ(F ′)| is odd, where the summation extends over all antipodal pairs {F, F ′} of faces.
(1) (It would be enough to sum over all antipodal pairs {F, F ′} with dim(F)+dim(F ′) = n−1.) To see that it suffices to prove (1), it is enough to apply a random projective transfor-mation close to the identity. To be more precise, assume that we have a polytope P that has parallel faces. For every pair (E, E′) of faces whose affine hulls intersect, choose a (fi-nite) point pEE′ in the intersection of the affine hulls. For every pair (E, E′) of faces whose projective hulls intersect, choose an infinite point qEE′ in the intersection of their projective hulls. Let H be a finite hyperplane having all the points pEE′ on one side, and avoiding all the points qFF ′. Apply a projective transformation that maps H onto the hyperplane at infinity, to get a new polytope P ′. It is clear that P ′ has no parallel faces, and it is easy to argue that every pair of faces that are antipodal in P ′ correspond to antipodal faces in P.
Hence (1) implies the theorem.
We now prove (1). Let P be a convex polytope in Rn having no parallel faces. For any 3 two faces F, F ′, denote F ≤F ′ if F ⊆F ′. Then: (i) if A and B are antipodal faces, then A −B is a face of P −P, with dim(A −B) = dim(A) + dim(B); (ii) if F is a face of P −P, then there exists a unique pair A, B of antipodal faces with A −B = F; (iii) for any two pairs A, B and A′, B′ of antipodal faces one has: A−B ≤A′−B′ if and only if A ≤A′ and B ≤B′.
(2) This gives the following observation: For every pair of faces A and B with dim(A) + dim(B) = n −2, the number of antipodal pairs {F, F ′} of faces with A ≤F and B ≤F ′ and dim(F)+dim(F ′) = n −1 is 0 or 2.
(3) To see (3), it is clear that if A and B are not antipodal, then this number is 0. Suppose that they are antipodal. Then the number is 2, since by (2), it is equal to the number of facets of P −P incident with the (n −2)-face A −B.
To prove (1), we use a “deformation” argument. The statement is true for the following mapping φ: pick a point q very near the center of gravity of some facet F (outside P), and project ∂P from q onto the hyperplane H of F. Then the only nontrivial intersection is that the image of the (unique) vertex of P farthest from H is contained in F.
Now we deform this map to φ. We may assume that the images of two faces E and E′ with dim(E)+dim(E′) ≤n−3 never meet; but we have to watch when φ(A) passes through φ(B), where A and B are faces with dim(A) + dim(B) = n −2. But then |φ(F) ∩φ(F ′)| changes exactly when A ⊆F and B ⊆F ′. By (3), this does not change the parity. This proves (1), and hence the theorem.
For any polytope P, let (P)k denote its k-skeleton. Two disjoint images A and B of (d −1)-spheres in R2d−1 are said to have an odd linking number if the image of A can be extended to the image of a d-ball with an odd number of transveral intersections with the image of B (and no other intersections). So having an odd linking number implies being linked.
Corollary 1.1.
Let P be a full-dimensional convex polytope in R2k+1 and let φ be an embedding of (P)k−1 into R2k−1. Then there exists a pair of antipodal k-faces F and F ′ such that φ(∂F) and φ(∂F ′) have an odd linking number.
Proof. First we extend φ with a last coordinate equal to 0, to obtain an embedding ψ of (P)k−1 into R2k. Next we extend ψ to a generic mapping ∂P →R2k, in such a way that ψ(x) has last coordinate positive if x ∈∂P \ (P)k−1.
By Theorem 1, P has two antipodal faces F and F ′ such that dim(F) + dim(F ′) = 2k and |ψ(F) ∩ψ(F ′)| is odd. If dim(F) ≤k −1, then the last coordinate of each point in ψ(F) is 0, while the last coordinate of each point in ψ(F ′) is positive (as dim(F ′) ≥k). So dim(F) ≥k and similarly dim(F ′) ≥k. Therefore, dim(F) = dim(F ′) = k.
Then the boundaries of F and F ′ are (k −1)-spheres S1 and S2, mapped disjointly into R2k−1, and the mappings extend to mappings of the k-balls into the “upper” halfspace of 4 R2k, so that the images of the balls intersect at an odd number of points. But this implies that the images of the spheres are linked. In fact, if they were not linked, then there exists an extension of the map of ∂F to a continuous mapping ψ′ of F into R2k such that the image of every point in the interior of F has last coordinate equal to 0, and ψ′(F) intersects ψ(∂F ′) transversally in an even number of points. We can extend the map of ∂F ′ to a continuous mapping ψ′ of F ′ into R2k such that the image of every point in the interior of F has a negative last coordinate 0. Then we get two maps of the k-sphere into R2k with an odd number of transversal intersection points, which is impossible. This contradiction completes the proof.
3. λ(G) ≤4 for linklessly embeddable graphs G In this section we focus on the graph parameter λ(G) introduced in , and show that λ(G) ≤4 for linklessly embeddable graphs. The proof method also serves as an introduction to the methods used in proving that µ(G) ≤4 for linklessly embeddable graphs. We gave the definition of λ(G) in Section 1.
Theorem 2. If G is linklessly embeddable, then λ(G) ≤4.
Proof. Let G be linklessly embedded in R3, and suppose that λ(G) ≥5. Then there is a 5-dimensional subspace L of RV such that G|supp+(x) is nonempty and connected for each nonzero x ∈L.
Call two elements x and x′ of L equivalent if supp+(x) = supp+(x′) and supp−(x) = supp−(x′). The equivalence classes decompose L into a centrally symmetric complex P of pointed polyhedral cones. Choose a sufficiently dense set of vectors of unit length from every cone in P, in a centrally symmetric fashion, and let P be the convex hull of these vectors. Then P is a 5-dimensional centrally symmetric convex polytope such that every face of P is contained in a cone of P.
We define an embedding φ of (P)1 in R3. For each vertex v of P, we choose a node v′ in supp+(v), end we let φ(x) be a point in R3 very near v′. For each edge e = uv of P, we choose a path e′ connecting u′ and v′ in G|supp+(x), where x is an interior point of e. (By our construction, supp+(x) is independent of the choice of x, and contains both supp+(u) P + -u v e x 0 + -+ Figure 2: Constructing an embedding of (P)1 5 and supp+(v).) Then we map e onto a Jordan curve connecting φ(u) and φ(v) very near e′.
Clearly we can choose the images of the vertices and edges so that this map φ is one-to-one (Fig. 2).
Then by Corollary 1.1, P has two antipodal 2-faces F and F ′ such that the images of their boundaries are linked. Since P is centrally symmetric, there is a facet D of P such that F ⊆D and F ′ ⊆−D. Let y be a vector in the interior of D. Then the images of ∂F and ∂F ′ are very near subgraphs spanned by supp+(y) and supp−(y), respectively, and hence some cycle of G spanned by supp+(y) must be linked with some cycle in supp−(y), a contradiction.
Corollary 2.1. If G is obtained from linklessly embedded graphs by taking clique sums and subgraphs, then λ(G) ≤4.
Proof. Directly from Theorem 2 and the fact that the class of graphs G with λ(G) ≤4 is closed under taking clique sums and subgraphs ().
4. The Colin de Verdiere parameter µ(G) We now go over to the Colin de Verdi ere parameter µ(G), to which we first give some background.
Let G = (V, E) be an undirected graph, which we assume without loss of generality to have vertex set {1, . . . , n}. Then µ(G) is the largest corank of any symmetric real-valued n × n matrix M = (mi,j) satisfying: (i) M has exactly one negative eigenvalue, of multiplicity 1, (ii) for all i, j with i ̸= j: mi,j < 0 if i and j are adjacent, and mi,j = 0 if i and j are nonadjacent, (iii) there is no nonzero symmetric n × n matrix X = (xi,j) such that MX = 0 and such that xi,j = 0 whenever i = j or mi,j ̸= 0.
(4) There is no condition on the diagonal entries mi,i. (The corank corank(M) of a matrix M is the dimension of its kernel.) Condition (4)(iii) is called the Strong Arnold Property (or Strong Arnold Hypothesis).
There exist matrices M satisfying (4), for which ker(M) is a not a representation of G; that is, for which there exist x ∈ker(M) with G|supp+(x) disconnected. (Otherwise µ(G) ≤λ(G) would follow, which is not true.) The Petersen graph provides an example.
Let A be the adjacency matrix of the Petersen graph and let M = I −A. Let e and e′ be two edges at distance 2, and define a vector qee′ ∈RV as 1 on the endnodes of e, −1 on the endnodes of e′, and 0 elsewhere. Then qee′ ∈ker(M), and it is easy to see that in fact ker(M) is generated by these vectors. Now if e, e′ and e′′ are three edges that are mutually at distance 2, then qee′ + qee′′ is a vector in ker(M) with supp−(q) having two components (Fig. 3). We shall see that this is as bad as it ever gets.
The following lemma extends a lemma in ; the methods we use to prove it are close to those used in .
For any graph G = (V, E) and U ⊆V , let N(U) be the set of vertices in V \ U that are adjacent to at least one vertex in U. For any V × V matrix and I, J ⊆V , let MI×J denote 6 2 2 -1 -1 -1 -1 -1 -1 1 1 Figure 3: A vector on the Petersen graph with disconnected positive support the submatrix induced by the rows in I and columns in J, and let MI := MI×I. For any vector z ∈RI and J ⊆I, let zJ be the subvector of z induced by the indices in J.
Lemma 1. Let G be a connected graph, let M be a matrix satisfying (4), and let x be a vector in ker(M) with G|supp+(x) disconnected. Then there are no edges connecting supp+(x) and supp−(x), and each component K of G|supp(x) satisfies N(K) = N(supp(x)).
Proof. Let z be a positive eigenvector of M (by the Perron-Frobenius theorem, z is unique up to scaling, and belongs to the smallest eigenvalue of M).
Let I and J be two components of G|supp+(x). Let L := supp−(x). As Mx = 0, we have MI×IxI + MI×LxL = 0, MJ×JxJ + MJ×LxL = 0.
(5) Let λ := zT I xI/zT J xJ. Define y ∈RV by: yi := xi if i ∈I, yi := −λxi if i ∈J, and yi := 0 if i ̸∈I ∪J. Then zTy = zT I xI −λzT J xJ = 0. Moreover, one has (since MI×J = 0): yT My = yT I MIyI + yT J MJyJ = xT I MIxI + λ2xT J MJxJ = −xT I MI×LxL −λ2xT J MJ×LxL ≤0 (6) (using (5)), since MI×L and MJ×L are nonpositive, and since xI > 0, xJ > 0 and xL < 0.
Now zTy = 0 and yT My ≤0 imply that My = 0 (as M is symmetric and has exactly one negative eigenvalue, with eigenvector z). Therefore, y ∈ker(M). By (4)(i), for distinct u, v, entry Mu,v of M is negative if and only if u and v are adjacent. This implies that each vertex in V \ supp(y) adjacent to supp+(y) = I is also adjacent to supp−(y) = J, and conversely; that is, N(I) = N(J).
Also x −y belongs to ker(M). Hence, again, any vertex v ∈V \ supp(x −y) is adjacent to supp−(x−y) if and only if v is adjacent to supp+(x−y). As supp+(x−y) = supp+(x)\I and as I is a component of G|supp+(x), no vertex in I is adjacent to supp+(x −y). Hence no vertex in I is adjacent to supp−(x −y) = supp−(x). As I is an arbitrary component of G|supp+(x), it follows that there is no edge connecting supp+(x) and supp−(x). So each component of G|supp(x) is a component of either G|supp+(x) or G|supp−(x).
7 Since N(I) = N(J) for any two components I, J of G|supp+(x), and similarly, N(I) = N(J) for any two components I, J of G|supp−(x), and since N(supp+(x)) = N(supp−(x)), we have that N(I) = N(supp(x)) for any component I of G|supp(x).
5. µ(G) ≤4 for linklessly embeddable graphs Theorem 3. A graph G is linklessly embeddable if and only if µ(G) ≤4.
Proof. By the results of and it suffices to show that µ(G) ≤4 if G is linklessly embeddable.
Let G be linklessly embeddable and suppose that µ(G) ≥5. By the results of we may assume that G is ‘flatly embedded’ in R3; that is, for each circuit C in G there exists an open disk (“panel”) D in R3 with the property that D is disjoint from G and has boundary C. (Any flat embedding is also linkless, but the reverse is generally not true. But in it has been shown that if G has a linkless embedding it also has a flat embedding.) We take a counterexample G with a minimum number of vertices.
Then G is 4-connected. For suppose that G has a minimum-size vertex cut U with |U| ≤3. Consider any component K of G −U. Then the graph G′ obtained from G −K by adding a clique on U is a linkless embeddable graph again, because, if |U| ≤2, G′ is a minor of G, and if |U| = 3, G′ can be obtained from a minor of G by a Y∆-operation. As G′ has fewer vertices than G, we have µ(G′) ≤4. As this is true for each component K, G is a a subgraph of a clique sum of graphs G′ with µ(G′) ≤4, with cliques of size at most 3, and hence by the results of , µ(G) ≤4.
Let M be a matrix satisfying (4) with corank(M) = 5. We proceed as in the proof of Theorem 2. Call two elements x and x′ of ker(M) equivalent if supp+(x) = supp+(x′) and supp−(x) = supp−(x′). The equivalence classes decompose ker(M) into a centrally symmetric complex P of pointed polyhedral cones. Call a cone f of P broken if G|supp+(x) is disconnected for any x ∈f.
To study broken cones, we first remark: for each x ∈ker(M) with G|supp+(x) disconnected, G|supp(x) has exactly three components, say K1, K2, and K3, with K1∪K2 = supp+(x) and K3 = supp−(x), and with N(Ki) = V \ supp(x) for i = 1, 2, 3.
(7) This follows directly from Lemma 1, using the 4-connectivity of G and the fact that G has no K4,4-minor (as K4,4 is not linklessly embeddable (cf. )).
Now (7) gives: any broken cone f is 2-dimensional.
(8) Indeed, choose x ∈f, and let K1, K2, and K3 be as in (7). Consider any y ∈f. As supp(y) = supp(x) we have that MKiyKi = 0 for i = 1, 2, 3. As MKixKi = 0 and as xKi is fully positive or fully negative, we know by the Perron-Frobenius theorem that yKi = λixKi for some λi > 0 (i = 1, 2, 3). Moreover, for the positive eigenvector z of M we have that zT y = zT x = 0. Conversely, any vector y ∈RV with zTy = 0 and supp(y) = supp(x) and for which there exist λ1, λ2, λ3 > 0 with yKi = λixKi for i = 1, 2, 3, belongs to f, since 8 it belongs to ker(M). This follows from the fact that zT y = 0 and yT My = 0. So f is 2-dimensional, proving (8).
Now choose a sufficiently dense set of vectors of unit length from every cone in P, in a centrally symmetric fashion, and let P be the convex hull of these vectors. Then P is a 5-dimensional centrally symmetric convex polytope such that every face of P is contained in a cone of P. We choose the vectors densely enough such that every face of P contains at most one edge that is part of a 2-dimensional cone in P. We call an edge of P broken if it is contained in a broken cone in P.
We define an embedding φ of the 1-skeleton (P)1 of P in R3. We map each vertex x of P to a point φ(x) near supp+(x), and we map any unbroken edge e = xy of P to a curve connecting φ(x) and φ(y) near G|supp+(z), where z ∈e. We do this in such a way that the mapping is one-to-one.
Consider next a broken edge e of P. Choose x ∈e, let K1, K2,and K3 be as in (7), and let T := N(supp(x)). Then there is a curve C in R3 \ G connecting K1 and K2 such that there is no pair of disjoint linked circuits A in G|(K1 ∪K2 ∪T) ∪C and B in G|(K3 ∪T).
(9) To see this, let H be the flatly embedded graph obtained from G by contracting Ki to one vertex vi (i = 1, 2, 3). It suffices to show that there is a curve C connecting v1 and v2 such that the graph H ∪C is linklessly embedded. (Indeed, having C with H ∪C linklessly embedded, we can decontract each Ki slightly, and make C connecting two arbitrary points in K1 and K2. Consider a circuit A in G|(K1 ∪K2 ∪T) ∪C and a circuit B in G|(K3 ∪T) disjoint from A. Contracting K1, K2, and K3, we obtain disjoint cycles A′ and B′ in H ∪C.
(A cycle is an edge-disjoint union of circuits.) As H ∪C is linklessly embedded, A′ and B′ are unlinked. Hence A and B are unlinked.) Now H|T is a Hamiltonian circuit on T, or part of it. Otherwise, H|T would contain, as a minor, a graph on four vertices that is either a K1,3 or a triangle with an isolated vertex.
In both cases, it implies that H has a minor in the Petersen family, which is not possible since H is linklessly embedded.
So H is isomorphic to the complete bipartite graph K3,|T|, with some edges on T added forming part (or whole) of a Hamiltonian circuit on T. As H is flatly embedded, for each edge t1t2 of H|T there is an open disk (“panel”) with boundary the triangle t1t2v3, in such a way that the panels are pairwise disjoint (by B¨ ohme’s lemma (cf. , )). Since the union of H|(K3 ∪T) with the panels is contractible, there is a curve C from v1 to v2 not intersecting any panel. This curve has the required properties, showing (9).
We now define φ on e close to a curve in G|(K1 ∪K2)∪C, again so that it is one-to-one.
We do this for each broken edge e, after which the construction of φ is finished.
Then by Corollary 1.1, there are two antipodal 2-faces F and F ′ such that the images of their boundaries are linked. Since P is centrally symmetric, there is a facet D of P such that F ⊆D and F ′ ⊆−D. Let y be a vector in the interior of D. Then ∂F and ∂F ′ have image in supp+(y) and supp−(y) respectively. If ∂F and ∂F ′ do not contain any broken edge, then it would follow that G has two disjoint linked circuits — a contradiction.
So we can assume that ∂F contains a broken edge e. Then it is the only broken edge in ∂F, since by our construction, ∂D contains at most one edge of P that is part of a 2-dimensional cone f in P. So f is broken. Moreover, ∂F ′ does not contain any broken 9 edge. For suppose that ∂F ′ contains broken edge e′ of P. Then e′ is part of a broken 2-dimensional cone f ′ in P, and hence f ′ = −f (since D is incident with at most one edge that is part of a 2-dimensional cone of P). However, as f is broken, −f is not broken, since G|supp−(x) is connected for any x ∈f (by (7)).
Choose x ∈f. and consider the partition of V into K1, K2, K3, and T as above, with supp+(x) = K1 ∪K2 and supp−(x) = K3. Then K1 ∪K2 ⊆supp+(y) and K3 ⊆supp−(y), and hence supp+(y) ⊆K1 ∪K2 ∪T and supp−(y) ⊆K3 ∪T. So the image of ∂F is close to G|(K1 ∪K2 ∪T) ∪C, where C is the curve constructed for the broken edge e of P, and the image of ∂F ′ is close to G|(K3 ∪T). This contradicts (9).
Note that in this proof, the Strong Arnold Property is hardly used. Also the Lemma remains true without the Strong Arnold Property.
In fact, the above shows that for a 4-connected linklessly embeddable graph G, each matrix M satisfying (4)(i) and (ii) has corank(M) ≤4.
6. Some open questions A first question that comes up is if one can prove that a graph G is linklessly embeddable if µ(G) ≤4 directly, that is, without using the Robertson-Seymour-Thomas theorem. In other words, does the nonexistence of a matrix M of corank 5 satisfying (4) imply, in a direct way, the existence of a linkless embedding?
Secondly, the above does not complete the characterization of graphs G satisfying λ(G) ≤4. Each graph G obtainable from linklessly embeddable graphs by taking clique sums and subgraphs, satisfies λ(G) ≤4, but it is an open question if the reverse also holds.
To answer this question, one might investigate the class of minor-minimal graphs that can-not be obtained from linklessly embeddable graphs by taking clique sums and subgraphs.
It is not known what the full list of these graphs is.
In the following graphs are shown to be minor-minimal with respect to the property λ(G) ≥5. First G = K6 (all other graphs G in the Petersen family satisfy λ(G) ≤4).
Next consider the graph V8 with vertices v1, . . . , v8, with vi and vj adjacent if and only if |i−j| ∈{1, 4, 7}. (It was shown by Wagner that a graph G can be obtained from planar graphs by taking clique sums and subgraphs if and only if G has no K5- or V8-minor. So K5 and V8 are the only minor-minimal graphs G with λ(G) ≥4.) Let V ′ 9 arise from V8 by adding an extra vertex v0, adjacent to v2, v4, v6, v7, v8. Similarly, let V ′′ 9 arise from V8 by adding an extra vertex v0 adjacent to v2, v3, v5, v7, v8. It is shown in that V ′ 9 and V ′′ 9 are minor-minimal graphs G with λ(G) ≥5.
The graphs V ′ 9 and V ′′ 9 are also minor-minimal graphs not obtainable from linklessly embeddable graphs by taking clique sums and subgraphs.
This can be seen as follows.
Since λ(V ′ 9) = λ(V ′′ 9 ) = 5, it follows from Corollary 2.1 that these two graphs indeed are not obtainable in such a way. Moreover, to see that they are minor-minimal, observe that deleting or contracting any edge of V ′ 9 or V ′′ 9 produces a graph that has a vertex whose deletion makes the graph a clique sum of planar graphs.
Since the class of graphs G with λ(G) ≤4 is closed under taking ∆Y operations (not under Y∆), we can obtain other graphs with λ(G) ≥5 by applying a Y∆operation to V ′ 9 or 10 V ′′ 9 . Any of them contains a K6-minor, except if we apply Y∆to vertex v1 (or equivalently, to v5) of V ′ 9.
So it could be true that λ(G) ≤4 if and only if G is obtainable from linklessly embeddable graphs by taking clique sums and subgraphs.
Another open question is if λ(G) ≤µ(G) for each graph G. That is, if for any represen-tation L of G there is a matrix M satisfying (4) with dim(L) ≤corank(M). This is true if µ(G) ≤4.
In fact, a tempting, more general speculation is that for any natural number t: (???) a graph G satisfies λ(G) ≤t if and only if G is obtainable from graphs H satisfying µ(H) ≤t by taking clique sums and subgraphs (???) (10) This has been proved for t ≤3, and the ‘if’ part for t ≤4.
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11 |
8167 | https://projecteuclid.org/journals/sut-journal-of-mathematics/volume-57/issue-2/Tate-Hochschild-cohomology-rings-for-eventually-periodic-Gorenstein-algebras/10.55937/sut/1641859464.pdf | SUT Journal of Mathematics Vol. 57, No. 2 (2021), 133–146 Tate-Hochschild cohomology rings for eventually periodic Gorenstein algebras Satoshi Usui (Received March 1, 2021) Abstract. Tate-Hochschild cohomology of an algebra is a generalization of or-dinary Hochschild cohomology, which is defined on positive and negative degrees and has a ring structure. Our purpose of this paper is to study the eventual periodicity of an algebra by using the Tate-Hochschild cohomology ring. First, we deal with eventually periodic algebras and show that they are not necessar-ily Gorenstein algebras. Secondly, we characterize the eventual periodicity of a Gorenstein algebra as the existence of an invertible homogeneous element of the Tate-Hochschild cohomology ring of the algebra, which is our main result.
Finally, we use tensor algebras to establish a way of constructing eventually periodic Gorenstein algebras.
AMS 2010 Mathematics Subject Classification. 16E40, 16E05, 16G50.
Key words and phrases. Eventually periodic algebras, Gorenstein algebras, sin-gularity categories, Tate-Hochschild cohomology, complete resolutions.
§1.
Introduction The Tate-Hochschild cohomology of an algebra was introduced by Wang based on the notion of Tate cohomology defined by Buchweitz .
It was proved in that the Tate-Hochschild cohomology carries a structure of a graded commutative algebra. There are studies on the ring structure of the Tate-Hochschild cohomology, such as [10, 17, 18, 19].
Recently, Dotsenko, G é linas and Tamaroffproved in [10, Corollary 6.4] that, for a monomial Gorenstein algebra Λ, the Tate-Hochschild cohomology ring d HH •(Λ) is iso-morphic to d HH ≥0(Λ)[χ−1], where d HH ≥0(Λ) stands for the subring consisting of the non-negative part of d HH •(Λ) and χ is an invertible homogeneous ele-ment of positive degree. Moreover, the author also showed in [18, Corollary 3.4] that the same isomorphism holds for a periodic algebra. In both cases, the 133 134 S. USUI invertible element χ was obtained from the fact that any minimal projective resolution of the given algebra eventually becomes periodic.
In this paper, we first deal with eventually periodic algebras (i.e., algebras Λ with the n-th syzygy Ωn Λe(Λ) periodic for some n ≥0). It will be revealed that eventually periodic algebras are not necessarily Gorenstein (see Example 3.2), although it is known that periodic algebras are all Gorenstein. Secondly, we will study the relationship between the eventual periodicity of a Gorenstein algebra and the Tate-Hochschild cohomology ring of the algebra. The following is the main result of this paper: Main Result (see Theorem 3.5). Let Λ be a Gorenstein algebra. Then the following are equivalent.
(1) Λ is an eventually periodic algebra.
(2) The Tate-Hochschild cohomology ring d HH •(Λ) has an invertible homo-geneous element of positive degree.
In this case, there exists an isomorphism d HH •(Λ) ∼ = d HH ≥0(Λ)[χ−1] of graded algebras, where the degree of an invertible homogeneous element χ equals the period of the periodic syzygy Ωn Λe(Λ) of Λ for some n ≥0.
Our main result requires only the eventual periodicity of a minimal pro-jective resolution of a given Gorenstein algebra. Hence it turns out that [10, Corollary 6.4] and [18, Corollary 3.4] can be obtained from our main result, because monomial Gorenstein algebras and periodic algebras are both even-tually periodic Gorenstein algebras. Finally, using tensor algebras, we will provide one of the constructions of eventually periodic Gorenstein algebras.
This paper is organized as follows. In Section 2, we recall basic facts on Tate cohomology and Gorenstein algebras. In Section 3, we give examples of eventually periodic algebras and prove our main result. In Section 4, we establish a way to construct eventually periodic Gorenstein algebras.
§2.
Preliminaries Throughout this paper, let k be an algebraically closed field. We write ⊗k as ⊗. By an algebra Λ, we mean a finite dimensional associative unital k-algebra.
All modules are assumed to be finitely generated left modules. For an algebra Λ, we denote by Λ-mod the category of Λ-modules, by Λ-proj the category of projective Λ-modules, by gl.dim Λ the global dimension of Λ and by Λe the enveloping algebra Λ ⊗Λop. Remark that we can identify Λe-modules with Λ-bimodules. For a Λ-module M, we denote by inj.dimΛM (resp. proj.dimΛM) TATE-HOCHSCHILD COHOMOLOGY RINGS 135 the injective (resp. projective) dimension of M. By a complex X•, we mean a chain complex X• = · · · →Xi+1 dX i+1 − − − →Xi →· · · .
For a complex X• and an integer i, we denote by Ωi(X•) the cokernel Cok dX i+1 of the differential dX i+1 and by X•[i] the complex given by (X•[i])j = Xj−i and dX[i] = (−1)idX.
2.1.
Tate cohomology rings In this subsection, we recall some facts on Tate cohomology rings and Tate-Hochschild cohomology rings. Let Λ be an algebra. Recall that the singularity category Dsg(Λ) of Λ is defined to be the Verdier quotient of the bounded derived category Db(Λ-mod) of Λ-mod by the bounded homotopy category Kb(Λ-proj) of Λ-proj. Let M and N be Λ-modules and i an integer. Following , we define the i-th Tate cohomology group of M with coefficients in N by d Ext i Λ(M, N) := HomDsg(Λ)(M, N[i]), where M and N are viewed as complexes concentrated in degree 0. We call d Ext i Λe(Λ, Λ) the i-th Tate-Hochschild cohomology group of Λ and denote it by d HH i(Λ).
Let T be a triangulated category with shift functor . For an object X of T , one can endow End• T (X) := ⊕ i∈Z HomT (X, X[i]) with a structure of a graded ring. The multiplication is given by the Yoneda product ⌣: HomT (X, X[i]) ⊗HomT (X, X[j]) →HomT (X, X[i + j]) sending α⊗β to α[j]◦β. If T = Dsg(Λ) and X = M ∈Λ-mod, then we obtain a graded algebra d Ext • Λ(M, M) := End• Dsg(Λ)(M) and call it the Tate cohomology ring of M, which is called the stabilized Yoneda Ext algebra of M by Buchweitz . It was proved by Wang that the Tate-Hochschild cohomology ring d HH •(Λ) := d Ext • Λe(Λ, Λ) of any algebra Λ is a graded commutative algebra.
2.2.
Singularity categories of Gorenstein algebras The aim of this subsection is to recall facts on the singularity category of a Gorenstein algebra from . Let Λ be an algebra. Recall that the stable category Λ-mod of Λ-modules is the category whose objects are the same as Λ-mod and morphisms are given by HomΛ(M, N) := HomΛ(M, N)/P(M, N), 136 S. USUI where P(M, N) is the space of morphisms factoring through a projective mod-ule. We denote by [f] the element of HomΛ(M, N) represented by a morphism f : M →N. There exists a canonical functor F : Λ-mod →Dsg(Λ) making the following square commute: Λ-mod / Db(Λ-mod) Λ-mod F / Dsg(Λ) where the two vertical functors are the canonical ones, and the upper horizon-tal functor is the one sending a module M to the complex M concentrated in degree 0. Further, the functor F satisfies F ◦ΩΛ ∼ = [−1] ◦F, where ΩΛ is the syzygy functor on Λ-mod (i.e., the functor sending a module M to the kernel of a projective cover of M). On the other hand, let APC(Λ) be the homo-topy category of acyclic complexes of projective Λ-modules. Then taking the cokernel Ω0(X•) = Cok dX 1 of the differential dX 1 for a complex X• defines a functor Ω0 : APC(Λ) →Λ-mod satisfying Ω0 ◦[−1] ∼ = ΩΛ ◦Ω0.
Recall that an algebra Λ is Gorenstein if inj.dimΛΛ < ∞and inj.dimΛopΛ < ∞. Since the two dimensions coincide (see [21, Lemma A]), we call a Goren-stein algebra Λ with inj.dimΛΛ = d a d-Gorenstein algebra. In the rest of this subsection, let Λ denote a Gorenstein algebra. We call a Λ-module M Cohen-Macaulay if Exti Λ(M, Λ) = 0 for all i > 0. It is clear that projective Λ-modules are Cohen-Macaulay. We denote by CM(Λ) the category of Cohen-Macaulay Λ-modules. It is well known that CM(Λ) is a Frobenius exact category whose projective objects are precisely projective Λ-modules, so that the stable cat-egory CM(Λ), the full subcategory of Λ-mod consisting of Cohen-Macaulay Λ-modules, carries a structure of a triangulated category (see [7, 13]). In par-ticular, the syzygy functor ΩΛ on Λ-mod gives rise to the inverse of the shift functor Σ on CM(Λ). We end this subsection with the following result due to Buchweitz.
Theorem 2.1 ([7, Theorem 4.4.1]). Let Λ be a Gorenstein algebra. Then there exist equivalences of triangulated categories APC(Λ) Ω0 / CM(Λ) ιΛ / Dsg(Λ), where the equivalence ιΛ is given by the restriction of F : Λ-mod →Dsg(Λ) to CM(Λ).
2.3.
Tate cohomology over Gorenstein algebras This subsection is devoted to recalling another description of Tate cohomology over a Gorenstein algebra. Throughout, let Λ denote a d-Gorenstein algebra TATE-HOCHSCHILD COHOMOLOGY RINGS 137 unless otherwise stated. Thanks to Theorem 2.1, we can associate to any Λ-module M an object T• = T M • in APC(Λ), uniquely determined up to isomor-phism, satisfying that Ω0(T•) ∼ = M in Dsg(Λ). Thus the triangle equivalence ιΛ : CM(Λ) →Dsg(Λ) induces an isomorphism d Ext i Λ(M, M) ∼ = HomΛ(Ω0(T•), ΣiΩ0(T•)) for all i ∈Z. We identify d Ext • Λ(M, M) with End• CM(Λ)(Ω0(T•)) via this iso-morphism.
Recall that, for an algebra Λ, the Gorenstein dimension G-dimΛM of a Λ-module M is defined by the shortest length of a resolution of M by Λ-modules X with X ∼ = X∗∗and Exti Λ(X, Λ) = 0 = Exti Λop(X∗, Λ) for all i > 0, where we set (−)∗:= HomΛ(−, Λ) (see for its original definition). The next proposition is easily obtained from the results in applied to the case of Gorenstein algebras: (1), (2) and (3) follow from [3, Theorems 3.1 and 3.2], [3, Lemma 2.4 and Theorem 3.1] and [3, Theorem 5.2], respectively.
Proposition 2.2. The following hold for a module M over a d-Gorenstein algebra Λ.
(1) The Gorenstein dimension G-dimΛM of M satisfies G-dimΛM ≤d and is equal to the smallest integer r ≥0 for which Ωr Λ(M) is Cohen-Macaulay.
(2) There exists a diagram T• θ − →P• ε − →M satisfying the following conditions: (i) T• ∈APC(Λ) and P• ε − →M is a projective resolution of M.
(ii) θ : T• →P• is a chain map with θi an isomorphism for any i ≫0.
(3) We have that Exti Λ(M, M) ∼ = d Ext i Λ(M, M) for all i > G-dimΛM.
We call such a diagram as in Proposition 2.2 (2) a complete resolution of M (see for its definition in a general setting). A complete resolution is unique in the sense of [3, Lemma 5.3] (when it exists).
Finally, we explain how we find the corresponding object T M • in APC(Λ) for any Λ-module M. Let T• →P• →M be a complete resolution of M. Then the complex T• in APC(Λ) is the object corresponding to M via the triangle equivalence ι ◦Ω0 : APC(Λ) →Dsg(Λ). Indeed, the morphism Ω0(T•) →M induced by the chain map θ≥0 : T≥0 →P• is an isomorphism in Dsg(Λ). Here, T≥0 stands for the following truncated complex of T•: T≥0 = · · · →T2 dT 2 − →T1 dT 1 − →T0 →0 →0 →· · · .
Thus we conclude that constructing a complete resolution of M is equivalent to finding the corresponding object T M • of APC(Λ).
138 S. USUI §3.
Tate-Hochschild cohomology for eventually periodic Gorenstein algebras In this section, we first define eventually periodic algebras and provide exam-ples of them. We then prove our main result.
3.1.
Eventually periodic algebras As mentioned above, let us first define the eventual periodicity of algebras and provide examples of eventually periodic algebras.
Definition 3.1. Let Λ be an algebra. A Λ-module M is called periodic if Ωp Λ(M) ∼ = M in Λ-mod for some p > 0. The smallest such p is said to be the period of M. We say that M ∈Λ-mod is eventually periodic if Ωn Λ(M) is periodic for some n ≥0. An algebra Λ is called periodic (resp. eventually periodic) if Λ ∈Λe-mod is periodic (resp. eventually periodic).
From the definition, periodic algebras are eventually periodic algebras. Pe-riodic algebras have been studied for a long time (see ). We know from [12, Lemma 1.5] that periodic algebras are self-injective algebras (i.e., 0-Gorenstein algebras). On the other hand, it follows from the proof of [10, Corollary 6.4] that monomial Gorenstein algebras are eventually periodic algebras. It also follows from the formula gl.dim Λ = proj.dimΛeΛ (see [14, Section 1.5]) that algebras of finite global dimension are eventually periodic algebras. As will be seen in Example 3.2 below, not all eventually periodic algebras are Gorenstein algebras.
Example 3.2.
(1) Let Λ1 be the algebra given by a quiver with relation 1 2 α β αβα = 0.
Then Λ1 is a monomial algebra that is not Gorenstein (since inj.dimΛΛe1 = ∞, where e1 is the primitive idempotent corresponding to the vertex 1). Using Bardzell’s minimal projective resolution of a monomial alge-bra (see ), we have that Λ1 is an eventually periodic algebra having Ω2 Λe 1(Λ1) as its first periodic syzygy.
(2) Let Λ2 be the algebra given by a quiver with relation 1 2 α β α2 = 0.
TATE-HOCHSCHILD COHOMOLOGY RINGS 139 Then the algebra Λ2 is monomial 1-Gorenstein and hence eventually periodic. Bardzell’s minimal projective resolution allows us to see that Ω2 Λe 2(Λ2) is the first periodic syzygy of Λ2.
Moreover, one can see that the algebras in [8, Example 4.3] are eventually periodic algebras.
3.2.
Main result This subsection is devoted to showing our main result. We prove it after two propositions below. Before the first one, we prepare some terminology. Recall that we write Ωi(X•) = Cok dX i+1 for a complex X• and i ∈Z. For a module M over a Gorenstein algebra Λ, its complete resolution T• →P• →M is called periodic if there exists an integer p > 0 such that Ωi(T•) ∼ = Ωi+p(T•) in Λ-mod for all i ∈Z. We call the least such p the period of the complete resolution.
Proposition 3.3. Let Λ be a Gorenstein algebra and M a Λ-module. If there exists an integer n ≥0 such that Ωn Λ(M) is periodic of period p, then M admits a periodic complete resolution of period p. Further, the period of the periodic complete resolution is independent of the choice of periodic syzygies.
Proof. Assume that there exists a minimal projective resolution P• →M satisfying that Ωn Λ(M) is periodic of period p. Then, by using the periodicity of Ωn Λ(M), we can extend the truncated complex P≥n to an (unbounded) complex T• in APC(Λ) having the following properties: (i) T≥n = P≥n.
(ii) For each i ∈Z, there exists an integer 0 ≤j < p such that Ωi(T•) ∼ = Ωn+j Λ (M).
In particular, one sees that Ωi(T•) ∼ = Ωi+p(T•) for all i ∈Z. Note that one may take T• = 0 if proj.dimΛM < ∞.
It follows from Theorem 2.1 that Ωi(T•) = Σ−iΩ0(T•) is Cohen-Macaulay for each i ∈Z, where Σ denotes the shift functor on CM(Λ). Then it is easily checked that HomK(Λ)(T•, Λ[i]) = 0 for all i ∈Z, where K(Λ) is the homotopy category of Λ-modules. Hence, as in [9, Lemma 2.4], the family {idTj}j≥n can be extended uniquely up to homotopy to a chain map θ : T• →P• with θj the identity for all j ≥n.
Therefore, the chain map θ gives rise to the desired complete resolution. We remark that the period of the resulting complete resolution does not depend on the choice of n. Indeed, if we take the smallest integer r ≥0 such that Ωr Λ(M) is periodic, then, for each i ≥n, the module Ωi Λ(M) is periodic and has the same period as Ωr Λ(M).
140 S. USUI Recall that the Yoneda product of the Tate cohomology ring d Ext • Λ(M, M) is denoted by ⌣.
Proposition 3.4. Let Λ be a Gorenstein algebra and M a Λ-module. Then the following are equivalent.
(1) M is eventually periodic.
(2) The Tate cohomology ring d Ext • Λ(M, M) has an invertible homogeneous element of positive degree.
Proof. It suffices to prove the statement for M ∈Λ-mod with proj.dimΛM = ∞. First, we assume that a Λ-module M satisfies that Ωn Λ(M) is periodic of period p for some n ≥0. By Proposition 3.3, there exists a complete resolution T• →P• →M such that Ω0(T•) is periodic of period p, where p is the period of Ωn Λ(M). We fix this complete resolution. Then the shift functor Σ on CM(Λ) satisfies ΣiΩ0(T•) = Ω−i(T•) for all i ∈Z. Let f ∈HomΛ(Ωp(T•), Ω0(T•)) be an isomorphism and consider two homogeneous elements x := Σp[f] ∈d Ext p Λ(M, M) and y := [f−1] ∈d Ext −p Λ (M, M).
Then we have x ⌣y = (Σ−px) ◦y = [f] ◦[f−1] = 1 and similarly y ⌣x = 1, where we set 1 := [idΩ0(T•)].
Conversely, we let T• →P• →M be a complete resolution of M and assume that there exists an isomorphism x ∈HomΛ(Ω0(T•), ΣpΩ0(T•)) = d Ext p Λ(M, M) of degree p > 0. From the definition of complete resolutions, we have HomΛ(Ω0(T•), ΣpΩ0(T•)) ∼ = HomΛ(Σ−m−pΩ0(T•), Σ−mΩ0(T•)) ∼ = HomΛ(Ωm+p Λ (M), Ωp Λ(M)) for some sufficiently large m > 0. Hence we get Ωm+p Λ (M) ∼ = Ωm Λ (M) in Λ-mod. This implies that Ωm+p Λ (M) ⊕P ∼ = Ωm Λ (M) ⊕Q in Λ-mod for some P and Q ∈Λ-proj. By applying the syzygy functor ΩΛ to this isomorphism, we obtain an isomorphism Ωm+p+1 Λ (M) ∼ = Ωm+1 Λ (M) in Λ-mod. This completes the proof.
Using Proposition 3.4, we obtain our main result.
Theorem 3.5. Let Λ be a Gorenstein algebra. Then the following are equiv-alent.
(1) Λ is an eventually periodic algebra.
TATE-HOCHSCHILD COHOMOLOGY RINGS 141 (2) The Tate-Hochschild cohomology ring d HH •(Λ) has an invertible homo-geneous element of positive degree.
In this case, there exists an isomorphism d HH •(Λ) ∼ = d HH ≥0(Λ)[χ−1] of graded algebras, where the degree of an invertible homogeneous element χ equals the period of the periodic syzygy Ωn Λe(Λ) of Λ for some n ≥0.
Proof. We know from [2, Proposition 2.2] that if Λ is a Gorenstein algebra, then so is the enveloping algebra Λe. Hence the former statement follows from Proposition 3.4 applied to Λ ∈Λe-mod. On the other hand, suppose that the Gorenstein algebra Λ satisfies that Ωn Λe(Λ) is periodic for some n ≥0. By the proof of Proposition 3.4, there exists an invertible homogeneous element χ ∈d HH •(Λ) whose degree equals the period of the periodic Λe-module Ωn Λe(Λ).
Then the fact that d HH •(Λ) is a graded commutative algebra yields the desired isomorphism of graded algebras (cf. the proof of [18, Corollary 3.4]).
We end this subsection with the following three remarks.
Remark 3.6. From the definition of singularity categories, an algebra Λ has finite projective dimension as a Λe-module if and only if its Tate-Hochschild cohomology ring is the zero ring (cf. [7, Section 1]). Thus Theorem 3.5 provides a new result if and only if the projective dimension of a given Gorenstein algebra Λ over Λe is infinite.
Remark 3.7. Applying Theorem 3.5 to monomial Gorenstein algebras and to periodic algebras, one obtains [10, Corollary 6.4] and [18, Corollary 3.4], respectively.
Remark 3.8. For an eventually periodic Gorenstein algebra Λ, one can obtain all of dimk d HH ∗(Λ) by using Theorem 3.5 and the Hochschild cohomology HH•(Λ) := ⊕ i≥0 Exti Λe(Λ, Λ) of Λ (see Example 4.7). However, it is still open how we compute the ring structure of d HH ≥0(Λ) (cf. [18, Proposition 3.7]).
§4.
Construction of eventually periodic Gorenstein algebras In this section, we aim at establishing a way of constructing eventually periodic Gorenstein algebras. First, we show two propositions which will be used latter.
Let us start with the following.
Proposition 4.1. Any periodic Λ-module M over a d-Gorenstein algebra Λ is Cohen-Macaulay.
Proof. Assume that M is a periodic Λ-module of period p. Since Ωi Λ(M) ∈ CM(Λ) for i ≥d by [7, Lemma 4.2.2], we have that M ∼ = Ωjp Λ (M) ∈CM(Λ) for some j ≫0.
142 S. USUI We now show that, for an eventually periodic Gorenstein algebra Λ, the smallest integer n ≥0 satisfying that Ωn Λe(Λ) is periodic has a lower bound.
Proposition 4.2. Let Λ be a d-Gorenstein algebra. Assume that there exists an integer n ≥0 such that Ωn Λe(Λ) is periodic. Then the least such integer n satisfies n ≥d. In particular, the equality holds if and only if there exists a simple Λ-module S such that Extn Λ(S, Λ) ̸= 0.
Proof. Let Λ be an eventually periodic Gorenstein algebra and P• →Λ a minimal projective resolution of Λ over Λe satisfying that Ωn Λe(Λ) is the first periodic syzygy of period p. For any M ∈Λ-mod, an exact sequence P• ⊗Λ M →Λ ⊗Λ M = M is a projective resolution of M and has the property that Ωn(P• ⊗Λ M) = Ωn Λe(Λ) ⊗Λ M ∼ = Ωn+ip Λe (Λ) ⊗Λ M = Ωn+ip(P• ⊗Λ M) for all i ≥0. In particular, as in Proposition 4.1, one concludes that Ωn(P• ⊗Λ M) is Cohen-Macaulay. This implies that n ≥inj.dimΛΛ = d. Indeed, for any Λ-module M, we have Extn+1 Λ (M, Λ) ∼ = Ext1 Λ(Ωn(P• ⊗Λ M), Λ) = 0.
For the latter statement, we first suppose that n = d.
Then it follows from [10, Proposition 2.4] that we have n = G-dimΛ(Λ/r), where r denotes the Jacobson radical of Λ. This shows that Extn Λ(Λ/r, Λ) ̸= 0, so that one obtains the desired simple Λ-module. Conversely, assume that Extn Λ(S, Λ) ̸= 0 for some simple Λ-module S. Then one concludes that Ωn−1 Λ (S) ̸∈CM(Λ).
However, since we know that Ωn Λ(S) is Cohen-Macaulay, we have n = G-dimΛS and hence n ≤d. Then the proof is completed since n ≥d by the former statement.
Now, we recall some facts on projective resolutions for tensor algebras. Let Λ and Γ be algebras and P• εΛ − →Λ and Q• εΓ − →Γ projective resolutions as bimodules. Then the tensor product P• ⊗Q• εΛ⊗εΓ − − − − →Λ ⊗Γ is a projective resolution of the tensor algebra Λ ⊗Γ over (Λ ⊗Γ)e (see [16, Section X.7]).
Here, we identify (Λ ⊗Γ)e with Λe ⊗Γe. It also follows from [6, Lemma 6.2] that if both P• →Λ and Q• →Γ are minimal, then so is P• ⊗Q• →Λ ⊗Γ.
From now on, we assume that Λ is a periodic algebra of period p and that Γ is an algebra of finite global dimension n. Set A := Λ ⊗Γ. Since periodic algebras are self-injective algebras, it follows from [6, Lemma 6.1] that we have inj.dim A = inj.dim Λ + inj.dim Γ = 0 + n = n as one-sided modules.
Thus A is an n-Gorenstein algebra.
Note that the same lemma also implies that the enveloping algebra Ae is a (2n)-Gorenstein algebra. We now show that the algebra A has an eventually periodic minimal projective resolution.
Proposition 4.3. Let Λ and Γ be as above. Then A = Λ ⊗Γ is an eventually periodic n-Gorenstein algebra having Ωn Ae(A) as its first periodic syzygy.
TATE-HOCHSCHILD COHOMOLOGY RINGS 143 Proof. Let P• →Λ and Q• →Γ be minimal projective resolutions as bimod-ules. Recall that the r-th component of the total complex P• ⊗Q• with r ≥0 is given by (P• ⊗Q•)r = r ⊕ i=0 Pr−i ⊗Qi.
Since Qi = 0 for i > n = gl.dim Γ = proj.dimΓeΓ, we have (P• ⊗Q•)r = n ⊕ i=0 Pr−i ⊗Qi for all r ≥n. Moreover, the (r + 1)-th differential dP⊗Q r+1 : (P• ⊗Q•)r+1 →(P• ⊗Q•)r (r ≥n) can be written as the square matrix (∂ij r+1)ij of degree n + 1 whose (i, j)-th entry ∂ij r+1 : Pr+1−(j−1) ⊗Qj−1 →Pr−(i−1) ⊗Qi−1 (1 ≤i, j ≤n + 1) is given by ∂ij r+1 = dP r−i+2 ⊗idQi−1 if i = j; (−1)r−i+1 idPr−i+1 ⊗dQ i if j = i + 1; 0 otherwise.
We claim that Cok dP⊗Q n+p+1 ∼ = Cok dP⊗Q n+1 . First, suppose that p is even. Since ∂ij n+p+1 = ∂ij n+1 for all 1 ≤i, j ≤n + 1 because p is even and dP l = dP l+p for any l ≥0, we conclude that dP⊗Q n+p+1 = dP⊗Q n+1 , which implies the claim.
Now, assume that p is odd. Consider the isomorphism of Ae-modules between (P• ⊗Q•)r and (P• ⊗Q•)r+p with r ≥n induced by the diagonal matrix D of degree n+1 whose (i, i)-th entry is (−1)n+i. Together with the fact that p+1 is even, a direct calculation shows that there exists a commutative diagram of Ae-modules with exact rows (P• ⊗Q•)n+p+1 dP ⊗Q n+p+1 / D ∼ = (P• ⊗Q•)n+p / D ∼ = Cok dP⊗Q n+p+1 / 0 (P• ⊗Q•)n+1 dP ⊗Q n+1 / (P• ⊗Q•)n / Cok dP⊗Q n+1 / 0 This implies the claim. Since the projective resolution P• ⊗Q• →A is min-imal, we have that Ωn+p Ae (A) = Cok dP⊗Q n+p+1 ∼ = Cok dP⊗Q n+1 = Ωn Ae(A).
From Proposition 4.2 and this isomorphism, we see that the n-th syzygy Ωn Ae(A) is the first periodic syzygy of A.
144 S. USUI Remark 4.4. Proposition 2.2 allows us to get G-dimAeA ≤2n = inj.dimAeAe and hence HHi(A) ∼ = d HH i(A) for all i > 2n. On the other hand, the i-th syzygy Ωi Ae(A) of A is Cohen-Macaulay for any i ≥n by Propositions 4.1 and 4.3.
Again, Proposition 2.2 yields that G-dimAeA ≤n. One of the advantages of this observation is that there exists an isomorphism HHi(A) ∼ = d HH i(A) for all i > n.
Remark 4.5. It follows from Theorem 3.5 and the proof of Proposition 4.3 that the Tate-Hochschild cohomology ring d HH •(A) of A is of the form d HH ≥0(A)[χ−1], where the degree of χ divides the period p of Λ. We hope to address the degree of χ in a future paper.
We end this section with the following two examples. Note that the tensor algebra A in Example 4.7 can be found in [6, Example 6.3].
Example 4.6. For an integer n ≥0, let Γn be the algebra given by a quiver with relations 0 1 · · · n −1 n α0 αn−1 αi+1αi = 0 for i = 0, . . . , n −2.
Then we have gl.dim Γn = n. By Proposition 4.3, any periodic algebra Λ gives us an eventually periodic n-Gorenstein algebra A = Λ ⊗Γn with Ωn Ae(A) the first periodic syzygy of A.
Example 4.7. Let Λ = k[x]/(x2) and let Γ be the algebra Γ1 defined in Example 4.6. Thanks to Bardzell’s minimal projective resolution, we see that Λ is a periodic algebra whose period is equal to 1 if char k = 2 and to 2 otherwise. On the other hand, the tensor algebra A = Λ ⊗Γ is given by the following quiver with relations 1 2 α β γ α2 = 0 = γ2 and βα = γβ.
Thus we see that A is a (non-monomial) eventually periodic Gorenstein algebra whose first periodic syzygy is Ω1 Ae(A). Now, we compute dimk d HH i(A) for all i ∈Z. It follows from [14, Section 1.6] that the Hochschild cohomology ring HH•(Γ) is of the form HH•(Γ) = k.
According to [5, Section 5], the Hochschild cohomology ring HH•(Λ) is as follows: HH•(Λ) = k[a0, a1]/(a2 0) if char k = 2; k[a0, a1, a2]/(a2 0, a2 1, a0a1, a0a2) if char k ̸= 2, TATE-HOCHSCHILD COHOMOLOGY RINGS 145 where the index i of a homogeneous element ai denotes the degree of ai. On the other hand, by [15, Lemma 3.1], there exists an isomorphism of graded algebras HH•(A) ∼ = HH•(Λ) ⊗HH•(Γ) = HH•(Λ).
It follows from Remark 4.4 that HHi(A) ∼ = d HH i(A) for all i > 1. Hence, the fact that d HH ∗(A) ∼ = d HH ∗+p(A) with the period p of Λ (see Remark 4.5) implies that, for any integer i, we have dimk d HH i(A) = { 2 if char k = 2; 1 if char k ̸= 2.
Acknowledgments The author would like to express his appreciation to the referee(s) for valuable suggestions and comments and for pointing out an error in the manuscript.
The author also would like to thank Professor Katsunori Sanada, Professor Ayako Itaba and Professor Tomohiro Itagaki for their tremendous support for the improvement of the manuscript of the paper.
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E. L. Green, N. Snashall, and Ø. Solberg. The Hochschild cohomology ring of a selfinjective algebra of finite representation type.
Proc. Amer. Math. Soc., 131(11):3387–3393, 2003.
D. Happel.
Triangulated categories in the representation theory of finite-dimensional algebras, volume 119 of London Mathematical Society Lecture Note Series. Cambridge University Press, Cambridge, 1988.
D. Happel. Hochschild cohomology of finite-dimensional algebras. In S´ eminaire d’Alg` ebre Paul Dubreil et Marie-Paul Malliavin, volume 1404 of Lecture Notes in Math., pages 108–126. Springer, Berlin, 1989.
J. Le and G. Zhou. On the Hochschild cohomology ring of tensor products of algebras. J. Pure Appl. Algebra, 218(8):1463–1477, 2014.
S. Mac Lane. Homology. Classics in Mathematics. Springer-Verlag, Berlin, 1995.
V. C. Nguyen. The Tate-Hochschild cohomology ring of a group algebra. (2012).
arXiv:1212.0774.
S. Usui. Tate-Hochschild cohomology for periodic algebras. Arch. Math. (Basel), 116(6):647–657, 2021.
Z. Wang. Tate-Hochschild cohomology of radical square zero algebras. Algebr.
Represent. Theory, 23(1):169–192, 2020.
Z. Wang. Gerstenhaber algebra and Deligne’s conjecture on the Tate–Hochschild cohomology. Trans. Amer. Math. Soc., 374(7):4537–4577, 2021.
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Satoshi Usui Department of Mathematics, Tokyo University of Science 1-3 Kagurazaka, Shinjuku-ku, Tokyo, 162-8601, Japan E-mail: 1119702@ed.tus.ac.jp |
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Water Vapor Saturation Pressure Formulae and Calculator
Thermodynamics and Heat Transfer
Water Vapor Saturation Pressure
Increasing temperature of liquid (or any substance) enhances its evaporation that
results in the increase of vapor pressure over the liquid. By lowering temperature of the
vapor we can make it condense back to the liquid. These two phase transitions,
evaporation and condensation, are accompanied by consuming/evolving enthalpy of
transition and by a change in entropy of the material.
The water vapor saturation pressure is required to determine
a number of moist air properties, principally the saturation humidity
ratio. Saturated water vapor pressure is a function of temperature only and independent on the presence of other gases. The temperature dependence is exponential. For water vapor the semi empirical dependence reads as :
Equation 1
pw, s = eA + B/T + C lnT +Dt
Where:
pw, s = water vapor saturation pressure (Pa)
T = Temperature in Kelvin, K = °F + 255.927778
A = 77.34,
B = -7235,
C = - 8.2,
D = 0.005711,
e =
2.718281828,
t = saturation temperature
Reference:
Lecture Notes
Sampo Smolander
University of Helsinki
Alternative (recommended) formula given by IPAWS R7-97(2012) and R14-08 (2011). The saturation pressure over liquid water for the temperature range
of 32 to 392°F is given by:
Equation 2
ln pw, s = C8/T + C9 + C10T + C11T2 + C12T3 + C13 ln T
Where:
pw, s = saturation pressure, psia
T = absolute temperature, °R = °F + 459.67
C8 = –1.0440397 E+04
C9= –1.1294650 E+01
C10 = –2.7022355 E–02
C11 = 1.2890360 E–05
C12 = –2.478068 1 E–09
C13 = 6.5459673 E+00
The coefficients of Equations (2) were derived from the
Hyland-Wexler equations, which are given in SI units.
Above the surface of liquid water there always exists some amount of gaseous water and consequently there exists a vapor pressure. When a container containing water is open then the number of the escaping molecules is larger than the number of molecules coming back from the gaseous phase (Fig. 1). In this case vapor pressure is small and far from saturation. When the container is closed then the water vapor pressure above the surface increases (concentration of molecules increases) and therefore the number of molecules coming back increases too (Fig. 2). After some time, the number of molecules escaping the liquid and that coming back becomes equal. Such situation is called by dynamic equilibrium between the escaping and returning molecules (Fig. 3). In this case, it is said that the water vapor pressure over the liquid water is saturated.
Figues 1, 2 and 3
Table 1 - Derived from Equation 2
| | | | |
--- --- |
| Temperature | | Saturation Pressure | |
| ˚ F | ˚ R | psia | Pa |
| 32 | 491.7 | 0.0886 | 611.21 |
| 44 | 503.7 | 0.1420 | 979.27 |
| 56 | 515.7 | 0.2220 | 1530.79 |
| 68 | 527.7 | 0.3392 | 2338.80 |
| 80 | 539.7 | 0.5074 | 3498.08 |
| 92 | 551.7 | 0.7439 | 5129.31 |
| 104 | 563.7 | 1.0709 | 7383.46 |
| 116 | 575.7 | 1.5151 | 10446.37 |
| 128 | 587.7 | 2.1093 | 14543.35 |
| 140 | 599.7 | 2.8926 | 19943.75 |
| 152 | 611.7 | 3.9110 | 26965.43 |
| 164 | 623.7 | 5.2183 | 35978.96 |
| 176 | 635.7 | 6.8765 | 47411.59 |
| 188 | 647.7 | 8.9562 | 61750.80 |
| 200 | 659.7 | 11.5374 | 79547.49 |
| 212 | 671.7 | 14.7095 | 101418.67 |
| 224 | 683.7 | 18.5720 | 128049.71 |
| 236 | 695.7 | 23.2345 | 160196.16 |
| 248 | 707.7 | 28.8168 | 198685.05 |
| 260 | 719.7 | 35.4495 | 244415.77 |
| 272 | 731.7 | 43.2735 | 298360.56 |
| 284 | 743.7 | 52.4405 | 361564.68 |
| 296 | 755.7 | 63.1126 | 435146.16 |
| 308 | 767.7 | 75.4625 | 520295.41 |
| 320 | 779.7 | 89.6731 | 618274.54 |
| 332 | 791.7 | 105.9380 | 730416.57 |
| 344 | 803.7 | 124.4604 | 858124.52 |
| 356 | 815.7 | 145.4541 | 1002870.48 |
| 368 | 827.7 | 169.1422 | 1166194.58 |
| 380 | 839.7 | 195.7580 | 1349704.14 |
| 392 | 851.7 | 225.5442 | 1555072.74 |
Related:
Steam Tables - Thermodynamics - Thermodynamics
Water Saturation Pressure Temperature
Partial Pressure of Water Vapor in Saturated Air Table Chart
Saturation Thermodynamic
Saturated Steam Tables Imperial Units
Saturated Steam Table Chart Metric Units
Thermodynamic Properties Saturated Water Pressure Entry Tables
References:
Adapted from NASA (1976).
Link to this Webpage:
Copy Text
to clipboard
Click for Suggested Citation
© Copyright 2000 -
2025, by Engineers Edge, LLC
www.engineersedge.com
All rights reserved
Disclaimer |
Feedback
Advertising| Contact
Thermodynamics and Heat Transfer
Water Vapor Saturation Pressure
Increasing temperature of liquid (or any substance) enhances its evaporation that
results in the increase of vapor pressure over the liquid. By lowering temperature of the
vapor we can make it condense back to the liquid. These two phase transitions,
evaporation and condensation, are accompanied by consuming/evolving enthalpy of
transition and by a change in entropy of the material.
The water vapor saturation pressure is required to determine
a number of moist air properties, principally the saturation humidity
ratio. Saturated water vapor pressure is a function of temperature only and independent on the presence of other gases. The temperature dependence is exponential. For water vapor the semi empirical dependence reads as :
Equation 1
pw, s = eA + B/T + C lnT +Dt
Where:
pw, s = water vapor saturation pressure (Pa)
T = Temperature in Kelvin, K = °F + 255.927778
A = 77.34,
B = -7235,
C = - 8.2,
D = 0.005711,
e =
2.718281828,
t = saturation temperature
Reference:
Lecture Notes
Sampo Smolander
University of Helsinki
Alternative (recommended) formula given by IPAWS R7-97(2012) and R14-08 (2011). The saturation pressure over liquid water for the temperature range
of 32 to 392°F is given by:
Equation 2
ln pw, s = C8/T + C9 + C10T + C11T2 + C12T3 + C13 ln T
Where:
pw, s = saturation pressure, psia
T = absolute temperature, °R = °F + 459.67
C8 = –1.0440397 E+04
C9= –1.1294650 E+01
C10 = –2.7022355 E–02
C11 = 1.2890360 E–05
C12 = –2.478068 1 E–09
C13 = 6.5459673 E+00
The coefficients of Equations (2) were derived from the
Hyland-Wexler equations, which are given in SI units.
Above the surface of liquid water there always exists some amount of gaseous water and consequently there exists a vapor pressure. When a container containing water is open then the number of the escaping molecules is larger than the number of molecules coming back from the gaseous phase (Fig. 1). In this case vapor pressure is small and far from saturation. When the container is closed then the water vapor pressure above the surface increases (concentration of molecules increases) and therefore the number of molecules coming back increases too (Fig. 2). After some time, the number of molecules escaping the liquid and that coming back becomes equal. Such situation is called by dynamic equilibrium between the escaping and returning molecules (Fig. 3). In this case, it is said that the water vapor pressure over the liquid water is saturated.
Figues 1, 2 and 3
Table 1 - Derived from Equation 2
| | | | |
--- --- |
| Temperature | | Saturation Pressure | |
| ˚ F | ˚ R | psia | Pa |
| 32 | 491.7 | 0.0886 | 611.21 |
| 44 | 503.7 | 0.1420 | 979.27 |
| 56 | 515.7 | 0.2220 | 1530.79 |
| 68 | 527.7 | 0.3392 | 2338.80 |
| 80 | 539.7 | 0.5074 | 3498.08 |
| 92 | 551.7 | 0.7439 | 5129.31 |
| 104 | 563.7 | 1.0709 | 7383.46 |
| 116 | 575.7 | 1.5151 | 10446.37 |
| 128 | 587.7 | 2.1093 | 14543.35 |
| 140 | 599.7 | 2.8926 | 19943.75 |
| 152 | 611.7 | 3.9110 | 26965.43 |
| 164 | 623.7 | 5.2183 | 35978.96 |
| 176 | 635.7 | 6.8765 | 47411.59 |
| 188 | 647.7 | 8.9562 | 61750.80 |
| 200 | 659.7 | 11.5374 | 79547.49 |
| 212 | 671.7 | 14.7095 | 101418.67 |
| 224 | 683.7 | 18.5720 | 128049.71 |
| 236 | 695.7 | 23.2345 | 160196.16 |
| 248 | 707.7 | 28.8168 | 198685.05 |
| 260 | 719.7 | 35.4495 | 244415.77 |
| 272 | 731.7 | 43.2735 | 298360.56 |
| 284 | 743.7 | 52.4405 | 361564.68 |
| 296 | 755.7 | 63.1126 | 435146.16 |
| 308 | 767.7 | 75.4625 | 520295.41 |
| 320 | 779.7 | 89.6731 | 618274.54 |
| 332 | 791.7 | 105.9380 | 730416.57 |
| 344 | 803.7 | 124.4604 | 858124.52 |
| 356 | 815.7 | 145.4541 | 1002870.48 |
| 368 | 827.7 | 169.1422 | 1166194.58 |
| 380 | 839.7 | 195.7580 | 1349704.14 |
| 392 | 851.7 | 225.5442 | 1555072.74 |
Related:
Steam Tables - Thermodynamics - Thermodynamics
Water Saturation Pressure Temperature
Partial Pressure of Water Vapor in Saturated Air Table Chart
Saturation Thermodynamic
Saturated Steam Tables Imperial Units
Saturated Steam Table Chart Metric Units
Thermodynamic Properties Saturated Water Pressure Entry Tables
References:
Adapted from NASA (1976).
Link to this Webpage:
Copy Text
to clipboard
Click for Suggested Citation
© Copyright 2000 -
2025, by Engineers Edge, LLC
www.engineersedge.com
All rights reserved
Disclaimer |
Feedback
Advertising| Contact
Water Vapor Saturation Pressure
Increasing temperature of liquid (or any substance) enhances its evaporation that
results in the increase of vapor pressure over the liquid. By lowering temperature of the
vapor we can make it condense back to the liquid. These two phase transitions,
evaporation and condensation, are accompanied by consuming/evolving enthalpy of
transition and by a change in entropy of the material.
The water vapor saturation pressure is required to determine
a number of moist air properties, principally the saturation humidity
ratio. Saturated water vapor pressure is a function of temperature only and independent on the presence of other gases. The temperature dependence is exponential. For water vapor the semi empirical dependence reads as :
Equation 1
pw, s = eA + B/T + C lnT +Dt
Where:
pw, s = water vapor saturation pressure (Pa)
T = Temperature in Kelvin, K = °F + 255.927778
A = 77.34,
B = -7235,
C = - 8.2,
D = 0.005711,
e =
2.718281828,
t = saturation temperature
Reference:
Lecture Notes
Sampo Smolander
University of Helsinki
Alternative (recommended) formula given by IPAWS R7-97(2012) and R14-08 (2011). The saturation pressure over liquid water for the temperature range
of 32 to 392°F is given by:
Equation 2
ln pw, s = C8/T + C9 + C10T + C11T2 + C12T3 + C13 ln T
Where:
pw, s = saturation pressure, psia
T = absolute temperature, °R = °F + 459.67
C8 = –1.0440397 E+04
C9= –1.1294650 E+01
C10 = –2.7022355 E–02
C11 = 1.2890360 E–05
C12 = –2.478068 1 E–09
C13 = 6.5459673 E+00
The coefficients of Equations (2) were derived from the
Hyland-Wexler equations, which are given in SI units.
Above the surface of liquid water there always exists some amount of gaseous water and consequently there exists a vapor pressure. When a container containing water is open then the number of the escaping molecules is larger than the number of molecules coming back from the gaseous phase (Fig. 1). In this case vapor pressure is small and far from saturation. When the container is closed then the water vapor pressure above the surface increases (concentration of molecules increases) and therefore the number of molecules coming back increases too (Fig. 2). After some time, the number of molecules escaping the liquid and that coming back becomes equal. Such situation is called by dynamic equilibrium between the escaping and returning molecules (Fig. 3). In this case, it is said that the water vapor pressure over the liquid water is saturated.
Figues 1, 2 and 3
Table 1 - Derived from Equation 2
| | | | |
--- --- |
| Temperature | | Saturation Pressure | |
| ˚ F | ˚ R | psia | Pa |
| 32 | 491.7 | 0.0886 | 611.21 |
| 44 | 503.7 | 0.1420 | 979.27 |
| 56 | 515.7 | 0.2220 | 1530.79 |
| 68 | 527.7 | 0.3392 | 2338.80 |
| 80 | 539.7 | 0.5074 | 3498.08 |
| 92 | 551.7 | 0.7439 | 5129.31 |
| 104 | 563.7 | 1.0709 | 7383.46 |
| 116 | 575.7 | 1.5151 | 10446.37 |
| 128 | 587.7 | 2.1093 | 14543.35 |
| 140 | 599.7 | 2.8926 | 19943.75 |
| 152 | 611.7 | 3.9110 | 26965.43 |
| 164 | 623.7 | 5.2183 | 35978.96 |
| 176 | 635.7 | 6.8765 | 47411.59 |
| 188 | 647.7 | 8.9562 | 61750.80 |
| 200 | 659.7 | 11.5374 | 79547.49 |
| 212 | 671.7 | 14.7095 | 101418.67 |
| 224 | 683.7 | 18.5720 | 128049.71 |
| 236 | 695.7 | 23.2345 | 160196.16 |
| 248 | 707.7 | 28.8168 | 198685.05 |
| 260 | 719.7 | 35.4495 | 244415.77 |
| 272 | 731.7 | 43.2735 | 298360.56 |
| 284 | 743.7 | 52.4405 | 361564.68 |
| 296 | 755.7 | 63.1126 | 435146.16 |
| 308 | 767.7 | 75.4625 | 520295.41 |
| 320 | 779.7 | 89.6731 | 618274.54 |
| 332 | 791.7 | 105.9380 | 730416.57 |
| 344 | 803.7 | 124.4604 | 858124.52 |
| 356 | 815.7 | 145.4541 | 1002870.48 |
| 368 | 827.7 | 169.1422 | 1166194.58 |
| 380 | 839.7 | 195.7580 | 1349704.14 |
| 392 | 851.7 | 225.5442 | 1555072.74 |
Related:
References:
Link to this Webpage:
Copy Text
to clipboard
© Copyright 2000 -
2025, by Engineers Edge, LLC
www.engineersedge.com
All rights reserved
Disclaimer |
Feedback
Advertising| Contact
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| |
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8169 | https://indico.global/event/8261/contributions/75115/attachments/36184/67484/Bruschini-QWG.pdf | Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary RB, Phys. Rev. D 109, L031501 (2024), arXiv:2306.17120 Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed Roberto Bruschini The Ohio State University QWG 2024 IISER Mohali, February 27, 2024 Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 1 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary 1 Born-Oppenheimer Approximation 2 Quarkonium Hybrids 3 Decay Selection Rules Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 2 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Born-Oppenheimer Approximation for QCD K.J. Juge, J. Kuti and C.J. Morningstar 1999 Q ¯ Q ¯ q q Heavy degrees of freedom heavy quarks Light-QCD fields gluons light quarks 0 ΛQCD mg mu md ms mc mb Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 3 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary The Born-Oppenheimer Hamiltonian Expansion in powers of 1/m HBO(⃗ r, ⃗ p ) = Hstatic(⃗ r ) + p2 m + . . .
Leading order (m →∞): the static limit Hstatic(⃗ r ) = P n |ζn(⃗ r )⟩Vn(r) ⟨ζn(⃗ r )| n Born-Oppenheimer quantum numbers Vn(r) energy levels of light QCD with static Q, ¯ Q at distance r |ζn(⃗ r )⟩eigenstates of light QCD with static Q, ¯ Q at +⃗ r/2, −⃗ r/2 Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 4 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Matching with Lattice QCD Correlation matrix in light QCD with static Q, ¯ Q at +⃗ r/2, −⃗ r/2 Cij(r, τ, τ0) = ⟨0|Oi(⃗ r, τ)U(τ, τ0)O† j(⃗ r, τ0)|0⟩ The correlation matrix C can be calculated using lattice QCD.
QCD quantity that is determined B-O C eigenvalues at large τ static energy levels Vn(r) C eigenvectors at large τ mixing angles |ζn(⃗ r )⟩ Truncation to N channels N eigenvalues and eigenvectors →truncated B-O approximation Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 5 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Diabatic Born-Oppenheimer Approximation W. Lichten 1963; F.T. Smith 1969 Adiabatic Schrödinger equation −1 m ⃗ ∇+ ⃗ Π(⃗ r ) 2Ψ(⃗ r ) + Vdiag(r)Ψ(⃗ r ) = EΨ(⃗ r ) transitions proceed through nonadiabatic coupling matrix ⃗ Π(⃗ r ) unitary transformation Diabatic Schrödinger equation −∇2 m Ψ(⃗ r ) + V(⃗ r )Ψ(⃗ r ) = EΨ(⃗ r ) transitions proceed through diabatic potential matrix V(⃗ r ) Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 6 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary From Static Quarks to Quarkonium Hybrids RB 2023 The static energy levels with Born-Oppenheimer quantum numbers η, λ are the eigenvalues of a matrix Gη,λ(r) that solely depends on the distance r between Q and ¯ Q.
The diabatic potential matrix that depends on the relative position ⃗ r of Q and ¯ Q is a linear combination of the matrices Gη,λ(r) for different values of λ, V η i,σ;i′,σ′(⃗ r ) = P λ Dji σ,λ(φ, θ, ψ)Dji′ σ′,λ(φ, θ, ψ)∗Gη,λ i,i′ (r), where the angular dependence is governed by Wigner D-matrix elements.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 7 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Static Energy Levels of Pure SU(3) Gauge Theory K.J. Juge, J. Kuti and C.J. Morningstar 1999 S. Capitani, O. Philipsen, C. Reisinger, C. Riehl and M. Wagner 2019 r V Σ− u Πu Σ+ g Πu, Σ− u : hybrid potentials r →0: 1+−gluelump r →∞: N = 1, 3 string Σ+ g : quarkonium potential r →0: 0++ vacuum r →∞: N = 0 string Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 8 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Diabatic Schrödinger Equation for Quarkonium Hybrids The quarkonium-hybrid spectrum forms degenerate multiplets of heavy-quark spin symmetry.
The potential matrix for each hybrid multiplet is given by: ▶angular-momentum coefficients, ▶functions of r calculable using lattice QCD.
Potential matrix for H1 multiplet: JPC = 1−−, (0, 1, 2)−+ 1 3 2VΠu(r)+VΣ− u (r) √ 2 3 VΠu(r)−VΣ− u (r) q 1 3 g(r) √ 2 3 VΠu(r)−VΣ− u (r) 1 3 VΠu(r)+2VΣ− u (r) − q 2 3 g(r) q 1 3 g(r) − q 2 3 g(r) VB(∗) ¯ B(∗)(r) Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 9 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Spectrum and Decays of Quarkonium Hybrids Quarkonium Hybrids Are associated with poles of the S-matrix for heavy-meson pairs.
S-matrix Can be calculated nonperturbatively by solving the Schrödinger equation for coupled Q ¯ Q and heavy-meson-pair channels.
Decay selection rules Can be determined using Born-Oppenheimer symmetries.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 10 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Born-Oppenheimer Selection Rules for Hybrids RB 2024 1 The Born-Oppenheimer quantum numbers are: ▶Πu and Σ− u for quarkonium hybrids, ▶Σ+ g , Πg, and Σ− u for pairs of S-wave heavy mesons.
2 The Born-Oppenheimer quantum numbers are conserved.
3 Decays into pairs of S-wave heavy mesons are: ▶allowed for pure Σ− u or mixed Πu/Σ− u quarkonium hybrids, ▶forbidden for pure Πu quarkonium hybrids.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 11 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Decays of Lowest Hybrids into Two S-Wave Heavy Mesons RB 2024 Multiplet Potential H1 1−− (0, 1, 2)−+ Πu/Σ− u H2 1++ (0, 1, 2)+− Πu H3 0++ 1+− Σ− u H4 2++ (1, 2, 3)+− Πu/Σ− u H5 2−− (1, 2, 3)−+ Πu JPC forbidden forbidden allowed allowed See talk by Chunjiang Shi on Friday about the decays of the lowest 1−+ charmoniumlike hybrid in lattice QCD.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 12 Outline Born-Oppenheimer Approximation Quarkonium Hybrids Decay Selection Rules Summary Quarkonium hybrids can be studied ab initio using the Born-Oppenheimer approximation for QCD.
One can derive model-independent selection rules for decays into pairs of heavy mesons.
The Born-Oppenheimer selection rules allow decays of many quarkonium hybrids into pairs of S-wave heavy mesons.
This finding contradicts the conventional wisdom of the last 40 years from constituent models that hybrid mesons are forbidden to decay into pairs of S-wave heavy mesons.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 13 Born-Oppenheimer Symmetries The static Q ¯ Q break rotations, parity, charge-conjugation, down to cylindrical symmetries, combined CP symmetry.
The quantum numbers are not J angular momentum, P parity, C charge-conjugation, but rather λ angular momentum projection on the Q ¯ Q axis, η (g or u) CP = + or −.
Heavy-quark spin symmetry Static energy levels are independent of the heavy-quark spins.
Roberto Bruschini The Ohio State University Why Quarkonium Hybrid Coupling to Two S-Wave Heavy-Light Mesons is Not Suppressed 14 |
8170 | https://ocw.mit.edu/courses/3-091-introduction-to-solid-state-chemistry-fall-2018/4d44cb084ffa34989c4fe75e4f8d4c47_MIT3_091F18_REC3.pdf | 1 3.091: Introduction to Solid State Chemistry Maddie Sutula, Fall 2018 Recitation 3 Waves and photons Photons are quanta of light: the energy of a photon with frequency ν (or wavelength λ) is given by the Planck-Einstein relation. hc E = hν = λ Here, h is Planck’s constant, and c is the speed of light. The electromagnetic spectrum is shown here: Recall that the equality above came from the fact that frequency and wavelength are related through the speed of light as c = λν. It’s also important to remember that the energy of a photon is independent of the amplitude of the wave: The energy is set by the wavelength (or equivalently, frequency); the amplitude of the wave is related to its intensity: this can be thought of as the number of photons of a certain energy. Example: A radio station broadcasts at 100.7 MHz with a power output of 50 kW. How many pho tons are emitted each second? Recall that 1W = 1J/s. The energy per second that leaves the radio station is 1000 W 1 J/s 50 kW × × = 5 × 104 J/s 1 kW 1W We can calculate the energy per photon with the Planck-Einstein relation: E = hν = (6.626 × 10−34 Js) × (100.7 × 106 s −1) = 6.672 × 10−26 J/photon Finally, the number of photons per second is 1 photon N = (5 × 10−4 J/s) = 7 × 1029 photons/s 6.672 × 10−26 J 1 2 3.091: Introduction to Solid State Chemistry Maddie Sutula, Fall 2018 Recitation 3 Bohr model The photoelectric effect tells us that shining light with enough energy on an atom can cause the emission of electrons. In 3.091, we’ll use the Bohr model to model this process. In the Bohr model, an electron in a hydrogen atom can only live in discrete energy levels, which we label with n. The lowest energy level is the ground state, with n = 1. The change in energy corresponding to the transition from an initial energy level, ni to a final energy level, nf is given by 1 1 ΔE = −13.6 − [eV ] 2 2 n n f i If a photon with just the right energy impinges on the atom, the photon can be absorbed and the electron excited to a higher energy level. Conversely, if an electron relaxes from a higher energy level to a lower one within the atom, a photon with frequency ν = ΔE/h is released. Example: A power source emits 8 × 1018 photons/s at 10 W. Determine how much energy each pho ton has. Then, calculate what state an electron in the ground state of a hydrogen atom would end up in if excited by a photon with this energy. To calculate the energy of each photon, we can use dimensional analysis to get to Joules: 10 J/s 1 eV E = × = 12.5 eV 5 × 1018 photons/s 1.6 × 10−19 J If the electron starts in the ground state, we need to solve 1 1 ΔE = 12.5 eV = −13.6 eV 2 − nf = 3.51 nf 12 Since the electron must live in an integer state, it must either go to the third or fourth level. It would take more energy than we have to get to the fourth energy level, the highest accessible energy level is n = 3. 2 MIT OpenCourseWare 3.091 Introduction to Solid-State Chemistry Fall 2018 For information about citing these materials or our Terms of Use, visit: 3 |
8171 | https://www2.oberlin.edu/physics/dstyer/P103/Welcome.pdf | Welcome to Physics 103, “Elementary Physics I”. This document presents information about significant figures and about dimensions (or units) that I expect you to understand and to use on a daily basis throughout this course.
— Dan Styer; 1 November 2023 Significant Figures The impossibility of certainty What is the pattern of thought most characteristic of science? Is it knowing that “net force causes acceleration rather than velocity”? Is it knowing that“momentum is conserved in the absence of external forces”? Is it knowing that “an object starting at rest with constant acceleration has position change proportional to the time squared”? No, it is none of these three facts — important though they are. The pattern of thought most characteristic of science is knowing that “every observation is imperfect and thus every observed number comes with an uncertainty”.
This has an important philosophical consequence: Because all scientific conclusions are based on obser-vations, and all observations contain some uncertainty, no scientific conclusion can be absolutely certain.
Anyone worshiping at the alter of science is fooling himself.
This course is more interested in the day-to-day practicalities of uncertainty than in the grand philosoph-ical consequences. How do we express our lack of certainty? How do we work with (add, subtract, multiply) quantities that aren’t certain?
Expressing uncertainty As with any facet of science, the proper approach to uncertainty is not “plug into a formula for error propagation” but instead “think about the issues involved”. For example, in the course of building a tree house I measured a plank with a meter stick and found it to be 187.6 cm long.
But a more accurate measurement would of course provide a more accurate length: perhaps 187.64722031 cm. I don’t know the plank’s exact length, I only know an approximate value. In a math class, 187.6 cm means the same as 187.60000000 cm. But in a physics class, 187.6 cm means the same as 187.6??????? cm, where the question marks represent not zeros, but digits that you don’t know. The digits that you do know are called “significant digits” or “significant figures.” The convention for expressing uncertain quantities is simple: Any digit written down is a significant digit.
A plank measured to the nearest millimeter has a length expressed as, say, 103.7 cm or 91.5 cm or 135.0 cm.
Note particularly the trailing zero in 135.0 cm: this final digit is significant. The quantity “135.0 cm” is different from “135 cm”. The former means “135.0??????? cm”, the latter means “135.???????? cm”. In the former, the digit in the tenths place is 0, while in the latter, the digit in the tenths place is unknown.
1 This convention gives rise to a problem for representing large numbers. Suppose the distance between two stakes is 45.6 meters. What is this distance expressed in centimeters? The answer 4560 cm is unsatisfactory, because the trailing zero is not significant and so, according to our rule, should not be written down.
This quandary is resolved using exponential notation: 45.6 meters is the same as 4.56 × 103 cm. (This is, unfortunately, one of the world’s most widely violated rules.) Working with uncertainty Addition and subtraction. I measure a plank with a meter stick and find it to be 187.6 cm long. Then I measure a dowel with a micrometer and find it to be 2.3405 cm in diameter. If I place the dowel next to the plank, how long is the dowel plus plank assembly? You might be tempted to say 187.6 + 2.3405 --------189.9405 But no! This is treating the unknown digits in 187.6 cm as if they were zeros, when in fact they’re question marks. The proper way to perform the sum is 187.6?????
+ 2.3405??
----------189.9?????
So the correct answer, with only significant figures written down, is 189.9 cm.
2 Multiplication and division. The same question mark technique works for multiplication and division, too. For example, a board measuring 124.3 cm by 5.2 cm has an area given through 1243?
x 52?
--------?????
2486?
6215?
--------64????
Adjusting the decimal point gives an answer of 6.4 × 102 cm2 But while the question mark technique works, it’s very tedious. (It’s even more tedious for division.) Fortunately, the following rule of thumb works as well as the question mark technique and is a lot easier to apply: When multiplying or dividing two numbers, round the answer down to the number of significant digits in the least certain of the two numbers.
For example, when multiplying a number with four significant digits by a number with two significant digits, the result should be rounded to two significant digits (as in the example above).
Doing the rounding Most of the rules for rounding are straightforward.
If your result has two significant figures and your calculator gives you the result 1.2317431, you round down to 1.2. If your calculator gives 1.2617431, you round up to 1.3. But what if your calculator gives 1.2517431, given that 1.25 is exactly halfway between 1.20 and 1.30?
Many teachers use the “round to even rule”: If the digit to be rounded offis a 5, and the digit that’s retained is even, round down. If the digit that’s retained is odd, round up. Under this rule 1.2517431 is rounded down to 1.2 (with last digit even), whereas 1.3517431 is rounded up to 1.4 (with last digit even).
Because 0 is an even number, 1.0517431 is rounded down to 1.0. Under this rule you round up half the time and you round down half the time, with pleasing symmetry.
I don’t insist on the “round to even rule”. It gives the misimpression that for any scientific problem there is a single correct answer and a single correct way to find it, that science doesn’t involve creativity or ingenuity. In this course, you may round 1.2517431 to either 1.2 or 1.3, I don’t care which. But do not round it to 1.25, giving you an illegitimate extra digit.
3 Examples of rounding Suppose you’re working a problem to find a length. You’ve already determined that the result has three significant digits and the units of meters. This table shows what comes out of your calculator, and then the correct and incorrect results: Calculator: 1.7432987 1.7032987 1743.2987 1703.2987 Correct: 1.74 m 1.70 m 1.74 × 103 m 1.70 × 103 m Correct: 1.74 km 1.70 km Wrong: 1.7 m 1740 m 1700 m The first wrong answer errs in not reporting a trailing zero that is significant. The second two wrong answers err in reporting a digit (the rightmost zero) that is not significant. In 1700 m, the zero to the left is significant so must be reported, the zero to the right is not significant so must not be reported.
Numbers that are certain Any measured number is uncertain, but counted and defined quantities can be certain. If there are seven people in a room, there are 7.0000000 . . . people. There are never 7.00395 people in a room. And the inch is defined to be exactly 2.5400000 . . . centimeters — there’s no uncertainty in this conversion factor, either.
4 Dimensions What does “dimensions” mean?
Suppose I say that a table is six feet long or, in symbols, ℓT = 6 ft, where ℓT represents the length of the table. This means that the table is six times as long as the length of the standard foot: ℓT = 6 ft means ℓT = 6 × (the length of the standard foot).
In other words, the symbol “ft” used in the equations above stands for “the length of the standard foot”.
The symbol “ℓT ” stands for “6 ft”. That is, it stands for the number “six” times the length of the standard foot, or in other words, for the number “six” times the unit “ft”. If I wrote “ℓT = 6” instead of “ℓT = 6 ft”, I’d be dead wrong. . . just as wrong as if the solution to an algebra problem were “y = 6x” and I wrote “y = 6”, or if the solution to an arithmetic problem were “6 × 7” and I wrote “6”. In all three cases, my answer would be wrong because it omitted a factor. (These are, respectively, the factor of “the length of the standard foot”, the factor of x, and the factor of 7.) The length of the table is not 6 — rather, the ratio of the length of the table to the length of the standard foot is 6.
Ignoring the units of a measurement results in practical as well as conceptual error. On 23 September 1999 the “Mars Climate Orbiter” spaceprobe crashed into the surface of Mars, dashing the hopes and dreams of dozens of scientists and resulting in the waste of $125 million. This spacecraft had survived perfectly the long and perilous trip from Earth to Mars. How could it have failed so spectacularly the final, relatively easy, phase of its journey? The manufacturer, Lockheed Martin Corporation, had told the the spacecraft controllers, at NASA’s Jet Propulsion Laboratory, the thrust that the probe’s rockets could produce. But the Lockheed engineers gave the thruster performance data in pounds (the English unit of force), and they didn’t specify which units they used. The NASA controllers assumed that the data were in newtons (the metric unit of force).
5 New York Times, 1 October 1999, page 1, an embarrassing place to have your blunders reported.
Modern information technology actually encourages mistakes like this.
When you use a calculator, a spreadsheet, or a computer program, you enter pure numbers like “1.79”, rather than quantities like “1.79 feet”. So it’s especially important to be on your guard and document your units when using computers.
Keep the units in your mind, even if you can’t keep them in your calculator!
A nitpicky distinction is that the length of the table has the units of “feet” and the dimensions of “length”.
If the length of the table were measured in yards or meters it would still have the dimensions of length. But in everyday language the terms “units” and “dimensions” are often used interchangeably.
6 Unit conversions It is standard usage to refer to the length of the standard foot by the symbol “ft”. But in this document I’ll also refer to the length of the standard foot by the symbol ℓF . Similarly I’ll call the length of the standard yard either “yd” or ℓY .
You know that if a table is 6 feet long it is also 2 yards long: ℓT = 6 ft = 6ℓF = 2 yd = 2ℓY .
How can this be? It’s certainly not true that 6 = 2! It’s true instead that 6 ft = 2 yd because the length of a yardstick is three times the length of a foot ruler: ℓY = 3ℓF .
This tells us that 2 yd = 2ℓY = 2 × (3ℓF ) = 6ℓF = 6 ft, or, in the opposite direction, 6 ft = 6ℓF = 6ℓF (1) = 6ℓF ℓY 3ℓF = 6ℓF \ ℓY 3ℓF \ = 2ℓY = 2 yd.
In short, the symbol “ft” can be manipulated exactly like the symbol “ℓF ”, because that’s exactly what it means!
Incompatible dimensions If I walk for 4 yd, and then for 2 ft, how far did I go? The answer is 14 ft or 4 2 3 yd, but not 4 + 2 = 6 of anything!
If I walk for 4 yd, and then pause for 30 seconds, how far did I go? Certainly not 4 yd + 30 sec. The number 34 has no significance in this problem. For example, it can’t be converted into minutes.
In general, you can’t add quantities with different units.
This rule can be quite helpful. Suppose you’re working a problem that involves a speed v and a time t, and you’re asked to find a distance d. Someone approaches you and whispers: “Here’s a hint: use the equation d = vt+ 1 2vt2.” You know that this hint is wrong: The quantity vt has the dimensions of [distance], but the quantity 1 2vt2 has the dimensions of [distance×time]. You can’t add a quantity with the dimensions of [distance×time] to a quantity with the dimensions of [distance], any more than you could add 30 sec to 4 yd.
7 Dimensions aren’t a cure-all It’s not meaningful to add quantities with different units, but just because two quantities do have the same units doesn’t mean it must be meaningful to add them. For example, two quantities from mechanics are work and torque. Both have the dimensions of [force×distance], but the entities are quite distinct and it never makes sense to add a quantity of work to a quantity of torque. Why? Work is defined as a force times a distance parallel to that force, while torque is defined as a force times a distance perpendicular to that force. They are different types of entity, even though they share the same dimensions.
Another example: The rate at which a conveyor belt delivers gravel is measured in kilograms/second.
And if a frictional force is proportional to velocity, Ffriction = −bv, then the friction coefficient b has the units of kilograms/second. But these are clearly entities of different character!
A famous use of dimensional analysis The following story about the use of dimensional analysis comes from David L. Goodstein, States of Matter (Prentice-Hall, 1975, page 436). (See also Physics Today, May 2000, page 35; and G.I. Barenblatt, Scaling Phenomena in Fluid Mechanics (Cambridge University Press, 1994).) “Dimensional analysis is a technique by means of which it is possible to learn a great deal about very complicated situations if you can put your finger on the essential features of the problem. An example is the well-known story of how G.I. Taylor was able to deduce the yield of the first nuclear explosion from a series of photographs of the expanding fireball in Life magazine. He realized that he was seeing a strong shock expanding into an undisturbed medium. The pictures gave him the radius as a function of time, r(t). All that could be important in determining r(t) was the initial energy release, E, and the density of the undisturbed medium, ρ. The radius, with the dimension of length, depended on E, ρ, and t, and he constructed a distance out of these quantities. E and ρ had to come in as E/ρ to cancel the mass. E/ρ has the dimensions [length]5/[time]2, so the only possible combination was r(t) ∝ E ρ t2 1/5 .
A log-log plot of r versus t (measured from the pictures) gave a slope of 2 5, which checked the theory, and E/ρ could be obtained from extrapolation to the value of log r when log t = 0. Since ρ, the density of undisturbed air, was known, E was determined to within a [dimensionless] factor of order one. For the practitioner of the art of dimensional analysis, the nation’s deepest secret had been published in Life magazine.” Problem Sound speed. The speed of sound vs in air, in a given room, could reasonably depend on three things: the air density ρ, the air pressure p, and the room volume V . In other words vs = CρxpyV z, where C is some dimensionless number. Use dimensional analysis to determine the exponents x, y, and z.
8 |
8172 | https://www.milan-museum.com/leonardo-last-supper-cenacolo.php | The Last Supper - Leonardo Da Vinci - Useful Information
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A bit of history about the Leonardo's Last Supper
The Last Supper of Leonardo da Vinci (Cenacolo Vinciano) is one of the most famous paintings in the world. This artwork was painted between 1494 and 1498 under the government of Ludovico il Moro and represents the last "dinner" between Jesus and his disciples.
In order to create this unique work, Leonardo carried out an exhaustive research creating an infinity of preparatory sketches. Leonardo abandons the traditional method of fresco painting, painting the scene "dry" on the wall of the refectory. Traces of gold and silver foils have been found which testify to the artist's willingness to make the figures in a much more realistic manner, including precious details. After completion, his technique and environmental factor had contributed to the eventual deterioration of the fresco, which had undergone numerous restorations.
The most recent restoration was completed in 1999 where several scientific methods were used to restore the original colors as close as possible, and to eliminate traces of paint applied in previous attempts to restore the fresco.
What is the Leonardo's Last Supper nowadays?
Leonardo’s Last Supper is located in its original place, on the wall of the dining room of the former Dominican convent of Santa Maria delle Grazie, exactly in the refectory of the convent and is one of the most celebrated and well known artworks in the world.
Leonardo Da Vinci's "Last Supper", a huge painting of 4.60 meters high and 8.80 meters wide was made with tempera and oil on a gypsum preparation instead of the technique commonly used in the fresco period.
Several measures have now been implemented to protect the paint from deterioration. To ensure that the fresco is maintained at room temperature, since the last restoration, visitor's' entry has been restricted to a group of 25 people every 15 minutes.
Curiosities about the Leonardo's Last Supper
Did you know that the great fame of this masterpiece has awaken the interest of many historians, researchers and novelists who seek to solve the supposed mysteries and enigmas that surround this painting. For example, in the books "The Templar Revelation" by Clive Prince and Lynn Picknett and in the novel Dan Brown's Da Vinci Code, it is affirmed that the figure to the right of Jesus is not the apostle John, but a female figure. The truth is that these mysteries and curiosities have not yet been solved.
Did you know that during the French Revolutionary War Napoleon's troops used the wall of the refectory to make target practice and during the Second World War in 1943 the bombings managed to tear off the roof of the old Dominican dining room leaving the paint in the open for several years.
Why visit the Leonardo's Last Supper?
The Last Supper of Leonardo da Vinci is undoubtedly one of the most important works of art of all times, both for its innovative approach and for the impact it has had on artists of all ages. This magnificent work of art has been seeing by Leonardo's contemporary artists as the "painting that speaks," something that had never happened before.
How to visit the Last Supper of Leonardo da Vinci?
The Last Supper of Leonardo da Vinci is undoubtedly one of the most interesting attractions in the city of Milan. The availability of tickets is very limited, so the advance reservation is considered "mandatory". Tickets to see this masterpiece can be booked online but must be sold as part of a package, so it is recommended to combine them with a Milan Audioguide, or with entries for the Brera Gallery or for the Pinacoteca Ambrosiana.
Naturally, it is also possible to book a guided tour that may also include visiting other churches or attractions located in the center of Milan.
It is also possible to request a visit with a private guide that includes, in addition to the entrance to the Last Supper, a visit to the adjacent Church of Santa Maria delle Grazie.
Other attractions in the area
Milan is a big city full of surprises. On a short distance from Leonardo da Vinci's Last Supper is the Duomo Square in the historic center of the city. In this square is located the famous gothic Cathedral of Milan and the statue of Victor Emanuel II erected in 1896 in honor to the king of Italy. Here you will also find the wonderful Vittorio Emanuele II Gallery, a shopping center full of cafes, restaurants and shops. After crossing this gallery you will arrive at the luxurious theater of the city of Milan, La Scala. A walk through a historic center simply unforgettable!
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8173 | https://leanprover-community.github.io/logic_and_proof/natural_deduction_for_first_order_logic.html | Natural Deduction for First Order Logic — Logic and Proof 3.18.4 documentation
8. Natural Deduction for First Order Logic¶
8.1. Rules of Inference¶
In the last chapter, we discussed the language of first-order logic, and the rules that govern their use. We summarize them here:
The universal quantifier:
In the introduction rule, x should not be free in any uncanceled hypothesis. In the elimination rule, t can be any term that does not clash with any of the bound variables in A.
The existential quantifier:
In the introduction rule, t can be any term that does not clash with any of the bound variables in A. In the elimination rule, y should not be free in B or any uncanceled hypothesis.
Equality:
Strictly speaking, only r e f l and the second substitution rule are necessary. The others can be derived from them.
8.2. The Universal Quantifier¶
The following example of a proof in natural deduction shows that if, for every x, A(x) holds, and for every x, B(x) holds, then for every x, they both hold:
Notice that neither of the assumptions 1 or 2 mention y, so that y is really “arbitrary” at the point where the universal quantifiers are introduced.
Here is another example:
As an exercise, try proving the following:
∀x(A(x)→B(x))→(∀x A(x)→∀x B(x)).
Here is a more challenging exercise. Suppose I tell you that, in a town, there is a (male) barber that shaves all and only the men who do not shave themselves. You can show that this is a contradiction, arguing informally, as follows:
By the assumption, the barber shaves himself if and only if he does not shave himself. Call this statement ().
Suppose the barber shaves himself. By (), this implies that he does not shave himself, a contradiction. So, the barber does not shave himself.
But using () again, this implies that the barber shaves himself, which contradicts the fact we just showed, namely, that the barber does not shave himself.
Try to turn this into a formal argument in natural deduction.
Let us return to the example of the natural numbers, to see how deductive notions play out there. Suppose we have defined e v e n and o d d in such a way that we can prove:
∀n(¬e v e n(n)→o d d(n))
∀n(o d d(n)→¬e v e n(n))
Then we can go on to derive ∀n(e v e n(n)∨o d d(n)) as follows:
We can also prove and ∀n¬(e v e n(n)∧o d d(n)):
As we move from modeling basic rules of inference to modeling actual mathematical proofs, we will tend to shift focus from natural deduction to formal proofs in Lean. Natural deduction has its uses: as a model of logical reasoning, it provides us with a convenient means to study metatheoretic properties such as soundness and completeness. For working within the system, however, proof languages like Lean’s tend to scale better, and produce more readable proofs.
8.3. The Existential Quantifier¶
Remember that the intuition behind the elimination rule for the existential quantifier is that if we know ∃x A(x), we can temporarily reason about an arbitrary element y satisfying A(y) in order to prove a conclusion that doesn’t depend on y. Here is an example of how it can be used. The next proof says that if we know there is something satisfying both A and B, then we know, in particular, that there is something satisfying A.
The following proof shows that if there is something satisfying either A or B, then either there is something satisfying A, or there is something satisfying B.
The following example is more involved:
In this proof, the existential elimination rule (the line labeled 3) is used to cancel two hypotheses at the same time. Note that when this rule is applied, the hypothesis ∀x(A(x)→¬B(x)) has not yet been canceled. So we have to make sure that this formula doesn’t contain the variable x freely. But this is o.k., since this hypothesis contains x only as a bound variable.
Another example is that if x does not occur in P, then ∃x P is equivalent to P:
This is short but tricky, so let us go through it carefully. On the left, we assume ∃x P to conclude P. We assume P, and now we can immediately cancel this assumption by existential elimination, since x does not occur in P, so it doesn’t occur freely in any assumption or in the conclusion. On the right we use existential introduction to conclude ∃x P from P.
8.4. Equality¶
Recall the natural deduction rules for equality:
Keep in mind that we have implicitly fixed some first-order language, and r, s, and t are any terms in that language. Recall also that we have adopted the practice of using functional notation with terms. For example, if we think of r(x) as the term (x+y)×(z+0) in the language of arithmetic, then r(0) is the term (0+y)×(z+0) and r(u+v) is ((u+v)+y)×(z+0). So one example of the first inference on the second line is this:
The second axiom on that line is similar, except now P(x) stands for any formula, as in the following inference:
Notice that we have written the reflexivity axiom, t=t, as a rule with no premises. If you use it in a proof, it does not count as a hypothesis; it is built into the logic.
In fact, we can think of the first inference on the second line as a special case of the second one. Consider, for example, the formula ((u+v)+y)×(z+0)=(x+y)×(z+0). If we plug u+v in for x, we get an instance of reflexivity. If we plug in 0, we get the conclusion of the first example above. The following is therefore a derivation of the first inference, using only reflexivity and the second substitution rule above:
Roughly speaking, we are replacing the second instance of u+v in an instance of reflexivity with 0 to get the conclusion we want.
Equality rules let us carry out calculations in symbolic logic. This typically amounts to using the equality rules we have already discussed, together with a list of general identities. For example, the following identities hold for any real numbers x, y, and z:
commutativity of addition: x+y=y+x
associativity of addition: (x+y)+z=x+(y+z)
additive identity: x+0=0+x=x
additive inverse: −x+x=x+−x=0
multiplicative identity: x⋅1=1⋅x=x
commutativity of multiplication: x⋅y=y⋅x
associativity of multiplication: (x⋅y)⋅z=x⋅(y⋅z)
distributivity: x⋅(y+z)=x⋅y+x⋅z,(x+y)⋅z=x⋅z+y⋅z
You should imagine that there are implicit universal quantifiers in front of each statement, asserting that the statement holds for any values of x, y, and z. Note that x, y, and z can, in particular, be integers or rational numbers as well. Calculations involving real numbers, rational numbers, or integers generally involve identities like this.
The strategy is to use the elimination rule for the universal quantifier to instantiate general identities, use symmetry, if necessary, to orient an equation in the right direction, and then using the substitution rule for equality to change something in a previous result. For example, here is a natural deduction proof of a simple identity, ∀x,y,z((x+y)+z=(x+z)+y), using only commutativity and associativity of addition. We have taken the liberty of using a brief name to denote the relevant identities, and combining multiple instances of the universal quantifier introduction and elimination rules into a single step.
There is generally nothing interesting to be learned from carrying out such calculations in natural deduction, but you should try one or two examples to get the hang of it, and then take pleasure in knowing that it is possible.
8.5. Counterexamples and Relativized Quantifiers¶
Consider the statement:
Every prime number is odd.
In first-order logic, we could formulate this as ∀p(p r i m e(p)→o d d(p)). This statement is false, because there is a prime number that is even, namely the number 2. This is called a counterexample to the statement.
More generally, given a formula ∀x A(x), a counterexample is a value t such that ¬A(t) holds. Such a counterexample shows that the original formula is false, because we have the following equivalence: ¬∀x A(x)↔∃x¬A(x). So if we find a value t such that ¬A(t) holds, then by the existential introduction rule we can conclude that ∃x¬A(x), and then by the above equivalence we have ¬∀x A(x). Here is a proof of the equivalence:
One remark about the proof: at the step marked by 4 we cannot use the existential introduction rule, because at that point our only assumption is ¬∀x A(x), and from that assumption we cannot prove ¬A(t) for a particular term t. So we use a proof by contradiction there.
As an exercise, prove the “dual” equivalence yourself: ¬∃x A(x)↔∀x¬A(x). This can be done without using proof by contradiction.
In Chapter 7 we saw examples of how to use relativization to restrict the scope of a universal quantifier. Suppose we want to say “every prime number is greater than 1”. In first order logic this can be written as ∀n(p r i m e(n)→n>1). The reason is that the original statement is equivalent to the statement “for every natural number, if it is prime, then it is greater than 1”. Similarly, suppose we want to say “there exists a prime number greater than 100.” This is equivalent to saying “there exists a natural number which is prime and greater than 100,” which can be expressed as ∃n(p r i m e(n)∧n>100).
As an exercise you can prove the above results about negations of quantifiers also for relativized quantifiers. Specifically, prove the following statements:
¬∃x(A(x)∧B(x))↔∀x(A(x)→¬B(x))
¬∀x(A(x)→B(x))↔∃x(A(x)∧¬B(x))
For reference, here is a list of valid sentences involving quantifiers:
∀x A↔A if x is not free in A
∃x A↔A if x is not free in A
∀x(A(x)∧B(x))↔∀x A(x)∧∀x B(x)
∃x(A(x)∧B)↔∃x A(x)∧B if x is not free in B
∃x(A(x)∨B(x))↔∃x A(x)∨∃x B(x)
∀x(A(x)∨B)↔∀x A(x)∨B if x is not free in B
∀x(A(x)→B)↔(∃x A(x)→B) if x is not free in B
∃x(A(x)→B)↔(∀x A(x)→B) if x is not free in B
∀x(A→B(x))↔(A→∀x B(x)) if x is not free in A
∃x(A(x)→B)↔(A(x)→∃x B) if x is not free in B
∃x A(x)↔¬∀x¬A(x)
∀x A(x)↔¬∃x¬A(x)
¬∃x A(x)↔∀x¬A(x)
¬∀x A(x)↔∃x¬A(x)
All of these can be derived in natural deduction. The last two allow us to push negations inwards, so we can continue to put first-order formulas in negation normal form. Other rules allow us to bring quantifiers to the front of any formula, though, in general, there will be multiple ways of doing this. For example, the formula
∀x A(x)→∃y∀z B(y,z)
is equivalent to both
∃x,y∀z(A(x)→B(y,z))
and
∃y∀z∃x(A(x)→B(y,z)).
A formula with all the quantifiers in front is said to be in prenex form.
8.6. Exercises¶
Give a natural deduction proof of
∀x(A(x)→B(x))→(∀x A(x)→∀x B(x)).
2. Give a natural deduction proof of ∀x B(x) from hypotheses ∀x(A(x)∨B(x)) and ∀y¬A(y).
From hypotheses ∀x(e v e n(x)∨o d d(x)) and ∀x(o d d(x)→e v e n(s(x))) give a natural deduction proof ∀x(e v e n(x)∨e v e n(s(x))). (It might help to think of s(x) as the function defined by s(x)=x+1.)
Give a natural deduction proof of ∃x A(x)∨∃x B(x)→∃x(A(x)∨B(x)).
Give a natural deduction proof of ∃x(A(x)∧C(x)) from the assumptions ∃x(A(x)∧B(x)) and ∀x(A(x)∧B(x)→C(x)).
Prove some of the other equivalences in the last section.
Consider some of the various ways of expressing “nobody trusts a politician” in first-order logic:
∀x(p o l i t i c i a n(x)→∀y(¬t r u s t s(y,x)))
∀x,y(p o l i t i c i a n(x)→¬t r u s t s(y,x))
¬∃x,y(p o l i t i c i a n(x)∧t r u s t s(y,x))
∀x,y(t r u s t s(y,x)→¬p o l i t i c i a n(x))
They are all logically equivalent. Show this for the second and the fourth, by giving natural deduction proofs of each from the other. (As a shortcut, in the ∀ introduction and elimination rules, you can introduce / eliminate both variables in one step.)
Formalize the following statements, and give a natural deduction proof in which the first three statements appear as (uncancelled) hypotheses, and the last line is the conclusion:
Every young and healthy person likes baseball.
Every active person is healthy.
Someone is young and active.
Therefore, someone likes baseball.
Use Y(x) for “is young,” H(x) for “is healthy,” A(x) for “is active,” and B(x) for “likes baseball.”
Give a natural deduction proof of ∀x,y,z(x=z→(y=z→x=y)) using the equality rules in Section 8.4.
Give a natural deduction proof of ∀x,y(x=y→y=x) using only these two hypotheses (and none of the new equality rules):
∀x(x=x)
∀u,v,w(u=w→(v=w→u=v))
(Hint: Choose instantiations of u, v, and w carefully. You can instantiate all the universal quantifiers in one step, as on the last homework assignment.)
Give a natural deduction proof of ¬∃x(A(x)∧B(x))↔∀x(A(x)→¬B(x))
Give a natural deduction proof of ¬∀x(A(x)→B(x))↔∃x(A(x)∧¬B(x))
Remember that both the following express ∃!x A(x), that is, the statement that there is a unique x satisfying A(x):
∃x(A(x)∧∀y(A(y)→y=x))
∃x A(x)∧∀y∀y′(A(y)∧A(y′)→y=y′)
Do the following:
Give a natural deduction proof of the second, assuming the first as a hypothesis.
Give a natural deduction proof of the first, assuming the second as a hypothesis.
(Warning: these are long.)
Logic and Proof
1. Introduction
2. Propositional Logic
3. Natural Deduction for Propositional Logic
4. Propositional Logic in Lean
5. Classical Reasoning
6. Semantics of Propositional Logic
7. First Order Logic
8. Natural Deduction for First Order Logic
8.1. Rules of Inference
8.2. The Universal Quantifier
8.3. The Existential Quantifier
8.4. Equality
8.5. Counterexamples and Relativized Quantifiers
8.6. Exercises
9. First Order Logic in Lean
10. Semantics of First Order Logic
11. Sets
12. Sets in Lean
13. Relations
14. Relations in Lean
15. Functions
16. Functions in Lean
17. The Natural Numbers and Induction
18. The Natural Numbers and Induction in Lean
19. Elementary Number Theory
20. Combinatorics
21. The Real Numbers
22. The Infinite
23. Axiomatic Foundations
24. Appendix: Natural Deduction Rules
Lean Home
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8174 | https://brilliant.org/wiki/hockey-stick-identity/ | Hockey Stick Identity
Andy Hayes,
Rohit Udaiwal,
Mateo Matijasevick,
and
Worranat Pakornrat
Eli Ross
Jimin Khim
contributed
The hockey stick identity is an identity regarding sums of binomial coefficients.
For whole numbers n and r (n≥r),
k=r∑n(rk)=(r+1n+1). □
The hockey stick identity gets its name by how it is represented in Pascal's triangle.
In Pascal's triangle, the sum of the elements in a diagonal line starting with 1 is equal to the next element down diagonally in the opposite direction. Circling these elements creates a "hockey stick" shape:
1+3+6+10=20.
The hockey stick identity is a special case of Vandermonde's identity. It is useful when a problem requires you to count the number of ways to select the same number of objects from different-sized groups. It is also useful in some problems involving sums of powers of natural numbers.
Contents
Counting Problems
Triangular Numbers
Sums of Powers of Natural Numbers
Proofs
Counting Problems
The hockey stick identity is often used in counting problems in which the same amount of objects is selected from different-sized groups.
Treating each of the balls in the figure below as distinct, how many ways are there to select 3 balls from the same horizontal row?
The smallest row has 3 balls and the largest row has 9 balls. The number of ways to select 3 balls from the same row can be expressed as a sum of binomial coefficients. This can then be computed with the hockey stick identity:
k=3∑9(3k)=(410)=210.
So, there are 210 ways to select 3 balls from the same row. □
Mabel is playing a carnival game in which she throws balls into a triangular array of cups.
She will throw three balls, and she will win the game if the three balls are in a straight line (not necessarily adjacent) and in different cups.
Disregarding the order the balls go into the cups, how many ways can she win the game?
The correct answer is: 114
In how many different ways can 4 squares be chosen on a chessboard such that the 4 squares lie in a diagonal line?
Image Credit: Wikipedia Commons
The correct answer is: 364
Triangular Numbers
You might have noticed that Pascal's triangle contains all of the positive integers in a diagonal line.
Each of these elements corresponds to the binomial coefficient (1n), where n is the row of Pascal's triangle. The sum of all positive integers up to n is called the nth triangular number. It can be represented as
k=1∑nk=k=1∑n(1k).
These two expressions are equivalent because k=(1k). As this sum can be expressed as the sum of binomial coefficients, it can be computed with the hockey stick identity:
The sum of the first n positive integers is
k=1∑nk=k=1∑n(1k)=(2n+1). □
This leads to the more well-known formula for triangular numbers.
The sum of the first n positive integers is
k=1∑nk=(2n+1)=(n−1)!(2)!(n+1)!=2n(n+1). □
Since each triangular number can be represented with a binomial coefficient, the hockey stick identity can be used to calculate the sum of triangular numbers.
The sum of the first n triangular numbers is
k=1∑nj=1∑kj=k=1∑n(2k+1)=(3n+2). □
This can also be represented algebraically.
The sum of the first n triangular numbers is
k=1∑nj=1∑kj=(3n+2)=(n−1)!(3)!(n+2)!=6n(n+1)(n+2). □
The oranges are arranged such that there is 1 top orange; the second top layer has 2 more oranges than the top; the third has 3 more oranges than the second, and so on. Forming a tetrahedron of oranges, these "tetrahedral" numbers of oranges run as a series, as shown above.
What is the value of the 100th term of this series?
The correct answer is: 171700
I drew rectangular grids with one more row than their column. The number of dots in the first four grids are 2, 6, 12, and 20, as shown in the diagram below:
What is the total number of dots used in the first eleven grids?
The correct answer is: 572
Sums of Powers of Natural Numbers
The hockey stick identity can be used to develop the identities for sums of powers of natural numbers.
The sum of the squares of the first n positive integers is
k=1∑nk2=6n(n+1)(2n+1)=2(3n+2)−(2n+1). □
Recall from the previous example that the identities for the sum of the first n positive integers is
k=1∑nk=(2n+1)=2n(n+1).
The sum of the first n triangular numbers can be expressed as
k=1∑n2k(k+1)=21k=1∑nk2+21k=1∑nk.
The sum of the first n triangular numbers was found previously using the hockey stick identity:
k=1∑n2k(k+1)=6n(n+1)(n+2).
Substituting these identities, the identity for the sum of squares of the first n positive integers can be developed:
6n(n+1)(n+2)21k=1∑nk2k=1∑nk2=21k=1∑nk2+21(2n(n+1))=6n(n+1)(n+2)−4n(n+1)=6n(n+1)(2n+1).
This can also be written in terms of the binomial coefficient:
k=1∑nk2=2(3n+2)−(2n+1). □
This method can be continued indefinitely to develop an identity for the sum of any power of natural numbers.
The sum of the cubes of the first n natural numbers is
k=1∑nk3=4n2(n+1)2=6(4n+3)−6(3n+2)+(2n+1). □
Consider the previous identity for the sum of squares of positive integers:
k=1∑nk2=6n(n+1)(2n+1)=2(3n+2)−(2n+1).
Now consider the sum of the sum of squares of positive integers:
k=1∑nj=1∑kk2=k=1∑n6k(k+1)(2k+1)=31k=1∑nk3+21k=1∑nk2+61k=1∑nk.
This sum can be alternatively computed using binomial coefficients and the hockey stick identity:
k=1∑nj=1∑kk2=k=1∑n[2(3k+2)−(2k+1)]=2(4n+3)−(3n+2)=12n(n+1)(n+2)(n+3)−6n(n+1)(n+2)=12n(n+1)2(n+2).
Substituting these identities gives
12n(n+1)2(n+2)k=1∑nk3=31k=1∑nk3+12n(n+1)(2n+1)+12n(n+1)=31k=1∑nk3+122n(n+1)2=4n2(n+1)2.
This can also be expressed with binomial coefficients:
k=1∑nk3=6(4n+3)−6(3n+2)+(2n+1). □
For the past decade, the king of Mathlandia has forced his subjects to build a pyramid in his honor. The king decreed the pyramid to be constructed with cubic stone slabs. The king also decreed the pyramid to be built in 100 square levels, with each subsequent level 2 units less on a side than the previous level. The top level was to be constructed with a single cube. (See the picture for an example of a pyramid 3 levels high constructed in the same way).
Moments after the final cube was placed, the king changed his mind. He ordered the pyramid to be taken down, and in its place, a cubic monolith was to be built. He ordered the monolith to be built as large as possible with the same stone slabs the pyramid was made of.
The king would tolerate no waste, so he ordered one of his subjects to be sacrificed for each leftover slab of stone.
How many of the king's subjects will be sacrificed?
The correct answer is: 2300
Proofs
Inductive Proof of Hockey Stick Identity:
Base Case: r=n
We have
k=n∑n(nk)=(nn)(n+1n+1)=1=1.
Inductive Step:
Suppose that for whole numbers n and r (n≥r),
k=r∑n(rk)=(r+1n+1).
k=r∑n+1(rk)=(r+1n+1)+(rn+1)=(n−r)!(r+1)!(n+1)!+(n−r+1)!r!(n+1)!=(n−r+1)!(r+1)!(n−r+1)(n+1)!+(n−r+1)!(r+1)!(r+1)(n+1)!=(n−r+1)!(r+1)!(n+2)!=(r+1n+2). □
Combinatorial Proof using Identical Objects into Distinct Bins
Imagine that there are m identical objects to be distributed into q distinct bins such that some bins can be left empty. Using the stars and bars approach outlined on the linked wiki page above, this can be done in (q−1m+q−1) ways.
Now consider a slightly different approach to compute this same result. Distribute j objects among the first q−1 bins, and then distribute the remaining m−j objects into the last bin. This can be done in (q−2j+q−2) ways for each value of j. Count all of the distributions among all possible values of j up to m:
j=0∑m(q−2j+q−2).
These two methods for counting the distributions of m identical objects into q bins are equivalent, so the expressions which give the results are equal:
j=0∑m(q−2j+q−2)=(q−1m+q−1).
Let k=j+q−2, let r=q−2, and let n=m+q−2. Substituting these variables in the identity above gives the hockey stick identity:
k=r∑n(rk)=(r+1n+1). □
Cite as:
Hockey Stick Identity.
Brilliant.org.
Retrieved
02:34, August 22, 2025,
from |
8175 | https://artofproblemsolving.com/wiki/index.php/Double_angle_identities?srsltid=AfmBOoqECKB3gOCV7PVIcqhmutsiQMC8n5MtNtn9Hgwin8P9kRFOSUl6 | Art of Problem Solving
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Double angle identities
The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting . Doing so yields:
This article is a stub. Help us out by expanding it.
Proof
Please try to prove it on your own.
See Also
Trigonometric identities
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8176 | https://math-master.org/tutorials/finding-the-area-of-a-parallelogram | MathMaster Blog
Home
tutorials
finding the area of a parallelogram
A parallelogram is a four-sided, two-dimensional figure with the following properties:
two equal and opposing sides
two intersecting but non-equal diagonals, and
two equal and opposite angles
1:
A parallelogram's area is calculated by multiplying its base by its altitude. The base and altitude of a parallelogram are perpendicular to each other.
2:
If the length of neighboring sides and the angle between them are known, the area of a parallelogram can be determined without the height. In this case, we can use the area of the triangle formula from the trigonometry concept:
Area = ab sin θ
Where:
a and b = length of parallel sides, and,
θ = angle between the sides of the parallelogram
3:
The length of a parallelogram's diagonals can also be used to calculate its area.
Area = ½ x d1x d2sinx
Where:
d1 and d2 = length of the diagonals, and,
x = angle between the diagonals
Example 1:
Calculate the area of a parallelogram with 18 cm and 15 cm diagonals and a 43° angle of intersection between them.
Solution:
Let d1 = 18 cm and d2 = 15 cm.
β = 43°
A = ½ x d1 x d2 sine β
= ½ x 18 x 15 sine 43°
= 135 sine 43°
= 92.07 cm2
Answer: The area of the parallelogram is 92.07 cm^2
Example 2:
Calculate the height of a parallelogram with parallel sides of 30 cm and 40 cm and a 36-degree angle between them. Assume that the parallelogram's base is 40 cm.
Solution:
Area = ab sine a = bh
30 x 40 sine36 = 40 x h
1.200 sine36 = 40 x h
Divide both sides by 40.
h = 1200/40 sine 36
= 30 sine 36
h = 1200/40 sine 36
= 30 sine 36
h = 17.63 cm
Answer: The height of the parallelogram is 17.63 cm.
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\r\n\u003Cp>Where:\u003C/p>\r\n\u003Cp>a and b = length of parallel sides, and,\u003C/p>\r\n\u003Cp>θ = angle between the sides of the parallelogram\u003C/p>\r\n\u003Cspan class=\"step-tutor\">#3:\u003C/span>\r\n\u003Cp>The length of a parallelogram's diagonals can also be used to calculate its area.\u003C/p>\r\n\u003Cp>Area = ½ x d1x d2sinx\u003C/p>\r\n\u003Cp>Where:\u003C/p>\r\n\u003Cp>d1 and d2 = length of the diagonals, and,\u003C/p>\r\n\u003Cp>x = angle between the diagonals\u003C/p>\r\n\u003Ch2>Example 1:\u003C/h2>\r\n\u003Cp>Calculate the area of a parallelogram with 18 cm and 15 cm diagonals and a 43° angle of intersection between them.\u003C/p>\r\n\u003Ch3>Solution:\u003C/h3>\r\n\u003Cp>Let d1 = 18 cm and d2 = 15 cm.\u003C/p>\r\n\u003Cp>β = 43°\u003C/p>\r\n\u003Cp>A = ½ x d1 x d2 sine β\u003C/p>\r\n\u003Cp>= ½ x 18 x 15 sine 43°\u003C/p>\r\n\u003Cp>= 135 sine 43°\u003C/p>\r\n\u003Cp>= 92.07 cm2\u003C/p>\r\n\u003Cp>Answer: The area of the parallelogram is 92.07 cm^2\u003C/p>\r\n\u003Ch2>Example 2:\u003C/h2>\r\n\u003Cp>Calculate the height of a parallelogram with parallel sides of 30 cm and 40 cm and a 36-degree angle between them. Assume that the parallelogram's base is 40 cm.\u003C/p>\r\n\u003Ch3>Solution:\u003C/h3>\r\n\u003Cp>Area = ab sine a = bh\u003C/p>\r\n\u003Cp>30 x 40 sine36 = 40 x h\u003C/p>\r\n\u003Cp>1.200 sine36 = 40 x h\u003C/p>\r\n\u003Cp>Divide both sides by 40.\u003C/p>\r\n\u003Cp>h = 1200/40 sine 36\u003C/p>\r\n\u003Cp>= 30 sine 36\u003C/p>\r\n\u003Cp>h = 1200/40 sine 36\u003C/p>\r\n\u003Cp>= 30 sine 36\u003C/p>\r\n\u003Cp>h = 17.63 cm\u003C/p>\r\n\u003Cp>Answer: The height of the parallelogram is 17.63 cm.\u003C/p>",{"data":157},[158,162,166,170,174,178,182,186,190,194],{"id":159,"post_title":160,"post_slug":161},213,"Top 5 Hardest ACT Math Questions","top-5-hardest-act-math-questions",{"id":163,"post_title":164,"post_slug":165},234,"What Are Composite Numbers?","what-are-composite-numbers",{"id":167,"post_title":168,"post_slug":169},220,"5 Best Math Apps For Learning Math","5-best-math-apps-for-learning-math",{"id":171,"post_title":172,"post_slug":173},222,"7 Tricky Fun Math Questions","7-tricky-fun-math-questions",{"id":175,"post_title":176,"post_slug":177},231,"What Is Mean Average?","what-is-mean-average",{"id":179,"post_title":180,"post_slug":181},229,"What Are Vertices, Faces, and Edges? 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Curve fitting: Find the smoothest function that satisfies a list of constraints
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Consider the set of non-decreasing surjective (onto) functions from (-inf,inf) to [0,1].
(Typical CDFs satisfy this property.)
In other words, for any real number x, 0 <= f(x) <= 1.
The logistic function is perhaps the most well-known example.
We are now given some constraints in the form of a list of x-values and for each x-value, a pair of y-values that the function must lie between.
We can represent that as a list of {x,ymin,ymax} triples such as
```
constraints = {{0, 0, 0}, {1, 0.00311936, 0.00416369}, {2, 0.0847077, 0.109064},
{3, 0.272142, 0.354692}, {4, 0.53198, 0.646113}, {5, 0.623413, 0.743102},
{6, 0.744714, 0.905966}}
```
Graphically that looks like this:
(source: yootles.com)
We now seek a curve that respects those constraints.
For example:
(source: yootles.com)
Let's first try a simple interpolation through the midpoints of the constraints:
```
mids = ({#1, Mean[{#2,#3}]}&) @@@ constraints
f = Interpolation[mids, InterpolationOrder->0]
```
Plotted, f looks like this:
(source: yootles.com)
That function is not surjective. Also, we'd like it to be smoother.
We can increase the interpolation order but now it violates the constraint that its range is [0,1]:
(source: yootles.com)
The goal, then, is to find the smoothest function that satisfies the constraints:
Non-decreasing.
Tends to 0 as x approaches negative infinity and tends to 1 as x approaches infinity.
Passes through a given list of y-error-bars.
The first example I plotted above seems to be a good candidate but I did that with Mathematica's FindFit function assuming a lognormal CDF.
That works well in this specific example but in general there need not be a lognormal CDF that satisfies the constraints.
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edited Jul 24, 2019 at 7:34
Glorfindel
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asked Apr 23, 2010 at 22:52
dreevesdreeves
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I guess a CDF is a Cumulative Distribution Function.
– brainjam
Commented
Apr 24, 2010 at 0:36
2
Because you have ranges I feel there is a way to do this with a finite sum of arctan functions (which would of course be infinitely differentiable). If someone else wants to run with this answer feel free; otherwise I'll see if inspiration strikes tomorrow. While such an answer would strictly speaking be smoother than the cubic splines, it may satisfy the word of smoothness and not the spirit. That is it will look crassly stepwise.
– Steven Noble
Commented
Apr 24, 2010 at 6:45
1
I think you are missing an important point. The error bars are indicators of certainty and by just taking the mean value A as the point to be fit you are throwing away a lot of information. The fitting algorithm will treat deviation from A and B (another mean value) on equal basis, even if the error for B is a much larger than for A. E.g. your points at x=0,1,2 should be given much higher priority to be "on the curve" than the other points. Most Mathematica fitting functions have the option Weights which I would use to assign the inverse of the error interval to each mean point as weight.
– Timo
Commented
Apr 24, 2010 at 10:04
1
Damns! Read your post again and they are not error bars but constraints.
– Timo
Commented
Apr 24, 2010 at 10:07
3
I think I'll pass on providing a full write up of my arctan solution simply because I quite like ~unutbu's interpolation solution (particularly with the transform). You may want to check out blog.noblemail.ca/2009/03/… where I've implemented some locally monotonic splines like this in js. Perhaps it will inspire some mathematica code.
– Steven Noble
Commented
Apr 24, 2010 at 21:17
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I don't think you've specified enough criteria to make the desired CDF unique.
If the only criteria that must hold is:
CDF must be "fairly smooth" (see below)
CDF must be non-decreasing
CDF must pass through the "error bar" y-intervals
CDF must tend toward 0 as x --> -Infinity
CDF must tend toward 1 as x --> Infinity.
then perhaps you could use Monotone Cubic Interpolation.
This will give you a C^2 (twice continously differentiable) function which,
unlike cubic splines, is guaranteed to be monotone when given monotone data.
This leaves open the question, exactly what data should you use to generate the
monotone cubic interpolation. If you take the center point (mean) of each error
bar, are you guaranteed that the resulting data points are monotonically
increasing? If not, you might as well make some arbitrary choice to guarantee
that the points you select are monotonically increasing (because the criteria does not force our solution to be unique).
Now what to do about the last data point? Is there an X which is guaranteed to
be larger than any x in the constraints data set? Perhaps you can again make an
arbitrary choice of convenience and pick some very large X and put (X,1) as the
final data point.
Comment 1: Your problem can be broken into 2 sub-problems:
Given exact points (x_i,y_i) through which the CDF must pass, how do you generate CDF? I suspect there are infinitely many possible solutions, even with the infinite-smoothness constraint.
Given y-errorbars, how should you pick (x_i,y_i)? Again, there infinitely many possible solutions. Some additional criteria may need to be added to force a unique choice. Additional criteria would also probably make the problem even harder than it currently is.
Comment 2: Here is a way to use monotonic cubic interpolation, and satisfy criteria 4 and 5:
The monotonic cubic interpolation (let's call it f) maps R --> R.
Let CDF(x) = exp(-exp(f(x))). Then CDF: R --> (0,1). If we could find the appropriate f, then by defining CDF this way, we could satisfy criteria 4 and 5.
To find f, transform the CDF constraints (x_0,y_0),...,(x_n,y_n) using the transformation xhat_i = x_i, yhat_i = log(-log(y_i)). This is the inverse of the CDF transformation. If the y_i's were increasing, then the yhat_i's are decreasing.
Now apply monotone cubic interpolation to the (x_hat,y_hat) data points to generate f. Then finally, define CDF(x) = exp(-exp(f(x))). This will be a monotonically increasing function from R --> (0,1), which passes through the points (x_i,y_i).
This, I think, satisfies all the criteria 2--5. Criteria 1 is somewhat satisfied, though there certainly could exist smoother solutions.
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edited Apr 24, 2010 at 18:07
answered Apr 24, 2010 at 2:12
unutbuunutbu
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Thanks for this answer! I didn't know about Monotone Cubic Interpolation. If anyone knows of a Mathematica implementation of that, please post it as an answer. You're right that the criteria do not uniquely determine a CDF; thanks for clarifying what the criteria are! (Btw, your #4 is actually subsumed by #3 -- note the first degenerate error bar in my list of example constraints.)
– dreeves
Commented
Apr 24, 2010 at 3:38
As for your proposed hack for criterion 5: I tried that with Mathematica's normal cubic interpolation (not enforcing monotonicity, though that didn't seem to be an issue) and didn't find it satisfying. I don't want the interpolating function to exceed 1 for any X. True, X can be made big enough that it doesn't matter in practice, but then the function doesn't go to 1 quickly enough.
– dreeves
Commented
Apr 24, 2010 at 3:38
What if we stated the problem as "find the smoothest possible non-decreasing function from (-inf,inf) to [0,1] such that it passes through a given list of error bars"? Maybe I'll edit the question to put it that way.
– dreeves
Commented
Apr 24, 2010 at 3:45
I've now reworked the question, thanks in part to your answer.
– dreeves
Commented
Apr 24, 2010 at 4:31
Wow, your transformation idea in Comment 2 is really clever. I hadn't seen that when I wrote my solution (now posted as an answer and perhaps sufficient for my immediate purposes). Thanks again for all the help on this!
– dreeves
Commented
Apr 24, 2010 at 21:38
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I have found a solution that gives reasonable results for a variety of inputs.
I start by fitting a model -- once to the low ends of the constraints, and again to the high ends.
I'll refer to the mean of these two fitted functions as the "ideal function".
I use this ideal function to extrapolate to the left and to the right of where the constraints end, as well as to interpolate between any gaps in the constraints.
I compute values for the ideal function at regular intervals, including all the constraints, from where the function is nearly zero on the left to where it's nearly one on the right.
At the constraints, I clip these values as necessary to satisfy the constraints.
Finally, I construct an interpolating function that goes through these values.
My Mathematica implementation follows.
First, a couple helper functions:
```
( Distance from x to the nearest member of list l. )
listdist[x_, l_List] := Min[Abs[x - #] & /@ l]
( Return a value x for the variable var such that expr/.var->x is at least (or
at most, if dir is -1) t. )
invertish[expr_, var_, t_, dir_:1] := Module[{x = dir},
While[dir(expr /. var -> x) < dirt, x = 2];
x]
```
And here's the main function:
```
( Return a non-decreasing interpolating function that maps from the
reals to [0,1] and that is as close as possible to expr[var] without
violating the given constraints (a list of {x,ymin,ymax} triples).
The model, expr, will have free parameters, params, so first do a
model fit to choose the parameters to satisfy the constraints as well
as possible. )
cfit[constraints_, expr_, params_, var_] :=
Block[{xlist,bots,tops,loparams,hiparams,lofit,hifit,xmin,xmax,gap,aug,bests},
xlist = First /@ constraints;
bots = Most /@ constraints; ( bottom points of the constraints )
tops = constraints /. {x_, , ymax} -> {x, ymax};
( fit a model to the lower bounds of the constraints, and
to the upper bounds )
loparams = FindFit[bots, expr, params, var];
hiparams = FindFit[tops, expr, params, var];
lofit[z_] = (expr /. loparams /. var -> z);
hifit[z_] = (expr /. hiparams /. var -> z);
( find x-values where the fitted function is very close to 0 and to 1 )
{xmin, xmax} = {
Min@Append[xlist, invertish[expr /. hiparams, var, 10^-6, -1]],
Max@Append[xlist, invertish[expr /. loparams, var, 1-10^-6]]};
( the smallest gap between x-values in constraints )
gap = Min[(#2 - #1 &) @@@ Partition[Sort[xlist], 2, 1]];
( augment the constraints to fill in any gaps and extrapolate so there are
constraints everywhere from where the function is almost 0 to where it's
almost 1 )
aug = SortBy[Join[constraints, Select[Table[{x, lofit[x], hifit[x]},
{x, xmin,xmax, gap}],
listdist[#,xlist]>gap&]], First];
( pick a y-value from each constraint that is as close as possible to
the mean of lofit and hifit )
bests = ({#1, Clip[(lofit[#1] + hifit[#1])/2, {#2, #3}]} &) @@@ aug;
Interpolation[bests, InterpolationOrder -> 3]]
```
For example, we can fit to a lognormal, normal, or logistic function:
```
g1 = cfit[constraints, CDF[LogNormalDistribution[mu,sigma], z], {mu,sigma}, z]
g2 = cfit[constraints, CDF[NormalDistribution[mu,sigma], z], {mu,sigma}, z]
g3 = cfit[constraints, 1/(1 + cExp[-kz]), {c,k}, z]
```
Here's what those look like for my original list of example constraints:
(source: yootles.com)
The normal and logistic are nearly on top of each other and the lognormal is the blue curve.
These are not quite perfect.
In particular, they aren't quite monotone.
Here's a plot of the derivatives:
```
Plot[{g1'[x], g2'[x], g3'[x]}, {x, 0, 10}]
```
(source: yootles.com)
That reveals some lack of smoothness as well as the slight non-monotonicity near zero.
I welcome improvements on this solution!
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edited Jun 20, 2020 at 9:12
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answered Apr 24, 2010 at 21:24
dreevesdreeves
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I noticed the obvious smoothness problems go away when I use the option Method->"Spline" in the Interpolation call. I don't yet know how to fix the monotonicity problem though.
– dreeves
Commented
Apr 24, 2010 at 21:58
Actually, just setting InterpolationOrder->1 (linear interpolation) in the Interpolation call solves the monotonicity problem but of course makes the curve much less smooth. It's still roughly the same shape though and could be improved to an arbitrary degree by increasing the sampling granularity of the fitted function before interpolating.
– dreeves
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Apr 24, 2010 at 22:36
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You can try to fit a Bezier curve through the midpoints. Specifically I think you want a C2 continuous curve.
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answered Apr 24, 2010 at 2:04
i_am_jorfi_am_jorf
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Thanks jeff. I think the tricky part is what I've now listed as criterion 2 in my revised version of the question, or what unutbu lists as criterion 5: the function should tend to 1 as x goes to infinity.
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Le Chatelier's Principle
Equilibrium Temperature
Haber Process
Chemical Processes
Chemical Equilibrium
Physical Chemistry
Chemical Reactions
5
How is the haber process an example to explain that temperature affects the position of the equilibrium?
·
The Haber process is a key industrial method for synthesizing ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. The reaction can be represented as follows:
N2(g)+3H2(g)⇌2NH3(g)
This reaction is exothermic, meaning it releases heat. According to Le Chatelier's principle, if a system at equilibrium experiences a change in temperature, pressure, or concentration, the equilibrium will shift in a direction that counteracts the change.
Effect of Temperature on Equilibrium
Exothermic Reaction: Since the Haber process is exothermic, increasing the temperature shifts the equ
The Haber process is a key industrial method for synthesizing ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. The reaction can be represented as follows:
N2(g)+3H2(g)⇌2NH3(g)
This reaction is exothermic, meaning it releases heat. According to Le Chatelier's principle, if a system at equilibrium experiences a change in temperature, pressure, or concentration, the equilibrium will shift in a direction that counteracts the change.
Effect of Temperature on Equilibrium
Exothermic Reaction: Since the Haber process is exothermic, increasing the temperature shifts the equilibrium to the left (toward the reactants, N₂ and H₂). This is because the system tries to absorb the added heat by favoring the endothermic direction (the reactants).
Lowering Temperature: Conversely, lowering the temperature shifts the equilibrium to the right (toward the products, NH₃). This is because the system releases heat, thereby favoring the exothermic reaction that produces ammonia.
Practical Implications
Optimal Conditions: In industrial applications, the Haber process is typically carried out at moderate temperatures (around 450°C) and high pressures (about 200 atm). This balance allows for a reasonable rate of reaction while maximizing the yield of ammonia.
Yield vs. Rate: While lower temperatures favor ammonia production, they also slow down the reaction rate. Therefore, a compromise is made to achieve a good yield of ammonia in a practical timeframe.
The Haber process illustrates how temperature affects the position of equilibrium in chemical reactions. By understanding this principle, chemists can optimize conditions for industrial processes to maximize product yield while maintaining efficiency.
Samuel Tella
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In the
Haber process
, every 3 moles of hydrogen gas will react with 1 mole of nitrigen gas to yield 2 moles of ammonia . The forward reaction is exothermic, while the reverse is endothermic.
Since the forward reaction is exothermic, decreasing the temperature will cause the equilibrium position to shift to the right, which will give a high yield of ammonia. An increase in temperature will cause the equilibrium position to shift to the left, i.e. it favours reactants formation.
In the
Haber process
, every 3 moles of hydrogen gas will react with 1 mole of nitrigen gas to yield 2 moles of ammonia . The forward reaction is exothermic, while the reverse is endothermic.
Since the forward reaction is exothermic, decreasing the temperature will cause the equilibrium position to shift to the right, which will give a high yield of ammonia. An increase in temperature will cause the equilibrium position to shift to the left, i.e. it favours reactants formation.
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The Haber process is a very important chemical reaction because it allows mankind to make fertilizer (and explosives) to grow more food (and reshape our environment).
The reaction in words is
reactants, nitrogen and hydrogen yield the products ammonia and heat
as an equilibrium reaction.
Normally to make a chemical reaction proceed faster, a chemist would raise the temperature.
However, now speaking anthropomorphically, the equilibrium reaction would not like the addition of heat by raising the temperature because it disturbs it's balance at any temperature, so the reaction tries to get rid of the
The Haber process is a very important chemical reaction because it allows mankind to make fertilizer (and explosives) to grow more food (and reshape our environment).
The reaction in words is
reactants, nitrogen and hydrogen yield the products ammonia and heat
as an equilibrium reaction.
Normally to make a chemical reaction proceed faster, a chemist would raise the temperature.
However, now speaking anthropomorphically, the equilibrium reaction would not like the addition of heat by raising the temperature because it disturbs it's balance at any temperature, so the reaction tries to get rid of the added heat (energy) . The equilibrium reaction is able to get rid of the heat by going in the reverse direction from the products which is exactly what we as humans do not want. We want the ammonia.
Fritz Haber was able to make the reaction work to produce ammonia at higher temperatures by increasing the pressure at which the reaction is operated at and he also added a catalyst to make the reaction go by a different route. The higher pressure of the reactants in this reaction causes the reaction, again speaking anthropomorphically, to get rid of the higher pressure of the 4 moles of reactant gases ( 1 N2 + 3 H2) by making 2 moles of product ammonia (2 NH3).
For this important work, Fritz Haber received the 1918 Nobel Prize in Chemistry, although not with some controversy.
Related questions
What is the quantitative effect of temperature on equilibrium?
At equilibrium, the concentration of the reactants and product is [NO] = 0.0542 M, =0.127 M and [N02] =15.5 M. What is the equilibrium constant Kc at this temperature?
If the temperature higher than 450c was used in the Haber process, what would happen to the rate of the reaction?
Why does the Haber process not reach equilibrium?
What is the effect of temperature on equilibrium?
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Why is equilibrium constant affected by the temperature but not affected by the concentration of reactants?
When a system is at equilibrium it looks as if nothing is going on: no change in macroscopic properties. At a microscopic level there is still a lot going on. Reactants are becoming Products in the forward reaction and Products are becoming Reactants in the reverse reaction.
The simple version of this system can be represented as:
Reactants = Products
Equilibrium is defined as the point in time when the rate of the forward reaction equals the rate of the reverse reaction:
Rate(f) = Rate(r).
The equilibrium constant is calculated by taking the ratio of the concentration of the participants:
Keq = [pr
When a system is at equilibrium it looks as if nothing is going on: no change in macroscopic properties. At a microscopic level there is still a lot going on. Reactants are becoming Products in the forward reaction and Products are becoming Reactants in the reverse reaction.
The simple version of this system can be represented as:
Reactants = Products
Equilibrium is defined as the point in time when the rate of the forward reaction equals the rate of the reverse reaction:
Rate(f) = Rate(r).
The equilibrium constant is calculated by taking the ratio of the concentration of the participants:
Keq = [products]/[reactants]
Messing with a system at equilibrium introduces what chemistry is all about: the ability to predict the outcome of an event before it happens. A French chemist named Henry Louis Le Chatelier (1850 - 1936) figured this out. His principle states:
If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change and re-establish equilibrium.
The change in condition can be all the usual suspects: concentration, temperature and pressure.
Take concentration for example: if you increase the concentration of a reactant or product participant in an equilibrium, the system will absorb some of the added material and given some time the system will reestabish equilibrium. The interesting part here is that when you recalculate Keq, you get the same value. Hence the term Equlibrium Constant. Knowing this allows us to predict changes in concentration of the players in a dynamic equlibrium when one of the participants changes.
Missing in most simple chemical equations is the heat term. All chemical reactions are either endo or exothermic. This means heat is a player in the system.
Reactants + heat = Products… Endothermic
Reactants = Products + heat… Exothermic
Le Chatelier predicts that increasing temperature will drive the reaction to absorb the added heat. So, in an endothermic system, when you add heat the reaction will absorb some of the heat, and shift to the right to reestabilsh equilibrium. That, of course, changes the concentration of the reactants and products and will produce a different (larger) Keq. So… Keq is temperature dependent.
The reverse is true for exothermic reactions. Adding heat to an exothermic system at equilibrium will cause the system to absorb some of the heat and shift to the reactant side. When concentrations are measured at the new equilibrium, we find a different (smaller) Keq.
Let’s talk about gases and solids. Solids don’t have a concentration, they have density. Density is a state property, it belongs to the substance and can not be changed. Therefore, solids do not play in Keq systems. Gases do have concentration and therefore belong in the Keq equation. You can increase the concentration of a gas by adding more of the gas or by changing the pressure of the system. Le Chatelier will predict the outcome. Changing pressure becomes a predictable thing and using Le Chatelier will allow you to predict the outcome.
Numerical values of Keq are always given at a specific T & P.
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N2(g)+3H2(g) 4volumes…..>2NH3(g)2volumes
Gay lussac's law 4 volumes produces 2 volumes of ammonia; the equilibrium will be favoured by high pressure and low temperature.
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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What is the effect of temperature on equilibrium?
The effect of temperature on a system at equilibrium is more rigorously discussed by the 1st Chemistry Nobel Laureate Van’t Hoff.
The pertinent equation is
log〖K_2/K_1 〗= H/2.303R ([T_2-T_1])/(T_1 T_2 )
Where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively whereas
is the Heat of Reaction (assumed to be constant over the temperature range T1 and T2) and R is the Gas Constant. Note that the temperatures shown are in Kelvin scale and are always positive.
The above equation is known as the Van’t Hoff’s Reaction Isochore. Note that how this is different from the Van’t H
The effect of temperature on a system at equilibrium is more rigorously discussed by the 1st Chemistry Nobel Laureate Van’t Hoff.
The pertinent equation is
log〖K_2/K_1 〗= H/2.303R ([T_2-T_1])/(T_1 T_2 )
Where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively whereas
is the Heat of Reaction (assumed to be constant over the temperature range T1 and T2) and R is the Gas Constant. Note that the temperatures shown are in Kelvin scale and are always positive.
The above equation is known as the Van’t Hoff’s Reaction Isochore. Note that how this is different from the Van’t Hoff’s Reaction Isotherm (which relates the effect of activities, or say concentrations or pressures for ideal solutions and ideal gases respectively) of the reactants and products on the Gibbs’ Free Change, DG of the process.
If T2 is the changed temperature from the existing temperature T1, then T2 > T1 means heating and T2 < T1 means cooling.
Also it is to be noted that DH positive means an endothermic process and DH negative means an exothermic process.
The second term in the above equation becomes positive when T2 > T1 and the process is endothermic (DH is positive). Then, necessarily K2 has to larger than K1. In other words, a rise in temperature of an endothermic equilibrium process has the effect of shifting the position of equilibrium in favour of the final state (product, if the process is a chemical reaction). Or the same endothermic reaction upon cooling (T2 < T1) renders the K2 < K1. Or the process shifts towards the initial state (reactants).
In case of exothermic process, wherein DH is negative, a rise in temperature (T2 > T1) means a negative value on the right side of the above equation and hence K2 < K1. So, heating of an exothermic reaction pushes the process towards the initial state and cooling the process betters the final state (products).
Related questions
What are the factors affecting equilibrium?
What examples can you give temperature affecting solubility?
Why does catalyst not affect equilibrium?
How does temperature affect the equilibrium?
What is the temperature in the Haber process?
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What is the effect of temperature on equilibrium?
Dependency of temperature on equilibrium depends on whether the forward reaction or the backward reaction is exothermic or endothermic.
Let's, say we have a reaction:
A+B -> C +D
And let's say that the forward reaction is exothermic which means the backward is endothermic.
So, in this case if we increase temperature, then as the forward reaction is exothermic (heat is released which subsequently increases temperature), thus by Le-Chatelier's Principle, the equilibrium should shift to the reaction which decreases the temperature i.e. the backward reaction.
Similarly, if we decrease the temperature,
Dependency of temperature on equilibrium depends on whether the forward reaction or the backward reaction is exothermic or endothermic.
Let's, say we have a reaction:
A+B -> C +D
And let's say that the forward reaction is exothermic which means the backward is endothermic.
So, in this case if we increase temperature, then as the forward reaction is exothermic (heat is released which subsequently increases temperature), thus by Le-Chatelier's Principle, the equilibrium should shift to the reaction which decreases the temperature i.e. the backward reaction.
Similarly, if we decrease the temperature, then the equilibrium will shift in the reaction where heat is released(so as to counterbalance the said effect) and thus will shift to the forward reaction.
In conclusion,
Increasing temperature favours endothermic reaction
And, Decreasing temperature favours exothermic reaction.
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Why is equilibrium constant affected by the temperature but not affected by the concentration of reactants?
Equilibrium constants are ratios of the rate constants for the forward and reverse reactions, once the concentrations of reactants and products no longer change.
If you change the concentrations, the rate constants do not change once equilibrium has been re-established, because rate constants are what’s left of the observed rates after concentrations have been accounted for.
But changing the temperature can change rate constants, which are dependent on temperature; furthermore, it may (and often does) change the two different rate constants in different ways. (Since equilibrium constants are pro
Equilibrium constants are ratios of the rate constants for the forward and reverse reactions, once the concentrations of reactants and products no longer change.
If you change the concentrations, the rate constants do not change once equilibrium has been re-established, because rate constants are what’s left of the observed rates after concentrations have been accounted for.
But changing the temperature can change rate constants, which are dependent on temperature; furthermore, it may (and often does) change the two different rate constants in different ways. (Since equilibrium constants are proportional to the logarithm of the temperature, you are guaranteed that the forward and reverse rate constants will be affected differently.)
Nissim A.
Author has 595 answers and 4.7M answer views
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In the Haber process why does an increase in the pressure skew the equilibrium to the right?
When you increase the pressure, you decrease the volume since they are inversely proportional to each other. When you decrease the volume, you increase the concentration of the gases because you are forcing the molecules to occupy less space than they used to. Recall that concentration is the amount of molecules per unit volume. Since volume decreases when you increase the pressure, the concentration increases.
Now this change in concentration affects all molecules.
Take the reaction:
2N2(g)+3H2(g)⇌2NH3(g)
There are more number of molecules on the left hand side tha
When you increase the pressure, you decrease the volume since they are inversely proportional to each other. When you decrease the volume, you increase the concentration of the gases because you are forcing the molecules to occupy less space than they used to. Recall that concentration is the amount of molecules per unit volume. Since volume decreases when you increase the pressure, the concentration increases.
Now this change in concentration affects all molecules.
Take the reaction:
2N2(g)+3H2(g)⇌2NH3(g)
There are more number of molecules on the left hand side than on the right. When you increase the pressure, you increase the concentration on the both sides of the equilibrium. Since there are more number of molecules on the left hand side, that means the concentration of the left hand side increased more than it did in the right hand side. According to Le Chatalier's principle, when this happens the equilibrium shifts where this increase in concentration is neutralized i.e. in this case, the right. So, the equilibrium shifts right.
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If the temperature higher than 450c was used in the Haber process, what would happen to the rate of the reaction?
If the temperature higher than 450 ᵒC was used in the Haber process, what would happen to the rate of the reaction?
For every chemical reaction, the rate speeds up when temperature increases.
The reaction of Haber process is represented by the following equation:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
When temperature is higher than 450 ᵒC, both the rate of the forward reaction and the rate of the backward reaction (reverse reaction) will be higher than those at 450 ᵒC respectively.
(Prevent using such high temperature. This is because the equilibrium position will shift to the left as the backwa
If the temperature higher than 450 ᵒC was used in the Haber process, what would happen to the rate of the reaction?
For every chemical reaction, the rate speeds up when temperature increases.
The reaction of Haber process is represented by the following equation:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
When temperature is higher than 450 ᵒC, both the rate of the forward reaction and the rate of the backward reaction (reverse reaction) will be higher than those at 450 ᵒC respectively.
(Prevent using such high temperature. This is because the equilibrium position will shift to the left as the backward reaction is endothermic, and this leads to the decrease of yield.)
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How does high temperature and pressure affect chemical equilibrium?
How does high temperature and pressure affect chemical equilibrium?
Le Chatelier’s Principle states that…
“A chemical system that is in a dynamic equilibrium will shift in order to annul the effect of any change on that system.”
If a reaction is exothermic as it is written down and thermal energy is applied so as to raise the temperature of the system, then the system will shift to the left, so as to reform some reactants and decompose some products and achieve a new dynamic equilibrium.
If a reaction is endothermic as it is written, then the exact opposite occurs and the dynamic equilibrium shift
How does high temperature and pressure affect chemical equilibrium?
Le Chatelier’s Principle states that…
“A chemical system that is in a dynamic equilibrium will shift in order to annul the effect of any change on that system.”
If a reaction is exothermic as it is written down and thermal energy is applied so as to raise the temperature of the system, then the system will shift to the left, so as to reform some reactants and decompose some products and achieve a new dynamic equilibrium.
If a reaction is endothermic as it is written, then the exact opposite occurs and the dynamic equilibrium shifts to the right.
If a reaction occurs entirely in solution or in the solid phase, then there is, effectively, no change in volume and therefore pressure changes have no effect.
If a gas is involved in a reaction and the volume of reactant equals that of the product, then pressure has no effect on the equilibrium.
If a reaction as it is written down has an increase in the volume of a gaseous product, then increasing the pressure will tend to shift the dynamic equilibrium to the left, forming a new equilibrium with higher concentrations of reactant and smaller concentrations of product.
If a reaction, as it is written down, has a decrease in the volume of a gaseous product, then the exact opposite of the above paragraph will occur and the dynamic equilibrium will shift to the right.
NOTE: Le Chatelier’s Principle is just that - a PRINCIPLE - and not a Law. Other factors (such as kinetics) may come into play that change the general rule that Le Chatelier noted.
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Why is it that only temperature can affect the equilibrium constant?
For a bit of theory:
The equilibrium constant is the ratio of the two rate constants for the forward and reverse reactions of the equilibrium.
Kc = kf / kr
Rate constants are described by the Arrhenius equation as:
k = A e^(-Ea/RT)
From this, you see that the value of the rate constant for a reaction depends on the values of the “frequency factor” (A), the activation energy for the reaction (Ea) and the temperature (T). Also, note that concentrations of reactants does not affect the value of the rate constant.
Since the activation energies of the forward and reverse reactions in an equilibrium are d
For a bit of theory:
The equilibrium constant is the ratio of the two rate constants for the forward and reverse reactions of the equilibrium.
Kc = kf / kr
Rate constants are described by the Arrhenius equation as:
k = A e^(-Ea/RT)
From this, you see that the value of the rate constant for a reaction depends on the values of the “frequency factor” (A), the activation energy for the reaction (Ea) and the temperature (T). Also, note that concentrations of reactants does not affect the value of the rate constant.
Since the activation energies of the forward and reverse reactions in an equilibrium are different, as the temperature changes, the values of the forward and reverse rate constants will be affected to different extents. This will lead to a change in the value of the equilibrium constant with changes in temperature.
Antonio Daniels
Former Learner 2020- Chemist
·
Author has 7.9K answers and 4.4M answer views
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1y
Related
How does temperature affect the amount of products at equilibrium?
Endothermic reaction??
An increase in temperature favours the endothermic reaction (equilibrium shift to the right) .
If an endothermic reaction is heated, the reaction shifts right; if the reaction is cooled, it shifts to the left.
Example: That is why we store food in refrigerators.
The food needs heat to get rotten. It is an endothermic reaction.
We cool down the food and the reaction will go to the left. (Food will last more time)
If we increase the temperature the reaction will go to the right and the food will rotten faster.
Endothermic reaction??
An increase in temperature favours the endothermic reaction (equilibrium shift to the right) .
If an endothermic reaction is heated, the reaction shifts right; if the reaction is cooled, it shifts to the left.
Example: That is why we store food in refrigerators.
The food needs heat to get rotten. It is an endothermic reaction.
We cool down the food and the reaction will go to the left. (Food will last more time)
If we increase the temperature the reaction will go to the right and the food will rotten faster.
exothermic reaction??
By increasing the temperature of an exothermic reaction at equilibrium will shift to the left.
Related questions
What is the quantitative effect of temperature on equilibrium?
At equilibrium, the concentration of the reactants and product is [NO] = 0.0542 M, =0.127 M and [N02] =15.5 M. What is the equilibrium constant Kc at this temperature?
If the temperature higher than 450c was used in the Haber process, what would happen to the rate of the reaction?
Why does the Haber process not reach equilibrium?
What is the effect of temperature on equilibrium?
What are the factors affecting equilibrium?
What examples can you give temperature affecting solubility?
Why does catalyst not affect equilibrium?
How does temperature affect the equilibrium?
What is the temperature in the Haber process?
Why does the equilibrium constant not change with a change in pressure?
What is Haber process?
In the Haber process why does an increase in the pressure skew the equilibrium to the right?
How is ammonia manufactured by Haber's process?
What is equilibrium and examples of equilibrium?
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8179 | https://achievethecore.org/coherence-map/7/33/363/363 | Expressions And Equations
Represent And Analyze Quantitative Relationships Between Dependent And Independent Variables.
Major Cluster
6.EE.C.9
Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.
Ratios And Proportional Relationships
Understand Ratio Concepts And Use Ratio Reasoning To Solve Problems.
Major Cluster
6.RP.A.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams), double number line diagrams), or equations.
Expressions And Equations
Reason About And Solve One-Variable Equations And Inequalities.
Major Cluster
6.EE.B.7
Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers).
Ratios And Proportional Relationships
Analyze Proportional Relationships And Use Them To Solve Real-World And Mathematical Problems.
Major Cluster
7.RP.A.2
Recognize and represent proportional relationships between quantities.
Expressions And Equations
Solve Real-Life And Mathematical Problems Using Numerical And Algebraic Expressions And Equations.
Major Cluster
7.EE.B.4
Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.
Ratios And Proportional Relationships
Analyze Proportional Relationships And Use Them To Solve Real-World And Mathematical Problems.
Major Cluster
7.RP.A.3
Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.
Example Task
7.RP and 7.G Sand Under the Swing Set
Provided by Illustrative Mathematics
Task
The 7th graders at Sunview Middle School were helping to renovate a playground for the kindergartners at a nearby elementary school. City regulations require that the sand underneath the swings be at least 15 inches deep. The sand under both swing sets was only 12 inches deep when they started.
The rectangular area under the small swing set measures 9 feet by 12 feet and required 40 bags of sand to increase the depth by 3 inches. How many bags of sand will the students need to cover the rectangular area under the large swing set if it is 1.5 times as long and 1.5 times as wide as the area under the small swing set?
Solutions
Solution: Finding the scale factor the hard way
3 inches is foot, so the volume of sand that was used is
cubic feet. The amount of sand needed for an area that is 1.5 times as long and 1.5 times as wide would becubic feet.
We know that 40 bags covers 27 cubic feet. Since the amount of sand for the large swing set is
times as large, they will need 2.25 times as many bags. Since , they will need 90 bags of sand for the large swing set.
Solution: Finding the scale factor the easy way
Since we have to multiply both the length and the width by 1.5, the area that needs to be covered is
times as large. Since the depth of sand is the same, the amount of sand needed for the large swing set is 2.25 times as much as is needed for the small swing set, and they will need 2.25 times as many bags. Since , they will need 90 bags of sand for the large swing set.
Solution: Using a unit rate
The area they cover under the small swing set is square feet. Since the depth is the same everywhere, and we know that 40 bags covers 108 square feet, they can cover square feet per bag.
The area they need to cover under the large swing set is
times as big as the area under the small swing set, which issquare feet. If we divide the number of square feet we need to cover by the area covered per bag, we will get the total number of bags we need:So they will need 90 bags of sand for the large swing set.
Solution: The other unit rate
The area they cover under the small swing set is square feet. Since the depth is the same everywhere, and we know that 40 bags covers 108 square feet, they can cover bags per square foot.
The area they need to cover under the large swing set is
times as big as the area under the small swing set, which issquare feet. If we multiply the number of square feet they need to cover by the number of bags needed per square foot, we will get the total number of bags we need:So they will need 90 bags of sand for the large swing set.
Download Example Task
Progressions
Consider the difference in these two percent decrease and percent increase problems:
Skateboard problem 1. After a 20% discount, the price of a SuperSick skateboard is 140 now, but its price will go up by 20%. What will the new price be after the increase?
The solutions to these two problems are different because the 20% refers to different wholes or 100% amounts. In the first problem, the 20% is 20% of the larger pre-discount amount, whereas in the second problem, the 20% is 20% of the smaller pre-increase amount. Notice that the distributve property is implicitly involved in working with percent decrease and increase. For example, in the first problem, if x is the original price of the skateboard (in dollars), then after the 20% discount, the new price is x - 20% x. The distributive property shows that the new price is 80% x
x - 20% x = 100% x - 20% x = (100% - 20%)x = 80% x
Please reference page 10 in the Progression document
Download Progressions PDF
Tasks
Lincoln's math problem Comparing Years Tax and Tip The Price of Bread
Assessments
Proportional Relationships Mini-assessment
Prior Grade-Level Tasks
The tasks below align to standards that are critical building blocks to this standard. Students do not necessarily need to be successful with these tasks to be able to engage with this standard, but these tasks can help teachers ground themselves in the skills and knowledge that students are bringing to the current unit of study and gather valuable information to use when teaching this standard. These tasks can be used as homework, warm-ups, in one-on-one work with students, or in PLC discussions. Because of their close ties to the work of this standard, the standards and the tasks below are a good place to start to:
review just before the current unit of study,
surface content that deserves additional support while addressing this standard,
focus the time on readiness for the current unit of study, and/or
provide visibility into a student’s thinking about a related task and determine how best to bring the student into the current unit of study.
7.RP.A.2 Buying Bananas, Assessment Variation 6.RP.A.3 SBAC Assessment Item 6.EE.B.7 Fruit Salad 6.RP.A.3 Fizzy Juice 4.OA.A.2 SBAC Assessment Item
Assessment Items
NWEA Assessment Item Illustrating 7.RP.A.3 Smarter Balanced Assessment Item Illustrating 7.RP.A.3 - Option 1 Smarter Balanced Assessment Item Illustrating 7.RP.A.3 - Option 2
Focus
Focus in Grade 7
Statistics And Probability
Investigate Chance Processes And Develop, Use, And Evaluate Probability Models.
Supporting Cluster
7.SP.C.6
Approximate the probability) of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability. For example, when rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200 times, but probably not exactly 200 times.
Statistics And Probability
Investigate Chance Processes And Develop, Use, And Evaluate Probability Models.
Supporting Cluster
7.SP.C.7
Develop a probability model) and use it to find probabilities) of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy.
Statistics And Probability
Investigate Chance Processes And Develop, Use, And Evaluate Probability Models.
Supporting Cluster
7.SP.C.8
Find probabilities) of compound events using organized lists, tables, tree diagrams, and simulation.
Seeing Structure In Expressions
Write Expressions In Equivalent Forms To Solve Problems
Widely Applicable Prerequisite
Modeling Standard ★
HS.A-SSE.B.3.c
Use the properties of exponents to transform expressions for exponential functions. For example the expression 1.15t can be rewritten as (1.151/12)12t ≈ 1.01212t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%.
Interpreting Functions
Analyze Functions Using Different Representations
Widely Applicable Prerequisite
HS.F-IF.C.8.b
Use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change) in functions such as y = (1.02)t, y = (0.97)t, y = (1.01)12t, y = (1.2)t/10, and classify them as representing exponential growth or decay.
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List oftrigonometricidentities
In trigonometry,
trigonometric identities
are equalities that involve trigonometric
functions and are true for every value of the occurring variables for which both sides
of the equality are defined. Geometrically, these are identities involving certainfunctions of one or more angles. They are distinct from triangle identities, which are
identities potentially involving angles but also involving side lengths or other lengthsof a triangle.These identities are useful whenever expressions involving trigonometric functionsneed to be simplified. An important application is the integration of non-trigonometricfunctions: a common technique involves first using the substitution rule with atrigonometric function, and then simplifying the resulting integral with a trigonometricidentity.
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Trigonometric functions and their reciprocals on the unit circle. All ofthe right-angled triangles are similar, i.e. the ratios between theircorresponding sides are the same. For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them.The reciprocal identities arise as ratios of sides in the triangles wherethis unit line is no longer the hypotenuse. The triangle shaded blueillustrates the identity , and the red triangleshows that .
The basic relationship between the sine and cosine is given by the Pythagoreanidentity:where means and means This can be viewed as a version of the Pythagorean theorem, and follows from theequation for the unit circle. This equation can be solved for either thesine or the cosine:where the sign depends on the quadrant of
Pythagorean identities
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Dividing this identity by , , or both yields the following identities:Using these identities, it is possible to express any trigonometric function in terms ofany other (up to a plus or minus sign):
Each trigonometric function in terms of each of the other five.
in termsof
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By examining the unit circle, one can establish the following properties of thetrigonometric functions.
Transformation of coordinates (
) when shifting thereflection angle in increments of
When the direction of a Euclidean vector is represented by an angle this is theangle determined by the free vector (starting at the origin) and the positive -unitvector. The same concept may also be applied to lines in a Euclidean space, wherethe angle is that determined by a parallel to the given line through the origin and thepositive -axis. If a line (vector) with direction is reflected about a line with direction then the direction angle of this reflected line (vector) has the value
Reflections, shifts, andperiodicity
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Consensus statement on best practice management regarding the use of intravesical immunotherapy with BCG for bladder cancer
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Download PDF
Consensus Statement
Open access
Published:
Expert consensus document
Consensus statement on best practice management regarding the use of intravesical immunotherapy with BCG for bladder cancer
Ashish M. Kamat1,
Thomas W. Flaig2,
H. Barton Grossman1,
Badrinath Konety3,
Donald Lamm4,
Michael A. O'Donnell5,
Edward Uchio6,
Jason A. Efstathiou7 &
¦
John A. Taylor III8
Nature Reviews Urology volume 12, pages 225235 (2015)Cite this article
33k Accesses
159 Citations
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Subjects
Bladder cancer
Drug therapy
Immunotherapy
Abstract
Multiple clinical trials have demonstrated that intravesical Bacillus CalmetteGuérin (BCG) treatment reduces recurrences and progression in patients with non-muscle-invasive bladder cancer (NMIBC). However, although BCG has been in use for almost 40 years, this agent is often underutilized and practice patterns of administration vary. This neglect is most likely caused by uncertainties about the optimal use of BCG, including unawareness of optimal treatment schedules and about patient populations that most benefit from BCG treatment. To address this deficit, a focus group of specialized urologic oncologists (urologists, medical oncologists and radiation oncologists) reviewed the current guidelines and clinical evidence, discussed their experiences and formed a consensus regarding the optimal use of BCG in the management of patients with NIMBC. The experts concluded that continuing therapy with 3-week BCG maintenance is superior to induction treatment only and is the single most important factor in improving outcomes in patients with NMIBC. They also concluded that a reliable alternative to radical cystectomy in truly BCG-refractory disease remains the subject of clinical trials. In addition, definitions for common terms of BCG failure, such as BCG-refractory and BCG-intolerant, have been formulated.
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Introduction
Over the past decade, interest in the use of immunotherapy for cancer treatment has been steadily growing. Early results from clinical trials investigating immunomodulation with the programmed death ligand 1 (PD-L1) have contributed to this enthusiasm, particularly in the field of bladder cancer research. Experts will recognize, however, that immunotherapy for bladder cancer has been successfully used for almost 40 years. BCG was first introduced into clinical use in urology in 1976 and remains the most effective form of intravesical treatment for non-muscle-invasive bladder cancer (NMIBC). Despite this successful history, BCG therapy is internationally underutilized and practice patterns of BCG administration vary widely, probably because of misconceptions about its optimal use.1 This Expert Consensus Document provides a comprehensive review of immunomodulatory therapy with BCG, clarifies topics that can create confusion and recommends best practice guidelines to improve overall use and patient outcomes (Box 1).
Box 1: Recommendations for intravesical BCG
¢ RCTs and practice pattern research demonstrate that BCG immunotherapy in NMIBC reduces recurrences and progression, and affects mortality
¢ 3-week BCG maintenance is confirmed to reduce recurrence rates compared with induction alone, as well as metastasis and mortality compared with maintenance chemotherapy; thus, it is the optimal regimen for current practice
¢ BCG maintenance schedules other than the 3-week schedule show no significant benefit in RCTs
¢ In the period of around 1.52 years after the identification of high-grade NMIBC, nonradical alternative treatments for patients experiencing BCG-failure can be explored
¢ After the first BCG failure, patients (who have not progressed) have several treatment options, including repeated BCG (or continued maintenance), BCG plus interferon, single-agent intravesical chemotherapy (for example, mitomycin, gemcitabine, or valrubicin), sequential chemotherapy (for example, gemcitabinedocetaxel) or device-assisted chemotherapy
¢ After the second BCG failure, or if the disease is BCG-refractory, radical cystectomy should be considered with alternatives (listed above) considered a matter of investigation by clinical trials
¢ Patients with BCG-refractory disease who are not candidates for cystectomy can be considered for chemoradiation
Abbreviations: NMIBC, non-muscle-invasive bladder cancer; RCTs, randomized controlled trials.
Methods
Contributing authors were invited by the Leo & Anne Albert Institute for Bladder Cancer Care and Research to partake in a symposium designed to evaluate and comment on current practices in bladder cancer. Participants were selected based on their expertise and knowledge in the area of interest. Global literature searches were performed for each section and presentations made with subsequent group discussion and consensus development based on the strength of the data.
Historical perspective
BCG is a live, attenuated form of the slow-growing, aerobic bovine tuberculosis bacterium, Mycobacterium bovis. In 1900, Albert Léon Charles Calmette and Camille Guérin of the Pasteur Institute in Lille, France, began work on a vaccine against Mycobacterium tuberculosis, one of the species that can cause tuberculosis in humans. They discovered that M. bovis is less virulent after being subcultured in a bile-containing medium. Now Middlebrook 7H9 medium is usually used. From 1908 to 1913 repeated subculturing of M. bovis resulted in a nonvirulent strain and vaccine trials in humans began in 1921. Before lyophilized storage in 1960, BCG was passaged continually (over 1,000 times), which led to the formation of multiple daughter strainseach strain a genetic variant of the original. Many of these mutants are used today, leading to different possible product characteristics and strengths, as well as varying dosing requirements. Remarkably, as demonstrated by the high complete response rates observed with BCG preparations containing sufficient numbers of live bacteria, all strains are effective and clearly superior to standard intravesical chemotherapy. Although controlled clinical trials might confirm the superiority of one preparation over another, current shortages of BCG require that we assure both patients and government agencies that current experience suggests that all current BCG preparations are highly effective and superior to non-BCG alternatives.
The first report of BCG as an immunomodulator in cancer therapy was published in 1959.2 In 1976, Morales and colleagues3 published a landmark paper on the favourable effect of intravesical BCG on outcomes in recurrent superficial bladder cancer in nine patients. The first controlled trial showing similar results was published in 1980 and BCG received FDA approval for the treatment of superficial bladder cancer in 1990.4
Review of current guidelines on BCG use
Several guidelines exist regarding the use of BCG in NMIBC, such as those by the American Urological Association (AUA),5 the European Association of Urology (EAU),6 the National Comprehensive Cancer Network (NCCN),7 the International Bladder Cancer Group (IBCG)8 and the International Consultation on Urological Diseases (ICUD).9 Even though each of these guidelines is based on large-scale literature reviews on similar available data, the guidelines are somewhat disparate, making implementation of the recommendations into routine practice difficult for the practicing urologist.
Low-grade tumours
Consensus exists for small, solitary, superficial low-grade tumours (Ta) with all groups agreeing that BCG is not indicated in this setting given the very low risk of disease progression. For multiple and/or large or recurrent low-grade tumours (the intermediate-risk category), BCG therapy with or without maintenance is considered optional by EAU and AUA guidelines and the IBCG.5,6,8 The AUA's meta-analysis suggests 24% and 31% reduction in recurrence rates for BCG induction and maintenance, respectively.5 The EAU recommendation is based predominantly on the recent meta-analysis by Sylvester and colleagues,10 which demonstrated a 32% relative risk reduction with the use of BCG maintenance compared with intravesical chemotherapy. The IBCG recommendation is based on a risk assessment model.11 By contrast, the NCCN guidelines do not specifically address these intermediate-risk tumours.7
High-grade tumours (including carcinoma in situ)
The recommendations for use of BCG in high-grade tumours are relatively consistent across guidelines; variations mainly exist in the recommended duration of therapy. The AUA, EUA and IBCG advise BCG induction with 13 years of maintenance for all high-grade tumours.5,6,8 ICUD guidelines for Ta high-grade tumours do not include maintenance BCG, remarking a lack of conclusive evidence about the effect of maintenance BCG on disease progression in these tumours. In addition, the value of maintenance therapy in T1 disease is questioned.9 Carcinoma in situ (CIS) is the only pathology for which maintenance is advised by ICUD and the guideline also includes reinduction of BCG instillations if no response is detected at first evaluation.9 The NCCN recommends BCG as standard treatment for CIS only, but lists BCG as the favoured option for high-grade Ta and T1 tumours and offers that maintenance should be considered.7
Optimal schedule and duration of therapy
The 6-week course of intravesical and percutaneous BCG, which was the first reported schedule,3 is very effective. Most subsequent studies have failed to demonstrate that alternative schedules provide significant benefits. Early modifications, such as omission of percutaneous vaccination, and quarterly or monthly maintenance, though accepted, only marginally improve convenience or efficacy. Although randomized clinical trials (RCTs) failed to demonstrate improved efficacy of percutaneous BCG against bladder cancer, systemic immune responses are higher after percutaneous vaccination than after intravesical administration alone. This heightened response after immunization is promising and should prompt new RCTs to investigate this application method.
RCTs of quarterly, monthly and repeated 6-week BCG instillations, which aim to maintain the immune response, also failed to show a significant improvement of efficacy (Figure 1). These results led some experts to question the value of maintenance BCG and to abandon its use. By contrast, the 3-week BCG regimen, designed by the Southwest Oncology Group with immunological principles in mind,12 is sufficiently different that it should not be considered in the same category as other maintenance BCG regimens.
The potential benefit of 3-week BCG maintenance deserves emphasis. Specifically, the schedule is initiated with weekly intravesical and percutaneous BCG treatment for 6 weeks, followed by further weekly combination treatment for 3 weeks after routine cystoscopy at 3, 6, 12, 18, 24, 30 and 36 months.12 Only 14% of patients completed treatment in the original study, but compliance in modern studies is high. In the population with CIS, complete response at 6 months in patients randomized to receive the initial 6-week BCG treatment only (induction arm) was 69% in comparison with 84% (P <0.01) in patients randomized to receive the additional 3-week maintenance treatment (maintenance arm).12 In the maintenance arm, complete response increased from 55% at 3 months to 84% at 6 months (64% complete response in patients with treatment failure at 3 months). Even without additional BCG, 26% of patients with residual disease at 3 months in the induction arm went on to have complete response by 6 months, illustrating that 6 months is a preferred time to evaluate response.
Significant improvement in the treatment of CIS was confirmed by other RCTs with 3-week maintenance BCG, but no other treatment has level one evidence of superiority over induction BCG. The most commonly used treatment, repeated 6-week instillations, has been found to be ineffective in one RCT.13 Five RCTs compared non-3-week maintenance schedules to BCG induction only (Table 1).14,15,16,17,18 None of these alternative maintenance schedules showed a statistically significant reduction in recurrence (range -4% to 22%, average 7%). By contrast, two studies that compared 3-week maintenance to induction treatment alone showed a statistically significant 28% reduction in tumour recurrence (Table 2).12,19 Moreover, only the 3-week BCG maintenance schedule reduced disease progression and metastasis, as well as overall and cancer-specific mortality in RCTs.10,12
Two other trials compared 3-week maintenance BCG with alternative agents (Table 2).10,20 EORTC 30911 compared the 3-week maintenance BCG with 3-week maintenance epirubicin chemotherapy. BCG significantly reduced recurrence by 15%, metastasis by 45% (P = 0.046), and both overall (P = 0.023) and cancer-specific mortality (P = 0.026).10 No other intravesical treatment, BCG induction only, or non-3-week maintenance BCG schedule has achieved such success.
The life-saving benefit of BCG immunotherapy in RCTs is confirmed in real-world practice pattern research. In a review of data from nearly 24,000 patients with bladder cancer from the Surveillance, Epidemiology, and End Results (SEER)Medicare database, use of BCG significantly reduced mortality (HR 0.87), and bladder cancer deaths were reduced by 23% in patients with high-grade tumours.21 Education might now be the most important next step in improving care in bladder cancer. The review found that only 22% of eligible patients received BCG, and only a fraction of those received the critically important 3-week maintenance schedule.21
Practical issues of BCG administration
Some practical points need to be considered when using BCG for bladder cancer treatment. BCG manufacturers recommend evaluating the tuberculosis status of the patient with a PPD (purified protein derivative) tuberculosis skin test before initiation of therapy, with some practitioners obtaining chest radiographs in all patients.22]
, (2013).") This procedure has never proven necessary, probably owing to the exceedingly low incidence of tuberculosis in developed nations, where most bladder cancer patients do not meet the criteria for disease screening as recommended by the Centers of Disease Control and Prevention.23.") In fact, patients with a positive PPD test without active disease should not be excluded from BCG therapy, as the presence of a systemic immune response might help augment the antitumour response as noted above.24.") In addition, patients with a positive PPD result were shown to display adverse effect profiles during BCG treatment that are similar to patients with a negative test.25.")
In general, BCG use in patients with significant immunosuppression should be avoided, owing to the infectious risk of BCG. However, other patients with mild immune impairment (for example, due to steroid use for chronic obstructive pulmonary disease), well controlled HIV, or mature transplants, as well as healthy elderly patients, can be treated successfully with minimal adverse effects and good efficacy.26,27 In addition, medications such as statins or antiplatelet agents are not a contraindication for BCG therapy. Although these drugs have a potential negative effect on the immune response to BCG treatment, neither had an effect on the clinical efficacy of BCG therapy in multiple clinical studies.27,28,29,30,31,32 Other relative contraindications are prosthetic valves or orthopaedic hardware. However, a large phase II study with BCG combination therapy reported no infectious complications in the absence of prophylactic antibiotics, suggesting this population represents a low-risk setting.33 This recommendation is in concordance with the American Heart Association's guideline against antibiotic prophylaxis for genitourinary procedures such as urethral catheterization utilized in BCG administration.34
The technique used for the administration of BCG therapy must optimize the contact duration of the mycobacterium to the bladder urothelium to achieve maximum treatment effect. These measures include minimizing fluid intake by the patient before treatment and assuring complete bladder drainage via a lubricated, atraumatically placed catheter immediately before BCG instillation under gravity. Additional rolling and positional manoeuvers by the patient after instillation are unnecessary for complete contact with the urothelium and distribution of BCG. Local symptoms of bladder irritability are quite common with 60% of patients in clinical trials experiencing dysuria.22]
, (2013)."),35.") Symptom management, such as use of single-dose, short-term quinolones 6 h after BCG instillation and/or use of anticholinergics, has been shown to decrease moderate to severe adverse events in some published reports,36."),37.") and enables the majority (90%) of patients to tolerate full-dose BCG without discontinuation of therapy.38.")
Definition of BCG failure
Recognizing when BCG has failed is another area that can lead to confusion. Although the assumption seems logical that any tumour recurrence after therapy is a 'BCG failure', not all patients in this population have a similar prognosis. This situation has immense implications on options for alternative therapy, including the need to proceed to cystectomy. Definitions to clarify what constitutes BCG failure have been published.39 Herr & Dalbagni40 and O'Donnell & Boehle41 suggested definitions of the terms BCG relapse, BCG-refractory and BCG-intolerant:
¢ BCG relapse is considered a recurrence of tumour after a period of disease-free status. Most experts agree that the time point for evaluation should be at 3 months for papillary tumours and 6 months for CIS (except when disease progression was observed at 3 months). Relapse can be further stratified as early (<1 year after treatment), intermediate (12 years) or late (>2 years), as the disease-free interval is a prognostic variable; early-relapsing patients are more likely to progress and late-relapsing patients can possibly derive some benefit from reinduction with BCG.42
¢ BCG-refractory is the persistence of disease after adequate induction and one maintenance course of BCG. Of note, this category includes any progression in stage or grade by 3 months if patients received induction BCG only.
¢ BCG-intolerant is defined as the inability to tolerate at least one full induction course of BCG. The tumour recurs largely because of inadequate therapy, which does not have the same negative prognostic implications as a true BCG failure.
The above definitions highlight a major problem with the design of current clinical trials for bladder cancer using intravesical BCG therapy: lack of a standard definition of treatment failure and the complexity and ambiguity associated with related terminology. The situation becomes even more complex when the adequacy of BCG therapy is considered, as induction BCG alone is considered suboptimal therapy. Thus, when designing clinical trials for intravesical therapy, most clinicians recognize that to be included in a BGC-failure trial patients should receive at least five of six doses of induction BCG and at least two of three doses of maintenance BCG. This requirement brings about its own challenge: the lack of a good control group (other than cystectomy) for true BCG failures. If patients have recurrent high-grade tumours after BCG therapy as listed above, the salvage rates with intravesical therapy remain low. Thus, no clinical trial that randomizes BCG-refractory patients to alternative conservative measures compared to the much more invasive treatment of radical cystectomy has been proposed or even felt to be feasible to date.
Improving enrolment in clinical trials of new intravesical therapies will require changes in study design and approach. One such proposal is based on early identification of patients who experience 'cytogenetic failure' of BCG therapy. This suggestion is based on a prospective clinical trial, in which fluorescence in situ hybridization (FISH) assays that detect aneuploidy and gene mutation in urine samples (UroVysion, Abbott Molecular) were performed serially during BCG therapy. The data suggested that patients with a positive result in the FISH assay and a negative 3-month cystoscopy are at such high risk of recurrence and progression that their disease can be classified as a cytogenetic BCG failure, even though they are clinically disease-free.43 Thus, a proposed clinical trial design might include randomizing such patients to further BCG (as is standard of care) or alternative therapy that has previously demonstrated robust activity in other phase II trials.44.")
Predictors of response to BCG
Markers of response to intravesical BCG fall into three broad categories: determination of response (presence of cancer), use of surrogate endpoint biomarkers and prediction of response. BCG is most frequently used to treat high-grade disease and assessing the response of CIS specifically is challenging because it is often difficult to detect. A commonly used design is to determine response by cystoscopy, cytology and possibly bladder biopsy at 3 months and 6 months following a BCG induction regimen. However, the 3-month assessment is problematic because of the well documented conversion of positive cytology at 3 months to negative cytology at 6 months, particularly after maintenance BCG.12 Fluorescence cystoscopy can improve detection of both CIS and papillary bladder cancer, but fluorescence can also be induced by inflammation.45 It remains to be seen whether fluorescence cystoscopy can reliably improve the determination of BCG response.
Cytology is a subjective test with variable performance;46 molecular tests that are more objective might yield better and more consistent performance in detecting visually occult bladder cancer. Patients who have abnormal results in the UroVysion assay in a urine sample obtained immediately before the last of the 6-week induction instillations have increased risk for tumour recurrence and progression.43
Cytokine production has been explored as a source of potential surrogate endpoint biomarkers. Increased levels of IL-8 within 6 h and increased levels of IL-18 within 12 h of the first BCG dose correlate with response to BCG treatment.47 Detection of IL-2 in the patient's urine during induction and the first maintenance treatment is associated with a favourable response to BCG.48 Similarly, patients receiving BCG whose peripheral blood mononuclear cells produce IL-2 in response to challenge with phytohaemagglutinin are likely to have a favourable response to BCG.49 A team from the MD Anderson Cancer Center has developed a nomogram based on the measurement of multiple cytokines that predicts response with an area under a receiver operating characteristics curve value of 0.85.50
Response to immunotherapy with BCG depends on complex molecular interactions between the patient and their cancer. Genomic variations in oxidative stress pathway genes might affect response to BCG.51 Similarly, polymorphisms in immune response genes in combination with clinicopathological features can be used to distinguish patient groups with varying risk of relapse-free survival after BCG treatment.52 This approach is supported by evidence of a specific gene signature that is associated with response to BCG.53.") Molecular analysis has defined clinically relevant subtypes of bladder cancer.54.") These tumours differ in their response to chemotherapy, but it is not yet clear if these molecular phenotypes significantly differ in their response to BCG. Increasing knowledge of the multiple factors that influence what is clinically recognized as a BCG response is likely to result in improved instruments to assess this response.
Options after BCG failure
BCG treatment eventually fails in up to 50% of patients and in about half of those within the first 6 months.55 In tumours that do not respond to BCG anymore, conventional intravesical chemotherapy agents, such as doxorubicin, thiotepa and mitomycin, have limited activity. In addition, valrubicin, the only FDA-approved drug for BCG-refractory CIS, is only effective in <10% of patients at 2 years and none with coincident stage T1 disease.56 Thus, the traditional approach after BCG treatment failure has been to advocate early cystectomy.57 However, several lines of evidence suggest a reasonably safe window of opportunity, probably up to 1.52 years after onset of the index high-grade tumour, in which other conservative, nonradical therapies might be tried without undue risk.58,59,60,61
Repeated BCG induction is of limited value. About one-third of patients who do not respond to one course of BCG will eventually achieve a durable response. However, after further courses, the typical success rate of repeated BCG is <10-20%.41,62,63 After recognizing that BCG exerts its effects via cytokine stimulation, as well as response of type 1 T helper cells (TH1) and T helper cell recruitment, the addition of cytokine therapy to BCG treatment has been tried. For example, IFN-α2b functions by enhancing protein TRAIL release, augmenting the TH1 response, increasing IFN-γ release and reducing the expression of inflammation-inhibiting proteins such as IL-10.64,65 Although intravesical IFN-α provides a 2-year disease-free rate of only 12% for BCG-refractory CIS,66 several phase I and phase II studies indicate a robust response of around 50% for the combination of BCG with IFN-α2b in patients who have not responded to BCG alone. However, not all patients have the optimum salvage response that is observed in those who initially responded to BCG and developed recurrent tumours in a delayed fashion (>1 years later).67,68,69,70 Long-term studies of up to 5 years show that, although early responses to the BCGinterferon treatment were excellent, 64% of patients had relapsed at 5 years with 20% having evidence of disease progression.71 The combination of reduced-dose BCG with IFN-α has shown favourable long-term results in first-time BCG failures.70,72 By contrast, the only randomized study of the combination of BCG with IFN-α2b in BCG-naive patients failed to demonstrate any superiority of the combination over BCG alone, even though the tolerability of reduced dose BCGIFN-α2b seemed to be better than that of full-dose BCG alone.73 Nevertheless, the lack of RCTs and lack of superiority of BCGIFN versus BCG alone in BCG-naive patients has limited the widespread use of this combination treatment. Furthermore, even this therapy has unsatisfactory activity in BCG-refractory patients. The combination of BCG with IL-12, which is known to promote TH1 response and interferon release, has been studied in mice. Intravesical administration of a combination of BCG and IL-12 resulted in higher urine and serum levels of IFN-γ, especially after the third instillation, in comparison to either agent alone.74 Long-term outcome data for this treatment combination are not available.
Other alternative strategies have been developed for patients who experience a failure of BCG therapy. For example, the use of initial chemotherapy followed by BCG, the reverse version or alternating application, as well as other novel chemotherapy regimens, have been investigated in small clinical studies.55 In addition, several original approaches using electromotive acceleration have been investigated. Kaasinen et al.75 conducted a study in 236 patients with recurrent Ta or T1 bladder tumours. Eligible patients received five weekly instillations of mitomycin C and were subsequently randomized to monthly BCG only or alternating BCG and IFN-α2b. They found a significantly lower recurrence rate in the arm treated with mitomycin C followed by BCG only, indicating no benefit of IFN-α2b in this situation. The use of hyperthermic conditions with intravesical mitomycin C can increase the response rate compared with euthermic conditions. Hyperthermic conditions resulted in a significant improvement in long-term disease-free survival.76 In another combination approach, Di Stasi et al.77 administered electromotive mitomycin C alternating with BCG compared with BCG alone in a complicated schedule that also included maintenance therapy in the complete responders. In this randomized trial, patients receiving the combination therapy demonstrated a significantly lower recurrence rate (42% versus 58%, P = 0.0012). The randomized EORTC-30993 trial compared sequential mitomycin C and BCG to BCG alone, but did not demonstrate any benefit of the combination treatment in terms of recurrence or progression.78 Results from an aggregate meta-analysis did not show any difference in response rates or progression between administering chemotherapy and BCG concomitantly or in sequence.79 However, the combination approach did trend towards reduced recurrence (relative risk 0.75) and progression (relative risk 0.45) for patients with Ta or T1 tumours, while not having much benefit in those with CIS.
Several approved chemotherapeutics that are active in advanced disease, including cisplatin, gemcitabine and taxanes, are being studied for intravesical use. Unfortunately, the most active single systemic agent, cisplatin, is associated with unpredictable and sometimes fatal anaphylaxis when used intravesically.80 By contrast, intravesical gemcitabine is particularly active in patients with intermediate risk papillary disease in whom BCG treatment has failed, even demonstrating superiority in comparison with mitomycin.81,82 However, two studies of gemcitabine in BCG-refractory disease resulted in 2-year durable responses of only 7% and 21%.83,84 Intravesical docetaxel has shown good activity, even in BCG-refractory NMIBC. Among 54 patients, 59% experienced a complete response, 40% were disease-free at 12 months, and 25% at 3 years. Cystectomy was required in 24% with an overall 5-year disease-specific survival of 85%.85
Intravesical multi-agent chemotherapy in NMIBC has only recently been investigated. Past combinations of harsher, vesicant drugs, such as doxorubicin and mitomycin, although highly active, resulted in considerable local chemical cystitis.86 By contrast, newer nonvesicant drugs, such as gemcitabine and docetaxel, are very well tolerated.87,88 Sequential gemcitabinemitomycin C in 47 patients with high-risk disease, primarily after BCG failure, resulted in a complete response in 68% of patients with 1-year and 2-year disease-free survival of 48% and 38%, respectively.89 Alternatively, sequenced gemcitabinedocetaxel in 45 patients with high-risk disease yielded a complete response rate of 66%, as well as 1-year and 2-year recurrence-free survival of 54% and 34%, respectively.90 Reassuringly, although the cystectomy rate for both regimens was around 20%, metastasis was rarely reported.
Radiation as an option for recurrent disease
Cystectomy remains the standard for recurrent BCG-refractory, high-grade T1 bladder tumours. However, many patients (especially the elderly) are not undergoing surgery, possibly owing to comorbidities and/or a desire to avoid a large operation.91 Furthermore, the high rates of clinicopathological stage discordance noted after surgery (around 45% of T1 tumours are being upstaged) might contribute to the poor long-term results of second-line intravesical agents following BCG failure.92
Radiation therapy presents a non-surgical option for BCG failures. Early studies showed reasonable response rates (4869%),93,94 and one randomized trial in patients with high-grade T1 tumours comparing conservative treatment with radiation alone showed no difference between treatments in terms of recurrence.95
There is increasing data on combined radiochemotherapy for high-risk NMIBC following maximum transurethral resection of the bladder tumour (TURBT). One study compared radiotherapy or radiochemotherapy in 141 patients and reported a complete response rate of 88%; at 5 years and 10 years, overall progression rates were 19% and 30% (13% and 29% for high-grade T1 tumours) and overall disease-specific survival was 82% and 73% (89% and 79% for complete responders, 80% and 71% for high-grade T1 tumours), respectively.96 Over 80% of survivors could retain their native bladder and around 70% were delighted or pleased with their urinary function. A second study reported on 17 patients undergoing trimodality therapy (TURBT, radiation and chemotherapy) following T2 recurrence after failure of BCG for non-invasive disease.97 With 7 years follow-up time, only one patient required cystectomy, 10 patients (59%) were free of any bladder tumour recurrence and disease-specific survival was 70%.
Based on current evidence, the new 2015 NCCN bladder cancer guideline states that external-beam radiotherapy is rarely appropriate for patients with stage Ta, T1, or Tis disease. Radiochemotherapy might be a potentially curative alternative to cystectomy in patients with recurrent TaT1 disease (usually following BCG therapy but without extensive Tis).98]
, (2015).") Currently, the North American multicentre, cooperative Radiation Therapy Oncology Group (RTOG) protocol RTOG 0926 is evaluating the role of radiochemotherapy (61.2 Gy with concurrent cisplatin or 5-fluorouracil plus mitomycin C) after maximum TURBT for patients with high-risk T1 bladder cancer following BCG failure for whom the next therapy would have been radical cystectomy.99.")
Novel agents
Currently, several strategies to enhance BCG response are being explored. Testing of combinations of BCG with various new investigational agents that have independently demonstrated activity in early phase trials is ongoing and some of these combinations have been applied clinically with variable effect. In addition, the use of entirely novel agents based on emerging preclinical findings presents an important area for potential clinical investigation.
Some bladder cancer cells have a high level of VEGF expression. Tyrosine kinase inhibitors, such as sunitinib, can inhibit proliferation of such cells. In vitro experiments using T24 bladder cancer cells have demonstrated that the combination of sunitinib and BCG can enhance cell death and reduce cell migration compared with sequential administration.100 It remains to be seen whether similar responses can be obtained in vivo or in humans. Smac mimetics degrade inhibitors of apoptosis (IAP) proteins, which can block caspase-induced apoptotic death, and have been studied for augmenting response to BCG.101 One of the mechanisms of action of BCG is to enhance release of protein TRAIL from neutrophils, which can be inhibited by IAP proteins.102,103 In one in vitro study, application of BCG-conditioned neutrophil medium along with a small molecule Smac mimetic resulted in enhanced cytotoxicity compared with the use of BCG alone.104 Finally, trapping of BCG by exogenously introduced fibrinogen can enhance local concentration of BCG and uptake via fibronectin. This strategy has been shown to enhance BCG response in vitro.105
Other developments in oncology might also affect the future of intravesical treatment in bladder cancer. The immune checkpoint inhibitors ipilimumab and pembrolizumab (antibodies that target CTLA-4 and PD-1, respectively) have been approved for use in advanced melanoma. The receptor PD-1 is one checkpoint of T-cell function and is activated by PD-L1 and PD-L2. Anti-CTLA-4 and anti-PD-1 antibodies can prolong and enhance the T-cell response. In bladder tumours, PD-L1 expression has been correlated with bladder cancer T-stage and is highest in T3/T4 and CIS.106 In addition, immunohistochemical data suggest that PD-L1 expression is increased in tumours from patients with BCG failure.106 PD-L1 and PD-L2 can also be inhibited and could provide an alternative strategy, as some evidence exists that inhibition of PD-L1 might avoid autoimmune responses, which are usually modulated by PD-L2. IDO-1 inhibitors can work in a fashion similar to PD-1 inhibitors to enhance TH activity. One such inhibitor, indoximod is currently being studied in a clinical trial for advanced prostate cancer and could potentially be used together with BCG to enhance response.107.") Currently, immune checkpoint inhibitors for the treatment of bladder cancer are being evaluated in the advanced-disease setting, but they might also have a role in early-stage disease in combination with BCG.
Intravesical therapy with a conditionally replicating adenovirus that was genetically engineered to express GM-CSF was well tolerated and induced a local immune response.108 Instillation of a nonreplicating adenoviral vector encoding IFN-α2b together with the exipient Syn3 to enhance gene transfer also seemed to be well tolerated and could yield responses.109 Photodynamic therapy is another novel approach to bladder cancer therapy. Several early studies suggested a benefit and one small study in 34 patients reported a favourable disease-free recurrence rate.110,111,112 However, despite the novelty in these and other approaches, none have been evaluated in a definitive, practice-changing study or are currently considered standard therapy.
Another treatment for NMIBC are toxic protein conjugates. These agents contain both a targeting and a treatment component to increase specificity for cancer cells. One clinical example of this approach is denileukin diftitox, a fusion of the diphtheria toxin and IL-2, which targets a membrane receptor that is differentially expressed in target cancer cells compared with the normal cells.113 In bladder cancer, oportuzumab monatox, a recombinant fusion protein of an anti-Ep-CAM antibody with the pseudomonas exotoxin, has been evaluated. In a phase II study in 46 BCG-refractory patients with CIS, 44% achieved a complete response with this therapy.114 Other preclinical evaluations of this approach include targeting diphtheria toxin to the epidermal growth factor receptor (EGFR), which is frequently overexpressed in bladder cancer but not the normal urothelium.115 Diphtheria toxin linked to EGF has shown encouraging results in murine orthotopic models of bladder cancer without any clear systemic or urothelial toxicity.116
Nanotechnology can be directed to specific cells and then heated for focal thermal ablation with the application of external light energy. In bladder cancer therapy, direct intravesical delivery can avoid problems caused by systemic clearance. Gold nanorods conjugated to anti-EGFR antibodies have been tested in bladder cancer cell lines using near-infrared light.117 A similar approach has been developed using gold nanorods targeting the fibroblast growth factor receptor, highlighting the generalizability of this approach based on the expression of target molecules on tumour cells.118
One approach to address the toxicity observed with BCG treatment is the use of nonlive-BCG immunotherapy. For example, BCG cell wall components have been incorporated into octaarginine-modified liposomes, which have similarities to an envelope-type virus, for cellular delivery.119 In fact, a similar strategy was used in the MCNA agent, in which mycobacterial cell wall is complexed to nuclear DNA with the aim of achieving both immunological and direct cytotoxic effects. A recent study in 129 patients with high-risk disease after BCG failure suggested a high activity of this agent.120 Overall disease-free rate was 25.0% at 1 year and 19.0% at 2 years. In patients with papillary-only tumours, MCNA was even more effective with rates of 35.1% and 32.2% at 1 year and 2 years, respectively.
Conclusion
Immunotherapy for patients with NMIBC remains the gold standard, with efficacy of intravesical BCG surpassing that of chemotherapy. In order for BCG immunotherapy to be effective, however, practitioners must learn (or re-learn) how to use it in an optimal manner. First, patient selection is paramount and complete resection of all visible tumours is important, not only for staging, but also to improve outcomes. BCG induction with 6-week instillations must be followed by 3-week maintenance instillations, as recommended in the SWOG protocol, as no other maintenance schedule has been shown to work reliably. If a patient develops recurrence on BCG therapy, the timing and tumour characteristics of this recurrence should be considered when classifying the patient in the subgroups within the BCG-failure category. Although radical cystectomy remains the mainstay for patients with high-risk BCG failure, for those who are unfit or who refuse cystectomy, several alternative options exist as outlined in this Expert Consensus Document. Future strategies aimed at improving the immune response of patients (with or without BCG), device-assisted chemotherapy delivery and combination chemotherapy all seem to be promising alternatives.
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The University of Texas MD Anderson Cancer Center, 1515 Holcombe Boulevard, Unit 1373, Houston, 77030, TX, USA
Ashish M. Kamat & H. Barton Grossman
2. University of Colorado, MS 8117, 12801 E. 17th Avenue, Room 8117, Aurora, 80045, CO, USA
Thomas W. Flaig
3. University of Minnesota, 420 Delaware Street Southeast, Minneapolis, 55455, MN, USA
Badrinath Konety
4. BCG Oncology, 3815 East Bell Road, Suite 1210, Phoenix, 85032, AZ, USA
Donald Lamm
5. University of Iowa, 200 Hawkins Drive, Iowa City, 52242, IA, USA
Michael A. O'Donnell
6. University of California Irvine, 333 City Boulevard, Suite 2100, Orange, 92868, CA, USA
Edward Uchio
7. Harvard Medical School, 100 Blossom Street, Cox 3, Boston, 02114, MA, USA
Jason A. Efstathiou
8. University of Connecticut Health Center, 263 Farmington Avenue, Farmington, 06030, CT, USA
John A. Taylor III
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Ashish M. Kamat
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Kamat, A., Flaig, T., Grossman, H. et al. Consensus statement on best practice management regarding the use of intravesical immunotherapy with BCG for bladder cancer. Nat Rev Urol 12, 225235 (2015).
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Neutrinos
Feature
Neutrino production moves to an industrial scale
27 April 2012
The promise and challenges of the Neutrino Factory.
or smaller than m1 and m2 (the \"inverted hierarchy\"). The fractional contribution of each flavour to the mass eigenstates is indicted by the coloured bars. (After a figure by S Parke.)")Fig. 1. Diagram of the relationship between the mass eigenstates (labelled 1, 2 and 3) for neutrinos and the flavour eigenstates (ν e, ν μ and ν τ). Neutrinos from the Sun have been used to show that m 2> m 1; m 3 may be greater or less than m 1 and m 2 (the “normal hierarchy”) or smaller than m 1 and m 2 (the “inverted hierarchy”). The fractional contribution of each flavour to the mass eigenstates is indicted by the coloured bars. (After a figure by S Parke.) or smaller than m1 and m2 (the \"inverted hierarchy\"). The fractional contribution of each flavour to the mass eigenstates is indicted by the coloured bars. (After a figure by S Parke.)")
The measurements of the electron- and muon-neutrino fluxes published by the Super-Kamiokande collaboration in 1998 marked a turning point in the history of particle physics. This team showed that fewer muon-neutrinos arrive at the surface of the Earth than are produced by cosmic-ray interactions in the upper atmosphere (atmospheric neutrinos). This in turn indicated evidence for neutrino oscillations, the phenomenon in which the flavour of the neutrino changes (oscillates) as the neutrino propagates through space and time. Since the publication of Super-Kamiokande’s seminal paper, the phenomenon of neutrino oscillations has been established through further measurements of atmospheric neutrinos, as well as of neutrinos and antineutrinos produced in the Sun, by nuclear reactors and by high-energy particle accelerators. It is arguably the most significant advance in particle physics of the past decade.
Extending the Standard Model
Neutrino oscillations imply that the Standard Model is incomplete and must be extended to include neutrino mass as well as mixing among the three neutrino flavours. The mechanism by which neutrino mass is generated is not known. An intriguing possibility is that the tiny neutrino mass is the result of physics at extremely high energy scales. Such a “see-saw” mechanism might also help to explain why neutrino mixing is so much stronger than the mixing among quarks. Mixing among three massive neutrinos admits the possibility that symmetry between matter and antimatter (CP-symmetry) is violated via the neutrino mixing matrix. Nonzero neutrino mass implies that lepton number must be used to distinguish a neutrino from an antineutrino. If lepton number is not conserved then a neutrino is indistinguishable from an antineutrino, i.e. the neutrino is a Majorana particle – a completely new state of matter. The determination of the properties of the neutrino, therefore, is fundamental to the development of particle physics.
Oscillation processes at the Neutrino Factory.
These exciting new measurements imply that it may be possible to observe CP-violation in neutrino oscillations
Neutrino oscillations are readily described by extending the Standard Model to include three neutrino-mass eigenstates, ν 1, ν 2 and ν 3, such that the neutrino-flavour eigenstates, ν e, ν μ and ν τ, are quantum-mechanical mixtures of the mass eigenstates (figure 1). Neutrino oscillations arise from the “beating” of the phase of the neutrino-mass eigenstates as a neutrino produced as an eigenstate of flavour propagates through space and time. The matrix by which the mass-basis is rotated into the flavour-basis is parameterized in terms of three mixing angles (θ 12, θ 23 and θ 13) and one phase parameter (δ). If δ is nonzero (and not equal to π), then CP-violation in the neutrino sector will occur so long as θ 13> 0. Measurements of neutrino oscillations in vacuum can be used to determine the moduli of the mass-squared differences Δm 2 31 = m 2 3 – m 2 1 and Δm 2 21 = m 2 2 – m 2 1 and, with the aid of interactions with matter, also the sign.
The bulk of the measurements of neutrino oscillations to date have been collected using the dominant “disappearance” channels ν e → ν e and ν μ → ν μ. These data have yielded values for the three mixing angles, as well as for the magnitude of the mass-squared differences Δm 2 31 and Δm 2 21, and have shown that m 2> m 1 (i.e. that Δm 2 21> 0). Last year, the T2K, MINOS and Double Chooz experiments presented evidence that θ 13 may be greater than zero. Then, in March this year, the Daya Bay collaboration reported that sin 2 2θ 13 = 0.092 ± 0.016(stat.) ± 0.005(syst.), i.e. that sin 2 2θ 13 = 0 is excluded at 5.2 σ. The announcement was soon followed by the report of a similar result from the RENO experiment. These exciting new measurements imply that it may be possible to observe CP-violation in neutrino oscillations. The challenge for the neutrino community, therefore, is to refine the measurement of θ 13 to determine the sign of Δm 2 31 (the “mass hierarchy”), to discover CP-violation (if, indeed, it does occur) by measuring δ and to improve the accuracy with which θ 23 is known.
Over the next few years, several experiments – MINOS, T2K, NOνA, Double Chooz, Daya Bay and RENO – will exploit the ν μ→ ν e and ν e → ν x channels to improve significantly the precision with which θ 13 is known. The NOνA long-baseline experiment might also be able to determine the mass hierarchy. However, it is unlikely that either T2K or NOνA will be able to discover CP-violation, i.e. that δ ≠ 0 or π.
The Neutrino Factory
Neutrino oscillations also have implications well beyond the confines of particle physics. The possibility of CP-violation through the neutrino mixing matrix, combined with the possibility that the neutrino is a Majorana particle, makes it conceivable that the interactions of the neutrino led to the observed domination of matter over antimatter in the universe. The abundance of neutrinos in the universe is second only to that of photons. Even with a tiny mass, the neutrino may make a significant contribution to dark matter and thereby play an important role in determining the structure of the universe.
Fig. 2. Left to right: The discovery potential at 3 σ for CP-violation, the mass hierarchy and sin 2 2θ 13. The discovery reach is plotted in terms of the “CP fraction” (the fraction of all possible values of the mixing phase, δ) as a function of sin 2 2θ 13. The red solid line shows the performance of the IDS-NF baseline Neutrino Factory, while the blue line refers to the “low-energy” option, optimized for large sin 2 2θ 13.
Such a breadth of impact justifies an ambitious, far-reaching experimental programme. Determining the nature of the neutrino – whether Majorana or Dirac – through the search for neutrinoless double-beta decay (2β0ν) is an important part of this programme. The absolute neutrino mass must also be determined either through observations of 2β0ν decay or from the measurement of the end-point of the electron spectrum in beta decay. Equally important is the accurate determination of the parameters that determine the properties of the neutrino. This requires intense, high-energy neutrino and antineutrino beams – precisely what the Neutrino Factory is designed to produce.
In the Neutrino Factory, beams of ν e and ν μ (ν e,ν μ) are produced from the decays of μ+ (μ–) circulating in a storage ring. High neutrino-energies can readily be achieved because the neutrinos carry away a substantial fraction of the energy of the muon. Time-dilation is beneficial, allowing sufficient time to produce a pure, collimated beam. The table above lists the oscillation channels that are available at the Neutrino Factory. Charged-current interactions induced by ν e → ν μ oscillations – the “gold channel” – produce muons that are opposite in charge to those produced by the ν μ in the beam, so a magnetized detector is required. The additional capability to investigate the “silver” (ν e → ν τ) and “platinum” (ν μ → ν e) channels also makes the Neutrino Factory an excellent place to look for oscillation phenomena that are outside the standard three-neutrino mixing paradigm. It would be the ideal facility to serve the precision-era of neutrino-oscillation measurements.
In 2011, the International Design Study for the Neutrino Factory (the IDS-NF) collaboration presented two options for the facility in its Interim Design Report (IDR) (Choubey et al. 2011). The first, optimized for discovery reach at small θ 13 (sin 2 2θ 13< 10–2), calls for two distant detectors, with baselines of 2500–5000 km and 7000–8000 km, and a stored-muon energy of 25 GeV. The second option, optimized for sensitivity at large θ 13, requires a single detector at a distance of around 2000 km and a stored-muon beam with an energy of only 10 GeV. Figure 2 shows the discovery reach of the facility presented in terms of the fraction of all possible values of δ (the “CP fraction”) and plotted as a function of sin 2 2θ 13.
In the past few weeks, the Daya Bay and RENO collaborations have announced the first measurements of sin 2 2θ 13 with a value around 0.1. Figure 2 shows that at such a large value of θ 13, excellent performance can be achieved using the “low-energy” option. At such a large value of θ 13, the precision and discovery reach of a “low energy” Neutrino Factory is significantly better than the realistic alternatives (IDS-NF 2011).
Novel techniques
The IDS-NF baseline accelerator facility sketched in figure 3 provides a total of 10 21 muon decays per year, split between the two distant neutrino detectors. The process of creating the muon beam begins with the bombardment of a pion-production target with a pulsed proton beam. The pions are captured in a solenoidal channel in which they decay to produce the muon beam. A sequence of accelerators is then used to manipulate and reduce (cool) the muon-beam phase space and to accelerate the muons to their final energy.
Fig. 3. Schematic diagram of the IDS-NF baseline for the Neutrino Factory accelerator complex (Choubey et al. 2011). The various systems have been drawn to scale.
The muon’s short lifetime has required novel techniques to be developed to carry out these steps. Ionization cooling, the technique by which it is proposed to cool the muons, involves passing the beam through a material in which it loses energy through ionization and then re-accelerating it in the longitudinal direction to replace the lost energy. Muon acceleration will be carried out in a series of superconducting linear and recirculating linear accelerators. The final stage of acceleration, from 12.6 GeV to the stored-muon energy of 25 GeV, is provided by a fixed-field alternating-gradient (FFAG) accelerator. The baseline neutrino detector is a MINOS-like iron-scintillator sandwich calorimeter with a sampling fraction optimized for the Neutrino Factory beam. The baseline calls for a fiducial mass of 100 kilotonnes to be placed at the intermediate baseline and a detector of 50 kilotonnes at the magic baseline.
Fig. 5. The MICE Hall in June 2011. The ISIS synchrotron lies behind the wall at the back. The downstream quadrupoles (blue) are visible in front of the concrete shielding (white), with a TOF hodoscope for particle identification immediately after the final quadrupole. This hodoscope is followed by the lead-scintillator section and the prototype electron-muon ranger. In the foreground are the prototype liquid-hydrogen delivery system and R&D cryostat. Image credit: S Kill/STFC.
Much of the Neutrino Factory facility, the accelerator complex and the neutrino detectors exploit state-of-the-art technologies. To achieve the ultimate performance (10 21 muon decays per year) the IDS-NF baseline calls for: a proton-beam power of 4 MW, delivered at a repetition rate of 50 Hz in short (around 2 ns) bunches; a pion-production target capable of accepting the high proton-beam power; an ionization-cooling channel that increases the useful muon flux by a factor of around 2; and an FFAG to boost the beam energy rapidly to 25 GeV. R&D programmes that address each of these issues are underway. CERN, along with other proton-accelerator laboratories, is actively developing the technologies necessary to deliver multimega-watt, pulsed proton beams. The principle of a mercury-jet pion-production target was demonstrated by the MERIT experiment in 2008 that ran in the beamline of n_TOF, the neutron time-of-flight facility at CERN. The nonscaling FFAG accelerator EMMA (the Electron Model of Muon Acceleration, also known as the Electron Model of Many Applications) has been commissioned at the Daresbury Laboratory in the UK and used to demonstrate the “serpentine acceleration” characteristic of the nonscaling FFAG. The international Muon Ionization Cooling Experiment (MICE) at the Rutherford Appleton Laboratory will provide the engineering demonstration of the ionization-cooling technique (see box, previous page).
The Neutrino Factory is the facility of choice for the study of neutrino oscillations. It has excellent discovery reach and offers the best precision on the mixing parameters. The ability to vary the stored-muon energy and, perhaps the detector technology, gives the necessary flexibility to respond to developments in understanding neutrino physics and in the discovery of new phenomena. The R&D programme required to make the Neutrino Factory a reality will directly benefit the development of a muon collider and experiments that seek to discover charged lepton-flavour violation. The case for the Neutrino Factory as part of a comprehensive muon-physics programme is compelling indeed.
I gratefully acknowledge the help, advice, and support of my many colleagues within the IDS-NF, EUROnu and MICE collaborations and the Neutrino Factory community who have freely discussed their results with me and from whose work and results I have drawn freely.
BOX INSET
Cooling at MICE
Fig. 4. Cutaway rendering of the international Muon Ionization Cooling Experiment. The muon beam enters from the bottom left of the figure. Image credit: A DeMello/LBNL.
MICE is a single-particle experiment in which the position and momentum of each muon is measured before it enters the MICE cooling channel and is measured again after it has left (Gregoire et al. 2003 and 2005). Muons with momenta between 140 MeV/c and 240 MeV/c, with normalized emittance between 2 πmm and 10 πmm, will be provided by a purpose-built beamline at the 800 MeV proton synchrotron, ISIS, at the Rutherford Appleton Laboratory.
The MICE cooling channel, a single lattice cell, comprises three 20-l volumes of liquid hydrogen and two short linac modules each consisting of four 201 MHz cavities. Beam transport is achieved by a series of superconducting solenoids: the “focus coils” focus the beam into the liquid-hydrogen absorbers, while a “coupling coil” surrounds each of the linac modules. A particle-identification system, with scintillator time-of-flight (TOF) hodoscopes and threshold Cherenkov counters, upstream of the cooling channel allows a pure muon beam to be selected. Downstream of the cooling channel, a final hodoscope and a calorimeter system allow muon decays to be identified. The calorimeter is composed of a lead-scintillator section, of a similar design to that of the KLOE detector at DAΦNE but with thinner lead foils, followed by a fully active scintillator detector (the electron-muon ranger) in which the muons are brought to rest.
Charged-particle tracking in MICE is provided by two solenoidal spectrometers that together determine the relative change in transverse emittance of the beam, which is expected to be approximately 10%, with a precision of ±1% (i.e. a 0.1% measurement of the change in absolute emittance). The trackers themselves are required to have high track-finding efficiency in the presence of background that is induced by X-rays produced in the RF cavities.
In the first “step” of the experiment, the muon beam for MICE has been characterized using the beamline instrumentation and the TOF, Cherenkov and lead-scintillator systems (figure 5). The results, which are being prepared for publication, show that the muon beam can provide the range of momentum and emittance required by MICE. The trackers and a prototype of the electron-muon ranger have been tested and shown to perform to specification. The cavities that make up the two short linac sections have been manufactured by Lawrence Berkeley National Laboratory (LBNL). The superconducting magnets required for the cooling channel are all under construction. By the end of 2012, the collaboration will commission the two spectrometer modules and the first liquid-hydrogen absorber and focus-coil module. This will allow preliminary studies of the ionization-cooling effect to be performed. The full MICE cooling cell will be constructed once the initial cooling studies are complete.
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8184 | https://www.chegg.com/homework-help/questions-and-answers/04-prove-equation-tangent-line-point-t-xo-yo-circle-s-x-y-x2-y2-2gx-2fy-c-0-given-s-x-xo-y-q76505577 | Your solution’s ready to go!
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Question: 04 (a) Prove that the equation of the tangent line at the point T(xo, Yo) on the circle S(x, y) = x2 + y2 + 2gx + 2fy+c= 0 is given by S(x; Xo, y; yo) = xxo + yyo + g(x + xo) + f(y + yo) + c = 0. (b) If a point A(x', y') is any point, prove that A is inside, on or outside the circle S(x,y) = 0 if and only if S(x',y') ŞO. (c) Find the equation of the tangent
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8185 | https://www.wyzant.com/resources/answers/749384/ladder-problem-please-help | WYZANT TUTORING
Jeff A.
Ladder problem please help
A ladder is placed against a wall with small wheels at its top. The ladder starts at 8 ft in length and can be electronically extended to 15 ft in length. The ladder is at 8 ft in length when placed against the wall and the base is 3 ft from the wall. If the ladder extends by 1 inch per second, how fast is the tip of the ladder moving up the side of the wall when it is 8 ft in length?
1 Expert Answer
Lois C.
answered • 04/01/20
BA in secondary math ed with 20+ years of classroom experience
This is a related rate problem in which the key equation is the Pythagorean Theorem and where x is fixed at 3 feet ( 96 inches) for the distance from the bottom of the ladder to the base of the wall, y represents the height on the wall at the point where the ladder makes contact with the wall, and h is the length of the ladder.
We must first identify the known and unknown rates for all quantities involved. Measurements must be consistent in the unit of measure used, so we should convert everything either to feet or to inches. I prefer inches to avoid working with fractions in the key equation. For x, there is no rate of change as the distance from the bottom of the ladder to the base of the wall does not change throughout the problem. Regarding y, dy/dt is the rate we seek since we want to know how fast the height along the wall where the ladder makes contact with the wall is changing as the ladder is extended. Regarding h, dh/dt is given to us as 1 in/second.
We now set up the key equation based on the Pythagorean Theorem, were x 2 + y2 = h2. We then take the derivative of each side of the equation. Since the x quantity is fixed, its derivative is 0. So the new equation becomes 0 + 2y dy/dt = 2h dh/dt ( we must use the Chain Rule to take each derivative with respect to time). At the moment in question, we need the y value. By the Pythagorean Theorem, x2 + y2 = h2 becomes 362 + y2 = 962. Solving for y, we see that y is approximately equal to 88.9944. Inserting this into the derivative version of the equation, we have 2(88.9944) dy/dt = 2(96)(1). Solving this equation for dy/dt, we see that the rate of change of y at the moment in question is approximately 1.08 in/sec.
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样本方差的快速计算-遍历一遍样本集
最新推荐文章于 2024-07-13 22:36:43 发布
lxg0807于 2017-10-12 16:09:51 发布
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分类专栏：ML文章标签：方差统计学
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在统计学中，经常需要样本的方差计算，比较麻烦的方式是
（1）第一次遍历计算出样本的平均值
（2）第二次遍历才能计算出样本的方差
上述方法在样本较少的情况下，非常合适，但如果样本非常大，这种方式就不可取了，而且如果样本是持续增长的，就都不适用了。
先上代码
```
第一步：遍历一遍样本
样本e in 样本集：
self.n += 1
self.sum += e.val
self.sum_sq += e.val eval
第二步：计算样本的均值、均差和方差，
self.mean = self.sum / self.n
tmp = (self.sum_sq - (self.sum self.sum) / self.n)
self.var = tmp / (self</
```
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Variables as Elements of Sets
Suppose we have a set $A = {1, 2, 6}$. Let's also say we have a variable $x$. If you were asked if $x \in A$ is true, without knowing the value of $x$, how would you respond? Would the answer be false? Is there insufficient information so far?
Further, suppose we have another set $B = {x, y, 7}$. Is this notation even valid? Can we say that the variable $y \in B$ without knowing the value of $y$?
One last example: if we have a set $C ={x, y, z}$, to determine if $a \notin C$, would we need to know all of the following: $a \neq x$, $a \neq y$, $a \neq z$?
1 Answer 1
If someone asked you, "Is it true?" you'd need more information, and you'd probably reply with a question: "Is what true?" If you were asked, "Is he going to the party?", you couldn't answer without knowing whom "he" refers to.
Variables are like pronouns. If I were told that $A = {1,2,6}$ is and then asked, "Is $x\in A$?", without more information about $x$ no reply is possible. If I'm told that $x=4$ then I can say, No, $x\notin A$; if I'm told that $1\le x\le 2$ and that $x$ is an integer, then I know that, yes, $x\in A$.
If $x$ and $y$ have values, denote entities, then ${x,y,100}$ is well-defined; if they don't, it isn't.
Variables are not members of sets; sets (/mathematical entities) are, and variables denote them. Pronouns don't go to parties; people do, and pronouns denote them.
Re your last two examples:
For all $x,y$, if $B = {x,y,100}$ then $y\in B$. This is true. Notice that $x,y$ are bound variables here. In your first example, $x$ is free — it's not bound by a quantifier, and has no value.
If $C = {x,y,z}$, then $a\in C \iff (a=x \lor a=y \lor a=z)$, so if you negate both sides the results are equivalent too. If you know $a\notin C$ then you know that $a$ is not equal to any of $x,y,z$, and if you know the latter, then you know $a\notin C$. This holds for all $x,y,z,a$ (bound variables again).
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Difference Between pi and 22/7
Thread starter mathdad
Start date
mathdad
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#1
What is the difference (if there is one) between pi and 22/7?
Dr.Peterson
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Joined
: Nov 12, 2017
Messages
: 16,842
#2
22/7 is a rational approximation to pi, just as 3.14 is. That is, both 22/7 and 3.14 are rational numbers (fraction and terminating decimal, respectively) that are close enough to the exact value of pi for elementary work. Pi is an irrational number, namely 3.1415926535897932384626433832795... 22/7 is a rational number, namely 3.1428571428571428571428571428571... 3.14 is a rational number, namely 3.1400000000000000000000000000000... We could say that the difference between pi and 22/7 is 0.00126448926734961868021375957764
mathdad
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#3
Dr.Peterson said:
22/7 is a rational approximation to pi, just as 3.14 is. That is, both 22/7 and 3.14 are rational numbers (fraction and terminating decimal, respectively) that are close enough to the exact value of pi for elementary work. Pi is an irrational number, namely 3.1415926535897932384626433832795... 22/7 is a rational number, namely 3.1428571428571428571428571428571... 3.14 is a rational number, namely 3.1400000000000000000000000000000... We could say that the difference between pi and 22/7 is 0.00126448926734961868021375957764
Click to expand...
More interesting notes for my files. There are questions that specifically state to use 22/7 or 3.14 or to leave the answer in terms of pi. Why so?
Dr.Peterson
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: 16,842
#4
mathdad said:
More interesting notes for my files. There are questions that specifically state to use 22/7 or 3.14 or to leave the answer in terms of pi. Why so?
Click to expand...
When a book says to use 22/7 or 3.14, that is probably either because they don't want the student to use a calculator, and want to keep the manual work simple, or just to make sure everyone's answer agrees, so it's easier to check, and easier for students when they look in the back of the book. When it says to leave the answer in terms of pi, it may be for more or less the same reasons, or to emphasize that we have an exact answer in that form. Commonly we keep work in terms of named constants rather than specific numbers (until getting a numerical value at the end) for several reasons: to leave the decision how much precision to use for last; to allow simplification; to see better how the answer depends on various constants; or just to avoid having to copy lots of (rounded) decimal places and risk mistakes at every step. I suspect that many authors don't even think through why they do this; it's just tradition.
Steven G
Elite Member
Joined
: Dec 30, 2014
Messages
: 14,594
#5
mathdad said:
More interesting notes for my files. There are questions that specifically state to use 22/7 or 3.14 or to leave the answer in terms of pi. Why so?
Click to expand...
Seriously, ask the author.
mathdad
Full Member
Joined
: Apr 24, 2015
Messages
: 925
#6
Jomo said:
Seriously, ask the author.
Click to expand...
Sometimes teachers do the same thing in the classroom.
mathdad
Full Member
Joined
: Apr 24, 2015
Messages
: 925
#7
Dr.Peterson said:
When a book says to use 22/7 or 3.14, that is probably either because they don't want the student to use a calculator, and want to keep the manual work simple, or just to make sure everyone's answer agrees, so it's easier to check, and easier for students when they look in the back of the book. When it says to leave the answer in terms of pi, it may be for more or less the same reasons, or to emphasize that we have an exact answer in that form. Commonly we keep work in terms of named constants rather than specific numbers (until getting a numerical value at the end) for several reasons: to leave the decision how much precision to use for last; to allow simplification; to see better how the answer depends on various constants; or just to avoid having to copy lots of (rounded) decimal places and risk mistakes at every step. I suspect that many authors don't even think through why they do this; it's just tradition.
Click to expand...
Perhaps it is "just tradition" as you said. I hope my questions are not dull or boring for the site. Honestly, I think you would love to have me in your class because I sincerely love math. I am different than most students. Older people are better students by far. I find myself buried more in my math books than I do the Bible. I am a Christian. I love God. God is not too happy that right now math is my passion. However, God created math. Therefore, I am hoping that He will understand. BTW, God ALWAYS wants to be first.
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8189 | https://www.nagwa.com/en/videos/515178251514/ | Question Video: Finding the Intervals of Increasing and Decreasing of a Function Involving Logarithmic Functions Using the Chain Rule Mathematics • Third Year of Secondary School
Find the intervals on which the function 𝑓(𝑥) = 5 ln (−4 ln 𝑥 + 6) is increasing and decreasing.
Video Transcript
Find the intervals on which the function 𝑓 of 𝑥 is equal to five times the natural logarithm of negative four times the natural logarithm of 𝑥 plus six is increasing and decreasing.
We’re given a function 𝑓 of 𝑥. And we need to determine the intervals on which this function is increasing and the intervals on which this function is decreasing. And the first thing we should always do when we’re asked a question like this is to determine the domain of our function 𝑓 of 𝑥. The reason we want to find the domain is we can’t possibly say that our function 𝑓 of 𝑥 is increasing or decreasing on an interval if our function 𝑓 of 𝑥 is not defined across the entire interval. So let’s determine the domain of our function 𝑓 of 𝑥.
First, we can see that our function 𝑓 of 𝑥 contains the natural logarithm of 𝑥. And the natural logarithm of 𝑥 is only defined when 𝑥 is positive. So we know that 𝑥 must be greater than zero. Next, we can also see we’re taking the natural logarithm of negative four times the natural logarithm of 𝑥 plus six. And if we’re taking the natural logarithm of this, then it must be positive. Otherwise, 𝑓 of 𝑥 will not be defined. And for any other value of 𝑥, we’re just taking the natural logarithm of positive numbers. So this will be defined. So we need to find the values of 𝑥 where negative four times the natural logarithm of 𝑥 plus six is positive.
We can just solve this inequality. We’ll start by subtracting six from both sides. We get negative four times the natural logarithm of 𝑥 is greater than negative six. Next, we’ll divide our inequality through by negative four. Remember, because this is a negative number, we need to switch the direction of our inequality. This gives us the natural logarithm of 𝑥 should be less than negative six divided by negative four, which we simplify to give us three over two. To solve this inequality, we need to recall that the exponential function is an increasing function. In other words, if the natural logarithm of 𝑥 is less than three over two, then 𝑒 to the power of the natural logarithm of 𝑥 will be less than 𝑒 to the power of three over two. Finally, we know the exponential function and the natural logarithm functions are inverses. So this gives us that 𝑥 will be less than 𝑒 to the power of three over two.
So we found our domain 𝑓 of 𝑥. 𝑥 must be positive; however, 𝑥 must be less than three over two. And because in the question, we’re working with intervals, we’ll write this in interval notation. The domain of 𝑓 of 𝑥 is the open interval from zero to 𝑒 to the power of three over two. Now that we found the domain of 𝑓 of 𝑥, we’re ready to start finding the intervals on which 𝑓 of 𝑥 is increasing and decreasing.
Let’s start by clearing some space. To answer this question, we need to recall the following piece of information about increasing and decreasing functions. We know if 𝑓 prime of 𝑥 is greater than zero on an interval 𝐼, then 𝑓 of 𝑥 will be increasing on that interval 𝐼. However, if 𝑓 prime of 𝑥 is less than zero on our interval 𝐼, then we know that 𝑓 of 𝑥 will be decreasing on that interval 𝐼. In other words, to determine where our function 𝑓 of 𝑥 is increasing or decreasing, we can instead look at the sign of 𝑓 prime of 𝑥. And to do this, we’re going to need to differentiate our function 𝑓 of 𝑥. We can see that 𝑓 of 𝑥 is the composition of two functions. So we’re going to want to do this by using the chain rule.
So let’s start by recalling the following version of the chain rule. This tells us if 𝑢 and 𝑣 are differentiable functions, then the derivative of 𝑢 of 𝑣 of 𝑥 with respect to 𝑥 is equal to 𝑣 prime of 𝑥 times 𝑢 prime of 𝑥 evaluated at 𝑣 of 𝑥. To apply this, we’re first going to need to write 𝑓 of 𝑥 as the composition of two functions. We’ll need to set 𝑣 of 𝑥 to be our inner function. We’ll set 𝑣 of 𝑥 to be negative four times the natural logarithm of 𝑥 plus six. So, by setting 𝑓 of 𝑥 equal to this, we can see that 𝑓 of 𝑥 is five times the natural logarithm of 𝑣. Then all we need to do is set our function 𝑢 of 𝑣 to be the remaining function. 𝑢 of 𝑣 is five times the natural logarithm of 𝑣.
And now we can see we’ve successfully written 𝑓 of 𝑥 as 𝑢 evaluated at 𝑣 of 𝑥. So we can find 𝑓 prime of 𝑥 using the chain rule. To use the chain rule, we’re going to need to find the expressions for 𝑣 prime of 𝑥 and 𝑢 prime of 𝑣. Let’s start with 𝑣 prime of 𝑥. That’s the derivative of negative four times the natural logarithm of 𝑥 plus six. We can differentiate this term by term. Remember, the derivative of the natural logarithm function is the reciprocal function. We get 𝑣 prime of 𝑥 is negative four divided by 𝑥. And in fact we can do the same to find 𝑢 prime of 𝑣; we get 𝑢 prime of 𝑣 is five divided by 𝑣.
We can now find an expression for 𝑓 prime of 𝑥 by substituting our expressions for 𝑣 prime and 𝑢 prime into the chain rule. We get 𝑓 prime of 𝑥 is equal to negative four over 𝑥 multiplied by five divided by 𝑣 of 𝑥. And of course, we know that 𝑣 of 𝑥 is negative four times the natural logarithm of 𝑥 plus six. So we substitute this in. This gives us the following expression. Finally, we’ll multiply these two expressions together to give us that 𝑓 prime of 𝑥 is equal to negative 20 divided by 𝑥 times negative four multiplied by the natural logarithm of 𝑥 plus six.
We now need to determine on which intervals is this positive and on which intervals is this negative. And remember, we’ve shown the domain of our function 𝑓 of 𝑥 is the open interval from zero to 𝑒 to the power of three over two and in particular on this interval 𝑥 is positive. So let’s look at our expression for 𝑓 prime of 𝑥. We have 𝑥, and we know that this is positive. The next question we need to ask is, what happens to negative four times the natural logarithm of 𝑥 plus six on this interval? There’s a few different ways of answering this question. For example, we could use a graphical approach; however, we’re going to use the fact that the natural logarithm of 𝑥 is an increasing function.
If the natural logarithm of 𝑥 is an increasing function, then on the open interval from zero to 𝑒 to the power of three over two, we must have the natural logarithm of 𝑥 is less than the natural logarithm of 𝑒 to the power of three over two. And this is because the natural logarithm is an increasing function. The higher our input of 𝑥, the bigger the output. So we just inputted the largest value in our interval. But we can evaluate the natural logarithm of 𝑒 to the power of three over two because the natural logarithm and exponential functions are inverse functions. It’s just equal to three over two. What we’ve shown is on the domain of 𝑓 of 𝑥, the natural logarithm of 𝑥 is less than three over two. And if the natural logarithm of 𝑥 is less than three over two, what does this mean about negative four times the natural logarithm of 𝑥 plus six?
And there’s a few different ways of answering this question. We’re going to do this by recreating this from our inequality. We’ll start by multiplying both sides of our inequality through by negative four. And of course, because this is negative, this switches the direction of our inequality. We have negative four times the natural logarithm of 𝑥 must be bigger than negative six. Now all we need to do is add six to both sides of our inequality. This gives us that negative four times the natural logarithm of 𝑥 plus six is positive on the domain of 𝑓 of 𝑥. So on the entire domain of 𝑓 of 𝑥, negative four times the natural logarithm of 𝑥 plus six is positive. So, in fact, what we’ve shown is on the entire domain of 𝑓 of 𝑥, 𝑓 prime of 𝑥 is a negative number divided by the product of two positive numbers.
Therefore, 𝑓 prime of 𝑥 is negative on the entire domain of 𝑓 of 𝑥. And we know if 𝑓 prime of 𝑥 is negative on the entire domain of 𝑓 of 𝑥, then 𝑓 of 𝑥 is decreasing on the entire domain of 𝑓 of 𝑥. And of course, this is setting 𝐼 equal to the domain of 𝑓 𝑥, which we’ve shown is the open interval from zero to 𝑒 to the power of three over two.
Therefore, we were able to show for the function 𝑓 of 𝑥 is equal to five times the natural logarithm of negative four multiplied by the natural logarithm of 𝑥 plus six is decreasing on the open interval from zero to 𝑒 to the power of three over two.
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Antenna Design for Engineers
The value of distance by meter (m) wavelength (λ) 0.375 0.375 0.2λ 0.3 0.3λ 0.45 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 0.4λ 0.6 1. The docu…
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Study and Design "Yagi-Uda Crossed Antenna" using 4NEC2
Hesham Mokhtar Rasim Amer Ali
College of Engineering Sciences and Technology. Sebha University
Emails:- rasim63213@gmail.com Hma_trix2@yahoo.com
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ABSTRACT
Boils down to this project in the study of (Yagi-Ud a Crossed Antenna), and the study was in the bandwidth of (180-220MHz), using 4NEC2 " wired antennas simulator", and the simulation process shows that the characteristics of this antenna has been improved through automatic modification in the antenna amounts , and trough the optimization process the obtaining gain was having a maximum value of 14.17dBic.
.
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INTRODUCTION
Wireless communication is one of the most important technologies rapidly grow and spread, where the demand for data transmission without cables and by different distances in increasing every day ,therefo re become the development of wireless communi cation systems is absolutely imperative, it is also known that the communications system generally consists of, the source , which is asource of the information or data to be sent, the power converter is usually converts information from the source into electrical signals to be able to send it via system transmission, the sender is based processing electrical signals issued from the converter to be suitable for transmission through the communication channel, the transmission channel which is the medium that connects between the sender and the receiver, and finally the receiver to extraction the data from the signal transmitter and delivered to the power converter of output system which converts these signals to the original form as it was before sending it. The role of the antenna here in the sender and any future it an essential part and indispensable in the communications system as that in the case of sending it converts electrical energy from the transmission line which connected with to the electromagnetic energy represented in waves transmitted into the air or vice versa as in the case of reception depending on the type and characteristics of the antenna. And generally the Wired antenna characteri zed by simply inst alled ,Where any wire or metal surface have the ability to be antenna and be able to pick up electr omagnetic waves but p ractic al antennas ha ve its forms an d specific dimensions determined by the purpose for which it was made. And In this work was designed and study radiological characteristics to one of the most important applications of (Yagi-Uda) antenna, a (Yagy-Uda Crossed Antenna) at bandwidt h of (180-220 MHz),it’s a very practical antenna , Because the waves that radiate be circular or linear polarized
and This is appropriate in some applications such as satellite communi cation systems and some radars, The designed antenna has
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been simulated by using an a pplication software for the antennas wired design "4NEC2" , This simulation allows view detailed properties of the antenna, which provides high accuracy in the results And economize time and effort, The software also features the ability to improve the radiation characteristics of the antenna designed, This is happened by doing an automatic changing process to the input dimensions of the antenna for the optimization process and choose the best values of results after finishing the Automatic comparison process .
1.The Software ''2NEC4'' :-
(NEC) is the abbreviation to (Numerical Electric Code) which a way for simulate wired antennas, attributed to Gerald J. Burke and Andrew J. Poggio originally was created by using FORTRAN language in the mid-seventies, and to achieve this simulation the antenna must divided into a small sections linearly with a different values of current and voltage in each segment, the NEC is based on (Moment Methods theory), therefore the results are very accurate and typical, It also provides a lot of effort and time because of the difficulty and complexity of the mathematical analysis using this method, and over time increased the flaws and weaknesses in the simulation using NEC For example, the resulting simulation errors when wires are crossed in a very short distances or when using coated wire, In addition it was a highly confidential for a very long time were not allowed even marketed on the Internet, This has been overcome defects in 4NEC2,which programmed by (Arie Voors) and so named because it combines proportion between versions NEC2 and NEC4, In addition to it is free and also mentions in this field the simulation program"EZNEC", which is not as potential strong and effectiveness 4NEC2 .
Moment Methods :-
Method of moments technique known method to solve linear equations, The conversion is used integral equations for the electric field to the matrix equations or system of linear equations. Transactions can be found using the current fragmentation of matrixes (LU), or using
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| In the Bohr model of the hydrogen atom, the electron (–e) is in orbit about the nuclear proton (+e) at a radius of r=5.29×10–11 m, as Figure 18.11 shows. Determine the speed of the electron, assuming the orbit to be circular. | | | | --- | In the Bohr model of the hydrogen atom, the electron ( e) orbits the proton ( e) at a distance of r 5.29 10 11 m. The velocity of the electron is v. | | | | Figure 18.11 In the Bohr model of the hydrogen atom, the electron (– e) orbits the proton (+ e) at a distance of r = 5.29 × 10–11 m. The velocity of the electron is v. | | Reasoning Recall from Section 5.2 that any object moving with speed v on a circular path of radius r has a centripetal acceleration of ac=v2/r. This acceleration is directed toward the center of the circle. Newton’s second law specifies that the net force SF needed to create this acceleration is SF=mac=mv2/r, where m is the mass of the object. This equation can be solved for the speed: . Since the mass of the electron is m=9.11×–31 kg and the radius is given, we can calculate the speed, once a value for the net force is available. For the electron in the hydrogen atom, the net force is provided almost exclusively by the electrostatic force, as given by Coulomb’s law. This force points toward the center of the circle, since the electron and the proton have opposite signs. The electron is also pulled toward the proton by the gravitational force. However, the gravitational force is negligible in comparison to the electrostatic force. | | | --- | | | | | Problem solving insight When using Coulomb’s law (F=k |q1||q2|/r2), remember that the symbols |q1| and |q2| stand for the charge magnitudes. Do not substitute negative numbers for these symbols. | | Solution The electron experiences an electrostatic force of attraction because of the proton, and the magnitude of this force is | | | --- | | | | Using this value for the net force, we find | | | --- | | | | This orbital speed is almost five million miles per hour. |
| | |
In the Bohr model of the hydrogen atom, the electron (–e) is in orbit about the nuclear proton (+e) at a radius of r=5.29×10–11 m, as Figure 18.11 shows. Determine the speed of the electron, assuming the orbit to be circular.
| | | |
---
| In the Bohr model of the hydrogen atom, the electron ( e) orbits the proton ( e) at a distance of r 5.29 10 11 m. The velocity of the electron is v. | | | | Figure 18.11 In the Bohr model of the hydrogen atom, the electron (– e) orbits the proton (+ e) at a distance of r = 5.29 × 10–11 m. The velocity of the electron is v. | |
In the Bohr model of the hydrogen atom, the electron (
e) orbits the proton (
e) at a distance of r
5.29
10
11 m. The velocity of the electron is v.
| |
| Figure 18.11 In the Bohr model of the hydrogen atom, the electron (– e) orbits the proton (+ e) at a distance of r = 5.29 × 10–11 m. The velocity of the electron is v. |
Reasoning
Recall from Section 5.2 that any object moving with speed v on a circular path of radius r has a centripetal acceleration of ac=v2/r. This acceleration is directed toward the center of the circle. Newton’s second law specifies that the net force SF needed to create this acceleration is SF=mac=mv2/r, where m is the mass of the object. This equation can be solved for the speed: . Since the mass of the electron is m=9.11×–31 kg and the radius is given, we can calculate the speed, once a value for the net force is available. For the electron in the hydrogen atom, the net force is provided almost exclusively by the electrostatic force, as given by Coulomb’s law. This force points toward the center of the circle, since the electron and the proton have opposite signs. The electron is also pulled toward the proton by the gravitational force. However, the gravitational force is negligible in comparison to the electrostatic force.
| | |
--- |
| | | | Problem solving insight When using Coulomb’s law (F=k |q1||q2|/r2), remember that the symbols |q1| and |q2| stand for the charge magnitudes. Do not substitute negative numbers for these symbols. | |
| | |
--- |
| | | | Problem solving insight When using Coulomb’s law (F=k |q1||q2|/r2), remember that the symbols |q1| and |q2| stand for the charge magnitudes. Do not substitute negative numbers for these symbols. | |
| | |
--- |
| | | | Problem solving insight When using Coulomb’s law (F=k |q1||q2|/r2), remember that the symbols |q1| and |q2| stand for the charge magnitudes. Do not substitute negative numbers for these symbols. | |
| |
| Problem solving insight When using Coulomb’s law (F=k |q1||q2|/r2), remember that the symbols |q1| and |q2| stand for the charge magnitudes. Do not substitute negative numbers for these symbols. |
Solution
The electron experiences an electrostatic force of attraction because of the proton, and the magnitude of this force is
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--- |
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Using this value for the net force, we find
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This orbital speed is almost five million miles per hour.
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| Copyright © 2000-2003 by John Wiley & Sons, Inc. or related companies. All rights reserved. | |
8192 | https://www.youtube.com/watch?v=7c9KyjASnhQ | Master Solving for x given parallel lines and a transversal
Brian McLogan
1600000 subscribers
Description
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Posted: 15 Aug 2017
Subscribe! Want more math video lessons? Visit my website to view all of my math videos organized by course, chapter and section. The purpose of posting my free video tutorials is to not only help students learn math but allow teachers the resources to flip their classrooms and allow more time for teaching within the classroom. Please feel free to share my resources with those in need help.
30 comments
Transcript:
welcome ladies gentlemen so what i'd like to do is show you how to solve for x when we have parallel lines now it's very important when we have pair lines we have um our angle relationships or actually just when we have two lines in a transversal we have angle relationships however when we have parallel lines those angle relationships now are going to those angle relationships now actually have some you know meaning and i'll kind of discuss each one of those the first thing though i'd like to do for each one of these problems is first identify what the angle relationship is okay so if we can at least get to that start that therefore therefore that's at least starting with okay here's what i recognize so in this case um you can see that these two angles are like in the exact same position of the intersection they're in the top left quadrant so those are what we call corresponding angles okay now it's important for corresponding angles for us to remember that corresponding angles are exactly the same so if i had you know the measure of angle a is equal to the measure of angle b that is for two corresponding angles however that is only true you know if you look at the theorems and positives so that's only true when you have parallel lines in a transversal okay so it has to have pair of lines but you can see for this whole video all my two lines are parallel okay so um we have these two angles are not going to be equal so all i need to do them to solve is i just need to do i just need to set up an equation x minus 18 is equal to 115. add 18 add 18 to both sides x is going to equal that would be 133 okay now again they're exactly the same so you could replug in 133 in for x and it actually really just kind of depends on the problem you know you solve them for x are you trying to find the angle measure in this example we're trying to find uh we're actually just solving for the value of x so but you can see that these have to be the same 133 minus 15 or minus 18 is going to give you 115 anyways but um there you go that is uh that is at least our first one the next one here is you can see that we have two angles that are on the interior and they're on the same side as the um they're on the same side of the transversal line the transversal line is a line that's going to intersect your parallel lines so those are what we call consecutive interior now consecutive interior angles these two angles right here um base when you have parallel lines consecutive interior angles are supplementary and again that goes back to a theorem that you'll have in your geometry book so i know that these two angles since they're supplementary because we have pair of lines since they're supplementary they're going to add up to 180. so i'm going to write the equation x plus 1 plus x plus 4 equals 180 degrees now i didn't really need to put parentheses in there but i just wanted to put parentheses so you can see how i differentiated between those two angles now i can go and combine these this is going to be 2x plus 5 equals 180 degrees uh 3 minutes, 18 seconds subtract is 175 that's not divisible by two what the hell oh jesus |
8193 | https://www.youtube.com/watch?v=d93AarE0lKg | Arctangent of a complex number
Mostly Math
1540 subscribers
86 likes
Description
6631 views
Posted: 23 May 2021
I use the complex forms of the sine and cosine functions to write an expression for the arctangent of a complex variable. I solve the resulting linear equation and plug in a few known examples to confirm the formula's validity before evaluating it for a complex angle.
6 comments
Transcript:
hi everyone welcome to mostly math as you can see this is a special episode since we have a special guest a caricature that my wife drew of me for my son since he likes to draw he'll be with us for the remainder of this episode hope you enjoy and in this video we are going to be extending the arc tangent function to complex values so typically we can only evaluate um our tangent of one or arc tangent of uh pi over two or something like that well in this video we want to be considering our tangent of any complex number say uh square root of 2 e to the i pi over 7 or something like that we want to be able to evaluate the arc our tangent for complex values why just because we think it should be possible and let's let's go ahead and get started on that we're going to begin by letting our complex function w or omega it doesn't matter to be the arc tangent of z which of course tells us that tan of omega or w equals z and from here we just want to use the complex version of the sine and cosines and see where it gets us okay as we know if we let z equal to tangent omega well it's going to be sine omega of course over cosine omega which we know from elementary complex number theory would be one over two i e to the i omega minus e to the minus i omega over one half e to the i omega plus e to the minus psi omega and of course we see that the one halves cancel out and we just have an i down here fantastic so we have e to the i omega minus e to the minus omega over i times e to the i omega plus e to the minus i omega and here we want to do a pretty obvious trick we're going to multiply by e to the plus i omega and when we do this we get e to the 2 i omega minus 1 over i e to the 2 i omega plus one looks pretty symmetric and now we want to solve for omega or w in terms of z so the next step is to multiply both sides of the equation by the denominator which is i e to the two omega plus one excuse me so i z e to the two i omega plus one is equal to e to the 2 i omega minus 1 and basically we want to solve for e to the 2i omega that's pretty easy all we have to do is subtract this or yeah whichever way i'm i'm going to do it yeah okay so [Music] this tells us we have which one do i want to do it let's see yeah i want to move the left hand side over to the right hand side so i want 0 to be equal to factoring out uh i z here so we have one minus i z e to the two i omega we need to subtract an i z here so we get minus i z minus one or plus one ah let's see four minus one plus i z and now we have very interesting symmetry here we have one plus i z one minus i z we can obviously see that we're close to the right answer we just need to solve for e to the two i omega now [Applause] is simply equal to this add this to both sides 1 plus i z over 1 minus i z we can easily now take the log of both sides and divide by 1 over 2y to get the final answer we have omega which is actually equal to the arc tangent of z just equal to 1 over 2i natural log one plus i z over one minus i z that's our cool formula this is what we need to know for the rest of the video we have derived it should be correct there are several forms that you'll see on wikipedia and so forth but this is the one that i believe to be the most straightforward to derive and remember since it has the factor of one over two i up front uh there are other ways to go um for example here in the derivation instead of multiplying by e omega e to the omega you could call this something a for example and then this would be you know a squared minus one over i a squared plus one and you'd solve for a and there'd be a square root and you'd use the properties of the logarithm to take the one half out front which you need to do if you want to repeat the same derivation for other trigonometric functions uh like uh like the inverse sine of z you'd have to do that trick but for the arc tangent it works out nicely if you keep it like this i hope you enjoyed that derivation it's the slickest one i've seen now let's move on to some straightforward examples first we want to verify that the formula that we derived actually reduces to what we know about the arctangent formula for real variables before we go ahead and plug in complex variables into it so i'm just going to demonstrate one value here probably the easiest value we want to know that our tangent of one is equal to pi over four we want to know that because tan tan um pi over four is equal to one since uh we have the right triangle here pi over four the sine is actually equal to the cosine of both one over square root of two so when you divide them they should be equal to one we want to verify that this is true so we're going to start by evaluating inverse tangent of one is by our formula here one over two i log of one plus i over one minus i so all i've done is plugged z equals 1 into this formula here and now we just have to work with the 1 plus i over one minus psi number here we're gonna do the obvious thing multiply top and bottom by one plus i or to rationalize the denominator so we get 1 plus i over 1 minus i plus i or you can use the formula z z star is equal to magnitude of z squared which for a complex number z equals a plus b i it's just equal to a squared plus b squared that's how i always evaluate these sort of problems so just 1 squared plus 1 squared which is 2 so it's 1 plus i over 2. and this is squared of course evaluating the square one-half we get 1 squared plus 2i minus i sorry plus i squared which is 1 and these go away it just becomes i so interestingly 1 plus i over 1 minus i is actually just equal to i perhaps you knew that at some point i had forgotten that so this problem was a nice reminder that the strange identity is true so it's just one over two i natural log of i which and this is the trick to evaluating these sort of logarithms we want to write the complex number that we're trying to evaluate in polar form which is particularly easy for i 1 over 2 i log of e to the i pi over 2 at least one value the arc tangent is multi-valued we're going to be choosing the principal branch here and log e by pi over 2 is just 1 over 2 i i pi over two which is pi over four as we hoped so right yeah that's the same thing it's the same thing as what i wrote here obviously same thing just take the tangent to both sides so we have indeed showed that it holds for at least one value you can check other values perhaps by computer we're going to do one other value as well now that we have some indication that it should hold for real values we're going to look at a purely complex value to see how we can use this formula since the formula is really not useful unless you use it for something so to the end we are going to be evaluating the inverse tangent of 2 plus i so why did i choose 2 plus i instead of 1 plus i well you see that the math works out slightly better in terms of the polar coordinate description you might ask why i did not choose i or minus i well actually the arc tangent function has poles there in the complex planar or rather a branch point so it's not defined at these values otherwise i would have chosen those values next simplest one is one plus i which is messy and two plus is slightly more simpler so that's why i'm giving you this example all right let's begin 1 over 2 i log of 1 plus i times 2 plus i over 1 minus i times 2 plus i which just uh simplifying we have 2i here the i squared becomes minus 1. that cancels with that so it's just 2i up top and we have a minus 2i minus i squared is plus 1. so it becomes 2 minus 2i obviously the twos can factor out everywhere so it becomes one minus i here and now we're at the same situation that we were before we just want to write this in polar coordinates so let's go ahead and do that i over one minus i is i'm multiplying by one plus i on both sides over two again yeah yeah two just two all good [Applause] yep so far so good we get an i minus 1 minus 1 plus i over 2. now we want to write this in polar form we could use the straightforward techniques from elementary complex variable theory to just write it in r e to the i theta form by brute force but we want to recognize that this is closely related to the pi over four that we spoke about earlier so i'm going to write it as one over square root of 2 times yeah times minus 1 plus i over square roots of 2. and this we actually recognize as 1 over square root of 2 e to the something related to pi over 4. i'll show you in a second that it's 3 pi i over 4. since we have unit circle here we have pi pi over 4 here pi over 4 sorry 4 here and this has 1 over square root of 2 here for the sine and then 1 over square root of 2 for the cosine but this has the first quadrant negated so it must be over here and this tells us the angle we seek is actually three pi over four and this has that was a terrible horizontal vertical line sorry this has a minus one of square over two for the cosine and one over square root of two for this sign so it must be one over square root of two e to the minus three pi r over four now we can proceed by taking the logarithm more straightforwardly that was in the side and now we have our main result is equal to 1 over 2 i natural log of one over square root of two e to the minus three pi i over four that was on the side okay by the properties of logs if we take a log of a product we simply add the logs 1 over 2i log of 1 over square root of 2 well that's just gonna be minus one half log of two using a different property of logs the one over it negates it and the square root can come outside now we have a minus yeah log e three pi i over four so looking good so far blade shouldn't both be negative ah yes i made a mistake this is actually plus sorry about that the minus angle will be in this quadrant this is plus plus sorry about that and now we basically have our final answer here we're going to get a minus one quarter log two here and the i is going to cancel here so our final answer is now now let's see three pi over eight plus i over four log two that should be the simplest way to write it i typed it on mathematica so it should be right and this is how you use the formula to calculate log of a complex number maybe you don't need this particular one for anything but it's just an example the video was more to show you the proof of the formula i think the proof is more fun than applying it hopefully you thought so too and for homework you can feel free to try and derive the other formulas like uh sine inverse c cos inverse z etc wikipedia tells me that if you are smart enough you can write them in terms of a general formula and do one derivation for all the inverse trig functions i'm not smart enough at this time to do that but for homework i think you should be able to do these two if you apply the techniques and use the method of letting uh a be e to the i omega and then writing everything's in terms of a instead of a linear equation for these two you'll actually have a quadratic equation and there's a sample proof of this on wikipedia if you get stuck as well so if you enjoyed this want to see more please subscribe to my channel i'll see you next time |
8194 | https://arxiv.org/pdf/2507.16629 | Some classes of finite-dimensional ladder operators ∗
Fabio Bagarello †
, Antonino Faddetta ‡
, Francesco Oliveri §
July 23, 2025
Abstract
We introduce and study some special classes of ladder operators in finite-dimensional Hilbert spaces. In particular we consider a trun-cated version of quons, their psudo-version, and a third family of op-erators acting on a closed chain. In this latter situation, we discuss the existence of what could be considered discrete coherent states , as suitable eigenvectors of the annihilation operator of the chain. We see that, under reasonable assumptions, a resolution of the identity can be recovered, involving these states, together with a biorthogonal family of vectors, which turn out to be eigenstates of the raising operator of the chain.
Keywords : quons, ladder operators, closed chain, coherent states.
MSC classification : 46N50, 81Q12.
∗
To appear on Journal of Physics A: Mathematical and Theoretical, 2025.
†
Dipartimento di Ingegneria, Universit` a di Palermo, Viale delle Scienze, Edificio 8, I–90128 Palermo, Italy; I.N.F.N., Sezione di Catania, Italy. fabio.bagarello@unipa.it
‡
Dipartimento di Fisica e Chimica, Universit` a di Palermo, Via Archirafi 36, I–90128 Palermo, Italy. antonino.faddetta@community.unipa.it
§
Dipartimento di Scienze Matematiche e Informatiche, Scienze Fisiche e Scienze della Terra, Universit` a di Messina, Viale F. Stagno d’Alcontres 31, I–98166 Messina, Italy. francesco.oliveri@unime.it
1
arXiv:2507.16629v1 [math-ph] 22 Jul 2025
1 Introduction
The role of ladder operators is very useful in many fields of physics, from quantum mechanics to quantum field theory, passing through many body problem and complex systems. There are hundreds of books dealing with this class of operators, both in view of their applications but also because they produce a beautiful mathematics. We cite here only few monographs on this topics [1–4], but many others can be easily found, with focus on different aspects of ladder operators, different extensions, and different realms of applications. Bosons and fermions are the two classes of ladders which are likely the most known since the beginning of quantum mechanics, the reason being that they describe interactions between particles (the bosons), and the par-ticles themselves (the fermions). We could think of photons and electrons, respectively, just to give two concrete examples of bosons and fermions, re-spectively. Nevertheless, there are many other classes of ladder operators in the literature. Quons are one such class [5, 6]. Abstract ladder opera-tors produce another class. And there are more: those related to the angular momentum [1–3], or those arising from the generalized Heisenberg algebra . And more. Some of them live in finite-dimensional, others in infinite-dimensional Hilbert spaces. This aspect is related to the explicit definition of the operators. Moreover, some operators originally defined in infinite-dimensional Hilbert spaces admit a counterpart in finite-dimensional ones. This restriction can be motivated by physical reasons: for instance, bosons are used in the diagonalization of the quantum harmonic oscillator. Its Hamiltonian, Hqo , admits an infinite set of (equally spaced) eigenvalues and an infinite set of mutually orthogonal eigenvectors. Anyway, it is clear that in a realistic situation not all the energetic levels can be reached by the oscillator. Only a (possibly large) finite number of levels can be occupied and, therefore, the original Hilbert space L2(R) is too big ! It is sufficient to consider the subspace of L2(R) which is generated by a finite number of eigenvectors of Hqo , say those corresponding to the energetic levels which can really be occupied. This kind of restriction is discussed, for instance, in [9–11], where truncated bosons have been analyzed. We should also men-tion that all the ladder operators named so far can be extended or deformed in such a way that the raising operator is not necessarily the adjoint of the lowering operator, as it often happens in quantum mechanical systems. Arecent overview of many of these extensions is contained in . 2The main focus in this paper is to consider families of ladders which live in finite-dimensional Hilbert spaces different from those considered so far in the existing literature, and to deduce their main properties. In particular, we will concentrate on a finite-dimensional version of quons, and on their pseudo-deformations. Then, we will discuss also ladder operators on a closed chain. More in detail, the paper is organized as follows. In Section 2, we will introduce the truncated quons (TQs), whereas in Subsection 2.1 we will deform these TQs, so getting truncated pseudo-quons (TPQs) similar to those considered for infinite-dimensional Hilbert spaces in . Section 3 is devoted to the analysis of ladders on a closed chain, with a special focus on the possibility of defining some sort of coherent states for the chain. Section 4 contains our conclusions and plans for the future. To keep the paper self-contained, we list some results on quons in the Appendix.
2 From bosons to TQs
Our main starting ingredient is an operator a satisfying, together with its adjoint a†, the canonical commutation relation (CCR) [ a, a †] = 1 1. Here, 1 1is the identity operator on H, the Hilbert space where a and a† act. It is well known that (i) H is necessarily infinite-dimensional and that (ii) a and
a† are unbounded. An important consequence of the CCR is that, calling
Na = a†a, then [ Na, a ] = −a. This is one of the key identities needed to use a and a† to diagonalize, in a purely algebraic way, the Hamiltonian of a harmonic oscillator. In [9,10], it has been shown that it is possible to construct a sort of trun-cated version of the operators a, a † which live in finite dimensional Hilbert spaces. For that, a possible approach consists in replacing the CCR with a different commutation rule. Indeed, let us consider an operator A which satisfies, together with A†, the commutation relation [A, A †] = 1 1L − (L + 1) K0, (1) where 0 < L < ∞, 1 1L is the ( L + 1) × (L + 1) identity matrix, and K0 is an orthogonal projector ( K0 = K20 = K†
0
) satisfying K0A = 0. In this case it is possible to prove that A admits a representation as a matrix of order ( L + 1). In other words, A is an operator acting on HL, where dim( HL) = L + 1, and A† is the adjoint of A under the usual scalar product in HL. We stress 3once more that this finite-dimensional representation is not possible if the
correction of the CCR proportional to K0 is missing in (1). To be more explicit, we recall that the original bosonic operator a satisfies the lowering property on H
ae n = √n e n−1, n ≥ 1, where ae 0 = 0 , (2) with its standard representation
a =
∞
X
n=0
√n + 1 |en⟩⟨ en+1 |. (3) This series converges on Le = l.s {en}, the linear span of the vectors en,where Fe = {en, n ≥ 0} is the orthonormal (o.n.) basis of H constructed in the standard way: e0 is a vector in H satisfying ae 0 = 0, and en =
1√n!
(a†)ne0, n ≥ 1. It is well known that, computing [ a, a †] (in the sense of unbounded operators), we get the identity operator on H. Indeed, using a formal approach (which, however, could be made rigorous), we find that [a, a †] =
∞
X
n=0
|en⟩⟨ en| = 1 1,
since Fe is an o.n basis for H, and therefore resolves the identity. An inter-esting feature of a is that it admits, among the others, the following (infinite-dimensional) matrix representation:
a =
0 1 0 0 0 . . .
0 0 √2 0 0 . . .
0 0 0 √3 0 . . .
0 0 0 0 √4 . . .
... ... ... ... ... ...
.
Let us now consider a sort of cut-off on a, replacing the infinite series in (3) with the finite sum
A =
L−1
X
n=0
√n + 1 |en⟩⟨ en+1 |, (4) 4so that A† =
L−1
X
n=0
√n + 1 |en+1 ⟩⟨ en|, where en ∈ F e. It is possible to check that (4) produces an explicit realization of (1), and that A, A†, and other relevant operators, can all be represented in a Hilbert space HL ⊂ H for which Fe(L) = {en, n = 0 , 1, 2, . . . , L }, finite subset of Fe, is an o.n. basis.
HL can be seen as the set of all the vectors f of H for which, when expanded in terms of Fe, we get f =
∞
X
n=0
cnen with cn = 0 for all n > L . Now we can easily check that [A, A †] =
L
X
n=0
|en⟩⟨ en| − (L + 1) |eL⟩⟨ eL|,
which is exactly formula (1) with
L
X
n=0
|en⟩⟨ en| = 1 1L, the identity in HL (with dimension L + 1), and K0 = |eL⟩⟨ eL|. Notice that, as required, K0 = K†
0
=
K20 . The matrix expression of A is
A =
0 1 0 0 0 · · · · · · 00 0 √2 0 0 · · · · · · 00 0 0 √3 0 · · · · · · 00 0 0 0 √4 · · · · · · 0... ... ... ... ... ... ... 00 0 0 0 0 0 · · · √L
0 0 0 0 0 0 · · · 0
,
which looks as a truncated version of the matrix expression for a.Hereafter, what we are interested to is the possibility of repeating a similar construction for quons (see Appendix), in particular to understand if the approach described here for bosons is the only one, or if some similar, but possibly slightly different, technique can also be proposed. More explicitly, adopting the same notation as in Appendix, our main effort here is to understand if the operator c satisfying, as in (47), [ c, c †]q :=
cc † − qc †c = 1 1, q ∈ [−1, 1], can be somehow replaced by a different operator,
C, satisfying a sort of generalized version of (1): [C, C †]q = 1 1L − (L + 1) K, (5) 5where 0 < L < ∞ is again connected to the dimensionality of a suitable Hilbert space where the operators C and C† can be represented as ( L +1) × (L + 1) matrices, and K is some operator required to get this finite-dimensional representation, as for the bosonic case.
Remark 1 It is useful to prove that [c, c †]q = 1 1 can be defined on some finite-dimensional Hilbert space only in few very special cases. Indeed, if we assume that [c, c †]q = 1 1 can be implemented in some HL, with dimension L + 1 < ∞,then taking the trace of both sides of the identity [c, c †]q = 1 1, and using the properties of the trace, we would get (1 − q) tr (c†c) = L + 1 . Now, since c†c
is a positive operator, its trace is non negative, and it is strictly positive if we exclude the trivial case that c†c = 0 . Hence, q = 1 is not compatible with this result (in agreement with what we know for bosons). Now, if we assume first that c is described by a truncated (L + 1) × (L + 1) version of (49) with
βn given in (51), we see that tr (c†c) =
L−1
X
n=0
β2
n
, and therefore, using (51), we should have L−1X
n=0
(1 − qn+1 ) = L + 1 ,
which is in general not true for our values of q. The only cases where this equality can be satisfied is when L is odd and q = −1. For L = 1 , we have the fermionic case, and our conclusion is, therefore, expected. There is also another solution of [c, c †]q = 1 1: in fact, if q ∈ [−1, 1[ , the
(L + 1) × (L + 1) matrix
c = 1
√1 − q
0 1 0 0 . . . 00 0 1 0 . . . 00 0 0 1 . . . 0
... ... ... ... ... ...
0 0 0 0 . . . 11 0 0 0 . . . 0
.
satisfies [c, c †]q = 1 1L in HL, for all finite integer L. However, this situation is not so interesting for us, since the matrices for different values of q are simply proportional one to the other. In this sense, this particular solution is trivial.
6We assume here, in view of Remark 1 and of the expression (49) for c,that C admits the representation
C =
0 β0 0 0 0 · · · · · · 00 0 β1 0 0 · · · · · · 00 0 0 β2 0 · · · · · · 00 0 0 0 β3 · · · . 0... ... ... ... ... ... ... ...0 0 0 0 0 0 · · · βL−1
0 0 0 0 · · · · · · · · · 0
, (6) where βn here will be later compared with those given in (51); then, we look for the consequences of this ansatz. Before starting our analysis, it might be useful to stress that, for quons, the commutation rule [ Na, a ] = −a is replaced by a slightly different expression, [Nc, c ] = −c + (1 − q)Ncc, (7) where we have introduced the quonic number-like operator Nc = c†c. Notice that the one here is the standard commutator, while, as clearly stated before, [., . ]q is the q-mutator, which reduces to [ ., . ] only when q = 1, and to the anti-commutator {., . } when q = −1. For the truncated version C, we will be interested to consider the following two questions: 1. Find, if possible, conditions on βn which produce, in analogy with (7), [NC , C ] = −C + (1 − q)NC C, (8) where NC = C†C. Once these expressions of βn are found, if possible, we will use (5) to deduce the expression for K. In other words: is it possible to recover for the truncated quons the same commutation rule with the number-like operator as for ordinary quons? 2. Find, if possible, conditions for (8) to reduce to a boson-like rule [NC , C ] = −C, (9) while keeping q̸ = 1. This constraint on q makes the question non trivial. Indeed, if q = 1 formula (9) is exactly what we do expect, and there is not much to say. Different, and possibly more interesting, is the situation when q̸ = 1. 7We will show in the rest of this Section that the answers to both these questions are affirmative. Let us first consider condition (8), with C as in (6). Restricting to βk̸ = 0 for all k, it is easy to check that β2
k
= 1 + q + · · · qk.Notice that
k
X
j=0
qj ≥ 0 for all allowed values of q. This is indeed not different from what we have found in (51): the expression of βk remains the same both in H and in HL (and this is why we have used the same notation in both cases).
Remark 2 It may be interesting to stress that other solutions for βk also exist such that (8) is satisfied. In particular we could have β0 = β1 = . . . =
βj0 = 0 , for some j0 < L − 1. In this case C annihilates a set of orthogonal vectors, and therefore dim(ker( C)) > 1. We will not consider this other possibility further in this paper.
From now on, to fix the ideas, we only consider βk > 0: if βk =
qPkj=0 qj ,we can deduce the expression of K inverting (5). We get
K =
0 0 . . . 0 00 0 . . . 0 0... ... . . . ... ...0 0 . . . 0 00 0 . . . 0 β2
L
L+1
, (10) which, as we can see, has a single non vanishing entry in the ( L + 1 , L + 1) position. Incidentally we observe that KC = 0 and that K = K†. However,
K = K2 is possible only if q = 1. Otherwise K̸ = K2. This is similar, but not identical, to what we have found for the truncated bosons, and for K0.It is interesting to look at the same problem from a more abstract point of view, i.e. , without assuming that C has the form in (6), with βk as above. In particular, if we assume that C obeys (5), and that KC = 0, it is easy to check that [ NC , C ] is the one in (8), independently of the explicit representation of
C.We can rewrite the operators C and K in terms of rank one operators as follows:
K = β2
L
L + 1 |eL⟩⟨ eL| and C =
L−1
X
n=0
βn |en⟩⟨ en+1 |. (11) 8Here en is the n-th vector of Fe(L) ⊂ F e, the same sets we have introduced for bosons and truncated bosons. We observe that, for q = 1, K coincides with
K0 and the operator C becomes the analogous operator A for the truncated bosons. If q = −1 and L = 1 we recover fermionic operators. If L ≥ 2and q = −1 we get extended fermionic operators, similar to those considered in : in particular, the expression above for K shows that K = 0 for odd
L, while K̸ = 0 for even L.Formula (6) easily produces the following lowering and raising equations for the operators C and C†, respectively:
Ce n = βn−1 en−1, 0 ≤ n ≤ L, β−1 = 0 ,C†en = βn en+1 , 0 ≤ n ≤ L − 1,
(12) which can be extended to all of HL by linearity. This is not a problem, here, since L < ∞, and C and C† are obviously bounded. From (12) we also deduce the following eigenvalue equations for NC : NC en = β2
n−1
en for
n = 0 , 1, 2, . . . , L .The vectors en, n = 1 , 2, . . . , L , are related to e0 as follows:
en := (C†)n
βn−1! e0, n = 0 , . . . , L. (13) Hence, C† is a raising operator, satisfying ( C†)L+1 = 0. The main difference with respect to the truncated bosons relies, as it is clear, in the normalization factor βn−1!, which reduces to √n! when q = 1. We refer to the Appendix for the meaning of βn−1!, known sometimes as the q-factorial. Next we go to formula (9). We recall that what is interesting for us here is to check if (9) is compatible with (5) for q̸ = 1, and which are the properties of K in this case. The approach is the same as above. In this case, using the expression in (6) for C in (9), we find that the solution for C, restricting as before to βj > 0, reads
C =
0 1 0 0 . . . 00 0 √2 0 . . . 00 0 0 √3 . . . 0... ... ... ... . . . ...0 0 0 0 . . . √L
0 0 0 0 . . . 0
, (14) 9which is exactly the matrix of the truncated bosons A we have seen before. This does not fix q = 1 necessarily. Indeed, inserting now this solution in (5), we find that K must be
K =
0 0 0 . . . 0 00 q−1
L+1
0 · · · 0 00 0 2( q−1)
L+1
· · · 0 0... ... ... ... ... ...0 0 0 · · · (L−1)( q−1)
L+1
00 0 0 · · · 0 1+ Lq L+1
, (15) which is still diagonal but quite different from what we have found before, and which can be rewritten as
K = q − 1
L + 1
L−1
X
n=0
n |en⟩⟨ en| + 1 + qL L + 1 |eL⟩⟨ eL|. (16) It is particularly interesting to notice that KC ̸ = 0. In fact, it turns out that KC = q − 1
L + 1 NC C. This result can be somehow reversed: if we assume that C obeys (5) and that KC = q − 1
L + 1 NC C, then it is easy to verify that [NC , C ] is the one in (9), and the operators C and K have the expressions (14) and (15), respectively. Again, we have lowering and raising properties for the operators C and
C†, but with purely bosonic weights:
Ce n = √n e n−1, 1 ≤ n ≤ L, Ce 0 = 0 ,C†en = √n + 1 en+1 , 0 ≤ n ≤ L − 1,
(17) and we recover the eigenvalue equation NC en = n e n for n = 0 , 1, 2, . . . , L .Also in this case, the vectors en can be constructed out of e0 as follows:
en := (C†)n
√n! e0, n = 0 , . . . , L. (18) In conclusion, we are back to truncated bosons, but these truncated bosons obey also a truncated q-mutation relation, see (5), if K is chosen as in (15) and (16). 10 2.1 Deformed TQs
In view of some older results in the literature (see , and references therein), it might be interesting to sketch briefly if and how it is possible to deform TQs in order to obtain truncated pseudo-quons (TPQs), where the ladder operators C and C† are replaced by a pair D and G which are still ladder operators living in HL, but such that G†̸ = D. Hereafter, we will propose a very simple (but still non trivial) procedure, similar to what has been done in the literature for regular pseudo-bosons . The main ingredients are the operators C and C†, satisfying (5), and where (6) is the matrix form of C.Let R be an invertible operator on HL. We use R and R−1 to define D and
G as follows:
D = RCR −1, G = RC †R−1. (19) It is clear that, if R is not unitary, which is the case we will only consider here, D̸ = G†. Using (5) we see that D and G satisfy the following q-mutator: [D, G ]q = R[C, C †]qR−1 = R (1 1L − (L + 1) K) R−1 = 1 1L − (L + 1) Q, (20) where Q = RKR −1. It is clear that Q2 = Q if K2 = K. This is not true, however, neither for the choice (10) of K, nor for (15), except when q = 1. Also, in general, Q†̸ = Q. However, Q† = Q if [ K, R †R] = 0, since K = K†,at least for K given in (10) and (15). Moreover, whenever KC = 0 and K = K† (which is what we have de-duced out of (8)), we have QD = RKCR −1 = 0, so that D†Q† = 0. The same reasoning implies that Q†G† = 0, so that GQ = 0 as well. Similarly, whenever KC = q − 1
L + 1 NC C and K = K† (which is what we have deduced from (9)), introducing ˜N = GD we deduce that QD =
q − 1
L + 1 ˜N D , implying that D†Q† = q − 1
L + 1 D† ˜N †, and also Q†G† = q − 1
L + 1 ˜N †G†,leading to GQ = q − 1
L + 1 G ˜N .
Example 1 Here, we will show how the above operators look like, starting from fixed (but arbitrarily chosen) biorthogonal sets Fφ = {φn, n = 0 , 1, 2, 3}
and Fψ = {ψn, n = 0 , 1, 2, 3} for the Hilbert space H3 = C4, and using these
11 sets as the main ingredient of all our construction. Let us consider
φ0 =
1000
, φ1 =
1100
, φ2 =
0110
, φ3 =
0011
, (21)
and
ψ0 =
1
−11
−1
, ψ1 =
01
−11
, ψ2 =
001
−1
, ψ3 =
0001
.
(22)
It is clear that, in particular, ⟨φn, ψ m⟩ = δn,m . This implies that the four vectors in Fφ are linearly independent, and therefore that Fφ is a basis for
H3. The same is true for Fψ. Hence these two sets solve the identity:
3
X
n=0
|ψn⟩⟨ φn| =
3
X
n=0
|φn⟩⟨ ψn| = 1 13.
Let us now fix βn ∈ R, n = 0 , 1, 2, and define
E =
2
X
n=0
βn|φn⟩⟨ ψn+1 |, F =
2
X
n=0
βn|φn+1 ⟩⟨ ψn|. (23)
Given f ∈ C4, we have
EF f =
2
X
n=0
β2
n
⟨ψn, f ⟩φn, F Ef =
2
X
k=0
β2
k
⟨ψk+1 , f ⟩φk+1 , (24)
so that, after some straightforward computations,
[E, F ]qf = EF f − qF Ef = 1 13f − 4Qf, (25)
where we have introduced the operator Q, say
Q = 14 (1 − β20 )|φ0⟩⟨ ψ0| +
2
X
l=1
(1 − β2
l
qβ 2
l−1
)|φl⟩⟨ ψl| + (1 + qβ 22 )|φ3⟩⟨ ψ3|
!
,
12 or, in matrix form,
Q = 14
1 − β20 (q + 1) β20 − β21 β21 − (q + 1) β20 (q + 1) β20 − β21
0 1 + qβ 20 − β21 (q + 1) β21 − qβ 20 − β22 qβ 20 − (q + 1) β21 + β22
0 0 1 + qβ 21 − β22 (q + 1) β22 − qβ 21
0 0 0 1 + qβ 22
.
The operators E and F satisfy the commutation rule in (20), [E, F ]q =113 − 4Q. Next we want to show that E and F are related to an operator C
(and its adjoint), obeying (5), as in (19). For that, we first observe that Fφ
and Fψ are Riesz bases. Indeed, we have φn = R e n and ψn = ( R−1)† en,where Fe = {ek} is the canonical o.n. basis in H3, and R is the invertible operator
R =
1 1 0 00 1 1 00 0 1 10 0 0 1
. (26)
We can now determine the operator K inverting the original formula for
Q, Q = RKR −1:
K = R−1QR = 14
1 − β20 0 0 00 1 + qβ 20 − β21 0 00 0 1 + qβ 21 − β22 00 0 0 1 + qβ 22
.
(27)
Now, since
C =
2
X
n=0
βn|en⟩⟨ en+1 |, (28)
we can rewrite (27) as K = 14
113 − CC † + qC †C, which is exactly formula (5). Notice also that E = RCR −1 and F = RC †R−1.
Going back now to our definition of Q, Q = RKR −1, in the easiest case in which [ K, R ] = 0 then it follows that Q = K. Hence, [D, G ]q = 1 1L − (L + 1) K, (29) which also implies that [ G†, D †]q = 1 1L − (L + 1) K. We are interested in finding conditions such that the analogous of (8) and (9) hold true. In other 13 words, we want to understand here when the following commutators are satisfied: [ ˜N , D ] = −D + (1 − q) ˜N D, [ ˜N †, G †] = −G† + (1 − q) ˜N †G†, (30) and [ ˜N , D ] = −D, [ ˜N †, G †] = −G†. (31) The equations in (30) are satisfied if KD = 0 and KG † = 0, which are both true since KC = 0. For K as in (10), it is clear that R must admit the representation
R =
R1,1 R1,2 R1,3 . . . R1,L 0
R2,1 R2,2 R2,3 . . . R2,L 0
R3,1 R3,2 R3,3 . . . R3,L 0... ... ... . . . ... ...
RL, 1 RL, 2 RL, 3 . . . RL,L 00 0 0 . . . 0 RL+1 ,L +1
, (32) with det( R)̸ = 0, as R−1 must exist. If we rather consider (31), then a solution is given by imposing that KD =
q − 1
L + 1 ˜N D and KG † = q − 1
L + 1 ˜N †G†, which are true since KC = q − 1
L + 1 NC C.In this case, K has the expression in (15), and R must have the form
R =
R1,1 0 0 . . . 0 00 R2,2 . . . 0 00 0 R3,3 . . . 0 0... ... ... . . . ... ...0 0 0 . . . RL,L 00 0 0 . . . 0 RL+1 ,L +1
, (33) with all the elements Rj,j ̸ = 0, in order to have det( R)̸ = 0.
3 Ladders for a closed chain
In this Section, we will construct and study ladder operators which ap-pear to be very different from those considered above. In particular, we focus on a closed chain, i.e. , an ( N + 1)-levels system for which a ladder 14 (or shift) operator can be defined mapping the j-th to the ( j + 1)-th level,
j = 0 , 1, 2, . . . , N − 1, and mapping the N -th back into the zero level. For concreteness, we will choose N = 3 here, so that H3 = C4. The generaliza-tion to higher values of N is straightforward, although the formulas become rapidly longer. Let Fe = {ej , j = 0 , 1, 2, 3} be the canonical o.n. basis for H3, ⟨ej , e k⟩ =
δj,k , and let {γj , j = 0 , 1, 2, 3} be a set of real and positive numbers. Let us consider an operator a† on H3 acting on Fe as follows:
a†e0 = γ1e1, a†e1 = γ2e2, a†e2 = γ3e3, a†e3 = γ0e0. (34) Its matrix expression is
a† =
0 0 0 γ0
γ1 0 0 00 γ2 0 00 0 γ3 0
so that a =
0 γ1 0 00 0 γ2 00 0 0 γ3
γ0 0 0 0
. (35) Therefore
ae 0 = γ0e3, ae 1 = γ1e0, ae 2 = γ2e1, ae 3 = γ3e2. (36) The reader should be aware that the operator a here has nothing to do (except for its ladder nature) with the bosonic operator a introduced in Section 2, and that’s why we decided to keep this notation. We see that, with this choice, no vector ej is annihilated by either a or a†. This is different from what happens for bosons and quons, truncated or not, since they all have a vacuum (for instance, see (2)), or for fermions, where both the zero and the excited levels are annihilated by b and b†, respectively, where {b, b †} = 1 1. Our interest in the operators a and a† in (35) is based on the following aspects. First of all, we are interested in describing a system with no vacuum, which appears as a finite dimensional version of what has been recently considered in in connection with graphene and coherent states, or in for a quantum particle on a circle. Secondly, we want to check if it is possible to construct some sort of coherent state for the operator a, even in our finite-dimensional settings. This is well known to be impossible as long as a vacuum exists, since an equality aΦ( z) = zΦ( z) could never be satisfied by any non-zero Φ( z) and for all z ∈ C. The reason is simple 1. As Fe is an o.n. basis for
1We sketch the proof in our H3=C4.It is trivial to generalize it to any dimension
N < ∞.
15 H3, it is clear that we can expand Φ( z) as follows: Φ( z) = P3
k=0
ck(z)ek,where ck(z) are z-dependent coefficients. If A is a generic lowering operator for Fe, and if Ae 0 = 0, it is clear that AΦ( z) is a linear combination of e0, e 1
and e2, but not of e3. On the other hand, zΦ( z) =
3
X
k=0
z c k(z)ek. It is easy to check that the only solution, independently of the details of action of the lowering operator A, is that ck(z) = 0 for all k. Hence, Φ( z) = 0, which is not allowed 2. Last but not least, we would like to consider new operators that can be eventually used in some dynamical system, on the same lines it has been done in recent years in many applications where ladder operators of different nature appear in the analysis of macroscopic systems, [15–17]. Going back to the matrices a and a† in (35), we see that
N = a†a =
γ20 0 0 00 γ21 0 00 0 γ22 00 0 0 γ23
, (37) which is diagonal despite of the presence of the γ0 in a and a†. The commu-tator between these two operators is also diagonal, [a, a †] =
γ21 − γ20 0 0 00 γ22 − γ21 0 00 0 γ23 − γ22 00 0 0 γ20 − γ23
=: Γ (38) Incidentally, we observe that [ a, a †]̸ = 1 1 for any choice of the γj ’s. This is not a surprise, as it is in agreement with well known facts for bosonic operators: no operator X can satisfy [ X, X †] = 1 1 in any finite dimensional Hilbert space. However, these operators share some interesting features with ordinary bosons. Indeed, we have [ N, a ] = −Γ a and [ N, a †] = a†Γ. We also see that [ N, Γ] = [ M, Γ] = 0, where M = aa † = Γ + N . These commutators are similar to the analogous results we get for bosons, as we can check by replacing 1 1 with Γ. In particular, if we call a(t) = exp( iN t )a exp( −iN t ) (as a sort of Heisenberg dynamics for a, driven by the “Hamiltonian” H = N ),
2The obvious (and well known) way out is to work in an infinite dimensional Hilbert space. In this case, a non trivial solution can be usually found. This is the case, just to cite an example, of the bosonic lowering operator.
16 it is easy to see that
da (t)
dt = i exp( iN t )[ N, a ] exp( −iN t ) = −iΓa(t) ⇒ a(t) = exp( −iΓt)a,
as Γ( t) = exp( iN t )Γ exp( −iN t ) = Γ. This implies that a†(t)a(t) = a†a, as it is obvious because [ H, N ] = 0, if H = N .A simple way to see that our ladder operator a describes a closed chain consists in observing that (a†)4 = γ0γ1γ2γ311 = a4, (39) which implies that the fourth power of both a and a† are simply proportional to the identity operator: we have just gone all over the chain, going back to the initial state.
Remark 3 It is easy to understand that the operator a† can be expressed in terms of a family (bj , b †
j
) of fermionic operators as follows: let bj be such that
{bj , b †
k
} = δj,k 11, and b2
j
= 0 , j = 0 , 1, 2, 3, with all the other anti-commutators being zero. The operator bj annihilates a particle in the j-th level, while b†
j
creates such a particle. Hence, we can rewrite
a† =
3
X
j=0
bj b†
j+1
,
with the identification b†
4
≡ b†
0
; a† and P3
j=0
bj b†
j+1
behave effectively in the same way. Anyway, it is clear that the use of a is much more efficient than adopting the bj operators.
Remark 4 Another possible choice for a in (35) satisfying (39) was already considered in the Remark 1. In fact, setting
a = 1
√1 − q
0 1 0 00 0 1 00 0 0 11 0 0 0
,a satisfies (39) if q is such that γ0γ1γ2γ3 = 1(1 −q)2 . Notice that this is surely possible whenever γ0γ1γ2γ3 ≥ 14 .
It is clear that all the results we have deduced here can be easily general-ized to HL for L̸ = 3. Notice that, here, L + 1 is just the length of the closed chain we are interested to. 17 3.1 Discrete coherent states for a
We have already shown that, for a purely lowering operator in a finite dimen-sional vector space, it is not possible to find a vector which could be called acoherent state . Hereafter, we propose a general proof of this statement, to-gether with some related results. In the last part of this Section, we go back to our ladder operators a and a†, and analyze in some details what Theorem 1 below implies. Let A be a general n × n matrix (no other requirement: in particular, we are not assuming A is a lowering operator as we did before). Our general question is the following: is it possible to find a z-dependent non-zero vector Φ( z) ∈ H , H = Cn, such that AΦ( z) = zΦ( z) for all z ∈ C? Of course, we cannot expect the answer be positive in general, as it is well known that coherent states for fermionic operators ( n = 2 and A = σ+, one of the Pauli matrices) can only be defined if z is replaced by some Grassmann number . Indeed, this is a general aspect, which does not depend on the specific form of the matrix A.
Theorem 1 The only solution of the identity AΦ( z) = zΦ( z), ∀z ∈ C,is Φ( z) = 0 . However, it is possible to find z1, z 2, . . . , z n, not necessarily different, and n non-zero vectors Φ( zj ), j = 1 , 2, . . . , n , such that
AΦ( zj ) = zj Φ( zj ), (40)
j = 1 , 2, . . . , n . Moreover, if these zj are all different, FΦ = {Φ( zj ), j =1, 2, . . . , n } is a basis for H. Its biorthonormal basis, Fψ = {ψ(zj ), j =1, 2, . . . , n }, is a set of eigenstates of A† with eigenvalues zj :
A†ψ(zj ) = zj ψ(zj ), (41)
j = 1 , 2, . . . , n .
Proof. First we observe that, rewriting AΦ( z) = zΦ( z) as ( A−z11)Φ( z) = 0, this admits a non-trivial solution only if p(z) = det( A − z11) = 0. But p(z)is a polynomial in z of degree n. Hence, it has exactly n roots, z1, z 2 . . . , z n
not necessarily all different. For all ˜ z ∈ C not in σ(A) = {z1, z 2 . . . , z n}, the set of eigenvalues of A, p(˜ z)̸ = 0, and the only solution of AΦ(˜ z) = ˜ zΦ(˜ z) is Φ(˜ z) = 0: hence, AΦ( z) = zΦ( z) does not admit a non trivial solution Φ( z)for all z ∈ C.18 Now, let us assume that all the eigenvalues zj of A are distinct. Since we are not assuming that A = A†, each zj is in general complex. If they are all different, the n eigenvectors in FΦ are all linearly independent. Hence, they form a basis for H. Thus [18, 19], there exists a unique biorthonormal basis Fψ = {ψ(zj ), j = 1 , 2, . . . , n }. Then we have ⟨Φ( zj ), ψ (zk)⟩ = δj,k , and therefore
⟨Φ( zj ), A †ψ(zk)⟩ = ⟨AΦ( zj ), ψ (zk)⟩ = zj ⟨Φ( zj ), ψ (zk)⟩ = zj δj,k == zkδj,k = ⟨Φ( zj ), z kψ(zk)⟩, ∀j, k = 1 , 2, . . . , n.
This implies, in particular, that ⟨Φ( zj ), A †ψ(zk) − zkψ(zk)⟩ = 0, for all j.Formula (41) now follows from the fact that FΦ is complete.
□
From this theorem we conclude that each vector f ∈ H can be written as
f =
n
X
j=1
⟨Φ( zj ), f ⟩ψ(zj ) =
n
X
j=1
⟨ψ(zj ), f ⟩Φ( zj ), (42) or, equivalently, that
n
X
j=1
|ψ(zj )⟩⟨ Φ( zj )| =
n
X
j=1
|Φ( zj )⟩⟨ ψ(zj )| = 1 1. (43) The interpretation is that, under the assumption of the theorem and inde-pendently of the fact that A is a ladder operator or not, a sort of discrete set of coherent states can be found, and this set, together with its biorthogonal family, resolves the identity in H.
Remark 5 It might appear not so natural to call these vectors coherent states, since these are states with very specific properties . However, we like to keep this name, in view of the application to the chain, and to the fact that also ordinary coherent states obey a discrete resolution of the iden-tity .
Let us now go back to the operators a and a† in (35). The eigenvalues for a are
E0 = −γ, E1 = −iγ, E2 = iγ, E3 = γ, (44) 19 where γ = 4
√γ0γ1γ2γ3 > 0. The corresponding eigenvectors are Φ( z0) = 12
− γγ0
, γ2
γ0γ1
, −γ3
γ , 1
T
,
Φ( z1) = 12
−i γγ0
, − γ2
γ0γ1
, i γ3
γ , 1
T
,
Φ( z2) = 12
i γγ0
, − γ2
γ0γ1
, −iγ3
γ , 1
T
,
Φ( z3) = 12
γγ0
, γ2
γ0γ1
, γ3
γ , 1
T
.
(45) The biorthogonal set Fψ = {ψ(zj )} is the set of the following vectors:
ψ(z0) = 12
−γ0
γ , γ0γ1
γ2 , − γγ3
, 1
T
,ψ(z1) = 12
iγ0
γ , −γ0γ1
γ2 , −i γγ3
, 1
T
,ψ(z2) = 12
−iγ0
γ , −γ0γ1
γ2 , i γγ3
, 1
T
,ψ(z3) = 12
γ0
γ , γ0γ1
γ2 , γγ3
, 1
T
.
(46) It is possible to check that a†ψ(zj ) = Ej ψ(zj ), j = 0 , 1, 2, 3, in agreement with Theorem 1, and that FΦ and FΨ, together, resolve the identity, as in (43).
Remark 6 It is not difficult to extend this construction to higher dimen-sions, i.e. to operators a and a† that, despite of formulas in (35), are repre-sented by larger M × M matrices, as far as they still represent ladders on a closed chain. Of course, they will act on some o.n. basis which is made by
M elements, in a way which extends the one in (34).
We could expand these vectors in terms of the canonical orthonormal basis Fe = {ej , j = 0 , 1, 2, 3}. For instance, we have Φ( z0) = 12
− γγ0
e0 + γ2
γ0γ1
e1 − γγ3
e2 + e3
,
20 so that, using (37),
N Φ( z0) = 12
−γγ 0e0 + γ2γ1
γ0
e1 − γγ 22
γ3
e2 + γ23 e3
.
This result suggests that Φ( z0) is not stable under the time evolution pro-duced by N , i.e. , under the operator exp( iN t ). However, this does not exclude that exp( iN t )Φ( z0) can coincide, or be proportional, to some of the other vectors in (45) or in (46) for some specific time t. This is another (big) difference with what people usually call coherent states, since these are stable under time evolution . Another difference with the canonical coherent states (as those attached to the harmonic oscillator), is that, here, both the lowering and the raising operators, a and a†, admit eigenstates belonging to the Hilbert space where the operators act, and that these two families are biorthonormal. This is due to the fact that we are working with a simplified functional settings, i.e. ,in a finite dimensional Hilbert space. This, as already commented before, produces several interesting consequences.
4 Conclusions
In this paper, we have considered different classes of finite-dimensional ma-trices obeying interesting algebraic relations, and with interesting physical interpretations. In particular, we have considered first a truncated version of quons, and discussed some of their possible deformations, possibly useful for non-Hermitian Quantum Mechanics. Then, we have introduced and an-alyzed ladder operators which are defined on certain closed chains of finite length. These latter operators can be seen as discrete translation operators, moving a given particle from one site of the chain to another. We have further considered the general problem of the existence of a sort of coherent state for our lowering operators, and we have deduced that this is possible, but in a slightly different way with respect to what it is usually done for canonical coherent states. Indeed, we found a finite set of vectors which are eigenstates of the lowering operator A and which are (under mild assumptions) a basis for H. Its (unique) biorthonormal basis turns out to be a set of eigenstates of A†. Together, these two families resolve the identity. In a forthcoming paper, now in preparation, we plan to use the operators considered here in a dynamical context , i.e. , in the construction and analysis 21 of some specific dynamical systems described in terms of operators, on the same lines widely discussed in [15–17].
Acknowledgments
F. B. acknowledges partial financial support from Palermo University and by project ICON-Q, Partenariato Esteso NQSTI - PE00000023, Spoke 2.. F. O. acknowledges partial financial support from Messina University. F. B. and F. O. acknowledges partial financial support from the G.N.F.M. of the INdAM and by the PRIN grant Transport phenomena in low dimensional structures: models, simulations and theoretical aspects - project code 2022TMW2PY -CUP B53D23009500006. A. F. acknowledges partial financial support from Palermo University for his visit in Messina, and all the people in Messina University for their warm welcome during his days in the MIFT Department.
A Quons
In this appendix, we briefly review few facts on quons, starting with their definition. Quons are defined by q-mutator
[c, c †]q := cc † − qc †c = 1 1, q ∈ [−1, 1] , (47) where the creation and the annihilation operators are c† and c and of course the CCR are obtained for q = 1, while CAR (canonical anticommutation relations) for q = −1. Note that the condition c2 = 0 is not required here a priori, so that it should be added, when q = −1. For q in the interval ( −1, 1), we have that (47) describes particles which are neither bosons nor fermions. In , it is proved that the eigenstates of N0 = c† c are analogous to the bosonic ones, except that for the normalization. A simple concrete realization of (47) can be deduced as follows. Let Fe = {ek, k = 0 , 1, 2, . . . } be the canonical orthonormal basis of the Hilbert space H = l2(N0), with all zero entries except that in the ( k + 1)th position, which is equal to one:
⟨ek, e m⟩ = δk,m . (48) 22 If we take
c =
0 β0 0 0 0 0 · · ·
0 0 β1 0 0 0 · · ·
0 0 0 β2 0 0 · · ·
0 0 0 0 β3 0 · · ·
0 0 0 0 0 β4 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·
, (49) then (47) is satisfied if
β20 = 1 and β2
n
= 1 + qβ 2
n−1
, ∀n ≥ 1, (50) which implies that
β2
n
=
n + 1 , if q = 1 ,
1−qn+1
1−q
, if q̸ = 1 .
(51) In older papers the authors have often restricted to βn > 0 for all n ≥ 0. The above form of c shows that c e 0 = 0, and c† behaves as a raising operator. From (49) we deduce
en+1 = 1
βn
c†en = 1
βn!(c†)n+1 e0, (52) for all n ≥ 0. Here we have introduced the q-factorial :
βn! = βnβn−1 · · · β2β1, (53) which is often indicated as [ n]q in some literature. Of course, from (52) it follows that
c†en = βnen+1 . (54) Using (49), it is also easy to check that c acts as a lowering operator on Fe:
c e m = βm−1em−1 ∀m ≥ 0, (55) where it is also useful to introduce β−1 = 0, in order to ensure c e 0 = 0. Then we have
N0em = β2
m−1
em, ∀m ≥ 0. (56) 23 The operator N is formally defined in as
N = 1log( q) log(1 1 − N0(1 − q)) for 0 < q < 1, (57) and satisfies the eigenvalue equation
N e m = me m, ∀m ≥ 0. (58) It might be interesting to stress that there are other ways to represent the operator c in (49). We refer to for such an alternative representation in L2(R).
References References
E. Merzbacher, Quantum Mechanics , Wiley, New York, (1970) A. Messiah, Quantum mechanics , vol. 2, North Holland Publishing Com-pany, Amsterdam, (1962) C. Cohen-Tannoudji, B. Diu, F. Lalo¨ e, Quantum mechanics , vol. 1, Wiley-VCH, (1977) J-P. Gazeau, Coherent states in quantum physics , Wiley-VCH, Berlin (2009) R.N. Mohapatra, Infinite statistics and a possible small violation of the Pauli principle , Phys. Lett. B, 242 , 407-411, (1990) O.W. Greenberg, Particles with small violations of Fermi or Bose statis-tics , Phys. Rev. D, 43 , 4111-4120, (1991) F. M. Fernandez, Symmetric quadratic Hamiltonians with pseudo-Hermitian matrix representation , Ann. Phys. 369 , 168-176 (2016) E. M. F. Curado, Y. Hassouni, M. A. Rego-Monteiro, Ligia M.C.S. Ro-drigues, Generalized Heisenberg algebra and algebraic method: The ex-ample of an infinite square-well potential , Physics Letters A, 372 , 3350-3355 (2008) 24 H. A. Buchdahl, Concerning a kind of truncated quantized linear har-monic oscillator , Amer. Jour. of Phys., 35 , 210 (1967) B. Bagchi, S.N. Biswas, A. Khare, P.K. Roy, Truncated harmonic os-cillator and parasupersymmetric quantum mechanics , Pramana 49 , (2), 199-204 (1997) F. Bagarello, S. T. Ali, J. P. Gazeau, Extended pseudo-fermions from non commutative bosons , Journal of Math. Phys., 54 , 073516, (2013) F. Bagarello, Pseudo-Bosons and Their Coherent States , Springer, Mathematical Physics Studies, 2022 F. Bagarello, Ladder operators with no vacuum, their coherent states, and an application to graphene , Ann. of Phys., 462 , 169605 (2024) K. Kowalskiy, J. Rembielinski, L. C. Papaloucas, Coherent states for a quantum particle on a circle , J. Phys. A, 29 , 4149-4167 (1996) F. Bagarello, Quantum dynamics for classical systems: with applications of the Number operator , Wiley (2012) F. Bagarello, Quantum Concepts in the Social, Ecological and Biological Sciences , Cambridge University Press, 2019 F. Bagarello, F. Gargano, F. Oliveri, Quantum Tools for Macroscopic Systems , Springer. Synthesis Lectures on Mathematics & Statistics, 2023 O. Christensen, An Introduction to Frames and Riesz Bases , Birkh¨ auser, Boston, (2003) C. Heil, A basis theory primer: expanded edition , Springer, New York, (2010) H. Bacry, A. Grossmann, and J. Zak, Proof of the completeness of lattice states in kq representation , Phys. Rev. B, 12 , 1118–1120, (1975) V. V. Eremin, A. A. Meldianov, The q-deformed harmonic oscillator, coherent states and the uncertainty relation , Theor. and Math. Phys.,
147 (2): 709-715, (2006) 25 |
8195 | https://engineering.purdue.edu/~wassgren/teaching/ME30800/NotesAndReading/FlowVisualization_Reading.pdf | Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics 1.7. Flow Kinematics A good reference on the experimental aspects of this topic is: Merzkirch, W., Flow Visualization, Academic Press.
In 1986, the Chernobyl Nuclear Power Plant in the Soviet Union released radioactive fallout into the atmo-sphere as a result of an explosion in one of the reactors (Figure 1.25). The radioactive plume covered regions of the western Soviet Union, Europe, and even parts of eastern North America. If such an accident were to happen again, how would you know which communities would be a↵ected by the drifting radioactive cloud?
If one had predictions of wind velocity measurements as a function of location and time, i.e., u = u(x, t), could you figure out what areas would be covered by the cloud? In this section, we’ll present three forms of flow kinematics: streamlines, pathlines, and streaklines. Each of these lines provides di↵erent information on the movement of fluid. For the toxic cloud release, one would be most interested in determining the streakline passing through the location of the damaged reactor.
Figure 1.25. Photo of the damaged Chernobyl reactor.
Photo from
wikipedia.org/wiki/Chernobyl_disaster.
1.7.1. Streamlines Figure 1.26. Illustration of streamlines. Streamlines are everywhere tangent to the velocity vectors.
C. Wassgren 82 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics A streamline is a line that is everywhere tangent to the velocity field vectors (Figure 1.26). Experimentally, one can visualize streamlines using the Particle Image Velocimetry (PIV) technique. In PIV, fluid particles are “tagged”, usually by mixing in very small, neutrally buoyant bits of “paint,” and taking two photographs in rapid succession. Velocity vectors can then be produced by “connecting the dots” (actually this method is technically Particle Tracking Velocimetry (PTV), but PIV operates in a similar manner, but matching up regions of particles rather than individual particles) (Figure 1.27).
Figure 1.27. An illustration of the Particle Tracking Velocimetry technique for obtaining a velocity vector field. Particle Image Velocimetry works in a similar manner, but tracks regions of particles rather than individual particles.
Note that the approximate velocity vectors can be found using, ˙ x(t) ⇡x(t + δt) −x(t) δt .
(1.111) We can determine the equation of a streamline given a velocity field by simply using the definition of a streamline. Since the streamline is tangent to the velocity vector, the slope of the streamline will be equal to the slope of the velocity vector, dy dx |{z} slope of streamline = uy ux |{z} slope of velocity vector (1.112) Similarly, in the x - z and y - z planes we have, dz dx = uz ux dz dy = uz uy .
(1.113) We can combine these equations into a more compact form, dx ux = dy uy = dz uz .
(1.114) Notes: (1) There is no flow across a streamline since the velocity component normal to the streamline is zero.
(2) A stream tube is a tube made by all the streamlines passing through a closed curve (Figure 1.28).
There is no flow through a stream tube wall.
(3) A stream filament is a stream tube with infinitesimally small cross-sectional area.
C. Wassgren 83 2021-12-15 Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Figure 1.28. An illustration of a stream tube.
Figure 1.29. An illustration of a pathline.
1.7.2. Pathlines A pathline is the line traced out by a particular particle as it moves from one point to another (Figure 1.29).
It is the actual path a particle takes. Experimentally, a pathline can be visualized by “tagging” a particular fluid particle and taking a long-exposure photograph of the particle’s motion.
To determine the equation of a pathline at time, t, for a particle passing through the point (x0, y0, z0) at some previous time t0, we solve the di↵erential equation describing the particle’s position, u = dx dt , (1.115) where u is the velocity field subject to the initial condition that the particle passes through the point x0 = (x0, y0, z0) at time, t0, x(t = t0) = x0.
(1.116) For a pathline t0 will be a particular value. The solution of Eq. (1.115) subject to the initial condition in Eq. (1.116) will consist of a set of parametric equations in t (by varying t, we can trace out the location of the particle at various times).
1.7.3. Streaklines A streakline is a line that connects all fluid particles that have passed through the same point in space at a previous (or later) time. Experimentally a streamline can be visualized by injecting dye into a fluid flow at a particular point (Figure 1.30).
To determine the equation of a streakline at time, t, passing through the point (x0, y0, z0), we solve the di↵erential equation describing a particle’s position as a function of time, u = dx dt , (1.117) where u is the velocity field subject to the initial condition that a particle pass through the point x0 = (x0, y0, z0) at some previous (or later) time, t0, x(t = t0) = x0.
(1.118) C. Wassgren 84 2021-12-15 streakline Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Figure 1.30. An illustration of a streakline.
Note that t0 will be di↵erent for each particle, i.e., it varies, unlike a pathline where t0 is fixed. The solution of Eq. (1.117) subject to the initial condition in Eq. (1.118) will consist of a set of parametric equations in t0 (note that t is a known value since we want to know the streakline at a particular time, t).
Notes: (1) The streamline, streakline, and pathline passing through a particular location can be di↵erent in an unsteady flow, but will be identical in a steady flow.
(2) The quantity t0 is the time when a fluid particle passes through the point x0. Hence, for a pathline t0 is fixed since there is only one fluid particle. However, for a streakline t0 varies since there are many fluid particles passing through the point x0, each at a di↵erent t0.
(3) Why don’t we use a Lagrangian derivative (covered in Chapter 5) when solving Eq. (1.115) for a particle’s pathline (since the pathline is a Lagrangian concept)? It turns out that the Lagrangian derivative of a particle’s position is equal to its Eulerian derivative. Consider, for example, the change in the x position of the particle as we follow it. Note that the position, x, is an Eulerian quantity, Dx Dt = @x @t + (u · r)x = @x @t |{z} =0 +ux @x @x |{z} =1 +uy @x @y |{z} =0 +uz @x @z |{z} =0 = ux.
(1.119) Be sure to: (1) Understand the definitions for streamlines, streaklines, and pathlines.
(2) Understand what initial conditions to use when evaluating streaklines and pathlines.
(3) Draw the direction of flow on the streamlines, streaklines, and pathlines.
(4) It’s perfectly correct to represent the position of a fluid particle parametrically, i.e., x = x(t) and y = y(t).
C. Wassgren 85 2021-12-15 flowvis_03 Page 1 of 1 One technique for visualizing fluid flow over a surface is to attach short, lightweight pieces of thread or “tufts” to the surface. A photograph of the tufts on the surface of an automobile is shown in the figure below. Do tufts trace out the streamlines, streaklines, pathlines, or some other type of flow line? What if the flow is unsteady? Explain your answers. SOLUTION: The tufts show the local direction of the fluid velocity and, hence, are indicators of the local streamline slope. If the flow is steady, then the streamlines, streaklines, and pathlines are all the same. If the flow is unsteady, then the tufts will, in general, only indicate the local slope of the streamlines. Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 86 2021-12-15 flowvis_06 Page 1 of 1 Tiny hydrogen bubbles are being used as tracers to visualize a flow. All the bubbles are generated at the origin (x = 0, y = 0). The velocity field is unsteady and obeys the equations: u = 1 m/s v = 1 m/s 0 ≤ t < 2 s u = 0 v = 1.5 m/s 2 ≤ t ≤ 4 s Plot the pathlines of bubbles that leave the origin at t = 0, 1, 2, 3, and 4 s. Mark the locations of these five bubbles at t = 4 s. Use a dashed line to indicate the position of the streakline passing through (0, 0) at t = 4 s. What does the streamline passing through (0, 0) look like at t = 4 s? SOLUTION: One could solve the differential equations describing the particle pathlines and streakline using the velocities given above, or, more easily, simply plot the positions of the fluid particles at different times. The plot below shows the particle positions, pathlines, and streakline. The streamline passing through (0, 0) at t = 4 s (or any other point for that matter) will be a vertical line since the velocity at t = 4 s is purely vertical. x y 1 2 3 1 2 3 4 5 Position of the bubble released at t = 1 s at t = 4 s. Position of the bubble released at t = 0 s at t = 4 s. Position of the bubble released at t = 2 s at t = 4 s. Position of the bubble released at t = 3 s at t = 4 s. Position of the bubble released at t = 4 s at t = 4 s. Streakline at t = 4 s for the particles passing through (0, 0). Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 87 2021-12-15 flowvis_21 Page 1 of 1 The two-dimensional velocity field for an unsteady flow is given by, ! = # $̂ + '̂ 0 ≤ < 1 s $̂ −'̂ 1 s ≤ < 2 s $̂ 2 s ≤ < 3 s a. Write an equation for the streamline passing through the point (x, y) = (1, 1) for 0 ≤ t ≤ 3 s. b. Sketch the pathline for a fluid particle released from the origin at t = 0 s for 0 ≤ t ≤ 3 s. c. Sketch the streakline through the point (x, y) = (1, 1) at t = 3 s. SOLUTION: The slope of a streamline at a point is tangent to the velocity vector at that same point, !" !# = $!
$" = # 1 0 ≤ < 1 s −1 1 s ≤ < 2 s 0 2 s ≤ < 3 s , (1) ∫34 " "# = ∫35 # ## 0 ≤ < 1 s ∫34 " "# = −∫35 # ## 1 s ≤ < 2 s ∫34 " "# = 0 2 s ≤ < 3 s , (2) 4 −4% = 5 −5% 0 ≤ < 1 s 4 −4% = 5% −5 1 s ≤ < 2 s 4 −4% = 0 2 s ≤ < 3 s , (3) Using (x0, y0) = (1, 1), 4 −1 = 5 −1 0 ≤ < 1 s 4 −1 = 1 −5 1 s ≤ < 2 s 4 −1 = 0 2 s ≤ < 3 s , (4) 4 = 5 0 ≤ < 1 s 4 = 2 −5 1 s ≤ < 2 s 4 = 1 2 s ≤ < 3 s , (5) Sketches are shown below for (a) the pathline of a fluid particle released from the origin (x0, y0) = (0, 0) at t = 0 s, and (b) the streakline through the point (x0, y0) = (1, 1) at t = 3 s. (a) (b) x y 1 2 3 4 0 0 1 2 3 4 x y 1 2 3 4 0 0 1 2 3 4 t = 0 s t = 1 s t = 2 s t = 3 s particle released at t = 0 s particle released at t = 1 s particle released at t = 2 s particle released at t = 3 s streakline through (1, 1) at t = 3 s pathline for a particle released from (0, 0) at t = 0 s ! = #̂ + &̂ ! = #̂ −&̂ ! = #̂ ! = #̂ ! = #̂ ! = #̂ + &̂ ! = #̂ −&̂ ! = #̂ −&̂ ! = #̂ Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 88 2021-12-15 flowvis_08 Page 1 of 1 Show that for a steady flow, streamlines, streaklines, and pathlines are identical. SOLUTION: Streamlines are defined as lines that are everywhere tangent to the instantaneous velocity vectors. (The rest of the problem will be worked out in Cartesian coordinates for convenience.) (1) Re-writing: (2) where u is not a function of time since the flow is assumed steady but is, in general, a function of position, i.e. u = u(x). Streaklines are lines connecting all fluid particles that pass through the same point in space. where (3) where t0 is the time at which a fluid particle on the streamline passes through the point x0 on the streakline. Note that t0 will be different for each fluid particle on a given streakline. Pathlines trace the motion of individual fluid particles over time. where (4) where t0 is the time at which a fluid particle passes through the point x0 on the pathline. Note that t0 is a fixed quantity for a given pathline. We can re-write the differential equations for the streakline and pathline as: (5) Note that u is not a function of t (steady flow Þ u = u(x)) so that we needn’t worry about how the slope of the lines change with time. Thus, we can write: (6) Since Eqns. (6) and (2) are identical, we can conclude that streamlines, streaklines, and pathlines are identical for a steady flow. y x y x z x z x z y z y u dy dy dx dx u u u u dz dz dx dx u u u u dz dz dy dy u u u = Þ = = Þ = = Þ = x y z dx dy dz u u u = = d dt = x u ( ) 0 0 t t = = x x d dt = x u ( ) 0 0 t t = = x x x x y y z z dx dx u dt dt u dy dy u dt dt u dz dz u dt dt u = Þ = = Þ = = Þ = x y z dx dy dz u u u \ = = Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 89 2021-12-15 flowvis_01 Page 1 of 1 A velocity field is given by: where V0 is a positive constant, i.e. V0 > 0. Determine: a. where in the flow the speed is V0 b. the equation and sketch of the streamlines c. the equations for the streaklines and pathlines SOLUTION: The speed is given by: (1) where (2) (3) Substituting into Eqn. (1) gives: (4) The flow speed is everywhere equal to V0. The slope of the streamline is tangent to the slope of the velocity vector: (5) Substitute Eqns. (2) and (3) and solving the resulting differential equation. (where (x0, y0) is a point located on the streamline) (6) The streamlines are circles! Note that when x > 0 and y > 0, Eqns. (2) and (3) indicate that ux < 0 and uy > 0 (note that V0 > 0) so that the flow is moving in a counter-clockwise direction. Since the flow is steady, the streaklines and pathlines will be identical to the streamlines. 1 1 2 2 0 0 2 2 2 2 ˆ ˆ ( ) ( ) V y V x x y x y -= + + + u i j 2 2 x y u u = + u 12 0 2 2 ( ) x V y u x y -= + 12 0 2 2 ( ) y V x u x y = + 2 2 2 2 0 0 2 2 2 2 ( ) ( ) V y V x x y x y = + + + u 0 V \ = u y x u dy dx u = 12 12 0 2 2 0 2 2 ( ) ( ) V x dy x x y V y dx y x y + = = --+ 0 0 y x y x ydy xdx -= ò ò ( ) ( ) 2 2 2 2 1 1 0 0 2 2 y y x x --= -2 2 2 2 0 0 constant x y x y + = + = x y Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 90 2021-12-15 flowvis_02 Page 1 of 2 Consider a 2D flow with a velocity field given by: Determine the equations for the streamline, streakline, and pathline passing through the point (x,y)=(1,1) at time t=0. SOLUTION: The slope of a streamline is tangent to the velocity vector. (1) where (2) (3) Substitute Eqns. (2) and (3) into Eqn. (1) and solve the resulting differential equation. (where (x0, y0) is a point passing through the streamline) (4) For the streamline passing through the point (x0, y0) = (1, 1) at time t = 0: (5) A streakline is a line that connects all of the fluid particles that pass through the same point in space. The equation for the streakline can be found parametrically using Eqns. (2) and (3). (6) (7) Solve the previous two differential equations. Þ (8) Þ (9) where t0 is the time at which a fluid particle passes through the point (x0, y0) on the streakline. Hence, the streakline passing through the point (x0, y0) = (1, 1) at time t = 0 is given parametrically (in t0) as: Þ (10) Þ (11) Recall that t0 is the time when a fluid particle passes through the point (x0, y0). ˆ ˆ (1 2 ) x t y = + + u i j y x u dy dx u = (1 2 ) x u x t = + y u y = (1 2 ) dy y dx x t = + 0 0 (1 2 ) y x y x dy dx t y x + = ò ò 0 0 (1 2 )ln ln y x t y x æ ö æ ö + = ç ÷ ç ÷ è ø è ø (1 2 ) 0 0 t y x y x + æ ö æ ö = ç ÷ ç ÷ è ø è ø y x = (1 2 ) x dx u x t dt = = + y dy u y dt = = 0 0 (1 2 ) x t x t dx t dt x = + ò ò 2 2 0 0 0 ln x t t t t x æ ö = + --ç ÷ è ø 0 0 y t y t dy dt y = ò ò 0 0 ln y t t y æ ö = -ç ÷ è ø ( ) 2 0 0 ln x t t = --( ) 2 0 0 exp x t t = --( ) 0 ln y t = -( ) 0 exp y t = -Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 91 2021-12-15 flowvis_02 Page 2 of 2 A pathline is a line traced out by a particular fluid particle as it moves through space. The equation for the pathline can be found parametrically using Eqns. (2) and (3). (12) (13) Solve the previous two differential equations. Þ (14) Þ (15) where t0 is the time at which a fluid particle passes through the point (x0, y0) on the pathline. Hence, the pathline for a particle passing through the point (x0, y0) = (1, 1) at time t0 = 0 is given parametrically (in t) as: Þ (16) Þ (17) Note that the streamline, streakline, and pathline are all different. A plot of these lines through (1, 1) at t = 0 is shown below. (1 2 ) x dx u x t dt = = + y dy u y dt = = 0 0 (1 2 ) x t x t dx t dt x = + ò ò 2 2 0 0 0 ln x t t t t x æ ö = + --ç ÷ è ø 0 0 y t y t dy dt y = ò ò 0 0 ln y t t y æ ö = -ç ÷ è ø ( ) 2 ln x t t = + ( ) 2 exp x t t = + ( ) ln y t = ( ) exp y t = 0 2 4 6 8 10 12 14 16 0 2 4 6 8 10 x y streamline streakline pathline Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 92 2021-12-15 flowvis_07 Page 1 of 1 A tornado can be represented in polar coordinates by the velocity field, where and are unit vectors pointing in the radial (r) and tangential (q) directions, respectively, and a and b are constants. Show that the streamlines for this flow form logarithmic spirals, i.e. where c is a constant. SOLUTION: The slope of a streamline is tangent to the velocity vector. In polar coordinates, the streamline slope is given by: (1) so that the relation describing the streamline slope is: (2) where (3) (4) Substitute Eqns. (3) and (4) into Eqn. (2) and solve the resulting differential equation. (5) where the constants r0 and q0 have been incorporated into the constant c. ˆ ˆ r a b r r q = -+ u e e ˆr e ˆq e exp a r c bq æ ö = -ç ÷ è ø small displacement in -direction small displacement in -direction r dr rd q q = r u dr rd uq q = r a u r = -b u r q = a dr a r b rd b r q -= = -0 0 r r dr a d r b q q q = -ò ò ( ) 0 0 ln r a r b q q æ ö = --ç ÷ è ø ( ) 0 0 exp r a r b q q é ù = --ê ú ë û exp a r c bq é ù \ = -ê ú ë û Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics C. Wassgren 93 2021-12-15 flowvis_05 Page 1 of 2 Consider the 2D flow field defined by the following velocity: For this flow field, find the equation of: a. the streamline through the point (1,1) at t = 0, b. the pathline for a particle released at the point (1,1) at t = 0, and c. the streakline at t = 0 which passes through the point (1,1). SOLUTION: The slope of a streamline is tangent to the velocity vector. (1) where (2) (3) Substitute Eqns. (2) and (3) into Eqn. (1) and solve the resulting differential equation. (where (x0, y0) is a point passing through the streamline) (4) For the streamline passing through the point (x0, y0) = (1, 1) at time t = 0: (5) A streakline is a line that connects all of the fluid particles that pass through the same point in space. The equation for the streakline can be found parametrically using Eqns. (2) and (3). (6) (7) Solve the previous two differential equations. Þ (8) Þ (9) where t0 is the time at which a fluid particle passes through the point (x0, y0) on the streakline. Hence, the streakline passing through the point (x0, y0) = (1, 1) at time t = 0 is given parametrically (in t0) as: Þ (10) Þ (11) Recall that t0 is the time when a fluid particle passes through the point (x0, y0). 1 ˆ ˆ 1 t æ ö = + ç ÷ + è ø u i j y x u dy dx u = 1 1 x u t = + 1 y u = 1 1 11 dy t dx t = = + + ( ) 0 0 1 y x y x dy t dx = + ò ò ( )( ) 0 0 1 y y t x x -= + -y x = 1 1 x dx u dt t = = + 1 y dy u dt = = 0 0 1 x t x t dt dx t = + ò ò 0 0 1 ln 1 t x x t æ ö + -= ç ÷ + è ø 0 0 y t y t dy dt = ò ò 0 0 y y t t -= -0 1 1 ln 1 x t æ ö - = ç ÷ + è ø 0 1 ln 1 1 x t æ ö = + ç ÷ + è ø 0 1 y t - = -0 1 y t = -Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 94 2021-12-15 flowvis_05 Page 2 of 2 A pathline is a line traced out by a particular fluid particle as it moves through space. The equation for the pathline can be found parametrically using Eqns. (2) and (3). (12) (13) Solve the previous two differential equations. Þ (14) Þ (15) where t0 is the time at which a fluid particle passes through the point (x0, y0) on the pathline. Hence, the pathline for a particle passing through the point (x0, y0) = (1, 1) at time t0 = 0 is given parametrically (in t) as: Þ (16) Þ (17) Note that the streamline, streakline, and pathline are all different. A plot of these lines through (1, 1) at t = 0 is shown below. 1 1 x dx u dt t = = + 1 y dy u dt = = 0 0 1 x t x t dt dx t = + ò ò 0 0 1 ln 1 t x x t æ ö + -= ç ÷ + è ø 0 0 y t y t dy dt = ò ò 0 0 y y t t -= -( ) 1 ln 1 x t - = + ( ) ln 1 1 x t = + + 1 y t - = 1 y t = + -10 -8 -6 -4 -2 0 2 4 6 8 10 -5 0 5 10 x y streamline streakline pathline Notes on Thermodynamics, Fluid Mechanics, and Gas Dynamics Video solution: C. Wassgren 95 2021-12-15 |
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2 Simple Ways to Calculate the Angle Between Two Vectors
Learn how to take the dot or cross product of 2 vectors to find the angle between them
Dot Product Formula
| Cross Product Formula
| Understanding the Dot Product Formula
| Video
| Q&A
| Tips
If you’re learning about angles and vectors in math class, your teacher probably just assigned you problems to find the angle between 2 vectors. It can definitely seem a little confusing to get started, so that’s why we’re here to help! In this article, we’ll tell you about the 2 formulas that find the angle between 2 vectors and walk you through how to use them. Read on to get your math problems solved!
Things You Should Know
Use the formula ( • ) / (|||| ||||) to find the angle between vectors using the dot product.
To calculate the dot product, multiply the same direction coordinates of each vector and add the results together.
Then, find each vector’s magnitude using the Pythagorean Theorem, or √(u12 + u22). Plug the arccos, dot product, and magnitudes into a calculator to get the angle.
Or, use the cross product formula ( ) / (|||| ||||) to get the angle between the vectors.
Steps
Method 1
Method 1 of 3:
Using the Dot Product Formula
1
Use the formula: ( • ) / (|||| ||||). The angle between 2 vectors is where the tails of 2 vectors, or line segments, meet.
X
Research source Each vector has a magnitude, or length, and a direction that it’s heading. So, to find the angle between 2 vectors, you use the above formula where:
X
Research source
is the angle between the vectors.
is the inverse of cosine, or the arc cos.
• is the dot product of vector and .
|||| |||| is the magnitude of vector and .
2. 2
Identify the vectors’ coordinates in your math problem. Most math problems give you the dimensional coordinates of each vector, which are sometimes also called components. You use each vector’s coordinates to find their magnitudes and combined dot product. If your math problem already gives the vectors’ magnitudes, skip the magnitude step below.
X
Research source
For example, find the angle between vector and vector . Vector has coordinates at (2, 2) and vector has coordinates at (0, 3).
Sometimes, vectors are written as = 2i + 2j and = 0i + 3j.
While our example uses two-dimensional vectors, finding the angle between 3-dimensional vectors follows the same steps.
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3. 3
Calculate the magnitude of each vector. Picture a right triangle drawn from the vector's x-component, its y-component, and the vector itself. The vector forms the hypotenuse of the triangle, so to find its magnitude, simply use the Pythagorean theorem. Just plug each vector’s coordinates into the theorem.
X
Research source
In the Pythagorean theorem of a2 + b2 + c2, the vector’s magnitude is denoted by c. So, just rewrite the equation to isolate the magnitude on one side: ||u|| = √(u12 + u22) with u12 + u22 being the vector’s x and y coordinates.
Using our example, find the magnitudes for vector at (2, 2) and vector at (0, 3).
+ Insert coordinates into the theorem: √22 + 22 = √8 = 2√2. So, ||u|| = 2√2.
+ Find magnitude: √02 + 32 = √9. So, |||| = 3.
If a vector is 3-dimensional or has more than 2 components, simply continue adding +u32 + u42 + … to the Pythagorean Theorem.
4. 4
Calculate the dot product of the 2 vectors. The dot product is a way to multiply vectors, which is also commonly called the scalar product.
To calculate the dot product, multiply the same direction coordinates of the vectors, then add the results together.
For computer graphics programs, see Tips before you continue.
X
Research source
Using our example, = (2, 2) and = (0, 3). Find • .
+ Multiply the x-coordinates of and and the y-coordinates. uxvx + uyvy = (2)(0) + (2)(3) = 0 + 6 = 6.
+ 6 is the dot product of vector and .
Defining Dot Product
In mathematical terms, • = u1v1 + u2v2, where (u1, u2) are the coordinates for vector u. If your vector has more than 2 components, simply continue to add + u3v3 + u4v4...
5. 5
Plug the dot product and each vector’s magnitude into the formula. Remember, the formula is ( • ) / (|||| ||||) Now that you know both the dot product and the magnitudes of each vector, simply enter them into this formula.
X
Research source
Finding Cosine with Dot Product and Magnitude
In our example, θ = cos-16 / (2√2•3). Simplify to get θ = cos-1(√2 / 2).
6. 6
Use a scientific calculator to find the angle based on the cosine. On most calculators, use either the arccos or cos-1 function on your calculator to find the angle θ. Simply enter “arccos” and the dot product divided by the vectors’ magnitudes. For some results, use the unit circle to work out the angle.
Finding an Angle with Cosine
In our example, θ = cos-1(√2 / 2). Enter "arccos(√2 / 2)" in your calculator to get θ = 45º. Alternatively, find the angle θ on the unit circle where cosθ = √2 / 2.
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Method 2
Method 2 of 3:
Using the Cross Product Formula
1
Use the formula: ( ) / (|||| ||||). This formula uses sine and the cross product of vectors to find the angle between them. Unlike the dot product formula, which gives you a scalar answer, the cross product formula gives you an answer as a vector. In this formula:
X
Research source
is the angle between the vectors.
is the inverse of sine, or the arc sin.
is the cross product of vector and .
|||| |||| is the magnitude of vector and .
2. 2
Find the cross product using the vectors’ coordinates. In most math problems, you have the dimensional coordinates, or components, of each vector written as . To find the cross product, make a matrix with the first vector’s coordinates in the first row and the second vector’s coordinates in the second row. Calculate the i, j, and k values for each matrix section.
X
Research source
For example, find the angle between 2 vectors where is 1i - 2j + 3k and is 10i + 1j - 3k.
+ Draw a matrix for and : 1 2 3 is on the top row, and 10 1 -3 is on the bottom row.
+ Solve the matrix for i: i = (uj vk) - (vj uk)
i = (6 - 3) = 3
+ Solve the matrix for j: j = (ui vk) - (vi uk)
j = (-3 - 30) = -33
+ Solve the matrix for k: k = (ui vj) - (vi uj)
k = (1 - -20) = 21
+ Find the coordinates for i - j + k: 3i - -33j + 21k = 3i + 33j + 21k or <3, 33, 21>
3. 3
Calculate the magnitude of the cross product. The final step in finding the cross product of vectors is finding the magnitude of their coordinates. Remember, use the Pythagorean Theorem to find a vector’s magnitude.
X
Research source Just plug in the i, k, j coordinates of the cross product of the vectors to get their magnitude.
X
Research source
The cross product of x is 3i + 33j + 21k or <3, 33, 21>.
+ Use the Pythagorean Theorem to find the magnitude: ||u x v|| = √(i2 + j2 + k2)
+ Plug in 3i + 33j + 21k into the theorem: √((3)2 + (33)2 + (21)2)
+ Solve: √9 + 1089 + 441 = √1539
+ The cross product of vector x = √1539
4. 4
Find the magnitude of each vector. Now, calculate the magnitude of each vector using their dimensional coordinates.
X
Research source Just plug the coordinates into the Pythagorean Theorem like in the step above.
X
Research source
In the example, is 1i - 2j + 3k and is 10i + 1j - 3k.
+ Find the magnitude of : ||u|| = √i2 + j2 + k2 = √((1)2 + (-2)2 + (3)2) = √1 + 4 + 9 = √14
+ Find the magnitude of : ||v|| = √((10)2 + (1)2 + (-3)2) = √100 + 1 + 9 = √110
5. 5
Plug the cross product and the vectors’ magnitude into the formula. Now that you have the vectors’ cross product and magnitudes, simply enter them into the formula ( ) / (|||| ||||).
X
Research source
In our example, θ = sin-1(√1539 / √14 √110)
6. 6
Find the angle using a calculator. Simply take the inverse sine of the cross product and magnitudes to find the angle between the vectors. Using your calculator, find the arcsin or sin-1 function. Then, enter in the cross product and magnitude.
X
Research source
In our example, enter “arcsin(√1539 / √14 √110) into your calculator to get θ = 88.5º.
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Method 3
Method 3 of 3:
Understanding the Dot Product Formula
1
Understand the purpose of the angle formula. This formula was not derived from existing rules. Instead, it was created as a definition of 2 vectors' dot product and the angle between them. However, this decision was not arbitrary. With a look back to basic geometry, you see why this formula results in intuitive and useful definitions.
The examples below use 2-dimensional vectors because these are the most intuitive to use. Vectors with 3 or more components use the same formula.
2. 2
Review the Law of Cosines used in the formula. Take an ordinary triangle, with angle θ between sides a and b, and opposite side c. The Law of Cosines states that c2 = a2 + b2 -2abcos(θ). This is derived fairly easily from basic geometry.
X
Research source
3. 3
Connect 2 vectors to form a triangle. Sketch a pair of 2D vectors on paper, vectors and , with angle θ between them. Draw a third vector between them to make a triangle. In other words, draw vector such that + = . This vector = - .
X
Research source
4. 4
Write the Law of Cosines for the triangle. Insert the length of the "vector triangle" sides in our example into the Law of Cosines:
X
Research source
||(a - b)||2 = ||a||2 + ||b||2 - 2||a|| ||b||cos(θ)
5. 5
Write the Law of Cosines using the dot product of vector a and b. Remember, the dot product is the magnification of 1 vector projected onto another. A vector's dot product with itself doesn't require any projection, since there is no difference in direction. This means that • = ||a||2. Use this fact to rewrite the equation:
X
Research source
( - ) • ( - ) = • + • - 2||a|| ||b||cos(θ)
6. 6
Rewrite the dot product into the angle formula. Expand the left side of the formula, then simplify to reach the formula used to find angles.
X
Research source
• - • - • + • = • + • - 2||a|| ||b||cos(θ)
- • - • = -2||a|| ||b||cos(θ)
-2( • ) = -2||a|| ||b||cos(θ)
• = ||a|| ||b||cos(θ)
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How do I find the angle between two vectors? For example, vector A = 4i + 2j - 2k and vector B = 3i +2j + 3k?
wikiHow Staff Editor
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This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
Staff Answer
Use the formula with the dot product, θ = cos^-1 (a b) / ||a|| ||b||. To get the dot product, multiply Ai by Bi, Aj by Bj, and Ak by Bk then add the values together. To find the magnitude of A and B, use the Pythagorean Theorem (√(i^2 + j^2 + k^2). Then, use your calculator to take the inverse cosine of the dot product divided by the magnitudes and get the angle.
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If the cosine formula gives me 0, it means that the vector are perpendicular. But how do I know if it's 90° or -90° ?
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This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
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The angle between 2 vectors is always between 0° and 180°, so the angle is 90°.
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In the above example cosθ was 1/√2. But here cosθ can be 45 degrees or 315 degrees. Why is that the answer is not 315?
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When you find the angle between 2 vectors, the angle is always going to be between 0° and 180°.
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For a quick plug and solve, use this formula for any pair of two-dimensional vectors: cosθ = (u1 • v1 + u2 • v2) / (√(u12 • u22) • √(v12 • v22)).
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If you are working on a computer graphics program, you most likely only care about the direction of the vectors, not their length. Take these steps to simplify the equations and speed up your program:
+ Normalize each vector so the length becomes 1. To do this, divide each component of the vector by the vector's length.
+ Take the dot product of the normalized vectors instead of the original vectors.
+ Since the length equals 1, leave the length terms out of your equation. Your final equation for the angle is arccos( • ).
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The cosine formula tells you whether the angle between vectors is acute or obtuse. Start with cosθ = ( • ) / (|||| ||||):
+ The left side and right sides of the equation must have the same sign (positive or negative).
+ Since the lengths are always positive, cosθ must have the same sign as the dot product.
+ Therefore, if the dot product is positive, cosθ is positive. We are in the first quadrant of the unit circle, with θ < π / 2 or 90º. The angle is acute.
+ If the dot product is negative, cosθ is negative. We are in the second quadrant of the unit circle, with π / 2 < θ ≤ π or 90º < θ ≤ 180º. The angle is obtuse.
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About This Article
Reviewed by:
Grace Imson, MA
Math Teacher
This article was reviewed by Grace Imson, MA and by wikiHow staff writer, Devin McSween. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 2,887,103 times.
12 votes - 93%
Co-authors: 41
Updated: April 23, 2025
Views: 2,887,103
Categories: Coordinate Geometry
Article SummaryX
Calculate the length of each vector.
Calculate the dot product of the 2 vectors.
Calculate the angle between the 2 vectors with the cosine formula.
Use your calculator's arccos or cos^-1 to find the angle. For specific formulas and example problems, keep reading below!
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Reviewed by:
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Co-authors: 41
Updated: April 23, 2025
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Principle of Inclusion-Exclusion
The Principle of Inclusion-Exclusion (abbreviated PIE) provides an organized method/formula to find the number of elements in the union of a given group of sets, the size of each set, and the size of all possible intersections among the sets.
Contents
[hide]
1 Important Note(!)
2 Application
2.1 Two Set Example
2.2 Three Set Example
2.3 Four Set Example
2.3.1 Problem
2.3.2 Solution
2.4 Five Set Example
2.4.1 Problem
2.4.2 Solution
3 Statement
4 Proof
5 Remarks
6 Examples
7 See also
Important Note(!)
When using PIE, one should understand how to strategically overcount and undercount, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving.
Application
Here, we will illustrate how PIE is applied with various numbers of sets.
Two Set Example
Assume we are given the sizes of two sets, and , and the size of their intersection, . We wish to find the size of their union, .
To find the union, we can add and . In doing so, we know we have counted everything in at least once. However, some things were counted twice. The elements that were counted twice are precisely those in . Thus, we have that:
.
Three Set Example
Assume we are given the sizes of three sets, and , the size of their pairwise intersections, , and , and the size their overall intersection, . We wish to find the size of their union, .
Just like in the Two Set Example, we start with the sum of the sizes of the individual sets . We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out . Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times and then subtracted out three times. We have to add back in . Putting this all together gives:
.
Four Set Example
Problem
Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10)
Solution
Let be the event that the first, second, and third people are in ordered height, be the event that the second, third, and fourth people are in ordered height, be the event that the third, fourth, and fifth people are in ordered height, and be the event that the fourth, fifth and sixth people are in ordered height. By a combination of complementary counting and PIE, we have that our answer will be . Now for the daunting task of evaluating all of this.
For , we just choose people and there is only one way to put them in order, then ways to order the other three guys for . Same goes for , , and .
Now, for , that's just putting four guys in order. By the same logic as above, this is . Again, would be putting five guys in order, so . is just choosing guys out of , then guys out of for . Now, is just the same as , so , is so , and is so .
Moving on to the next set: is the same as which is , is ordering everybody so , is again ordering everybody which is , and is the same as so .
Finally, is ordering everybody so .
Now, lets substitute everything back in. We get a massive expression of .
Five Set Example
Problem
There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50 take language arts, social studies, and biology. 50 take language arts, social studies, and history. 50 take language arts, biology, and history. 50 take social studies, biology, and history. 20 take algebra, language arts, social studies, and biology. 15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history. 10 take algebra, social studies, biology, and history. 10 take language arts, social studies, biology, and history. 5 take all five. None take none.
How many people are in my school?
Solution
Let A be the subset of students who take Algebra, L-languages, S-Social Studies, B-biology, H-history, M-the set of all students. We have:
Thus, there are people in my school.
Statement
If are finite sets, then:
.
Proof
We prove that each element is counted once.
Say that some element is in sets. Without loss of generality, these sets are
We proceed by induction. This is obvious for
If this is true for we prove this is true for For every set of sets not containing with size there is a set of sets containing with size In PIE, the sum of how many times these sets are counted is There is also one additional set of sets so is counted exactly once.
Remarks
Sometimes it is also useful to know that, if you take into account only the first sums on the right, then you will get an overestimate if is odd and an underestimate if is even. So,
,
,
,
and so on.
Examples
2011 AMC 8 Problems/Problem 6
2017 AMC 10B Problems/Problem 13
2005 AMC 12A Problems/Problem 18
2001 AIME II Problems/Problem 9
2002 AIME I Problems/Problem 1
2020 AIME II Problems/Problem 9
2001 AIME II Problems/Problem 2
2017 AIME II Problems/Problem 1
See also
Combinatorics
Overcounting
Retrieved from "
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8199 | https://www.universalclass.com/articles/math/pre-algebra/how-to-perform-arithmetic-operations-on-radicals.htm | Browse Courses My Classes
How to Perform Arithmetic Operations on Radicals
Key Terms
o Associative
o Distributive
o Identify additional properties of multiplication and addition (associative and distributive properties)
o Be able to add, subtract, multiply, and divide radical expressions
o Simplify radical expressions when appropriate
Primer for Arithmetic Operations on Radicals
We will look carefully at how to work with radicals by way of arithmetic operations (addition, subtraction, division, and multiplication). First, we introduce a mathematical concept that may or may not be familiar to you, but that is crucial to properly understanding this topic. When we multiply two numbers, for instance, we can generally write the product as another number. For example, . But what if one of the factors is unknown or cannot be written in an exact form? Let's consider multiplication of 4 by some unknown number c.
We cannot simplify this expression further or write it as a single known number (because c is unknown). Likewise, consider multiplication of 4 by an irrational number . Because we cannot write as an exact decimal or as a fraction with integers in the numerator and denominator, we cannot write the product (in this case) as an exact decimal or a fraction.
Thus, the way we will treat radicals is very similar to the way we must treat unknown numbers. We will leave them (in many cases) in radical form, knowing that they correspond to a specific, known number, but in the interest of exactness, we will not attempt to evaluate the radical. The results of any arithmetic operations will, in such cases, be expressions that include radicals.
Some Additional Properties of Multiplication and Addition
Recall that we identified addition and multiplication as commutative operations. Thus, for any two numbers a and b,
To aid in our look at radical operations, we can also consider some additional properties of addition and multiplication (and, by implication, subtraction and division, since these operations can be rewritten as addition and multiplication, respectively). First, we note that addition and multiplication are both associative, meaning that we can group terms or factors in any way we wish. Given three numbers a, b, and c,
This property simply says that we can perform a series of additions or multiplications in any order. (Recall from our study of the order of operations that expressions in parentheses must be evaluated first-hence we express the associative property as we have above.)
Additionally, we can show that multiplication is distributive, meaning that if we multiply a number a by the sum b + c, then
We can illustrate this property in a more concrete manner by considering that the product of two factors x and y can be viewed as x sets of y objects. For instance, 7 sets of 8 objects is equal to 56 objects. But 7 sets of 8 is the same as 3 sets of 8 and 4 sets of 8, or 2 sets of 8 and 5 sets of 8. Consider a graphical illustration of multiplication for this example.
Note above that and that . Furthermore, according to the representation of multiplication that we have used above, not only is , but . Thus, we can see from this example that multiplication is indeed distributive.
We can now apply these concepts and properties to our understanding of radicals.
Radical Operations
Let's say we wanted to perform the following addition:
As we have discussed, the square root of a number that is not a perfect square is irrational; thus, we cannot express it exactly as either a finite decimal or a fraction containing an integer numerator and integer denominator. Note, however, that we can use the fact that multiplication is distributive: . We'll start by rewriting the radical expressions slightly-this does not change the value of the expression, however.
Note that we cannot simplify this expression any further and still keep it exact. Also, note that we can omit the multiplication symbol () whenever doing so does not affect the clarity of the expression. Thus, we will, for the most part, omit this symbol for the remainder of this article. Let's look at a couple other examples.
Note that we cannot perform these operations if the numbers under the radicals are different. Thus, for instance, the expressions below cannot be further simplified.
We can also multiply and divide radicals.
Practice Problem: Evaluate each expression, where possible.
a. b. c. d.
e. f.
Solution: In each case, simply follow the pattern demonstrated earlier. Until you get more familiar with the operations, you may need to very carefully follow the distributive property to find sums or differences of radical expressions. Note that the expression in part a cannot be evaluated any further; the expression in part f can be only partially evaluated.
a. b.
c. d.
e.
f.
Simplifying Radicals
To aid in performing the operations above, we must often simplify radicals. To illustrate, consider the following expression.
A cursory look at this expression might seem to indicate that the expression cannot be simplified any further. Nevertheless, if we simplify the radical in the second term, we will find that we can actually perform the addition. The manipulations below rely only on the rules we have discussed so far.
Thus,
Above, we simplified the radical expression ; this process is in some ways similar to reducing a fraction to lowest terms. In simplifying a radical, we are writing an equivalent expression that is easier to work with and that often leaves less room for error. The goal of this process is to write the radical expression such that the number under the radical is not divisible by a perfect square. For instance, in the radical expression , 8 is divisible by 4, which is a perfect square. As a result, we can simplify this expression by factoring out the 4, as we did above. We then calculate the square root of 4 (which is 2), leaving only a 2 under the radical. Since 2 is not divisible by any perfect squares, the radical form is in simplest form.
Practice Problem: Evaluate each radical expression, where possible.
a. b. c.
d. e. f.
Solution: Each expression requires some degree of simplification of the radical. In some cases, you may take a slightly different approach, but the answer should be the same nonetheless. Not every step of the process is shown; if you are unsure of how a particular answer is achieved, consult the examples discussed above.
a.
b.
c.
d.
e.
f.
Another case where simplification is beneficial is a radical occurring in the denominator of a fraction. Consider the quotient expression we encountered above:
One (perfectly legitimate) approach to evaluating this expression is to take the square root of 16 (which is 4) and then simplify the radical in the denominator.
We are left, however, with a square root in the denominator. Although this is not an incorrect result, preventing occurrences of radicals in the denominator is often beneficial. Thus, we must find a way to remove the radical from the denominator. First, let's take note of the following, where y is any number:
Of course, multiplying any number by 1 yields the original number; furthermore, if the numerator and denominator of a fraction are the same, then that fraction is equal to 1. Thus, we can remove a radical from the denominator of a fraction by calculating an equivalent fraction in the manner shown below.
This is the same result we obtained earlier using a different approach to evaluating this expression. Below is another example of simplifying a fraction with a radical in the denominator.
Finally, note that in such cases, we can write the expression with the radical in the numerator or with the radical as a factor multiplied by a fraction.
Practice Problem: Simplify each of the following fractions.
a. b. c. d.
Solution: In each case, multiply the numerator and denominator by the radical in the denominator and evaluate. Before doing so, however, performing other simplifications may make the process easier. In part c, the result can be written either as a single fraction or as a fraction minus an integer.
a.
b.
c.
d.
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